# 20.E: Electrochemistry (Exercises)

These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

## 20.1: Oxidation States

### Q20.1.1

Identify the oxidation state of the atoms in the following compounds:

1. $$PCl_3$$
2. $$CO_3^{2-}$$
3. $$H_2S$$
4. $$S_8$$
5. $$SCl_2$$
6. $$Na_2SO_3$$
7. $$SO_4^{2-}$$

### S20.1.1

1. An atom’s oxidation state is representative of the number of electrons that a specific atom has lost or gained when binding to another atom. This allows us to determine the changes that cause in redox reactions as well as balance redox reactions.

Below are some of the rules to determining an atom’s oxidation number:

2. The oxidation state of an element that is not combined with another element Fe, H2, O2, P4, S8 is zero (0).

3. The oxidation state of oxygen in a compounds is -2, except for when it is in peroxides like H2O2, and Na2O2, in this case the oxidation state for O is -1.

4. The oxidation state of hydrogen is +1 in its compounds, except for metal hydrides, such as NaH and LiH, where the oxidation state for H is -1.

5. The net charge on a molecule of ion is equal to the algebraic sum of the oxidation state of all the atoms present in the species.

6.

• +1 for alkali metals (Group 1): (Li, Na, K, Rb, Cs)

• +2 for alkaline earth metals (Group 2): (Be, Mg, Ca, Sr, Ba)

• The oxidation number of fluorine is always –1. Chlorine, bromine, and iodine usually have an oxidation number of –1, unless they are combined with an oxygen or fluorine atom

7. The oxidation number of a monatomic ion is equal to the charge on the ion

8. Solution:

a) PCl3

According to Rule #5 above, chlorine has an oxidation state of -1. There are 3 chlorine atoms present in this compound and so chlorine has a total oxidation state of (-1)3=-3. Because this is a neutral compound, we know that the net charge of the molecule must equal to 0 (Rule #4). Therefore we can find the oxidation state of phosphorous because :

O.S.(P)+O.S(Cl)=net charge

O.S.(P)+(-3)=0

O.S.(P)=+3

Oxidation state of chlorine atom=-1, total oxidation state of Chlorine atoms: -3

Oxidation state of phosphorous atom=+3

b) CO32−

According to Rule #2, oxygen has an oxidation number of -2. Since there are 3 oxygen atoms, the overall oxidation state of the oxygens is (-2)3=-6 Because the overall molecule has a net charge of 2- , carbon must have an oxidation state of (-2)-(-6)=+4.

Oxidation state of oxygen atom=-2

Oxidation state of carbon atom=+4

c) H2S

According to Rule #3, hydrogen has an oxidation state of +1. Since there are two hydrogen atoms, the total oxidation state of the hydrogens is (+1)2=+2. The overall molecule has a net charge of 0, and so the oxidation state of the sulfur atom is 0-(+2)=-2.

Oxidation state of hydrogen atom=+1

Oxidation state of sulfur atom=-2

d) S8

Since this is an elemental form of sulfur, according to Rule #1, the sulfur atom would have an oxidation state of 0.

Oxidation state of sulfur atom=0

e) SCl2

According to Rule #5 above, chlorine has an oxidation state of -1. There are 2 chlorine atoms present in this compound and so chlorine has a total oxidation state of (-1)2=-2. This is a neutral compound, so the net charge of the molecule is equal to 0. Therefore, the oxidation state of the sulfur atom is 0-(-2)-+2.

Oxidation state of chlorine atom=-1

Oxidation state of sulfur atom=+2

f) Na2SO3

Sodium is an alkali metal, and so one Na atom has an oxidation state of +1; the overall oxidation state of the two sodium atoms in this compound is equal to (+1)2=+2. Oxygen atom has an oxidation state of -2, and since there are 3 oxygen atoms, the total oxidation state of the oxygen atoms in this compound is (-2)3=-6. This is a neutral molecule with a net charge of 0, and so the oxidation state of the sulfur atom is 0-(+2)-(-6)=+4.

Oxidation state of sodium atom=+1

Oxidation state of sulfur atom=+4

Oxidation state of oxygen atom=-2

g) SO42−

According to Rule #2, the oxidation state of an oxygen atom is -2. The overall oxidation state of 4 atoms of oxygen is (-2)4=-8. The net charge of the compound is 2-, and so the oxidation state of the sulfur atom is -2-(-8)=+6.

Oxidation state of sulfur atom=+6

Oxidatin state of oxygen atom=-2

### A20.1.1

a)

Oxidation state of chlorine atom=-1

Oxidation state of phosphorous atom=+3

b)

Oxidation state of oxygen atom=-2

Oxidation state of carbon atom=+4

c)

Oxidation state of hydrogen atom=+1

Oxidation state of sulfur atom=-2

d)

Oxidation state of sulfur atom=0

e)

Oxidation state of chlorine atom=-1

Oxidation state of sulfur atom=+2

f)

Oxidation state of sodium atom=+1

Oxidation state of sulfur atom=+4

Oxidation state of oxygen atom=-2

g)

Oxidation state of sulfur atom=+6

Oxidatin state of oxygen atom=-2

## 20.2: Balanced Oxidation-Reduction Equations

### Q20.2.1

Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

### S20.2.1

A good oxidant is an element that easily oxidizes another species. This means this species takes an electron from the other species or is easily reduced. A good reductant is the opposite. A good reductant is an element that easily reduces another species. This means this species gives an electron to the other species or is easily oxidized.

Elements that are good oxidants or easily reduced are highly electronegative, meaning they want electrons. These elements tend to be the elements on the far right side of the periodic table(excluding the halogens, period 18) because their outermost electron shells are nearly full and being full is very "desirable." Therefore, they readily accept electrons to become reduced whenever they get the chance to.

Elements that are good reducers or are easily oxidized have very low electronegativities, meaning they would much rather lose electrons than gain any. These elements tend to be the elements on the far left side of the periodic table because their outermost electron shells have very few electrons in them. It is easier to obtain a full outermost electron shell by losing their few outer electrons and dropping down a shell than gaining all electrons needed to fill their current shell.

### Q20.2.2

If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why?

### S20.2.2

(1) The redox reaction would occur because a composition of a poor oxidant and poor reductant is spontaneous (i.e. ΔG is negative). For example the standard reduction potential of a poor oxidant such as such as $$Li^{+}$$ is -3.04V. The standard reduction potential of a poor reductant such as $$I_{2}/I^{-}$$ is +0.54. Therefore, the overall standard reduction potential of this reaction would be +3.58. Since the $$E^{o}_{cell}$$ value is positive, the reaction is spontaneous and will occur.

(2) A redox reaction would not occur because a composition of a strong oxidant and weak reductant is non-spontaneous (i.e. ΔG is positive). Look at a redox table to find the $$E^{o}_{cell}$$ of $$F_{2}/F^{-}$$, a strong oxidant, and $$Cu/Cu^{+}$$, a weak reductant. If we subtract the cathode and anode, the $$E^{o}_{cell}$$ value is negative so it a non-spontaneous reaction and will not occur.

### Q20.2.3

In each redox reaction, determine which species is oxidized and which is reduced:

1. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
2. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
3. BrO3(aq) + 2MnO2(s) + H2O(l) → Br(aq) + 2MnO4(aq) + 2H+(aq)

### S20.2.3

In order to determine which species is oxidized and which is reduced, we have to get the oxidation number of each species in the reactants and products and compare them. Oxidation is the loss of electrons, while reduction is the gain of electrons.  (OIL RIG: Oxidation Is Loss and Reduction Is Gain)

In the first problem, the charge of Zn goes from 0 to +2. The charge of H goes from +1 to 0.

ANSWER: Zn is oxidized while H is reduced.

In the second problem, the charge of Cu goes from 0 to 2. The charge of N goes from +5 to +4.

ANSWER: Cu is oxidized while N is reduced.

In the third problem, the charge of Br goes from +5 to -1. The charge of Mn goes from +4 to +7.

ANSWER: Mn is oxidized while Br is reduced.

### A20.2.3

a) Zn is oxidized while H is reduced.

b) Cu is oxidized while N is reduced.

c) Mn is oxidized while Br is reduced.

### Q20.2.4

Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is not a single-displacement reaction.

### S20.2.4

A redox reaction, or oxidation-reduction reaction, is a chemical reaction that involves the transferring of electrons between species as one is being oxidized, the reducing agent, and the other reduced, the oxidizing agent. In order to identify which of the species is the reducing or oxidizing agent, you must look at their oxidation numbers.

Oxidation or the one known as the reducing agent is the loss of electrons. Reduction or the one known as the oxidizing agent is the gaining of electrons.

A helpful mnemonic to remember this would be OIL RIG as Oxidation Is Loss Reduction Is Gain.

In single-displacement reactions, it involves “replacing” an element in the reactant with an element in the product. For example, in $$A + BC \rightarrow AB + C$$, C in the BC compound on the reactant side is being replaced by A to produce the AB compound.

Another example of this would be

$$Cl_2 + 2KBr \rightarrow Br_2+ 2KCl$$

as Br in KBr is being replaced by $$Cl_2$$, with K acting as a spectator ion. The half-reactions in this for this reaction are:

$$Cl_2+2e^- \rightarrow 2Cl^-$$
$$2Br^- \rightarrow Br_2+ 2e^-$$

In this, $$Cl_2$$ would be reduced from an oxidation number of (0) to (-1) while Br in KBr is oxidized from an oxidation number of (-1) to (0).

Redox reactions do not only have single-displacement reactions but also other types of redox reactions such as combination, combustion, decomposition, and disproportionation reactions.

An example of a redox reaction that is not a single-displacement reaction would be

$$CH_4 + 2O_2 \rightarrow 2H_2O + CO_2$$

which is a combustion reaction, another type of redox reaction. The balanced half-reactions are:

$$2O_2+ 8e^- \rightarrow 4O^{2-}$$
$$C^{-4} \rightarrow C^{+4}+ 8e^-$$

$$O_2$$ has been reduced from (0) to (-2), while C has been oxidized from (-4) to (+4).

### Q20.2.5

Of the following elements, which would you expect to have the greatest tendency to be oxidized: Zn, Li, or S? Explain your reasoning.

### S20.2.5

There are two ways to solve this problem:

When looking at this problem it is first important to know what oxidation is. Oxidation is the loss of electrons. If an element has a high tendency to be oxidized it means that the element easily gives up its electrons. This is related to electronegativity, a very electronegative element does not give its electrons up easily while a weak electronegative element does. The electronegativity trend decreases down a group and increases across a period.

The first way is to find the element that has the greatest tendency to be oxidized you have to find the least electronegative element by finding Zn, Li, and S on the periodic table. When you do that you will see that Li is the least electronegative element of the three and is more likely to be oxidized.

The second way to solve this problem is to look at the reduction potential of each element. Reduction potential measures the ability of a chemical species to be reduced (gain electrons). By looking at the reduction potential of each element on the activity series here you will see that Li has the lowest reduction potential of Zn, Li and S making it the element that has the greatest tendency to be oxidized.

### Q20.2.6

Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning.

### S20.2.6

Remember the pneumonic, OIL RIG. "Oxidation is loss, Reduction is gain." When an element is reduced, it gains electrons.

Elements want to have a full valence shell (outermost shell that accounts for the reactivity of the element). Since Se is a Group 6 element, it has 6 valence electrons. The most stable state has 8 valance electrons (a full shell). Se would be the easiest to reduce because it readily accepts electrons to complete its valence shell.

Sr is in Group 2, meaning it has 2 valence electrons. Elements in Group 1 or Group 2 are more likely to give up electrons to empty their outermost shell. In this case, those elements would be oxidized because they are giving up electrons.

Ni is part of the transition metals. Most transition metals have a lower number of valence electrons. In this case, Nickel has 2 valence electrons, putting it more likely to be oxidized like Sr.

### Q20.2.7

Which of these metals produces H2 in acidic solution?

1. Ag
2. Cd
3. Ca
4. Cu

### S20.2.7

In order to solve this problem we must consult the table of Standard Reduction Potentials. Using the table we cn determine that in order for $$H_2$$ to be produced, $$H^+$$ must be reduced by a reducing agent. The reduction of $$H^+$$ to $$H_2$$ is... $2H^{+}{(aq)} + 2e^− \rightarrow H_{2}{(g)} \;\;\; E° = 0\; V$ A spontaneous reaction (reaction will occur with no external energy source) has an E°>0. $$E°=E°_{cathode}-E°_{anode}$$. Since reduction occurs at the cathode and in an acidic solution $$H^+$$ gains electrons and is reduced to $$H_2$$, the $$E°_{cathode}=\; 0$$. $E°>0 \;\;\; and\; E°_{cathode}=0\; V$ $0>0-E°_{anode}$ Therefore, any of the 4 metals that has a standard reduction potential E°<0 will reduce $$H^+$$ to $$H_2$$ in an acidic solution. $Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)} \;\;\; E° = 0.7996\; V$ $Cd^{2+}{(aq)} + 2e^− \rightarrow Cd{(s)} \;\;\; E° = -0.403\; V$ $Ca^{2+}{(aq)} + 2e^− \rightarrow Ca{(s)} \;\;\; E° = -2.84\; V$ $Cu^{2+}{(aq)} + 2e^− \rightarrow Cu{(s)} \;\;\; E° = 0.3419\; V$E°<0 for Cd and Ca so Cd and Ca will produce $$H_2$$ in an acidic solution.

### Q20.2.8

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

1. Mg(s) + Cu2+(aq) →
2. Au(s) + Ag+(aq) →
3. Cr(s) + Pb2+(aq) →
4. K(s) + H2O(l) →
5. Hg(l) + Pb2+(aq) →

### S20.2.8

For activity series, we are using the oxidation activity series chart. The higher up the chart, the more likely oxidation will take place.

a. Cu(s) and Mg2+(aq) will be formed as products. Let's take a look at the reactants. We see that Mg has no charge and Cu2+ has a charge. If we look at the activity series in terms of oxidation, we see that Mg(s) is more likely to be oxidized than Cu(s). So, Cu2+ will gain electrons (be reduced) and Mg will lose electrons (be oxidized). We have the half reactions from the activity series chart:

Mg(s) → Mg2+(aq) + 2e-

Cu2+(aq) + 2e- → Cu(s)

Adding the half reactions together, we get the electrons canceled out and having the net ionic equation Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq)

b. Nothing will happen. Let's take a look at the reactants. We see that Au has no charge and Ag+ has a charge. If we look at the activity series in terms of oxidation, we see that Ag(s) is more likely to be oxidized than Au(s). Au being oxidized and Ag+ being reduced is less likely. The reaction is not spontaneous, and there needs to be an external power supply for a reaction to even happen.

c. Cr3+(aq) and Pb(s) will be formed as products. Let's take a look at the reactants. We see that Mg has no charge and Cu2+ has a charge. If we look at the activity series in terms of oxidation, we see that Cr(s) is more likely to be oxidized than Pb(s). Pb2+ will gain electrons (be reduced) and Cr will lose electrons (be oxidized). We have the half reactions from the activity series chart:

Cr(s) → Cr3+(aq) + 3e-

Pb2+(aq) + 2e- → Pb(s).

Adding the half reactions together after the oxidation half reaction is multiplied by a coefficient of 2 and the reduction half reaction is multiplied by a coefficient of 3, we get the electrons canceled out and having the net ionic equation 2 Cr(s) + 3 Pb2+(aq) → 3 Pb(s) + 2 Cr3+(aq).

d. KOH and H2 will be formed as products. Let's take a look at the reactants, K and H2O. We see that an alkali metal, which would be K, is reacting with water. When an alkali metal reacts with water, a strong base and hydrogen gas is formed. K is highly likely to be oxidized, as alkali metals are good reducing agents, and the H2O molecule is reduced.

We get the unbalanced equation: K(s) + H2O(l)→ KOH(aq) + H2(g)

We add the coefficient of 2 on K, H2O, and KOH to get the net ionic equation 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)

e. Nothing will happen. Let's take a look at the reactants. We see that Hg has no charge and Pb2+ has a charge. If we look at the activity series in terms of oxidation, we see that Pb(s) is more likely to be oxidized than Hg(l). The reaction is not spontaneous, and there needs to be an external power supply for a reaction to even happen.

### A20.2.8

a) Mg(s) + Cu2+(aq) → Cu(s) + Mg2+(aq)

b) No Reaction

c) 2 Cr(s) + 3 Pb2+(aq) → 3 Pb(s) + 2 Cr3+(aq)

d) 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)

e) No Reaction

### Q20.2.9

Balance each redox reaction under the conditions indicated.

1. CuS(s) + NO3(aq) → Cu2+(aq) + SO42−(aq) + NO(g); acidic solution
2. Ag(s) + HS(aq) + CrO42−(aq) → Ag2S(s) + Cr(OH)3(s); basic solution
3. Zn(s) + H2O(l) → Zn2+(aq) + H2(g); acidic solution
4. O2(g) + Sb(s) → H2O2(aq) + SbO2(aq); basic solution
5. UO22+(aq) + Te(s) → U4+(aq) + TeO42−(aq); acidic solution

### S20.2.9

1. CuS(s) + NO3(aq) → Cu2+(aq) + SO42−(aq) + NO(g); acidic solution

1. Write the oxidation and reduction half reactions.

Reduction: ${NO_3^-}_{(aq)}⟶ NO_(g)$

Oxidation: ${CuS}_{(s)}⟶ {Cu^{2+}}_{(aq)}+{SO_4^{2-}}_{(aq)}$

2. In each half reaction, balance all elements. Balance O with H2O and H with H+. Balance the charge on each side using e-.

Reduction: ${3e^-}+{4H^+}_{(aq)}+{NO_3^-}_{(aq)}⟶ {NO}_{(g)}+{2H_2O}_{(l)}$

Oxidation: ${4H_2O}_{(l)}+{CuS}_{(s)}⟶ {Cu^{2+}}_{(aq)}+{SO_4^{2-}}_{(aq)}+{8H^+}_{(aq)}+8e^-$

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: $8({3e^-}+{4H^+}_{(aq)}+{NO_3^-}_{(aq)}⟶ {NO}_{(g)}+{2H_2O}_{(l)})$

${24e^-}+{32H^+}_{(aq)}+{8NO_3^-}_{(aq)}⟶ {8NO}_{(g)}+{16H_2O}_{(l)}$

Oxidation: $3({4H_2O}_{(l)}+{CuS}_{(s)}⟶ {Cu^{2+}}_{(aq)}+{SO_4^{2-}}_{(aq)}+{8H^+}_{(aq)}+8e^-)$

${12H_2O}_{(l)}+{3CuS}_{(s)}⟶ {3Cu^{2+}}_{(aq)}+{3SO_4^{2-}}_{(aq)}+{24H^+}_{(aq)}+24e^-$

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: ${8H^+}_{(aq)}+{8NO_3^-}_{(aq)}+{3CuS}_{(s)}⟶ {8NO}_{(g)}+{4H_2O}_{(l)}+{3Cu^{2+}}_{(aq)}+{3SO_4^2-}_{(aq)}$

5. Check if number of atoms and charges balance.

The charges on both sides equal to 0, and the number of atoms are equal on both sides. ✓

2. Ag(s) + HS(aq) + CrO42−(aq) → Ag2S(s) + Cr(OH)3(s); basic solution

1. Write the oxidation and reduction half reactions.

Reduction: ${CrO_4^{2-}}_{(aq)}⟶ {Cr(OH)_3}_{(s)}$

Oxidation: ${Ag}_{(s)}+{HS^-}_{(aq)}⟶{Ag_2S}_{(s)}$

2. In each half reaction, balance all elements. Balance O with H2O and H with H+. Balance the charge on each side using e-.

Reduction: ${3e^-}+{5H^+}_{(aq)}+{CrO_4^{2-}}_{(aq)}⟶ {Cr(OH)_3}_{(s)}+{H_2O}_{(l)}$

Oxidation: ${2Ag}_{(s)}+{HS^-}_{(aq)}⟶ {Ag_2S}_{(s)}+{H^+}_{(aq)}+2e^-$

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: $2({3e^-}+{5H^+}_{(aq)}+{CrO_4^{2-}}_{(aq)}⟶ {Cr(OH)_3}_{(s)}+{H_2O}_{(l)})$

${6e^-}+{10H^+}_{(aq)}+{2CrO_4^{2-}}_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+{2H_2O}_{(l)}$

Oxidation: $3({2Ag}_{(s)}+{HS^-}_{(aq)}⟶ {Ag_2S}_{(s)}+{H^+}_{(aq)}+2e^-)$

${6Ag}_{(s)}+{3HS^-}_{(aq)}⟶ {3Ag_2S}_{(s)}+{3H^+}_{(aq)}+6e^-$

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: ${7H^+}_{(aq)}+{2CrO_4^{2-}}_{(aq)}+6Ag_{(s)}+3HS^-_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+{2H_2O}_{(l)}+3Ag_2S_{(s)}$

5. Basic solution: add a number of OH- ions equal to the number of H+ ions.

${7OH^-}_{(aq)}+{7H^+}_{(aq)}+{2CrO_4^{2-}}_{(aq)}+6Ag_{(s)}+3HS^-_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+{2H_2O}_{(l)}+3Ag_2S_{(s)}+{7OH^-}_{(aq)}$

6. Basic solution: Combine H+ and OH- ions to form water molecules. Cancel common water molecules, and leave a remainder H2O molecules on just one side.

${5H_2O}_{(l)}+{2CrO_4^{2-}}_{(aq)}+6Ag_{(s)}+3HS^-_{(aq)}⟶ {2Cr(OH)_3}_{(s)}+3Ag_2S_{(s)}+{7OH^-}_{(aq)}$

7. Check if number of atoms and charges balance.

The charges on both sides equal to -7, and the number of atoms are equal on both sides. ✓

3. Zn(s) + H2O(l) → Zn2+(aq) + H2(g); acidic solution

1. Write the oxidation and reduction half reactions.

Reduction: ${H_2O}_{(l)}⟶ {H_2}_{(g)}$

Oxidation: ${Zn}_{(s)}⟶ {Zn^{2+}}_{(aq)}$

2. In each half reaction, balance all elements. Balance O with H2O and H with H+. Balance the charge on each side using e-.

Reduction: ${2e^-}+{2H^+}_{(aq)}+{H_2O}_{(l)}⟶{H_2}_{(g)}+{H_2O}_{(l)}$

Oxidation: ${Zn}_{(s)}⟶{Zn^{2+}}_{(aq)}+2e^-$

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: ${2e^-}+{2H^+}_{(aq)}+{H_2O}_{(l)}⟶{H_2}_{(g)}+{H_2O}_{(l)}$

Oxidation: ${Zn}_{(s)}⟶{Zn^{2+}}_{(aq)}+2e^-$

They have the same number of electrons. No changes needed.

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: ${2H^+}_{(aq)}+{Zn}_{(s)}⟶ {H_2}_{(g)}+{Zn^{2+}}_{(aq)}$

5. Check if number of atoms and charges balance.

The charges on both sides equal to +2, and the number of atoms are equal on both sides. ✓

4. O2(g) + Sb(s) → H2O2(aq) + SbO2(aq); basic solution

1. Write the oxidation and reduction half reactions.

Reduction: ${O_2}_{(g)}⟶ {H_2O_2}_{(aq)}$

Oxidation: ${Sb}_{(s)}⟶{SbO_2^-}_{(aq)}$

2. In each half reaction, balance all elements. Balance O with H2O and H with H+. Balance the charge on each side using e-.

Reduction: ${2e^-}+{2H^+}_{(aq)}+{O_2}_{(g)}⟶ {H_2O_2}_{(aq)}$

Oxidation: ${2H_2O}_{(l)}+{Sb}_{(s)}⟶ {SbO_2^-}_{(aq)}+{4H^+}_{(aq)}+3e^-$

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: $3({2e^-}+{2H^+}_{(aq)}+{O_2}_{(g)}⟶ {H_2O_2}_{(aq)})$

${6e^-}+{6H^+}_{(aq)}+{3O_2}_{(g)}⟶ {3H_2O_2}_{(aq)}$

Oxidation: $2({2H_2O}_{(l)}+{Sb}_{(s)}⟶ {SbO_2^-}_{(aq)}+{4H^+}_{(aq)}+3e^-)$

${4H_2O}_{(l)}+{2Sb}_{(s)}⟶ {2SbO_2^-}_{(aq)}+{8H^+}_{(aq)}+6e^-$

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: ${3O_2}_{(g)}+{4H_2O}_{(l)}+2Sb_{(s)}⟶ {3H_2O_2}_{(aq)}+{2SbO_2^-}_{(aq)}+2H^{2+}_{(aq)}$

5. Basic solution: add a number of OH- ions equal to the number of H+ ions.

${2OH^-}_{(aq)}+{3O_2}_{(g)}+{4H_2O}_{(l)}+2Sb_{(s)}⟶ {3H_2O_2}_{(aq)}+{2SbO_2^-}_{(aq)}+2H^{2+}_{(aq)}+{2OH^-}_{(aq)}$

6. Basic solution: Combine H+ and OH- ions to form water molecules. Cancel common water molecules, and leave a remainder H2O molecules on just one side.

${2OH^-}_{(aq)}+{3O_2}_{(g)}+{2H_2O}_{(l)}+2Sb_{(s)}⟶ {3H_2O_2}_{(aq)}+{2SbO_2^-}_{(aq)}$

7. Check if number of atoms and charges balance.

The charges on both sides equal to -2, and the number of atoms are equal on both sides. ✓

5. UO22+(aq) + Te(s) → U4+(aq) + TeO42−(aq); acidic solution

1. Write the oxidation and reduction half reactions.

Reduction: ${UO_2^{2+}}_{(aq)}⟶ {U^{4+}}_{(aq)}$

Oxidation: ${Te}_{(s)}⟶ {TeO_4^{2-}}_{(aq)}$

2. In each half reaction, balance all elements. Balance O with H2O and H with H+. Balance the charge on each side using e-.

Reduction: ${2e^-}+{4H^+}_{(aq)}+{UO_2^{2+}}_{(aq)}⟶{U^{4+}}_{(aq)}+{2H_2O}_{(l)}$

Oxidation: ${4H_2O}_{(l)}+Te_{(s)}⟶{TeO_4^{2-}}_{(aq)}+8H^+_{(aq)}+6e^-$

3. Equalize the number of electrons in both reactions by multiplying by the appropriate integers.

Reduction: $3({2e^-}+{4H^+}_{(aq)}+{UO_2^{2+}}_{(aq)}⟶{U^{4+}}_{(aq)}+{2H_2O}_{(l)})$

${6e^-}+{12H^+}_{(aq)}+{3UO_2^{2+}}_{(aq)}⟶{3U^{4+}}_{(aq)}+{6H_2O}_{(l)}$

Oxidation: ${4H_2O}_{(l)}+Te_{(s)}⟶{TeO_4^{2-}}_{(aq)}+8H^+_{(aq)}+6e^-$

4. Add the half equations, cancel species common to both sides of the overall equation.

Overall: ${4H^+}_{(aq)}+{3UO_2^{2+}}_{(aq)}+{Te}_{(s)}⟶ {3U^{4+}}_{(aq)}+{2H_2O}_{(l)}+{TeO_4^{2-}}_{(aq)}$

5. Check if number of atoms and charges balance.

The charges on both sides equal to +10, and the number of atoms are equal on both sides. ✓

### Q20.2.10

Balance each redox reaction under the conditions indicated.

1. MnO4(aq) + S2O32−(aq) → Mn2+(aq) + SO42−(aq); acidic solution
2. Fe2+(aq) + Cr2O72−(aq) → Fe3+(aq) + Cr3+(aq); acidic solution
3. Fe(s) + CrO42−(aq) → Fe2O3(s) + Cr2O3(s); basic solution
4. Cl2(aq) → ClO3(aq) + Cl(aq); acidic solution
5. CO32−(aq) + N2H4(aq) → CO(g) + N2(g); basic solution

### Q20.2.11

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

1. Platinum wire is dipped in hydrochloric acid.
2. Manganese metal is added to a solution of iron(II) chloride.
3. Tin is heated with steam.
4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.

### Q20.2.12

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

1. A few drops of NiBr2 are dropped onto a piece of iron.
2. A strip of zinc is placed into a solution of HCl.
3. Copper is dipped into a solution of ZnCl2.
4. A solution of silver nitrate is dropped onto an aluminum plate.

### Q20.2.13

Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.

### Q20.2.14

Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu2+(aq) and nitric oxide gas.

1. What has been oxidized? What has been reduced?
2. Balance the chemical equation.

### Q20.2.15

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

1. Pt2+(aq) + Ag(s) →
2. HCN(aq) + NaOH(aq) →
3. Fe(NO3)3(aq) + NaOH(aq) →
4. CH4(g) + O2(g) →

### Q20.2.16

Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:

1. Zn(s) + HCl(aq) →
2. 3HNO3(aq) + AlCl3(aq) →
3. K2CrO4(aq) + Ba(NO3)2(aq) →
4. Zn(s) + Ni2+(aq) →

### A20.2.16

a) This is a precipitation reaction because Zinc Chloride (which would be a solid since zinc is insoluble according to solubility rules) is produced and Hydrogen gas.

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2 (s)

b) There is no reaction because there is not an acid and base, so it is not an acid-base reaction, no precipitate is formed (HCl is aq and AlNO3 is always soluble- again, this is due to solubility rules) so it's not a precipitation reaction, and the oxidation numbers of the elements do not change in the equation, so it is not a redox reaction.

3HNO3(aq) + AlCl3(aq) → Al(NO3)3 (aq) + 3HCl (aq)

c) This is a precipitation reaction because chromates like BaCrO4 are insoluble. Again, not dealing with acids and bases and not with changing oxidation numbers.

K2CrO4(aq) + Ba(NO3)2(aq) → BaCrO4 (s) + 2KNO3 (aq)

d) This is a redox reaction because there is a change in the oxidation number of Zn and Ni.

To find the products, split the reaction into two half reactions:

Zn(s) → Zn2+(aq) + 2e-

Ni2+(aq) + 2e- → Ni(s)

Since there are 2e- in the reactants side of the second half reaction and the same number of e- at the products side of the first half reaction, they cancel out and you do not have to multiply the reactions by anything. So, you get:

Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)

## 20.3: Voltaic Cells

### Q20.3.1

Is $$2NaOH_{(aq)} + H_2SO_{4(aq)} \rightarrow Na_2SO_{4(aq)} + 2H_2O_{(l)}$$ an oxidation–reduction reaction? Why or why not?

### Q20.3.2

If two half-reactions are physically separated, how is it possible for a redox reaction to occur? What is the name of the apparatus in which two half-reactions are carried out simultaneously?

S20.3.2

It's possible for a redox reaction to occur when two half-reactions are physically separated if there is a complete circuit made by an external electrical connection that helps the electrons flow from the oxidation reaction to the reduction reaction. The name of the apparatus in which two half-reactions are carried out simultaneously is called the electrochemical cell.

### Q20.3.3

What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?

### Q20.3.4

What is the purpose of a salt bridge in a galvanic cell? Is it always necessary to use a salt bridge in a galvanic cell?

### Q20.3.5

One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K+ and Cl ions do. What would happen if the diffusion rates of the anions and cations differed significantly?

### S20.3.5

Ions of similar rates of diffusion are needed to prevent a build-up of charge and reduction in the rate at which the electrodes can perform redox reactions. If ions with significantly different rates of diffusion were used, a charge will build up on one electrode more than the other. This would decrease the quantity of electrons that can flow through the wire as electrical resistance is generated from a build-up of charge.

### Q20.3.6

It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?

### S20.3.6

A large difference in cation/anion diffusion rates would cause charge to build up and increase the resistance of charge flow in the salt bridge and limit electron flow through the circuit. At a higher resistance, the favorability of the reaction decreases (Δ G decreases) the electric potential will decrease.

### Q20.3.7

Copper(II) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO4 solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of $$ZnCl_2$$?

### S20.3.7

1. Copper(II) Sulfate is soluble in water and forms Cu2+ ion and sulfate ion. If we look at the list of standard reduction potentials we can see from the values of Eo for each reaction taking place that, Zn2+: Eo=-0.76V and Cu2+: Eo=0.34V. Zn is the anode and is being oxidized and that Cu2+ is the cathode and is being reduced. This must be true as a redox reaction can only occur if the Eo of the cathode is bigger than that of the anode. This redox reaction causes the solution to change color and the zinc to erode and form a black solid Cu. The solution changing color represents the blue Cu2+ ions being reduced into a solid black Cu metal. The zinc strip eroding represents Zn metal being oxidized and forming a clear Zn2+ solution in the water.

2. These reaction can be found on the list of standard reduction potentials or derived from the problem.

Anode: Zn(s)⟶Zn2+(aq) + 2e-

Cathode:Cu2+(aq)+ 2e-⟶ Cu(s)

Net reaction: Zn(s) + Cu2+(aq) -> Cu(s) + Zn2+(aq)

3. Nothing will occur. Again if we look at the standard reduction potentials we will find that Cu being oxidized into Cu2+ can not occur with Zn2+ ion since then the cathode would have a lower reduction potential than the anode. In otherwords, the reverse reaction (Cu metal to Cu ion, Zn ion to Zn metal) is non-spontaneous, as the forward reaction is spontaneous.

### Q20.3.8

Consider the following spontaneous redox reaction: NO3(aq) + H+(aq) + SO32−(aq) → SO42−(aq) + HNO2(aq).

1. Write the two half-reactions for this overall reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which electrode corresponds to which half-reaction?
3. Which electrode is negatively charged, and which is positively charged?

### S20.3.8

First, we must identify which species is being oxidized and which is being reduced by assigning oxidation numbers to each element. The nitrogen from NO3- on the reactants side has an oxidation number of +5 and the nitrogen from HNO2 in the products side has an oxidation number of +1; therefore, the species containing nitrogen is being reduced because nitrogen gains 2 electrons. The sulfur from SO32- on the reactants side has an oxidation number of +4 and the sulfur from SO42- on the products side has an oxidation number of +6; therefore, the species containing sulfur is being oxidized because sulfur loses 2 electrons.

Next, we write out the two half-reactions, one for oxidation and one for reduction.

Reduction: NO3-(aq) -> HNO2 (aq)

Oxidation: SO32-(aq) -> SO42-(aq)

Then we balance each of these half reactions by first balancing all of the elements present in the half reactions except for hydrogen and oxygen. Next, add H2O to balance out the oxygens. Then use H+ to balance out the hydrogens. Lastly, we must balance the charges of each half reaction by adding the appropriate number of electrons, resulting in our 2 balanced half reactions.

Reduction: NO3-(aq) + 3H+ + 2e- -> HNO2(aq) + H2O(l)

Oxidation: SO32-(aq) + H2O(l) -> SO42-(aq) + 2H+ + 2e-

b) If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which electrode corresponds to which half-reaction?

Oxidation always occurs at the anode and reduction occurs at the cathode. Because NO3- reaction is being reduced, its half-reaction would correspond to the cathode electrode. The SO32- reaction is being oxidized, so that half-reaction would correspond to the anode electrode.

c) Which electrode is negatively charged, and which is positively charged?

The anode is negatively charged and the cathode is positively charged.

### Q20.3.9

The reaction $Pb(s) + 2VO^{2+}(aq) + 4H^+(aq) → Pb^{2+}(aq) + 2V^{3+}(aq) + 2H_2O(l)$ occurs spontaneously.

1. Write the two half-reactions for this redox reaction.
2. If the reaction is carried out in a galvanic cell using an inert electrode in each compartment, which reaction occurs at the cathode and which occurs at the anode?
3. Which electrode is positively charged, and which is negatively charged?

### S20.3.9

From the main reaction, we see that lead is reduced (it gains electrons), and Vanadium is oxidized (it loses electrons). The unbalanced half reactions are:

Reduction: $${Pb(s)} \rightarrow {Pb^{2+}(aq)}$$

Oxidation: $${VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}$$

To balance the reactions, we first add $$H_2O$$ to balance the oxygen, then add $$H^+$$ to balance the hydrogen. We then add electrons to balance the charge.

Reduction: $${Pb(s)} \rightarrow {Pb^{2+}(aq)}$$

$${Pb(s)} \rightarrow {Pb^{2+}(aq)}+{2e^-}$$

Oxidation: $${VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}$$

$${VO^{2+}(aq)} \rightarrow {V^{3+}(aq)}+{H_2O(l)}$$

$${VO^{2+}(aq)}+{2H^+} \rightarrow {V^{3+}(aq)}+{H_2O(l)}$$

Problem 2

The cathode is defined as the electrode where reduction takes place. This means the reduction reaction occurs at the cathode.

The anode is defined as the electrode where oxidation takes place. This means the oxidation reduction occurs at the anode.

Problem 3

The cathode is positively charged because electrons flow across the wire to this electrode.

The anode is negatively charged because electrons flow across the wire from this electrode.

It helps to remember this; regardless of whether you cell is a galvanic cell or an electrolytic cell, reduction always happens at the cathode, and oxidation at the anode. Most student attempt to memorize it using signs that; cathode is positively charged and anode is negatively charged.

### Q20.3.10

Phenolphthalein is an indicator that turns pink under basic conditions. When an iron nail is placed in a gel that contains [Fe(CN)6]3−, the gel around the nail begins to turn pink. What is occurring? Write the half-reactions and then write the overall redox reaction.

Corrosion of a iron nail with a coiled copper wire, in agar-agar medium with ferroxyl indicator solution (potassium hexacyanoferrate(III), indicator of iron ions, and phenolphthalein, indicator of hydroxide ions). Image used with permission (CC BY-SA 3.0; Ricardo Maçãs).

### S20.3.10

Phenolphthalein is used in many acid-base titrations because it is pH sensitive. The gel around the iron nail turns pink indicating that there is a reaction that produces a base occurring in the area surrounding the nail.

• Reduction:
• [Fe(CN)6]3−+e⟶[Fe(CN)6]4−
• Oxidation:
• Fe(s)+2OH-⟶Fe(OH)2+2e
• Overall reaction:
• 2[Fe(CN)6]3−+Fe(s)+2OH⟶Fe(OH)2+2[Fe(CN)6]4−

### Q20.3.11

Sulfate is reduced to HS in the presence of glucose, which is oxidized to bicarbonate. Write the two half-reactions corresponding to this process. What is the equation for the overall reaction?

### S20.3.11

We know that Sulfate (SO42−) is reduced, which means that it becomes more negatively charged, in this case, through the process of creating water. Follow the rules of Writing Equations for Redox Reactions, adding first water to balance oxygen, then H+ to balance Hydrogen, then electrons (e-) to balance charge for the first half reaction.

reduction half reaction: SO42−(aq) + 9H+(aq) + 8e → HS(aq) + 4H2O(l)

Now, complete the same process for the oxidation of glucose (C6H12O6) into bicarbonate (HCO3) to form the oxidation half reaction.

oxidation half reaction: C6H12O6(aq) + 12H2O(l) → 6HCO3(g) + 30H+(aq) + 24e

In order to combine these two half reactions into a single overall reaction, the electrons must be able to evenly cancel. In this case, the reduction reaction has 8 electrons and the oxidation reaction has 24, so multiply the first reaction by the common multiple of 4.

4(SO42−(aq) + 9H+(aq) + 8e → HS(aq) + 4H2O(l)

4SO42−(aq) + 36H+(aq) + 24e → 4HS(aq) + 16H2O(l)

Now we can line up both equations and cancel anything that appears on both sides of the equation to achieve the final overall equaiton.

reduction: 4SO42−(aq) + 36H+(aq) + 24e → 4HS(aq) + 16H2O(l)

Oxidation: C6H12O6(aq) + 12H2O(l) → 6HCO3(g) + 30H+(aq) + 24e

Overall: C6H12O6(aq) + 3SO42−(aq) → 6HCO3(g) + 3H+(aq) + 3HS(aq)

### A20.3.11

reduction: SO42−(aq) + 9H+(aq) + 8e → HS(aq) + 4H2O(l)

oxidation: C6H12O6(aq) + 12H2O(l) → 6HCO3(g) + 30H+(aq) + 24e

overall: C6H12O6(aq) + 3SO42−(aq) → 6HCO3(g) + 3H+(aq) + 3HS(aq)

### Q20.3.12

Write the spontaneous half-reactions and the overall reaction for each proposed cell diagram. State which half-reaction occurs at the anode and which occurs at the cathode.

1. Pb(s)∣PbSO4(s)∣SO42−(aq)∥Cu2+(aq)∣Cu(s)
2. Hg(l)∣Hg2Cl2(s)∣Cl(aq) ∥ Cd2+(aq)∣Cd(s)

### A20.3.12

To approach this problem, some important things to note are that:

• To the left of the cell diagram composes of the anode
• To the right of the cell diagram composes of the cathode
• the || portion of the cell diagram represents the salt bridge

Knowing this, we can use this information to write out both the half-reactions for the anode and cathode.

1. $$Pb(s) + SO_4^{2-}(aq) \to PbSO_4(s) + 2e^{-}$$ anode

$$Cu^{2+}(aq) + 2e^{-} \to Cu(s)$$ cathode

$$Cu^{2+}(aq) + Pb(s) + SO_4^{2-}(aq) \to PbSO_4(s) + Cu(s)$$ overall reaction

2. $$2Hg(l) + 2Cl^{-}(aq) \to Hg_{2}Cl_2(s) + 2e^{-}$$ anode

$$Cd^{2+} + 2e^{-} \to Cd(s)$$ cathode

$$2Hg(l) + 2Cl^{-}(aq) + Cd^{2+}(aq) \to Hg_{2}Cl_{2}(s) + Cd(s)$$ overall reaction

### Q20.3.13

For each galvanic cell represented by these cell diagrams, determine the spontaneous half-reactions and the overall reaction. Indicate which reaction occurs at the anode and which occurs at the cathode.

1. Zn(s)∣Zn2+(aq) ∥ H+(aq)∣H2(g), Pt(s)
2. Ag(s)∣AgCl(s)∣Cl(aq) ∥ H+(aq)∣H2(g)∣Pt(s)
3. Pt(s)∣H2(g)∣H+(aq) ∥ Fe2+(aq), Fe3+(aq)∣Pt(s)

### S20.3.13

In order to make it easier to describe a given electrochemical cell, a special cell notation has been adopted. The half-cells are separated by two bars or a slash, which represents a salt bridge. Single lines indicate different phases. In cell notation, the anode (oxidation) is written on the left and the cathode (reduction) is written on the right. Concentrations can be written in parentheses along with the state of each phase. When writing the half-reactions, it is important to balance the charges by putting electrons on either side. In these reactions, platinum is not included because it is simply what the electrode is made of.

### A20.3.13

1.

reduction: 2H+(aq) + 2e → H2(aq); cathode;

oxidation: Zn(s) → Zn2+(aq) + 2e; anode;

overall: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(aq)

1.

reduction: AgCl(s) + e → Ag(s) + Cl(aq); cathode;

oxidation: H2(g) → 2H+(aq) + 2e; anode;

overall: AgCl(s) + H2(g) → 2H+(aq) + Ag(s) + Cl(aq)

1.

reduction: Fe3+(aq) + e → Fe2+(aq); cathode;

oxidation: H2(g) → 2H+(aq) + 2e; anode;

overall: 2Fe3+(aq) + H2(g) → 2H+(aq) + 2Fe2+(aq)

### Q20.3.14

For each redox reaction, write the half-reactions and draw the cell diagram for a galvanic cell in which the overall reaction occurs spontaneously. Identify each electrode as either positive or negative.

1. Ag(s) + Fe3+(aq) → Ag+(aq) + Fe2+(aq)
2. Fe3+(aq) + 1/2H2(g) → Fe2+(aq) + H+(aq)

### S20.3.14

1. $Ag(s)+Fe^{_{3+}}(aq)\rightarrow Ag^{_{+}}(aq)+Fe^{_{2+}}$

• Oxidation is the loss of electrons and reduction is the gain of electrons. Therefore, we can break the given down into the following half reactions.
• Oxidation half reaction: $Ag(s)\rightarrow Ag^{_{+}}(aq)+e^{_{-}}$
• Reduction half reaction: $Fe^{_{3+}}(aq)+e^{_{-}}\rightarrow Fe^{_{2+}}(s)$
• Oxidation takes place in the anode (LHS of the cell) and reduction takes place in the cathode (RHS of the cell).
• Cell diagram: $Ag(s) | Ag^{_{+}}(aq) || Fe^{_{3+}}(aq), Fe^{_{2+}}(aq) | Pt(s)$
• Use Platinum as the electrode because it is inert, and their are no solids present in the reduction reaction.
• The anode (left hand side) is the negative electrode because the electrons are flowing from the anode to the cathode.

2. $Fe^{_{3+}}+1/2H_2(g)\rightarrow Fe^{_{2+}}(aq)+H^{_{+}}(s)$

• It may be helpful to rewrite the reaction as: $2Fe^{_{3+}}+1H_2(g)\rightarrow 2Fe^{_{2+}}(aq)+2H^{_{+}}(s)$
• Oxidation is the loss of electrons and reduction is the gain of electrons. Therefore, we can break the given down into the following half reactions.
• Oxidation half reaction: $H_2(g)\rightarrow 2H^{_{+}}(aq)+2e^{_{-}}$
• Reduction half reaction: $2Fe^{_{3+}}(aq)+2e^{_{-}}\rightarrow 2Fe^{_{2+}}(s)$
• Oxidation takes place in the anode (LHS of the cell) and reduction takes place in the cathode (RHS of the cell).
• Cell Diagram: $Pt(s)[H_2(g)] | 2H^{_{+}}(aq) || Fe^{_{3+}}(aq), Fe^{_{2+}}(aq) | Pt(s)$
• Use Platinum as the electrode in the cathode because it is inert, and their are no solids present in the reduction reaction.
• Use SHE (Standard Hydrogen Electrode) as the electrode in the anode.
• The anode (left hand side) is the negative electrode because the electrons are flowing from the anode to the cathode.

### Q20.3.15

Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.

1. 2Cl(aq) + Br2(l) → Cl2(g) + 2Br(aq)
2. $$N_2O_4 (g) + H_2O → HNO_2 (aq) + H^+ (aq) + NO_3^- (aq)$$ (at pH=0)
3. 2H2O(l) + 2Cl(aq) → H2(g) + Cl2(g) + 2OH(aq)
4. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

### S20.3.15

1. From the overall reaction, write oxidation and reduction half-reactions and balance the equations with the appropriate number of electrons:

$\text{ oxidation: } 2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g})+2\text{ e}^-$

$\text{ reduction: }\text{Br}_2 (\mathit{l}) +2\text{ e}^-\rightarrow 2\text{ Br}^- (\mathit{aq})$

The spontaneity of the overall reaction can be determined by calculating the standard cell potential using the equation

$\text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode}$

If $$\text{E}^°_\text{cell}$$ is positive then the reaction is spontaneous. If $$\text{E}^°_\text{cell}$$ is negative then the reaction is not spontaneous. The standard reduction potentials for this reaction are:

• anode: $2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g})+2\text{ e}^- \;\;\; \text{E}^° = 1.36\text{ V}$
• cathode: $\text{Br}_2 (\mathit{l}) +2\text{ e}^-\rightarrow 2\text{ Br}^- (\mathit{aq}) \;\;\; \text{E}^° = 1.07\text{ V}$

Then find the standard cell potential.

$\text{E}^°_\text{cell} = 1.07\text{ V} - 1.36\text{ V}=-0.29\text{ V}$

Since $$\text{E}^°_\text{cell}$$ is negative the reaction is not spontaneous in the forward direction.

The cell diagram for a galvanic cell in which this reaction is spontaneous is:

$\text{Pt }(\mathit{s}) | \text{Br}_2(\mathit{ l}) | \text{Br}^-(\mathit{ aq}) || \text{Cl}^-(\mathit{ aq}) | \text{Cl}_2(\mathit{ g}) | \text{Pt} (\mathit{s})$

1. From the overall reaction given we can write the incomplete oxidation and reduction half-reactions:

$\text{ oxidation: }\ce{N2O4} (\mathit{g}) \rightarrow \text{ NO}_3^- (\mathit{aq})$

$\text{ reduction: } \ce{N2O4} (\mathit{g}) \rightarrow \text{HNO}_2 (\mathit{aq})$

We will start with the oxidation half-reaction and balance the nitrogen atoms by scaling the appropriate side.

$\ce{N2O4} (\mathit{g}) \rightarrow 2\text{ NO}_3^- (\mathit{aq})$

Next is balancing the number of oxygen atoms by adding enough H2O molecules to the correct side.

$\ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq})$

Now we balance the number of hydrogen molecules by adding enough H+.

$\ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) + 4\text{ H}^+$

Finally, add electrons to balance charge.

$\ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) + 4\text{ H}^+ + 2\text{ e}^-$

Now move onto the reduction half-reaction. First, we balance elements in the equation other than O and H. In this case, balance nitrogen atoms.

$\ce{N2O4} (\mathit{g}) \rightarrow 2\text{ HNO}_2 (\mathit{aq})$

Oxygen is already balanced so move onto balancing hydrogen.

$\ce{N2O4} (\mathit{g}) + 2\text{ H}^+\rightarrow 2\text{ HNO}_2 (\mathit{aq})$

Finally, add electrons to balance charge and obtain the balanced reduction half-reaction.

$\ce{N2O4} (\mathit{g}) + 2\text{ H}^+ + 2\text{ e}^-\rightarrow 2\text{ HNO}_2 (\mathit{aq})$

Obtain and calculare the SRP:

• anode: $\ce{N2O4}(\mathit{g}) + \ce{2H2O}\rightarrow 2\text{ NO}_3^- (\mathit{aq}) + 4\text{ H}^+ + 2\text{ e}^- \;\;\; \text{E}^° = 0.79\text{ V}$
• cathode: $\ce{N2O4} (\mathit{g}) + 2\text{ H}^+ + 2\text{ e}^-\rightarrow 2\text{ HNO}_2 (\mathit{aq}) \;\;\; \text{E}^° = 1.07\text{ V}$

$\text{E}^°_\text{cell} = 1.07\text{ V} - 0.79\text{ V}=0.28\text{ V}$

Since $$\text{E}^°_\text{cell}$$ is positive the reaction is spontaneous in the forward direction.

The cell diagram is:

$\text{Pt }(\mathit{s}) | \ce{N2O4}(\mathit{g}) | \text{ NO}_3^- (\mathit{aq}) || \text{ HNO}_2 (\mathit{aq}) | \ce{N2O4}(\mathit{g}) | \text{Pt} (\mathit{s})$

1. From the overall reaction, write oxidation and reduction half-reactions and balance the equations with the appropriate number of electrons, H, O, and H2O molecules:

$2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g}) + 2\text{ e}^-$

$\ce{2H2O} (\mathit{ l}) + 2\text{ e}^-\rightarrow \ce{H2} (\mathit{g}) + 2\text{ OH}^-$

The SRP for this reaction are:

• anode: $2\text{ Cl}^- (\mathit{aq}) \rightarrow \text{ Cl}_2 (\mathit{g}) + 2\text{ e}^- \;\;\; \text{E}^° = 1.36\text{ V}$
• cathode: $\ce{2H2O} (\mathit{ l}) + 2\text{ e}^-\rightarrow \ce{H2} (\mathit{g}) + 2\text{ OH}^- \;\;\; \text{E}^° = -0.83\text{ V}$

$\text{E}^°_\text{cell} = -0.83\text{ V} - 1.36\text{ V}=-2.19\text{ V}$

Since $$\text{E}^°_\text{cell}$$ is negative the reaction is not spontaneous in the forward direction.

The cell diagram for a galvanic cell in which this reaction is spontaneous is:

$\text{Pt }(\mathit{s}) | \text{H}_2(\mathit{ g}) | \text{OH}^-(\mathit{ aq}) || \text{Cl}^-(\mathit{ aq})| \text{Cl}_2(\mathit{ g}) | \text{Pt} (\mathit{s})$

1. This is a combustion reaction so it will be spontaneous. To calculate $$\text{E}^°_\text{cell}$$ write the incomplete oxidation and reduction half-reactions:

$\text{ oxidation: }\ce{C3H8 }(\mathit{g}) \rightarrow 3\text{ CO}_2 (\mathit{g})$

$\text{ reduction: } \ce{5O2} (\mathit{ g}) \rightarrow \ce{4H2O} (\mathit{g})$

Add H2O molecules to balance oxygen atoms.

$\ce{C3H8 }(\mathit{g}) + \ce{6H2O}(\mathit{l})\rightarrow 3\text{ CO}_2 (\mathit{g})$

Balance the hydrogen atoms.

$\ce{C3H8 }(\mathit{g}) + \ce{6H2O}(\mathit{l}) \rightarrow 3\text{ CO}_2 (\mathit{g}) + 20\text{ H}^+$

Add electrons to balance charge and get the balanced oxidation half-reaction.

$\ce{C3H8 }(\mathit{g}) + \ce{6H2O} (\mathit{l}) \rightarrow 3\text{ CO}_2 (\mathit{g}) + 20\text{ H}^+ + 20\text{ e}^-$

For the reduction half-reaction we start by balancing oxygen atoms.

$\ce{5O2} (\mathit{ g}) \rightarrow \ce{4H2O} (\mathit{g}) + \ce{6H2O}$

$\ce{5O2} (\mathit{ g}) \rightarrow \ce{10H2O} (\mathit{g})$

Balance the hydrogen atoms.

$\ce{5O2} (\mathit{ g}) + 20\text{ H}^+\rightarrow \ce{10H2O} (\mathit{g})$

Add electrons to balance charge and simplify to obtain the balanced reduction half-reaction.

$\ce{5O2} (\mathit{ g}) + 20\text{ H}^+ + 20\text{ e}^-\rightarrow \ce{10H2O} (\mathit{g})$

$\ce{O2} (\mathit{ g}) + 4\text{ H}^+ + 4\text{ e}^-\rightarrow \ce{2H2O} (\mathit{g})$

To find the SRP, calculate ΔG from the equation

$\text{ΔG} = -\text{nF}\text{E}^°_\text{cell}$

To determine the specific $$\text{E}^°_\text{cell}$$ value for the overall reaction and conclude whether the reaction is spontaneous or not use the equation:

$\text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode}$

This process is spontaneous as it is combustion.

The cell diagram is:

$\text{Pt }(\mathit{s}) | \ce{C3H8 }(\mathit{g}) | \text{ CO}_2 (\mathit{g}) || \ce{O2} (\mathit{ g}) | \ce{H2O} (\mathit{g}) | \text{Pt} (\mathit{s})$

### A20.3.15

a)$\text{Pt }(\mathit{s}) | \text{Br}_2(\mathit{ l}) | \text{Br}^-(\mathit{ aq}) || \text{Cl}^-(\mathit{ aq}) | \text{Cl}_2(\mathit{ g}) | \text{Pt} (\mathit{s})$

The reaction is not spontaneous in the forward direction.

b)$\text{Pt }(\mathit{s}) | \ce{N2O4}(\mathit{g}) | \text{ NO}_3^- (\mathit{aq}) || \text{ HNO}_2 (\mathit{aq}) | \ce{N2O4}(\mathit{g}) | \text{Pt} (\mathit{s})$ The reaction is spontaneous in the forward direction.

c) $\text{Pt }(\mathit{s}) | \text{H}_2(\mathit{ g}) | \text{OH}^-(\mathit{ aq}) || \text{Cl}^-(\mathit{ aq})| \text{Cl}_2(\mathit{ g}) | \text{Pt} (\mathit{s})$

The reaction is not spontaneous in the forward direction.

d)$\text{Pt }(\mathit{s}) | \ce{C3H8 }(\mathit{g}) | \text{ CO}_2 (\mathit{g}) || \ce{O2} (\mathit{ g}) | \ce{H2O} (\mathit{g}) | \text{Pt} (\mathit{s})$

The reaction is spontaneous in the forward direction.

### Q20.3.16

Write the half-reactions for each overall reaction, decide whether the reaction will occur spontaneously, and construct a cell diagram for a galvanic cell in which a spontaneous reaction will occur.

1. Co(s) + Fe2+(aq) → Co2+(aq) + Fe(s)
2. O2(g) + 4H+(aq) + 4Fe2+(aq) → 2H2O(l) + 4Fe3+(aq)
3. 6Hg2+(aq) + 2NO3(aq) + 8H+ → 3Hg22+(aq) + 2NO(g) + 4H2O(l)
4. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

### S20.3.16

Graph of Standard Reduction Potentials, see P2: Standard Reduction Potentials by Value and P1: Standard Reduction Potentials by Element

1.)

$Co(s)+Fe^{_{2+}}(aq)\rightarrow Co^{_{2+}}(aq)+Fe(s)$

In the redox reaction, the oxidation half of the reaction loses electron, and the reduction half of th reaction gain electron.

Oxidation Half Reaction: $$Co(s)+2e^{_{-}}\rightarrow Co^{_{2+}}(aq)$$

Reduction Half Reaction: $$Fe^{_{2+}}(aq)+2e^{_{-}}\rightarrow Fe(s)$$

To decide whether the reaction will occur spontaneously, we need to check whether E°cell > 0 or not. If E°cell > 0, the reaction is spontaneous.

Note that $$E°cell=E°cathode−E°anode$$

Substitute the value of standard reduction potential of E°cathode (reduction half reaction) and the value of standard reduction potential of E°anode (oxidation half reaction) base on the Standard Reduction Potential graph.

$$E°cell= -0.44V - (-0.28V) = -0.16V$$

Thus, this reaction is not spontaneous.

Thus, the spontaneous reaction will be $$Co^{_{2+}}(aq)+Fe(s)\rightarrow Co(s)+Fe^{_{2+}}(aq)$$ Co2+(aq) + Fe(s) → Co(s) + Fe2+(aq)

Note: The anode (oxidation half reaction) always goes on the left and cathode (reduction half reaction) on the right. Separate changes in phase by | and indicate the the salt bridge with ||. The lack of concentrations indicates solutions are under standard conditions

In this problem, the oxidation half is $$Fe(s)+2e^{_{-}}\rightarrow Fe^{_{2+}}(aq)$$, and the reduction half is $$Co^{_{2+}}(aq)+2e^{_{-}}\rightarrow Co(s)$$.

The cell diagram is : Fe(s)│Fe2+(aq)║Co2++(aq)│Co(s)

2.) O2(g) + 4H+(aq) + 4Fe2+(aq) → 2H2O(l) + 4Fe3+(aq)

$O_2(g)+4H^{_{+}}(aq)+4Fe^{_{2+}}(aq)\rightarrow 2H_2O(l) +4Fe^{_{3+}}$

In the redox reaction, the oxidation half of the reaction loses electron, and the reduction half of th reaction gain electron.

Oxidation Half Reaction: $$Fe^{_{2+}}(aq)\rightarrow Fe^{_{3+}} +e^{_{-}}$$

Reduction Half Reaction:$$O_2(g)+4H^{_{+}}(aq)+4e^{_{-}}\rightarrow 2H_2O(l)$$

To decide whether the reaction will occur spontaneously, we need to check whether E°cell > 0 or not. If E°cell > 0, the reaction is spontaneous.

Note that $$E°cell=E°cathode−E°anode$$

Substitute the value of standard reduction potential of E°cathode (reduction half reaction) and the value of standard reduction potential of E°anode (oxidation half reaction) base on the Standard Reduction Potential graph.

$$E°cell=1.23V - 0.771V = 0.459V$$

Thus, this reaction is spontaneous.

Thus, the spontaneous reaction will be $$O_2(g)+4H^{_{+}}(aq)+4Fe^{_{2+}}(aq)\rightarrow 2H_2O(l) +4Fe^{_{3+}}$$

Note: The anode (oxidation half reaction) always goes on the left and cathode (reduction half reaction) on the right. Separate changes in phase by | and indicate the the salt bridge with ||. The lack of concentrations indicates solutions are under standard conditions

In this problem, as no solid reactants can be used as electrodes, we need to add Pt as electrodes.

The cell diagram is : Pt(s)│Fe2+(aq),Fe3+(aq)║O2(g)│H+(aq)│Pt(s)

3.)

$6Hg^{_{2+}}(aq)+2NO_3^{_{-}}(aq) + 8H^{_{+}}\rightarrow 3Hg_2^{_{2+}}(aq)+2NO(g) +4H_2O(l)$

In the redox reaction, the oxidation half of the reaction loses electron, and the reduction half of th reaction gain electron.

Oxidation Half Reaction: $$2Hg^{_{2+}}(aq) +2e^{_{-}}\rightarrow Hg_2^{_{2+}}(aq)$$

Reduction Half Reaction: $$NO_3^{_{-}}(aq) + 4H^{_{+}}+3e^{_{-}}\rightarrow NO(g) +2H_2O(l)$$

To decide whether the reaction will occur spontaneously, we need to check whether E°cell > 0 or not. If E°cell > 0, the reaction is spontaneous.

Note that $$E°cell=E°cathode−E°anode$$

Substitute the value of standard reduction potential of E°cathode (reduction half reaction) and the value of standard reduction potential of E°anode (oxidation half reaction) base on the Standard Reduction Potential graph.

$E°cell=0.96V - 0.911V = 0.059V$

Thus, this reaction is spontaneous.

Thus, the spontaneous reaction will be $$6Hg^{_{2+}}(aq)+2NO_3^{_{-}}(aq) + 8H^{_{+}}\rightarrow 3Hg_2^{_{2+}}(aq)+2NO(g) +4H_2O(l)$$ 6Hg2+(aq) + 2NO3(aq) + 8H+ → 3Hg22+(aq) + 2NO(g) + 4H2O(l)

Note: The anode (oxidation half reaction) always goes on the left and cathode (reduction half reaction) on the right. Separate changes in phase by | and indicate the the salt bridge with ||. The lack of concentrations indicates solutions are under standard conditions

In this problem, as no solid reactants can be used as electrodes, we need to add Pt as electrodes.

The cell diagram is : Pt(s)│Hg2+(aq) ,Hg22+(aq) ║2NO3(aq)│2NO(g)│Pt(s)

4.) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

$CH_4(g)+2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

In the redox reaction, the oxidation half of the reaction loses electron, and the reduction half of th reaction gain electron.

Oxidation : $$CH_4(g) \rightarrow CO_2(g)$$

Reduction: $$O_2(g) \rightarrow 2H_2O(l)$$

Balance each reaction:

Oxidation :$$CH_4(g)+ 2H_2O \rightarrow CO_2(g)+8H^{_{+}} +8e^{_{-}}$$

Reduction: $$O_2(g)+4H^{_{+}} +4e^{_{-}}\rightarrow 2H_2O(l)$$

To decide whether the reaction will occur spontaneously, we need to check whether E°cell > 0 or not. If E°cell > 0, the reaction is spontaneous.

Note that $$E°cell=E°cathode−E°anode$$

Substitute the value of standard reduction potential of E°cathode (reduction half reaction) and the value of standard reduction potential of E°anode (oxidation half reaction) base on the Standard Reduction Potential graph.

In this problem, we couldn't find the direct SRP value from the table.

But, because this reaction is a combustion reaction, we assume that this reaction is spontaneous

Thus, the spontaneous reaction will be $$CH_4(g)+2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$$

Note: The anode (oxidation half reaction) always goes on the left and cathode (reduction half reaction) on the right. Separate changes in phase by | and indicate the the salt bridge with ||. The lack of concentrations indicates solutions are under standard conditions

In this problem, as no solid reactants can be used as electrodes, we need to add C as electrodes.

The cell diagram is : C(s)│CH4(g)CO2(g)║O2(g),CO2(g),2H2O(g)│C(s)

## 20.4: Cell Potential Under Standard Conditions

### Q20.4.1

Is a hydrogen electrode chemically inert? What is the major disadvantage to using a hydrogen electrode?

### S20.4.1

A hydrogen electrode is chemically inert when properly set up with a specially prepared inert platinum electrode and when it does not participate in the oxidation or reduction half-reactions for that cell. A major disadvantage to using a hydrogen electrode is that it is extremely difficult to keep the pressure of hydrogen gas at exactly 1 bar and in general one needs to make a very advanced apparatus. It is also difficult because hydrogen can be extremely expensive and needs to constantly be used in the electrode for the life of the cell and therefore often called hypothetical because the major disadvantages are too much to get over.

### Q20.4.2

List two factors that affect the measured potential of an electrochemical cell and explain their impact on the measurements.

### Q20.4.3

What is the relationship between electron flow and the potential energy of valence electrons? If the valence electrons of substance A have a higher potential energy than those of substance B, what is the direction of electron flow between them in a galvanic cell?

### S20.4.3

Electrons tend to flow from the anode to the cathode. The differences in potential energy between the two electrodes make the elections flow from the side of higher potential energy to the side of lower potential energy. This comes from the potential in the anode to become oxidized and the potential for the cathode to become reduced. The electrons go from the anode, which has a higher potential to become oxidized, to the cathode, which has a lower potential to become oxidized. The electrons favor a lower potential energy because in nature, the lowest energy form is preferred in order to be more stable. By favoring a lower potential energy, the reaction is spontaneous and does not require energy to allow the reaction to be completed. So, using this information, you can see that the electrons would move from substance A to substance B, because it’ll move from more potential energy to less potential energy.

### A20.4.3

In a galvanic cell, the difference in potential energy of valence electrons cause the electrons to flow through the circuit from anode to cathode. This in turns produce a current. If the valence electrons of substance A have a higher potential energy than those of substance B, the electrons will flow from A to B.

### Q20.4.4

If the components of a galvanic cell include aluminum and bromine, what is the predicted direction of electron flow? Why?

### S20.4.4

A galvanic, or voltaic, cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place in a cell and involve the movement of electrons. Since we know that galvanic cells operate based on spontaneous redox reactions, we have to consider the spontaneity of the redox reactions associated with the given elements that we are working with. To find this information, we can refer to a table of standard reduction potentials, from which we find the following information for aluminum and bromine:

$\mathrm{Al^{3+}(aq)}+\mathrm{3e^-}\rightarrow\mathrm{Al(s)};\textrm{-1.676 V}$

$\mathrm{Br_2(l)}+\mathrm{2e^-}\rightarrow\mathrm{2Br^{-}(aq)};\textrm{+1.065 V}$

With the information above, we find that the reduction potential for bromine is higher than that of aluminum, which therefore tells us that the reduction of bromine is more spontaneous. This means that in a galvanic cell with aluminum and bromine, bromine will undergo reduction whilst aluminum will undergo oxidation. We are interested in finding the direction of electron flow, so we need to consider where electrons are being produced and where they are going. Also consider that oxidation always occurs at the anode whilst reduction always occurs at the cathode. Oxidation results in the loss of electrons, so in our case, electrons are being produced at the aluminum anode, and will therefore flow through the wire towards the bromine cathode.

### Q20.4.5

Write a cell diagram representing a cell that contains the Ni/Ni2+ couple in one compartment and the SHE in the other compartment. What are the values of E°cathode, E°anode, and E°cell?

### S20.4.5

To write the cell diagram for this cell, we need to find out the E°cathode, E°anode, and E°cell which can be worked out based off the standard electrode potentials. We need to find out the half cell reactions and potentials for Ni/Ni2+ and SHE.

NOTE: SHE is the Standard Hydrogen Electrode where hydrogen gas is used to compare all other values to. These reactions are always carried out at standard conditions of 1M H+, 1 atm and at 273K. Hydrogen in this case as a cell potential of 0.00V where H/H+ can be oxidized or reduced depending on the reaction.

 Element Half Cell Reaction Standard Reduction Potential (V) Nickel (Ni) Ni2+(aq)+2e−⇔Ni(s) -0.257V Hydrogen (H2) H+(aq)+2e−⇔H2(g) 0.00V

As Nickel has the more negative standard reduction potential (SRP) value than H2, Ni is less likely to be reduced than H2. Following the OIL RIG mnemonic, oxidation is loss and reduction is gain. We can see the H2 will be reduced and gain electrons whereas Ni will be oxidized and gain electrons. So the redox equations should be:

$Ni^{2+}(aq)+2e^-→Ni(s)$

$H+(aq)+2e^-→H_{2}(g)$

We should also remember that oxidation ALWAYS occurs at the anode and reduction ALWAYS occurs at the cathode so we can conclude that:

Anode: Ni2+(aq)+2e→ Ni(s) (-0.257V)

Cathode: H+(aq)+2e→H2(g) (0.000V)

NOTE: You can try to remember the mnemonic RED CAT where REDucition always occurs at the CAThode.

Since we know which reaction is at the cathode and anode, we can use their SRP values to calculate the Standard Electrode Potential with the equation:

cell = E°cathode - E°anode

cell = (0.000V) - (-0.257V) = 0.257V

The overall reaction will be: H+(aq)+Ni(s)→H2(g)+Ni2+(aq)

As we know everything about this reaction, we can construct the cell diagram which will always illustrate everything about the reaction, including what is being reduced and oxidized.

NOTE: In a cell diagram, the left side always indicates the half-cell reaction that is at the cathode (being reduced) and the right side always indicates the half-cell reaction at the anode (being oxidized)

• At the cathode, solid Ni is being converted into its Ni2+ ions (since oxidation occurs at the anode) where Ni will be used as the electrode so the phase boundary will be in between the solid and aqueous ions
• At the anode, hydrogen gas is formed however (since reduction occurs at the cathode) , the solid that is used in the SHE is always platinum (Pt)

NOTE: When writing down cell diagrams with hydrogen, always separate H+(aq) and H2(g) with a phase boundary before the Pt electrode

NOTE: Always write down all of the states of all of the ions and elements/ compounds

Ni(s)∣Ni2+(aq)∥H+(aq, 1 M)∣H2(g, 1 atm)∣Pt(s)

### S20.4.5

Ni(s)∣Ni2+(aq)∥H+(aq, 1 M)∣H2(g, 1 atm)∣Pt(s)

 $$E^\circ_{\textrm{anode}} \\ E^\circ_{\textrm{cathode}} \\ E^\circ_{\textrm{cell}}$$ $$\mathrm{Ni^{2+}(aq)}+\mathrm{2e^-}\rightarrow\mathrm{Ni(s)};\;-\textrm{0.257 V} \\ \mathrm{2H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{H_2(g)};\textrm{ 0.000 V} \\ \mathrm{2H^+(aq)}+\mathrm{Ni(s)}\rightarrow\mathrm{H_2(g)}+\mathrm{Ni^{2+}(aq)};\textrm{ 0.257 V}$$

### Q20.4.6

Explain why E° values are independent of the stoichiometric coefficients in the corresponding half-reaction.

### S20.4.6

E° is defined as the potential of a cell measured under standard conditions. Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties (meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system) and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.

### Q20.4.7

Identify the oxidants and the reductants in each redox reaction.

1. Cr(s) + Ni2+(aq) → Cr2+(aq) + Ni(s)
2. Cl2(g) + Sn2+(aq) → 2Cl(aq) + Sn4+(aq)
3. H3AsO4(aq) + 8H+(aq) + 4Zn(s) → AsH3(g) + 4H2O(l) + 4Zn2+(aq)
4. 2NO2(g) + 2OH(aq) → NO2(aq) + NO3(aq) + H2O(l)

### S20.4.7

oxidant/oxidizing agent: being reduced/gaining electrons

reductant/reducing agent: being oxidized/losing electrons

1. oxidant: $$\displaystyle Ni^{2+}(aq)$$

$$\displaystyle Ni^{2+}(aq) + 2e^{-} \rightarrow Ni(s)$$

Electrons on the reactant side indicates that $$\displaystyle Ni^{2+}(aq)$$ is "gaining electrons".

reductant: $$\displaystyle Cr(s)$$

$$\displaystyle Cr(s)\rightarrow Cr^{2+}+2e^{-}$$

Electrons on the product side indicates that $$\displaystyle Cr(s)$$ is "losing electrons".

2. oxidant: $$\displaystyle Cl_2(g)$$

$$\displaystyle Cl_2(g)\rightarrow 2Cl^{-}(aq)+2e^{-}$$

Electrons on the reactant side indicates that $$\displaystyle Cl_2(g)$$ is "gaining electrons".

reductant: $$\displaystyle Sn^{2+}(aq)$$

$$\displaystyle Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^{-}$$

Electrons on the product side indicates that $$\displaystyle Sn^{2+}(aq)$$ is "losing electrons".

3. oxidant: $$\displaystyle H_3AsO_4(aq)$$

For this redox reaction, it is best to examine the oxidation states of the elements.

Given that the oxidation state of the hydrogen atom is +1 and the oxidation state of the oxygen atom is -2 in this reaction, we can determine the oxidation of $$\displaystyle As$$,x, to be:

reactant(($$\displaystyle H_3AsO_4(aq)$$): $$\displaystyle 3(+1)+(x)+4(-2)=0$$

$$\displaystyle x=5$$

product(($$\displaystyle AsH_3(g)$$): $$\displaystyle (x)+3(+1)=0$$

$$\displaystyle x=-3$$

From reactants to products, the oxidation state of $$\displaystyle As$$ changes from 5 to -3, meaning that it has gained electrons. Therefore, $$\displaystyle H_3AsO_4(aq)$$ acts as the oxidant.

reductant: $$\displaystyle Zn(s)$$

$$\displaystyle 4Zn(s)\rightarrow 4Zn^{2+}+2e^{-}$$

Electrons on the product side indicates that $$\displaystyle 4Zn(s)\rightarrow 4Zn^{2+}+2e^{-}$$ is "losing electrons".

4. oxidant: $$\displaystyle NO_2(g)$$

reductant: $$\displaystyle NO_2(g)$$

For this redox reaction, the only oxidation state that changes from reactants to products is $$\displaystyle N$$. Therefore, $$\displaystyle NO_2(g)$$ acts as both the oxidant and reductant in this reaction.

### A20.4.7

1. oxidant: Ni2+(aq); reductant: Cr(s)
2. oxidant: Cl2(g); reductant: Sn2+(aq)
3. oxidant: H3AsO4(aq); reductant: Zn(s)
4. oxidant: NO2(g); reductant: NO2(g)

### Q20.4.8

Identify the oxidants and the reductants in each redox reaction.

1. Br2(l) + 2I(aq) → 2Br(aq) + I2(s)
2. Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
3. H+(aq) + 2MnO4(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4(aq)
4. IO3(aq) + 5I(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)

### S20.4.8

To begin, let us define an oxidant and reductant. An oxidant, *or oxidizing agent, gains electrons and is reduced. A reductant, *or reducing agent, loses electrons and is oxidized. *The oxidation states can also be used to determine if a reaction is an oxidant or reductant. *In oxidation reactions, the oxidation state increases. *In reduction reactions, the oxidation state decreases. One can either look at the half reactions, or just look at the species in the overall reaction (it is best to split it into half reactions if you are just learning how to do such a problem).

a. Br2(l)+e-→2Br-(aq)

Here Br goes from an oxidation state of 0 to 1-. It went down in its oxidation state by gaining electrons, thus, was reduced. Therefore, based on the definitions, Br is an oxidant.

2I-(aq)→ I2(s) +e-

Here the oxidation state of I- went up. It lost electrons and was oxidized. Therefore, it is a reductant.

b. Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)

Here Cu goes down in its oxidation state by gaining electrons. It is reduced and is therefore an oxidant.

Ag goes up in its oxidation state by losing electrons. It is oxidized and therefore is a reductant.

c. H+(aq) + 2MnO4(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4(aq)

Here H+ goes to HSO4. Its oxidation state doesn't change at all and therefore is neither an oxidant or reductant.

Mn goes down in its oxidation state from 7+ to 2+ by gaining electrons, thus, is reduced. Since it is reduced,it is an oxidant.

H2 goes up in its oxidation state by losing electrons, thus, is oxidized. Since it is oxidized, it is a reductant.

d. IO3(aq) + 5I(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)

Here Iodine's oxidation state in IO3-(aq) is 5+. It's oxidation state alone, 5I-(aq), is 1-. I2 has an oxidation state of zero. Since Iodine's oxidation states go down, they gain electrons, are oxidized, and therefore is a reductant. Hydrogen's oxidation state doesn't change at all and is neither oxidized nor reduced.

### A20.4.8

oxidants: Br2, Cu2+,MnO4,IO3

Reductants: I-,Ag,H2SO3, I

### Q20.4.9

All reference electrodes must conform to certain requirements. List the requirements and explain their significance.

### S20.4.9

An ideal reference electrode should be reversible and obey the Nernst equation, exhibit a potential that is constant with time, return to its original potential after being subjected to small currents, and exhibit little lag in response( also known as hysteresis) with temperature fluctuations. These properties are significant in maintaining the function of the reference electrode which is to maintain a constant electrical potential against any deviations. A reference electrode is an electrode that has a known half-cell potential, is constant, and is altogether insensitive to the composition of the solution.

### Q20.4.10

For each application, describe the reference electrode you would use and explain why. In each case, how would the measured potential compare with the corresponding E°?

1. measuring the potential of a Cl/Cl2 couple
2. measuring the pH of a solution
3. measuring the potential of a MnO4/Mn2+ couple

### S20.4.10

One Solution:

reference electrode is an electrode with an accurately maintained potential, that is used as a reference when measuring cell potentials in galvanic cells. The most common reference electrode that we use is the standard hydrogen electrode. On the standard reduction potential chart, the standard hydrogen electrode has a value of 0, and all other electrical potentials are written down compared to this one.

From the chart, the standard hydrogen electrode has the reaction $2H^+(aq)+2e^-\rightarrow H_2(g)$

1. I would choose the SHE as the reference electrode. The standard hydrogen electrode (SHE) has a measured voltage of 0, which makes it easier to compare with other reduction reactions on the standard reduction potential table. The Eo value for the Cl-/Cl2 reaction is 1.358, while the potential of the SHE is 0. This would be a much stronger reduction reaction than the reference electrode, since it is at a much higher standard reduction potential value. Is SCE an option?

2. When measuring the pH of a solution, I would choose a pH electrode. This would be a glass electrode sensitive to the measurement of hydrogen ions. This is commonly used in the measurement of pH of a solution. A glass electrode is typically thought of as a tube within a tube with the inner one containing an HCl solution at a constant molarity. This basically acts as a galvanic cell, and is effective in the measurement of pH of a solution. SCE reference electrode could be used to measure pH as well.

3. For the last one, I would also use the standard hydrogen electrode (SHE) as the reference electrode because it has a measured voltage of 0. This makes it easier to compare to the other values on the table. The Eo value for this reaction is 1.51 compared to the SHE, which is much higher. This would make this reaction a better and stronger reduction reaction than the reference electrode.

The measured potential in all of these cases will still differ slightly from the actual Eo value. The measured potential depends on the equation $E=E^o-\frac{RT}{n}lnQ$ This equation takes in to account Q, the reaction quotient, which is determined by dividing the molarities of the reactants by the molarities of the products. Because of this, the measured potential changes according to the different concentrations of the various compounds going through the oxidation/reduction reactions.

### Q20.4.11

Draw the cell diagram for a galvanic cell with an SHE and a copper electrode that carries out this overall reaction:

$H_2(g) + Cu^{2+}(aq) → 2H^+(aq) + Cu(s)$

### S20.4.10

Shorthand cell diagrams are constructed with solid electrodes on the outside, then liquid, gas, and aqueous. If there is no solid electrode present in the reaction, like there is in this equation, then you would use either Pt(s) or C(s) (graphite) as the electrode. There is a single straight line separating each of these phases. In the middle, the anode (left side) is separated from the cathode (right side) by a double bar that represents the salt bridge in between the galvanic cells.

The phases are written in parentheses, with 1 atm and 1M being used as the standard concentration for gases and aqueous/solid solutions, respectively.

$Pt(s)\vert H_2(g,1 atm)\vert H^+(aq,1M) \vert\vert Cu^{2+}(aq) \vert Cu(s)$

A20.4.10

Pt(s)∣H2(g, 1 atm) | H+(aq, 1M)∥Cu2+(aq)∣Cu(s)

### Q20.4.12

Draw the cell diagram for a galvanic cell with an SHE and a zinc electrode that carries out this overall reaction: Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g).

### A20.4.12

In this question, we are asked to write out the cell diagram of a certain reaction containing a standard hydrogen electrode. A standard hydrogen electrode (SHE) is a metal electrode that does not get used up in the reaction. In this example, the SHE that is used is solid platinum.

From the reaction given, we can determine the two half-reactions, which are

Zn(s) → Zn2+(aq)

2H+(aq) → + H2(g)

When observing these reactions we can tell that the reaction containing zinc is an oxidation reaction, so it occurs in the anode, and the reaction containing hydrogen is a reduction reaction, so it occurs in the cathode. With the two half reactions, we can determine the cell diagram of this cell. Following the structure for standard cell diagrams, the anode is written on the right and the cathode is written on the left. The cell diagram for this cell is

Zn(s) | Zn2+(aq, 1.0M) || H+ (aq, 1.0M) | H2(g, 1.0 atm) | Pt (s)

where | represents a phase change, || represents the salt bridge, and molarity and pressure are included for the aqueous and gaseous parts of the diagram. Since it is not explicitly stated, we assume this reaction occurs at standard conditions (1M and 1atm.)

### Q20.4.13

Balance each reaction and calculate the standard electrode potential for each. Be sure to include the physical state of each product and reactant.

1. Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq)
2. Br2(aq) + Fe2+(aq) → 2Br(aq) + Fe3+(aq)
3. Fe3+(aq) + Cd(s) → Fe2+(aq) + Cd2+(aq)

### S20.4.13

The standard cell potential for a reaction can be calculated by subtracting the standard reduction potential for half-reaction at the anode from the standard reduction potential for the half-reaction at the cathode.

$$E_{cell} = E_{cathode} - E_{anode}$$

If $$E_{cell}> 0$$, then the reaction is spontaneous

If $$E_{cell}< 0$$, then the reaction is nonspontaneous

We can find the standard reduction of each species using the Standard Reduction Potential Table found here.

1. Anode (oxidation): H2(g) →2H+(aq)+ $$2e^-$$

Cathode (reduction): Cl2(g) + $$2e^-$$→ 2Cl(aq)

Overall reaction: Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq)

$$E_{cathode}=1.358V$$

$$E_{anode}=0V$$

$$E_{cell}= 1.358V - 0V = 1.358V$$

2. Anode (oxidation): Fe2+(aq) → Fe3+(aq) + $$e^-$$

Cathode (reduction): Br2(aq) + $$2e^-$$ → 2Br(aq)

Overall reaction: Br2(aq) + 2Fe2+(aq) → 2Br(aq) + 2Fe3+(aq)

$$E_{cathode}= 1.087V$$

$$E_{anode}= 0.771V$$

$$E_{cell}= 1.087V - 0.771V = 0.316V$$

3. Anode (oxidation): Cd(s) → Cd2+(aq) + $$2e^-$$

Cathode (reduction): Fe3+(aq) + $$e^-$$→ Fe2+(aq)

Overall reaction: 2Fe3+(aq) + Cd(s) → 2Fe2+(aq) + Cd2+(aq)

$$E_{cathode}= -0.403V$$

$$E_{anode}= 0.771V$$

$$E_{cell}= -0.403V - 0.771V = -1.174V$$

### A20.4.13

1. Cl2(g) + H2(g) → 2Cl(aq) + 2H+(aq); E° = 1.358 V
2. Br2(l) + 2Fe2+(aq) → 2Br(aq) + 2Fe3+(aq); E° = 0.316 V
3. 2Fe3+(aq) + Cd(s) → 2Fe2+(aq) + Cd2+(aq); E° = 1.174 V

### Q20.4.14

Balance each reaction and calculate the standard reduction potential for each. Be sure to include the physical state of each product and reactant.

1. Cu+(aq) + Ag+(aq) → Cu2+(aq) + Ag(s)
2. Sn(s) + Fe3+(aq) → Sn2+(aq) + Fe2+(aq)
3. Mg(s) + Br2(l) → 2Br(aq) + Mg2+(aq)

### A20.4.14

a. $$Cu^+(aq) + Ag^+(aq) → Cu^{2+}(aq) + Ag(s)$$

Step 1: Check if it’s balanced (both sides of equation have the same # of each element and same overall charge)

1 Cu on each side, 1 Ag on each side, overall charge +2 on each side: BALANCED SHE; so there is no need to balance the equation using the half-reaction method.

Step 2: Determine the oxidation and reduction reactions

Oxidation: $$Cu^+(aq)→ Cu^{2+}(aq) + e^{-}$$ (oxidation is loss of electrons)

Reduction: $$Ag^+(aq) + e^{-} →Ag(s)$$ (reduction is gain of electrons)

Step 3: To find the standard reduction potential use E cell = E cathode - E anode

$$E \ cathode = E \ reduction = 0.7996$$

$$E \ anode = E \ oxidation = 0.159$$

Plug these values in the formula: $$0.7996-0.159 = 0.6406 V$$

b. $$Sn(s) + Fe^{3+}(aq) → Sn^{2+}(aq) + Fe^{2+}(aq)$$

Step 1: Check if it’s balanced (both sides of equation have the same # of each element and same overall charge)

1 Sn (yes), 1 Fe (yes), overall charge +3 on left +4 on right: NOT BALANCED

Step 2: Separate into half reactions, decide which one is oxidation and which one is reduction, and balance the reaction

Oxidation: $$Sn(s)→ Sn^{2+}(aq)+2e^-$$ (oxidation is loss of electrons)

Reduction: $$e^- + Fe^{3+}(aq) →Fe^{2+}(aq)$$ (reduction is gain of electrons)

Balance: multiply the reduction reaction by 2, so the e- will cancel when the equations are added

Net Reaction: $$Sn(s) + 2Fe^{3+}(aq) →2Fe^{2+}(aq) Sn^{2+}(aq)$$

Step 3: To find the standard reduction potential use /(E cell = E cathode - E anode/)

$$E \ cathode = E \ reduction = 0.771$$

$$E \ anode = E \ oxidation = -0.14$$

Plug these values in the formula: $$0.771-(-0.14) = 0.911 V$$

c. $$Mg(s) + Br_2(l) → 2Br^−(aq) + Mg^{2+}(aq)$$

Step 1 - Check if it’s balanced (both sides of equation have the same # of each element and same overall charge)

1 Mg on both sides, 2 Br on both sides, net charge 0 on both sides: BALANCED

Step 2 - Identify oxidation and reduction reactions

Oxidation: $$Mg(s)→ Mg^{2+}(aq) + 2e^{-}$$ (oxidation is loss of electrons)

Reduction: $$Br_2(l) + 2e^{-} → 2Br^−(aq)$$ (reduction is gain of electrons)

Step 3: To find the standard reduction potential use /(E cell = E cathode - E anode/)

$$E \ cathode = E \ reduction = 1.087$$

$$E \ anode = E \ oxidation = -2.356$$

Plug these values in the formula: $$1.087-(-2.356) = 3.443 V$$

### Q20.4.15

Write a balanced chemical equation for each redox reaction.

1. H2PO2(aq) + SbO2(aq) → HPO32−(aq) + Sb(s) in basic solution
2. HNO2(aq) + I(aq) → NO(g) + I2(s) in acidic solution
3. N2O(g) + ClO(aq) → Cl(aq) + NO2(aq) in basic solution
4. Br2(l) → Br(aq) + BrO3(aq) in basic solution
5. Cl(CH2)2OH(aq) + K2Cr2O7(aq) → ClCH2CO2H(aq) + Cr3+(aq) in acidic solution

### S20.4.15

General Rules to Follow:

STEP 1: To balance a redox reaction, you first must separate the half reactions. Focus on one at a time

STEP 2: Balance all elements that are not hydrogen and oxygen

STEP 3: Balance oxygen with water

STEP 4: Balance hydrogen with H+

STEP 5: Balance the charges by adding electrons to the more positive side

STEP 6: If balancing in basic solution, then balance H+ with OH- on both sides and combine to form water. Cancel water molecules accordingly

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms

1. H2PO2(aq) + SbO2(aq) → HPO32−(aq) + Sb(s) in basic solution

STEP 1: Separate the half reactions. Focus on one at a time

$H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)$

STEP 2: Balance all elements that are not hydrogen and oxygen. Notice that phosphorous is balanced

STEP 3: Balance oxygen with water

$H_{2}O(l)+H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)$

STEP 4: Balance hydrogen with H+

$H_{2}O(l)+H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)+3H^{+}(aq)$

STEP 5: Balance the charges by adding electrons to the more positive side

$H_{2}O(l)+H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)+3H^{+}(aq)+2e^{-}$

STEP 6: If balancing in basic solution, then balance H+ with OH- on both sides and combine to form water. Cancel water molecules accordingly

$3OH^{-}(aq)+ H_{2}O(l)+H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)+3H^{+}(aq)+2e^{-}+3OH^{-}(aq)$

$3OH^{-}(aq)+H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)+2H_{2}O(l)+2e^{-}$

Balance the other half reaction.

$SbO_{2}^{-}(aq)\rightarrow Sb(s)$

STEP 2: Balance all elements that are not hydrogen and oxygen. Notice that Antimony is balanced

STEP 3: Balance oxygen with water

$SbO_{2}^{-}(aq)\rightarrow Sb(s)+2H_{2}O(l)$

STEP 4: Balance hydrogen with H+

$4H^{+}(aq)+SbO_{2}^{-}(aq)\rightarrow Sb(s)+2H_{2}O(l)$

STEP 5: Balance the charges by adding electrons to the more positive side

$4H^{+}(aq)+SbO_{2}^{-}(aq)+3e^{-}\rightarrow Sb(s)+2H_{2}O(l)$

STEP 6: If balancing in basic solution, then balance H+ with OH- on both sides and combine to form water. Cancel water molecules accordingly

$4OH^{-}(aq)+4H^{+}(aq)+SbO_{2}^{-}(aq)+3e^{-}\rightarrow Sb(s)+2H_{2}O(l)+4OH^{-}(aq)$

$3e^{-}+2H_{2}O(l)+SbO_{2}^{-}(aq)\rightarrow Sb(s)+4OH^{-}(aq)$

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms

$3(3OH^{-}(aq)+H_{2}PO_{2}^{-}(aq)\rightarrow HPO_{3}^{2-}(aq)+2H_{2}O(l)+2e^{-})$

$2(3e^{-}+2H_{2}O(l)+SbO_{2}^{-}(aq)\rightarrow Sb(s)+4OH^{-}(aq))$

The overall reaction is

$OH^{-}(aq)+3H_{2}PO_{2}^{-}(aq)+2SbO_{2}^{-}(aq)\rightarrow 3HPO_{3}^{2-}(aq)+2Sb(s)+2H_{2}O(l)$

2. HNO2(aq) + I(aq) → NO(g) + I2(s) in acidic solution

Following STEPS 1-5, the half reactions are

$e^{-}+H^{+}(aq)+HNO_{2}(aq)\rightarrow NO(g)+H_{2}O(l)$

$2I^{-}(aq)\rightarrow I_{2}(s)+2e^{-}$

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

$2(e^{-}+H^{+}(aq)+HNO_{2}(aq)\rightarrow NO(g)+H_{2}O)(l)$

$2I^{-}(aq)\rightarrow I_{2}(s)+2e^{-}$

The overall reaction is

$2H^{+}(aq)+2HNO_{2}(aq)+2I^{-}(aq)\rightarrow 2NO(g)+2H_{2}O(l)+I_{2}(s)$

3. N2O(g) + ClO(aq) → Cl(aq) + NO2(aq) in basic solution

Following STEPS 1-6, the half reactions are

$6OH^{-}(aq)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)+3H_{2}O(l)+4e^{-}$

$2e^{-}+H_{2}O(l)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+2OH^{-}(aq)$

Following STEP 7:

$6OH^{-}(aq)+N_{2}O(g)\rightarrow 2NO_{2}^{-}(aq)+3H_{2}O(l)+4e^{-}$

$2(2e^{-}+H_{2}O(l)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+2OH^{-}(aq))$

The overall reaction is

$2OH^{-}(aq)+N_{2}O(g)+2ClO^{-}(aq)\rightarrow 2NO_{2}^{-}(aq)+H_{2}O(l)+2Cl^{-}(aq)$

4. Br2(l) → Br(aq) + BrO3(aq) in basic solution

Following STEPS 1-6, the half reactions are

$12OH^{-}(aq)+Br_{2}(l)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2}O(l)+11e^{-}$

$Br_{2}(l)+2e^{-}\rightarrow 2Br^{-}(aq)$

Following STEP 7:

$2(12OH^{-}(aq)+Br_{2}(l)\rightarrow 2BrO_{3}^{-}(aq)+6H_{2}O(l)+11e^{-})$

$11(Br_{2}(l)+2e^{-}\rightarrow 2Br^{-}(aq))$

The overall reaction is

$24OH^{-}(aq)+13Br_{2}(l)\rightarrow 4BrO_{3}^{-}(aq)+22Br^{-}(aq)+12H_{2}O(l)$

5. Cl(CH2)2OH(aq) + K2Cr2O7(aq) → ClCH2CO2H(aq) + Cr3+(aq) in acidic solution

Following STEPS 1-5, the half reactions are

$H_{2}O(l)+Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-}$

$6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{+3}(aq)+2K^{+}(aq)+7H_{2}O(l)$

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

$3(H_{2}O(l)+Cl(CH_{2})_{2}OH(aq)\rightarrow ClCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-})$

$2(6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{+3}(aq)+2K^{+}(aq)+7H_{2}O(l))$

The overall reaction is

$3Cl(CH_{2})_{2}OH(aq)+2K_{2}Cr_{2}O_{7}(aq)+16H^{+}(aq)\rightarrow 3ClCH_{2}CO_{2}H(aq)+4Cr^{+3}(aq)+4K^{+}(aq)+11H_{2}O(l)$

### A20.4.15

1) $OH^{-}(aq)+3H_{2}PO_{2}^{-}(aq)+2SbO_{2}^{-}(aq)\rightarrow 3HPO_{3}^{2-}(aq)+2Sb(s)+2H_{2}O(l)$

2) $2H^{+}(aq)+2HNO_{2}(aq)+2I^{-}(aq)\rightarrow 2NO(g)+2H_{2}O(l)+I_{2}(s)$

3) $2OH^{-}(aq)+N_{2}O(g)+2ClO^{-}(aq)\rightarrow 2NO_{2}^{-}(aq)+H_{2}O(l)+2Cl^{-}(aq)$

4) $24OH^{-}(aq)+13Br_{2}(l)\rightarrow 4BrO_{3}^{-}(aq)+22Br^{-}(aq)+12H_{2}O(l)$

5) $3Cl(CH_{2})_{2}OH(aq)+2K_{2}Cr_{2}O_{7}(aq)+16H^{+}(aq)\rightarrow 3ClCH_{2}CO_{2}H(aq)+4Cr^{+3}(aq)+4K^{+}(aq)+11H_{2}O(l)$

### Q20.4.16

Write a balanced chemical equation for each redox reaction.

1. I(aq) + HClO2(aq) → IO3(aq) + Cl2(g) in acidic solution
2. Cr2+(aq) + O2(g) → Cr3+(aq) + H2O(l) in acidic solution
3. CrO2(aq) + ClO(aq) → CrO42−(aq) + Cl(aq) in basic solution
4. S(s) + HNO2(aq) → H2SO3(aq) + N2O(g) in acidic solution
5. F(CH2)2OH(aq) + K2Cr2O7(aq) → FCH2CO2H(aq) + Cr3+(aq) in acidic solution

### S20.4.16

General Rules to Follow:

STEP 1: To balance a redox reaction, you first must separate the half reactions. Focus on one at a time

STEP 2: Balance all elements that are not hydrogen and oxygen

STEP 3: Balance oxygen with water

STEP 4: Balance hydrogen with H+

STEP 5: Balance the charges by adding electrons to the more positive side

STEP 6: If balancing in basic solution, then balance H+ with OH- on both sides and combine to form water. Cancel water molecules accordingly

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms

1. I(aq) + HClO2(aq) → IO3(aq) + Cl2(g) in acidic solution

STEP 1: Separate the half reactions. Focus on one at a time

$I^{-}(aq)\rightarrow IO_{3}^{-}(aq)$

STEP 2: Balance all elements that are not hydrogen and oxygen. Note that iodine is balanced.

STEP 3: Balance oxygen with water

$3H_{2}O(l)+ I^{-}(aq)\rightarrow IO_{3}^{-}(aq)$

STEP 4: Balance hydrogen with H+
$3H_{2}O(l)+ I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)$

STEP 5: Balance the charges by adding electrons to the more positive side
$3H_{2}O(l)+ I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}$

STEP 6: Skip. We are balancing in acidic solution

Balance the other half reaction.

$HClO_{2}(aq)\rightarrow Cl_{2}(g)$

STEP 2: Balance all elements that are not hydrogen and oxygen

$2HClO_{2}(aq)\rightarrow Cl_{2}(g)$

STEP 3: Balance oxygen with water

$2HClO_{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)$

STEP 4: Balance hydrogen with H+

$6H^{+}(aq)+2HClO_{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)$

STEP 5: Balance the charges by adding electrons to the more positive side

$6e^{-}+6H^{+}(aq)+2HClO_{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)$

STEP 6: Skip. We are balancing in acidic solution

STEP 7: Balance the electrons in each half reaction and combine unlike terms while cancelling like terms. Notice that the electrons are already equal.

$3H_{2}O(l)+ I^{-}(aq)\rightarrow IO_{3}^{-}(aq)+6H^{+}(aq)+6e^{-}$

$6e^{-}+6H^{+}(aq)+2HClO_{2}(aq)\rightarrow Cl_{2}(g)+4H_{2}O(l)$

The overall reaction is

$I^{-}(aq)+2HClO_{2}(aq)\rightarrow IO_{3}^{-}(aq)+Cl_{2}(g)+H_{2}O(l)$

2. Cr2+(aq) + O2(g) → Cr3+(aq) + H2O(l) in acidic solution

Following STEPS 1-5, the half reactions are

$Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)+e^{-}$

$4e^{-}+4H^{+}(aq)+O_{2}(g)\rightarrow 2H_{2}O(l)$

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

$4(Cr^{2+}(aq)\rightarrow Cr^{3+}(aq)+e^{-})$

$4e^{-}+4H^{+}(aq)+O_{2}(g)\rightarrow 2H_{2}O(l)$

The overall reaction is:

$4H^{+}(aq)+O_{2}(g)+4Cr^{2+}(aq)\rightarrow 2H_{2}O(l)+4Cr^{3+}(aq)$

3. CrO2(aq) + ClO(aq) → CrO42−(aq) + Cl(aq) in basic solution

Following STEPS 1-6, the half reactions are

$2e^{-}+H_{2}O(l)+ClO^{-}(aq)\rightarrow Cl^{-}(aq)+2OH^{-}(aq)$
$4OH^{-}(aq)+CrO_{2}^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+2H_{2}O(l)+2e^{-}$

Notice that the electrons are already balanced. Following STEP 7, and combining the two half reactions, the overall reaction is

$2OH^{-}(aq)+CrO_{2}^{-}(aq)+ClO^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+H_{2}O(l)+Cl^{-}(aq)$

4. S(s) + HNO2(aq) → H2SO3(aq) + N2O(g) in acidic solution

Notice that sulfur is balanced, while there is one more nitrogen in the products. There are also one more hydrogen and two more oxygens in the products. If we take all of these into account, we'll notice that adding one more HNO2 balances out these differences.

The overall reaction is

$S(s) + 2HNO_{2}(aq)\rightarrow H_{2}SO_{3}(aq) + N_{2}O(g)$

5. F(CH2)2OH(aq) + K2Cr2O7(aq) → FCH2CO2H(aq) + Cr3+(aq) in acidic solution

Following STEPS 1-5, the half reactions are

$H_{2}O(l)+F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-}$

$6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{+3}(aq)+2K^{+}(aq)+7H_{2}O(l)$

Note that we're balancing in acidic solution. STEP 6 can be ignored.

Following STEP 7:

$3(H_{2}O(l)+F(CH_{2})_{2}OH(aq)\rightarrow FCH_{2}CO_{2}H(aq)+4H^{+}(aq)+4e^{-})$

$2(6e^{-}+14H^{+}(aq)+K_{2}Cr_{2}O_{7}(aq)\rightarrow 2Cr^{+3}(aq)+2K^{+}(aq)+7H_{2}O)(l)$

The overall reaction is

$3F(CH_{2})_{2}OH(aq)+2K_{2}Cr_{2}O_{7}(aq)+16H^{+}(aq)\rightarrow 3FCH_{2}CO_{2}H(aq)+4Cr^{+3}(aq)+4K^{+}(aq)+11H_{2}O(l)$

### A20.4.16

a)$$I^{-}(aq)+2HClO_{2}(aq)\rightarrow IO_{3}^{-}(aq)+Cl_{2}(g)+H_{2}O(l)$$

b) $$4H^{+}(aq)+O_{2}(g)+4Cr^{2+}(aq)\rightarrow 2H_{2}O(l)+4Cr^{3+}(aq)$$

c) $$2OH^{-}(aq)+CrO_{2}^{-}(aq)+ClO^{-}(aq)\rightarrow CrO_{4}^{2-}(aq)+H_{2}O(l)+Cl^{-}(aq)$$

d) $$S(s) + 2HNO_{2}(aq)\rightarrow H_{2}SO_{3}(aq) + N_{2}O(g)$$

e) $$3F(CH_{2})_{2}OH(aq)+2K_{2}Cr_{2}O_{7}(aq)+16H^{+}(aq)\rightarrow 3FCH_{2}CO_{2}H(aq)+4Cr^{+3}(aq)+4K^{+}(aq)+11H_{2}O(l)$$

### Q20.4.17

The standard cell potential for the oxidation of Pb to Pb2+ with the concomitant reduction of Cu+ to Cu is 0.39 V. You know that E° for the Pb2+/Pb couple is −0.13 V. What is E° for the Cu+/Cu couple?

### S20.4.17

$Pb \rightarrow Pb^{2+} + 2e^{-}$

And we know the above equation is oxidation and its charge is -0.13V

$Cu^{+} + e^- \rightarrow Cu$

And we know this is a reduction because it gains electrons and its charge is unknown; Y.

Now we must use the following equation:

$E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode}$

Now we write what we know:

$E^o_{Cell}= 0.39V$

And the anode is the oxidation portion so: $E^o_{Red,Anode}= -0.13V$

Finally we solve:

$0.39V= E^o_{Red,Cathode} - -0.13V$

$0.39V= E^o_{Red,Cathode} + 0.13V$

$0.39V- 0.13V= E^o_{Red,Cathode}$

$E^o_{Red,Cathode}= 0.26V$

$\dfrac{Cu^{+}}{Cu}= +0.26V$

+0.26V

### Q20.4.18

You have built a galvanic cell using an iron nail, a solution of FeCl2, and an SHE. When the cell is connected, you notice that the iron nail begins to corrode. What else do you observe? Under standard conditions, what is Ecell?

### S20.4.18

In addition to the corrosion of the iron nail due to oxidation, other observations would include the solution to bubble because of the standard hydrogen electrode, where $H_{2}^{+}+2e^{-}\rightarrow H_{}2$. Thus, the solution would form hydrogen gas, which causes the solution to bubble. SHE has an reduction potential of 0.0V and so all the other half reactions have standard hydrogen reduction potentials relative to hydrogen. Also, the $FeCl_{2}$ would disassociate in the solution forming $Cl^{-}$ and $Fe^{2+}$ ions. Therefore another observation would be that the solution would turn green because $Fe^{2+}$ ions produce a green color in an aqueous solution.

The E°cell of this reaction would be calculated using the formula E°cell =E°cathode - E°​​​​​​​anode

The cathode side is the reduction: $H_{2}^{+}+2e^{-}\rightarrow H_{}2$ E°​​​​​​​cathode = 0.00 V

The anode side is the oxidation: Fe(s) → Fe2+ + 2e- E°​​​​​​​anode = -0.44 V

E°​​​​​​​cell = 0.00 V + 0.44 V = 0.44V

### Q20.4.19

Carbon is used to reduce iron ore to metallic iron. The overall reaction is as follows:

$2Fe_2O_3 \cdot xH_2O(s) + 3C(s) → 4Fe(l) + 3CO_2(g) + 2xH_2O(g)$

Write the two half-reactions for this overall reaction.

### S20.4.19

1. Separate the half-reactions. The first step in determining the two half-reactions for the overall reaction is to identify which elements is oxidized and which element is reduced. The overall equation should be separated into two equations as shown below:

Reduction Half-Reaction: $2Fe_{2}O_{3}\cdot xH_{2}O(s)\rightarrow4Fe(l)+2xH_{2}O(g)$

The oxidation number of iron on the left side of the equation should be +3 because oxygen is more electronegative than iron, making its oxidation number -2. Since the charge of the overall molecule is neutral, iron must be +3 to balance the charge of the oxygen atoms. The oxidation number of iron on the right side of the equation should be 0 since it is a free element (uncombined state). The iron in this half-reaction is reduced since the oxidation number of iron decreases from +3 in 2Fe2O3 (reactants side) to 0 in 4Fe(l) (products side).

Oxidation Half-Reaction: $3C(s)\rightarrow3CO_{2}(g)$

The oxidation number of carbon on the left side should be 0 since it is a free element (uncombined state). The oxidation number of carbon on the right side should be +4 since oxygen is more electronegative than carbon, making its oxidation number -2. The charge of the overall molecule is neutral, so carbon must be +4 to balance the charge of the oxygen atoms. The carbon in this half-reaction is oxidized as the oxidation number of carbon increases from 0 in 3C(s) (reactants side) to +4 in 3CO2(g) (products side).

2. Balance elements other than oxygen (O) and hydrogen (H). In this equation, all elements other than O and H are balanced.

$2Fe_{2}O_{3}\cdot xH_{2}O(s)\rightarrow4Fe(l)+2xH_{2}O(g)$

$3C(s)\rightarrow3CO_{2}(g)$

3. Add H2O to balance oxygen. In the reduction half-reaction, 6 H2O molecules need to be added to the right side of the reaction to balance the oxygen molecules on the left side of the reaction. In the oxidation half-reaction, 6 H2O molecules must be added to the left side of the reaction to balance the oxygen molecules on the right side of the reaction.

Reduction Half-Reaction: $2Fe_{2}O_{3}\cdot xH_{2}O(s)\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)$

Oxidation Half-Reaction: $3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)$

4. Balance hydrogen by adding protons (H+). 12 protons need to be added to the left side of the iron reaction to balance the 16 (2 per water molecule * 8 water molecules) hydrogens. 12 protons need to be added to the right side of the carbon reaction.

Reduction Half-Reaction: $2Fe_{2}O_{3}\cdot xH_{2}O(s)+12H^{+}(aq)\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)$

Oxidation Half-Reaction: $3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)+12H^{+}(aq)$

5. Balance the charge of each equation with electrons. The iron reaction has a +12 charge on the left side of the reaction due to the 12 protons, while the other side has a neutral (0) charge, so 12 electrons must be added to the left side of the reaction to balance the charge to 0. The carbon reaction has a +12 on the right side of the reaction due to the 12 protons, while the other side has a neutral (0) charge, so 12 electrons must be added to the right side of the reaction to balance the charge to 0.

This is the final form of the two half-reactions for the reduction of iron ore to metallic ore by carbon:

Reduction Half-Reaction: $2Fe_{2}O_{3}\cdot xH_{2}O(s)+12H^{+}(aq)+12e^{-}\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)$

Oxidation Half-Reaction: $3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)+12H^{+}(aq)+12e^{-}$

### A20.4.19

Reduction Half-Reaction: $2Fe_{2}O_{3}\cdot xH_{2}O(s)+12H^{+}(aq)+12e^{-}\rightarrow4Fe(l)+2xH_{2}O(g)+6H_{2}O(l)$

Oxidation Half-Reaction: $3C(s)+6H_{2}O(l)\rightarrow3CO_{2}(g)+12H^{+}(aq)+12e^{-}$

### Q20.4.20

Will each reaction occur spontaneously under standard conditions?

1. Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
2. Zn2+(aq) + Pb(s) → Zn(s) + Pb2+(aq)

### S20.4.20

1. $$\mathrm{Cu(s)+2H^{+}(aq)⟶Cu^{2+}(aq)+H_2(g)}$$

Split into half reactions and identify each reactions standard reduction potential:

$$\mathrm{Cu(s)⟶Cu^{2+}(aq)}$$ (+0.34) and $$\mathrm{2H^{+}(aq)⟶H_2(g)}$$ (+0.00)

Because copper is losing electrons, it is the oxidation half reaction and is the anode. Hydrogen is gaining electrons in the reduction half reaction, making it the cathode.

Plug in $$ε^o_{Cathode,2H^{+}}={+0.34}$$ and $$ε^o_{Anode,Cu}={+0.00}$$ to $$ε^o_{Cell}= ε^o_{Cathode} - ε^o_{Anode}$$ to get $$ε^o_{Cell}= {+0.34} - {+0.00}= {+0.34}$$

Because the solution is positive, the reaction is spontaneous.

2. $$\mathrm{Zn^{2+}(aq)+Pb(s)⟶Zn(s)+Pb^{2+}(aq)}$$

Split into half reactions and identify each reactions standard reduction potential:

$$\mathrm{Zn^{2+}(aq)⟶Zn(s)}$$ and $$\mathrm{Pb(s)⟶+Pb^{2+}(aq)}$$

Because lead is losing electrons, it is the oxidation half reaction and is the anode. Zinc is gaining electrons in the reduction half reaction, making it the cathode.

Plug in $$ε^o_{Cathode,Zn}={-0.76}$$ and $$ε^o_{Anode,Pb}={-0.13}$$ to $$ε^o_{Cell}= ε^o_{Cathode} - ε^o_{Anode}$$ to get $$ε^o_{Cell}= {-0.76} - {-0.13}= {-0.63}$$

Because the solution is negative, the reaction is nonspontaneous.

a)

Spontaneous

b)

Non-spontaneous

### Q20.4.21

Each reaction takes place in acidic solution. Balance each reaction and then determine whether it occurs spontaneously as written under standard conditions.

1. Se(s) + Br2(l) → H2SeO3(aq) + Br(aq)
2. NO3(aq) + S(s) → HNO2(aq) + H2SO3(aq)
3. Fe3+(aq) + Cr3+(aq) → Fe2+(aq) + Cr2O72−(aq)

### S20.4.21

a.) $$Se(s)+Br_2(l)→H_2SeO_3(aq)+Br^−(aq)$$

Start out by writing out both the half reactions.

$$Br_2(l)\rightarrow Br^-(aq)$$

$$Se(s)\rightarrow H_2SeO_3(aq)$$

The oxidation state of bromine on the left side is 0. The oxidation state of bromine on the right is -1. Because the oxidation state of bromine decreases from 0 to -1, this is the reduction half-reaction (cathode). That means the selenium half reaction is the oxidation half reaction (anode).

Now we can balance both the half reactions.

$$cathode:$$ $$Br_2(l)\rightarrow Br^-(aq)$$

$$anode:$$ $$Se(s)\rightarrow H_2SeO_3(aq)$$

1. Balance all the elements in the equations except O and H.

$$Br_2(l)\rightarrow 2Br^-(aq)$$

$$Se(s)\rightarrow H_2SeO_3(aq)$$

2. Balance the oxygen atoms by adding H2O molecules to the opposite side of the equation.

$$Br_2(l)\rightarrow 2Br^-(aq)$$

$$3H_2O(l)+Se(s)\rightarrow H_2SeO_3(aq)$$

3. Balance the hydrogen atoms by adding H+ ions to the opposite side of the equation.

$$Br_2(l)\rightarrow 2Br^-(aq)$$

$$3H_2O(l)+Se(s)\rightarrow H_2SeO_3(aq)+4H^+(aq)$$

4. Add up the charges on each side. Add electrons (e-) to the more positive side if needed.

$$2e^-+Br_2(l)\rightarrow 2Br^-(aq)$$

$$3H_2O(l)+Se(s)\rightarrow H_2SeO_3(aq)+4H^+(aq)+4e^-$$

5. The electrons on each side must be made equal. If they are not, they need to be multiplied the lowest common multiple to be made the same.

$$2(2e^-+Br_2(l)\rightarrow 2Br^-(aq))$$

$$4e^-+2Br_2(l)\rightarrow 4Br^-(aq)$$

$$3H_2O(l)+Se(s)\rightarrow H_2SeO_3(aq)+4H^+(aq)+4e^-$$

6. Add the half reactions equations together. Common terms should be canceled out.

$$3H_2O(l)+Se(s)+2Br_2(l)\rightarrow H_2SeO_3(aq)+4H^+(aq)+4Br^-(aq)$$

Now we need to figure out if the reaction occurs spontaneously as written under standard conditions. We need to calculate the standard cell potential ($$E^\circ_{cell}$$). If $$E^\circ_{cell}$$ is positive ($$E^\circ_{cell}$$>0) then the reaction is spontaneous. If $$E^\circ_{cell}$$ is negative ($$E^\circ_{cell}$$<0) then the reaction is not spontaneous. We already determined which reactions occurred in the anode and cathode above. Now we need to find the standard reduction potentials of the anode and cathode using the standard reduction potentials chart.

$$cathode:$$ $$2e^-+Br_2(l)\rightarrow 2Br^-(aq)$$ $$E^\circ_{cathode}=1.066 V$$

$$anode:$$ $$3H_2O(l)+Se(s)\rightarrow H_2SeO_3(aq)+4H^+(aq)+4e^-$$ $$E^\circ_{anode}=0.74 V$$

Now we can calculate $$E^\circ_{cell}$$

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$

$$E^\circ_{cell}=1.066 V-0.74 V$$

$$E^\circ_{cell}=0.33 V$$

$$E^\circ_{cell}$$ is greater than zero, so the reaction is spontaneous.

b.) $$NO^−_3(aq)+S(s)→HNO_2(aq)+H_2SO_3(aq)$$

Start out by writing out both the half reactions.

$$NO^-_3(aq)\rightarrow HNO_2(aq)$$

$$S(s)\rightarrow H_2SO_3(aq)$$

Balance both the half reactions.

1. Balance all the elements in the equations except O and H.

$$NO^-_3(aq)\rightarrow HNO_2(aq)$$

$$S(s)\rightarrow H_2SO_3(aq)$$

2. Balance the oxygen atoms by adding H2O molecules to the opposite side of the equation.

$$NO^-_3(aq)\rightarrow HNO_2(aq)+H_2O(l)$$

$$3H_2O(l)+S(s)\rightarrow H_2SO_3(aq)$$

3. Balance the hydrogen atoms by adding H+ ions to the opposite side of the equation.

$$3H^+(aq)+NO^-_3(aq)\rightarrow HNO_2(aq)+H_2O(l)$$

$$3H_2O(l)+S(s)\rightarrow H_2SO_3(aq)+4H^+(aq)$$

4. Add up the charges on each side. Add electrons (e-) to the more positive side if needed.

$$2e^-+3H^+(aq)+NO^-_3(aq)\rightarrow HNO_2(aq)+H_2O(l)$$

$$3H_2O(l)+S(s)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-$$

5. The electrons on each side must be made equal. If they are not, they need to be multiplied the lowest common multiple to be made the same.

$$2(2e^-+3H^+(aq)+NO^-_3(aq)\rightarrow HNO_2(aq)+H_2O(l))$$

$$4e^-+6H^+(aq)+2NO^-_3(aq)\rightarrow 2HNO_2(aq)+2H_2O(l)$$

$$3H_2O(l)+S(s)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-$$

6. Add the half reactions equations together. Common terms should be canceled out.

$$2NO^-_3(aq)+H_2O(l)+2H^+(aq)+S(s)\rightarrow H_2SO_3(aq)+2HNO_2(aq)$$

Now we need to figure out if the reaction occurs spontaneously as written under standard conditions. We need to calculate the standard cell potential ($$E^\circ_{cell}$$). If $$E^\circ_{cell}$$ is positive ($$E^\circ_{cell}$$>0) then the reaction is spontaneous. If $$E^\circ_{cell}$$ is negative ($$E^\circ_{cell}$$<0) then the reaction is not spontaneous. Cathodes will usually have electrons on the left side of the equation and anodes will usually have electrons on the right side of the equation. Use the standard reduction potentials chart to find the corresponding potentials.

$$cathode:$$ $$2e^-+3H^+(aq)+NO^-_3(aq)\rightarrow HNO_2(aq)+H_2O(l)$$ $$E^\circ_{cathode}=0.934 V$$

$$anode:$$ $$3H_2O(l)+S(s)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-$$ $$E^\circ_{anode}=0.449 V$$

Now we can calculate $$E^\circ_{cell}$$

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$

$$E^\circ_{cell}=0.934 V-0.449 V$$

$$E^\circ_{cell}=0.485 V$$

$$E^\circ_{cell}$$ is greater than zero, so the reaction is spontaneous.

c.) $$Fe^{3+}(aq)+Cr^{3+}(aq)→Fe^{2+}(aq)+Cr_2O^{2−}_7(aq)$$

Start out by writing out both the half reactions.

$$Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$

$$Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)$$

The oxidation state of iron on the left side is +3. The oxidation state of iron on the right is +2. Because the oxidation state of iron decreases from +3 to +2, this is the reduction half-reaction (cathode). That means the chromium half reaction is the oxidation half reaction (anode).

Now we can balance both the half reactions.

$$cathode:$$ $$Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$

$$anode:$$ $$Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)$$

1. Balance all the elements in the equations except O and H.

$$Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$

$$2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)$$

2. Balance the oxygen atoms by adding H2O molecules to the opposite side of the equation.

$$Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$

$$7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)$$

3. Balance the hydrogen atoms by adding H+ ions to the opposite side of the equation.

$$Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$

$$7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)$$

4. Add up the charges on each side. Add electrons (e-) to the more positive side if needed.

$$e^-+Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$

$$7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)+6e^-$$

5. The electrons on each side must be made equal. If they are not, they need to be multiplied the lowest common multiple to be made the same.

$$6(e^-+Fe^{3+}(aq)\rightarrow Fe^{2+}(aq))$$

$$6e^-+6Fe^{3+}(aq)\rightarrow 6Fe^{2+}(aq)$$

$$7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)+6e^-$$

6. Add the half reactions equations together. Common terms should be canceled out.

$$7H_2O(l)+2Cr^{3+}(aq)+6Fe^{3+}(aq)\rightarrow 6Fe^{2+}(aq)+Cr_2O^{2-}_7(aq)+14H^+(aq)$$

Now we need to figure out if the reaction occurs spontaneously as written under standard conditions. We need to calculate the standard cell potential ($$E^\circ_{cell}$$). If $$E^\circ_{cell}$$ is positive ($$E^\circ_{cell}$$>0) then the reaction is spontaneous. If $$E^\circ_{cell}$$ is negative ($$E^\circ_{cell}$$<0) then the reaction is not spontaneous. We already determined which reactions occurred in the anode and cathode above. Now we need to find the standard reduction potentials of the anode and cathode using the standard reduction potentials chart.

$$cathode:$$ $$e^-+Fe^{3+}(aq)\rightarrow Fe^{2+}(aq)$$ $$E^\circ_{cathode}=0.771 V$$

$$anode:$$ $$7H_2O(l)+2Cr^{3+}(aq)\rightarrow Cr_2O^{2-}_7(aq)+14H^+(aq)+6e^-$$ $$E^\circ_{anode}=1.36 V$$

Now we can calculate $$E^\circ_{cell}$$

$$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$$

$$E^\circ_{cell}=0.771 V-1.36 V$$

$$E^\circ_{cell}=-0.59 V$$

$$E^\circ_{cell}$$ is less than zero, so the reaction is not spontaneous.

### A20.4.21

a)

$$3H_2O(l)+Se(s)+2Br_2(l)\rightarrow H_2SeO_3(aq)+4H^+(aq)+4Br^-(aq)$$ ; Spontaneous

b)

$$2NO^-_3(aq)+H_2O(l)+2H^+(aq)+S(s)\rightarrow H_2SO_3(aq)+2HNO_2(aq)$$ ; Spontaneous

c)

$$7H_2O(l)+2Cr^{3+}(aq)+6Fe^{3+}(aq)\rightarrow 6Fe^{2+}(aq)+Cr_2O^{2-}_7(aq)+14H^+(aq)$$ ; Non-Spontaneous

### Q20.4.22

Calculate E°cell and ΔG° for the redox reaction represented by the cell diagram Pt(s)∣Cl2(g, 1 atm)∥ZnCl2(aq, 1 M)∣Zn(s). Will this reaction occur spontaneously?

### S20.4.22

The standard cell potential for a redox reaction, E°cell , measures the tendency of reactants in their standard states to form products while in standard states. It's a measure of the driving force of the reaction, also known as voltage. Using the two standard electrode potentials represented by the Zn/Cl cell from the cell diagram above we can find the standard cell potentials from Table P1 and calculate the standard cell potential for the cell overall.

Pt(s)∣Cl2(g, 1 atm)∥ZnCl2(aq, 1 M)∣Zn(s)

We know that oxidation occurs at the anode, by definition, and reduction at the cathode. The left hand side of the cell diagram represents what happens at the anode whereas the right hand side represent the cathode. Thus knowing the values of E°anode for the reduction of Cl2 and E°cathode for the reduction of ZnCl2 using Table P1 we can calculate E°cell .

cathode: $Zn^{2+}\textit{(aq)}+2e^{-}\rightarrow Zn\textit{(s)}\; \; \; \; -0.763V$

anode: $Cl_{2}\textit{(g)}+2e^{-}\rightarrow 2Cl^{-}\textit{(aq)}\; \; \; \; 1.358V$

Thus E°cell = -0.763 - 1.358

= -2.121 V

Since E°cell is negative the reaction must be spontaneous, this is proven by

With this value we can now calculate ΔG° using

$\Delta G^{\circ} = -nFE^{\circ}_{cell}$

where n = 2 because we can see from the half reactions above 2 electrons were transferred. Thus ΔG° = 409289.4

Since ΔG° is greater than 0, the reaction is not spontaneous.

cell = -2.121 V

ΔG° = 409289.4

Not spontaneous

### Q20.4.23

If you place Zn-coated (galvanized) tacks in a glass and add an aqueous solution of iodine, the brown color of the iodine solution fades to a pale yellow. What has happened? Write the two half-reactions and the overall balanced chemical equation for this reaction. What is E°cell?

### S20.4.23

When adding an aqueous iodine solution to the zinc-coated tacks in a glass, the brown color of the iodine solution turns to a pale yellow because the iodine is reduced due to the presence of zinc. Zinc holds onto its electrons very weakly and as a result, when iodine is added, it releases them to form zinc ions and iodide ions.

$$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$$

$$I_2(aq) + 2e^- \rightarrow 2I^-(aq)$$

Overall reaction: $$Zn(s) + I_2(aq) \rightarrow Zn^{2+}(aq) + 2I^-(aq)$$

$$I_2$$ is reduced while $$Zn$$ is oxidized

1. To determine $$E_\left(cell\right)^°$$, look up the standard reduction potentials (SRP) for the reduction half-reaction and oxidation half-reaction.

$$E_\left(reduction of I_2\right) ^° = +0.54$$

$$E_\left(reduction of Zn^{2+}\right) ^° = -0.76$$

As you can see in the SRP table, even though we are given the equation for the reduction of Zn, $$Zn^{2+} + 2e^- \rightarrow Zn$$, the equation for $$E_\left(cell\right) ^°$$, $$E_\left(cell\right)^° = E_\left(cathode\right)^°- E_\left(anode\right)^°$$, accounts for the oxidation potential with the "-" sign. You can therefore plug these values into the equation directly without changing the sign of any numbers.

2. Plug in the numbers into the equation $$E_\left(cell\right)^° = E_\left(cathode\right)^°- E_\left(anode\right)^°$$ to find $$E_\left(cell\right)^°$$, where the $$E_\left(cell\right)^°$$ value for the reduction of $$I_2$$ corresponds to $$E_\left(cathode\right)^°$$ and the $$E_\left(cell\right)^°$$ for $$Zn$$ corresponds to $$E_\left(anode\right)^°$$.

$$E_\left(cell\right)^°= 0.54 - (-0.76) = 1.30V$$

$$E_\left(cell\right)^°$$ is 1.30V.

### A20.4.23

$$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$$

$$I_2(aq) + 2e^- \rightarrow 2I^-(aq)$$

Overall reaction: $$Zn(s) + I_2(aq) \rightarrow Zn^{2+}(aq) + 2I^-(aq)$$

$$E_\left(cell\right)^°$$ is 1.30V.

### Q20.4.24

Your lab partner wants to recover solid silver from silver chloride by using a 1.0 M solution of HCl and 1 atm H2 under standard conditions. Will this plan work?

### S20.4.24

To solve this problem we first have to find the overall equation:

${2}\textrm{AgCl} +\textrm{H}_2\rightarrow{2}\textrm{HCl} + {2}\textrm{Ag}$

From this reaction we know that H2 has to "kick out" silver from silver chloride for this reaction to work. We can see if H2 will kick out silver by looking at standard reduction potentials:

$\textrm{Ag}^+ +\textrm{e}^-\rightarrow\textrm{Ag}$ Eo=0.80V

${2}\textrm{H}^+ + {2}\textrm{e}^-\rightarrow\textrm{H}_2$ Eo= 0.00V

Now we have to assign the oxidation and reduction states for each reaction:

We do this by looking at the overall reaction:

Ag gains an electron and H loses an electron. Because of this the silver reaction is reduction and the H2 reaction is oxidation.

Using this information we can use:

Eocell = Eocathode - Eoanode,

Eocell = 0.800V - 0V

Eocell = +0.800V

Because Eocell is positive (Nernst Equation) the reaction will be spontaneous so the lab partner's method of recovering solid silver will work.

## 20.5: Free Energy and Redox Reactions

### Q20.5.1

State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in Zn(s) → Zn2+(aq) + 2e—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.

### S20.5.1

Disagree: Standard electrode potentials remain constant regardless of the quantity electrons transferred - there is no direct connection or proportionality between standard electrode potentials and quantity of electrons transferred and therefore, the standard electron potential is independent of this value. All that matters is that the standard reduction potential correlates to the correct metal ion (i.e. $$Zn^{2+}$$ ).

### Q20.5.2

What is the relationship between the measured cell potential and the total charge that passes through a cell? Which of these is dependent on concentration? Which is dependent on the identity of the oxidant or the reductant? Which is dependent on the number of electrons transferred?

### S20.5.2

The product of cell potential and the total charge that passes through a cell is the maximum amount of work that can be produced by an electrochemical cell. Both potential and total charge are dependent on concentration. Cell potential is dependent on the identity of the oxidant or the reductant. The total charge that passes through a cell is dependent on the number of electrons transferred.

### Q20.5.3

In the equation wmax = −nFE°cell, which quantities are extensive properties and which are intensive properties?

### S20.5.3

1) Know the definition of extensive properties and intensive properties first.

--> According to IUPAC, an intensive property is one whose magnitude is independent of the size of the system. An extensive property is one whose magnitude is additive for subsystems.

2) Examples of Intensive Properties:

Molality, Temperature, Pressure, Melting point and Boiling point, Color, etc.

3) Examples of Extensive Properties:

Mole, Entropy, Enthalpy, Mass, Volume

4) Apply concepts into the question:

--> We get E0cell as an intensive property and both n (mole) and Wmax as extensive properties.

A20.5.3

extensive: wmax and n; intensive: E°cell

### Q20.5.4

For any spontaneous redox reaction, E is positive. Use thermodynamic arguments to explain why this is true.

### S20.5.4

For a redox reaction to be spontaneous at a constant temperature and pressure, ΔG must be negative. Based on the thermodynamic equation $\Delta G=-nFE_{cell}$ n is the number of electrons transferred in the reaction, which is always a non-negative number. F is Faraday’s constant (96,485 C/mol e-), which is also nonnegative. Therefore, the only contribution to the sign of ΔG is E. If E is positive, ΔG will be negative, resulting in a spontaneous reaction. If E is negative, ΔG will be positive, resulting in a non-spontaneous reaction. Therefore, E must be positive to give a negative ΔG.

### Q20.5.5

State whether you agree or disagree with this statement and explain your answer: Electrochemical methods are especially useful in determining the reversibility or irreversibility of reactions that take place in a cell.

### Q20.5.6

Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?

### Q20.5.7

Occasionally, you will find high-quality electronic equipment that has its electronic components plated in gold. What is the advantage of this?

### S20.5.7

Gold is highly resistant to corrosion because of its very positive reduction potential.

### Q20.5.8

Blood analyzers, which measure pH, $$P_\mathrm{CO_2}$$ , and $$P_\mathrm{O_2}$$ , are frequently used in clinical emergencies. For example, blood $$P_\mathrm{CO_2}$$ is measured with a pH electrode covered with a plastic membrane that is permeable to CO2. Based on your knowledge of how electrodes function, explain how such an electrode might work. Hint: CO2(g) + H2O(l) → HCO3(aq) + H+(aq).

### Q20.5.9

Concentration cells contain the same species in solution in two different compartments. Explain what produces a voltage in a concentration cell. When does V = 0 in such a cell?

### Q20.5.10

Describe how an electrochemical cell can be used to measure the solubility of a sparingly soluble salt.

### Q20.5.11

The chemical equation for the combustion of butane is as follows:

$$\mathrm{C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow4CO_2(g)+5H_2O(g)}$$

This reaction has ΔH° = −2877 kJ/mol. Calculate E°cell and then determine ΔG°. Is this a spontaneous process? What is the change in entropy that accompanies this process at 298 K?

### A20.5.11

Since there is a circle next to delta H, E cell, etc., it means "standard," in that you have to find these values at standard conditions.

K (which is used to find ΔG° with the equation ΔG°=-RT*ln Keq) is 1 at standard conditions (since gas pressures = 1 bar and K= concentrations/gas pressures of products/ concentrations/gas pressures of reactants).

Plugging in K=1 into the equation ΔG°=-RT*ln Keq, we find that ΔG°=0. This means that the reaction is at equilibrium and is neither spontaneous nor non-spontaneous.

To calculate E°cell, use the equation: ΔG°=-nFE°cell, where n is the number of electrons and F is Faraday's constant. Since ΔG°=0, E°cell=0.

To find the change in entropy, or ΔS°, use the equation: ΔG°= ΔH° - TΔS°, where T is temperature in kelvin.

So, the equation (after plugging in given ΔH°, ΔG°, and T values) would be 0 = −2877 kJ/mol - 298*ΔS°. Solving for ΔS°, we get that the change in entropy= -9.65 kJ/mol, which is equivalent to about -9650 J/mol (technically -9654 but sig figs). ΔS° is usually written as J/mol rather than kJ/mol.

### Q20.5.12

How many electrons are transferred during the reaction Pb(s) + Hg2Cl2(s) → PbCl2(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg2Cl2 spontaneous? Calculate ΔG° for this reaction.

### Q20.5.13

For the cell represented as Al(s)∣Al3+(aq)∥Sn2+(aq), Sn4+(aq)∣Pt(s), how many electrons are transferred in the redox reaction? What is the standard cell potential? Is this a spontaneous process? What is ΔG°?

### S20.5.13

6e; E°cell = 1.813 V; the reaction is spontaneous; ΔG° = −525 kJ/mol Al.

### Q20.5.14

Calculate the pH of this cell constructed with the following half reactions when the potential is 0 at 25 °C

$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$

$Au^{3+} + 3e^- \rightarrow Au(s)$

Under this condition, the concentrations of other species in the cell are:

• 0.36 M: $$MnO_4^-$$
• 0.004 M: $$Au^{3+}$$
• 0.001 M: $$Mn^{2+}$$

### Q20.5.15

Based on Table 19.2 and Table P2, do you agree with the proposed potentials for the following half-reactions? Why or why not?

1. Cu2+(aq) + 2e → Cu(s), E° = 0.68 V
2. Ce4+(aq) + 4e → Ce(s), E° = −0.62 V

### Q20.5.16

For each reaction, calculate E°cell and then determine ΔG°. Indicate whether each reaction is spontaneous.

1. 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
2. K2S2O6(aq) + I2(s) → 2KI(aq) + 2K2SO4(aq)
3. Sn(s) + CuSO4(aq) → Cu(s) + SnSO4(aq)

### S20.5.16

Step 1: Determine which half of the reaction is the anode (the oxidation half) and which half of the reaction is the cathode (the reduction half) with their corresponding standard reduction potentials using a redox table.

Step 2: Determine $$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$$

Step 3: Use $$\Delta G^{\circ}=-nFE_{cell}^{\circ}$$ where n is the number of moles of electrons transferred (which can determine by balancing our half reactions) and F is Faraday's constant. If $$\Delta G^{\circ}$$ is negative, the reaction is spontaneous; if it is positive, it is non-spontaneous.

a) Oxidation: $$2Na(s) \to 2NaOH$$

$$Na_{(s)}\rightarrow Na^+_{(aq)}$$

$$Na_{(s)}\rightarrow Na^+_{(aq)} + e^-$$

$$Na^{+} + e^- \to Na(s)= -2.71V = E^o_{Anode}$$

Reduction: $$2H_{2}O(l) \to H_{2}(g)$$

$$2H_{2}O(l) + 2e^{-}\to H_{2}(g) + OH^{-}(aq) = -0.83V = E^o_{Cathode}$$

$E^o_{cell} = E^o_{Cathode} - E^o_{Anode}$

$-0.83V - (-2.71V) = 1.88V$

$To\ determine\ \Delta G^o, we\ use\ the\ formula\ \Delta G^o=-nFE_{cell}.$

First, we find the number of electrons transferred.

$2(Na\rightarrow Na^++e^-)$

$2H_{2}O(l) + 2e^{-}\to H_{2}(g) + OH^{-}(aq)$

$$2Na(s) + 2H_2O(l) ⟶ 2NaOH(aq) + H_2(g)$$

$$\,n\,=\,2$$

We then plug in the values into the formula.

$-2(96,485\ \frac{C}{mol})(1.88V)$

$=-363 KJ \therefore the\ reaction\ is\ spontaneous\ since\ \Delta G^o\ is\ negative.$

b) Oxidation:

$K_2S_2O_6 \to 2K^+ + 2SO_3^{2-}$

$2SO_4^{2-}(aq) + 2H_2O + 4e^- \rightarrow 2SO_3^{2-}(aq)+ 4OH^-$

$=-0.93V=E^{\circ}_{anode}$

Reduction:

$I_2\rightarrow I^-$

$I_2\rightarrow 2I^-$

$I_2+2e^-\rightarrow 2I^-$

$=0.54V=E^{\circ}_{cathode}$

$$E^°_{cell} = \ 0.535 \ V \ - (-.93) \ V) =1.465V$$

After balancing our half reactions (multiply reduction half reaction by 2) we see n = 4 as four electrons are transferred.

$\Delta G^o=-nFE_{cell}$

$-4(96,485\ \frac{c}{mol})(1.465V)$

$= -565 KJ \therefore the\ reaction\ is\ spontaneous\ because \Delta G^o\ is\ negative.$

c) Reduction:

$Cu^{+2}_{(aq)}\rightarrow Cu_{(s)}$

$Cu^{+2}_{(aq)} + 2e^-\rightarrow Cu_{(s)}$

$=0.337V$

Oxidation:

$Sn_{(s)}\rightarrow Sn^{2+}_{(aq)} + 2e^-$

$=-0.14V$

$E^o_{cell} = E^o_{Cathode} - E^o_{Anode}$

$0.337V - (-0.14V) = 0.477V$

$\Delta G^o=-nFE_{cell}$

We then plug in the values in the formula given that we know that the number of electrons used for this redox reaction is 2 electrons.

$-2(96,485\ \frac{c}{mol})(0.477V)$

$= -93 KJ \therefore the\ reaction\ is\ spontaneous\ because \Delta G^o\ is\ negative.$

### Q20.5.17

What is the standard change in free energy for the reaction between Ca2+ and Na(s) to give Ca(s) and Na+? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?

### S20.5.17

The standard reduction potentials of calcium and sodium differ depending on the source, but based on the Libretext chart found here, the E° for Ca is -2.868 V and E° for Na is -2.71 V. To determine ΔG°, we need to first calculate $$E°_{cell}$$.

$$Ca^{2+} (aq) + 2e^- → Ca(s)$$ E° = -2.868 V

$$Na^{+} (aq) + e^- → Na(s)$$ E° = -2.71 V

Anode: $$Na (s) → Na^{+} + e^-$$ E° = -2.71 V (note that the sign of E° does not change because it is an intensive property)

Cathode: $$Ca^{2+} (aq) + 2e^- → Ca(s)$$ E° = -2.868 V

$$E°_{cell} = E°_{cathode} - E°_{anode}$$

$$E°_{cell} = -2.868 V - (-2.71 V)$$ = - 0.158 V

$$ΔG°= - nFE°$$

$$ΔG°= - {2 mol e^-} \times {96485 C \over mol e^-} {- 0.158 V}$$

*Note that $$1 V= 1 {J \over C}$$

$$ΔG°= 30489.26 J = 30.489 kJ$$

The sign of ΔG° is positive, which indicates that it is nonspontaneous. This makes sense because their standard reduction potentials are close in value, since they are close together on the periodic table (sodium is an alkali metal, calcium is an alkaline earth metal), so the cell voltage $$E°_{cell}$$ is very small in value as well.

### Q20.5.18

In acidic solution, permanganate (MnO4) oxidizes Cl to chlorine gas, and MnO4 is reduced to Mn2+(aq).

1. Write the balanced chemical equation for this reaction.
2. Determine E°cell.
3. Calculate the equilibrium constant.

### S20.5.18

a. To write the balanced equation of this redox reaction you have to understand what a redox reaction is.

An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species.

Oxidation is the loss of electrons while reduction is gain of electrons.

From the information given, 2 half reactions can be determined:

MnO4- → Mn2+

Cl- → Cl2

For each half reaction, follow these steps to balance a redox reaction under acidic conditions:

1. Balance elements in the equation other than O and H.

MnO4- → Mn2+

2. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.

MnO4- → Mn2+ + 4H2O

3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.

8H+ + MnO4- → Mn2+ + 4H2O

4. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.)

5e- + 8H+ + MnO4- → Mn2+ + 4H2O

5. Balance elements in the equation other than O and H.

2Cl- → Cl2

6. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.

2Cl- → Cl2

(there are no other elements besides Cl here, so it can be left alone for this step)

7. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H+ ions to the opposite side of the equation.

2Cl- → Cl2

(there are no other elements besides Cl here, so it can be left alone for this step)

8. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side. (Rule of thumb: e- and H+ are almost always on the same side.)

2Cl- → Cl2 + 2e-

Now, looking at both balanced half reactions:

9. The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.

(5e- + 8H+ + MnO4- → Mn2+ + 4H2O) x 2

(2Cl- → Cl2 + 2e-) x 5

———————————————————————————

10e- + 16H+ + 2MnO4- → 2Mn2+ + 8H2O

10Cl- → 5Cl2 + 10e-

10. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.

10e- + 16H+ + 2MnO4- + 10Cl- → 2Mn2+ + 8H2O + 5Cl2 + 10e-

16H+ + 2MnO4- + 10Cl- → 2Mn2+ + 8H2O + 5Cl2

MnO4+ 8H+ + 5e ⇌ Mn2+ + 4H2O(l)

Cl2(g) + 2e ⇌ 2Cl

b. The standard cell potential (Eocell) is the difference of the potentials of the half reactions at 2 electrodes, which forms the voltage of that cell:

Eocell = Eocathode - Eoanode

The cathode is where reduction happens and the anode is where oxidation happens, so

5e- + 8H+ + MnO4- → Mn2+ + 4H2O is where reduction is happening, so it is the cathode

2Cl- → Cl2 + 2e- is where oxidation is happening, so it is the anode

The standard reduction potential values of the two half reactions are given in a table where:

Eo for MnO4+ 8H+ + 5e ⇌ Mn2+ + 4H2O(l) = 1.51 V

Eo for Cl2(g) + 2e ⇌ 2Cl = 1.358 V

———————————————————————————————

Eocell = 1.51V - 1.358V = 0.152 V

c. To calculate the equilibrium constant from the standard cell potential, you use the Nernst equation:

$E^{0} _{cell} = \dfrac{RT}{nF} ln K$

Here, the problem assumes standard state, so we know that R=(8.3145 J)/(mol*K), T=298 K, F=96485 J or 96485 J/(V*mol). Using this data we get (RT/F)=.025693 and substitute the calculated value of (RT/F) into RT/nF to get .025693/n. Now we get the new Nernst equation:

$E^{0} _{cell} = \dfrac{0.025693}{n} ln K$

where: "n" is the total number of electrons being transferred and "K" is the equilibrium constant.

Plugging in the values, we get:

The number of electrons transferred is found by looking at the balanced half reactions as shown above in part a, and so we see that 2 electrons are transferred.

$0.152 V = \dfrac{0.025693}{10} ln K$

K= 4.93*1025

### Q20.5.19

Potentiometric titrations are an efficient method for determining the endpoint of a redox titration. In such a titration, the potential of the solution is monitored as measured volumes of an oxidant or a reductant are added. Data for a typical titration, the potentiometric titration of Fe(II) with a 0.1 M solution of Ce(IV), are given in the following table. The starting potential has been arbitrarily set equal to zero because it is the change in potential with the addition of the oxidant that is important.

Titrant (mL) E (mV)
2.00 50
6.00 100
9.00 255
10.00 960
11.00 1325
12.00 1625
14.00 1875
1. Write the balanced chemical equation for the oxidation of Fe2+ by Ce4+.
2. Plot the data and then locate the endpoint.
3. How many millimoles of Fe2+ did the solution being titrated originally contain?

### S20.5.19

1. $\ce{Fe^2+}+ \ce{Ce^4+} \rightarrow \ce{Fe^3+} + \ce{Ce^3+}$

2. From the graph below, the endpoint can be determined. In potentiometric titrations, the endpoint is characterized by a steep increase in voltage. In this case, the greatest increase is seen between 9 and 10 mL, so the endpoint of this titration is approximately 9.5 mL.

3. This reaction took approximately 9.5 mL to reach the endpoint. Since the solution was 0.1 M in Ce4+, approximately 0.95 millimoles of Ce4+ were consumed during this reaction. From the balanced equation in part 1, it is known that the reaction of Fe2+ and Ce4+ occurs in a one-to-one ratio. So, therefore there were approximately 0.95 millimoles of Fe2+ in the original solution.

### Q20.5.20

The standard electrode potential (E°) for the half-reaction Ni2+(aq) + 2e → Ni(s) is −0.257 V. What pH is needed for this reaction to take place in the presence of 1.00 atm H2(g) as the reductant if [Ni2+] is 1.00 M?

### S20.5.20

We are given the reduction half reaction:

$$Ni^{2+}(aq)+{2e^-} \rightarrow {Ni(s)}$$

The question states that hydrogen gas is the reductant and that the pH of the solution is involved. This means our second half reaction is the oxidation reaction between hydrogen gas and hydronium ions, so the oxidation half reaction must be:

$${H_2(g)} \rightarrow {2H^+(aq)}+{2e^-}$$

The same amount of electrons are in each half reaction, so they cancel when we combine the two half reaction equations.

$${Ni^{2+}(aq)}+{H_2} \rightarrow {Ni(s)}+{2H^+}$$

The standard reduction potential for the reduction (cathode) half reaction is $$−0.257 V$$. By looking up the standard reduction potential for our oxidation (anode) half reaction in a chart of standard reduction potentials, we get a value of $$0.00 V$$. We can find our standard cell potential by subtracting the standard cell potential of the anode from the standard cell potential of the cathode:

$E^o_{cell}=E^o_{cathode}−E^o_{anode}$

$E^o_{cell}=−0.257 V − 0.00 V= −0.257 V$

The question asks "What pH is needed for this reaction to take place...?", which means our reaction will not be under standard conditions. We must use the Nernst equation to find the concentration of hydronium ions required for the reaction to be spontaneous. Since the reaction occurs at 298K, we can use the simplified version of the Nernst equation where $$R$$, $$T$$ and $$F$$ are simplified to $$0.0592 V$$ and $$\ln{Q}$$ becomes $$\log{Q}$$. A reaction is spontaneous when $$E_{cell} > 0$$.

$E_{cell}={E^o_{cell}} − \frac{0.0592 V}{n} \log{Q}$

$0<{−0.257} − \frac{0.0592}{n} \log{Q}$

In our reaction, $$n=2$$ because there are two mols of electrons transferred for every mol of a product consumed (see half reactions). Based on the law of mass action, $$Q$$ for our reaction will be:

$Q=\frac{[H^+]^2}{(1 atm) [1 M]}$

$Q={[H^+]^2}$

The Nernst equation becomes:

$0<{−0.257} − \frac{0.0592}{2} \log{[H^+]^2}$

$0<{−0.257} − 0.0296 \log{[H^+]^2}$

We can now solve for the concentration of hydronium ions:

$\frac{−0.257}{0.0296} > \log{[H^+]^2}$

$exp(-8.68) > {[H^+]^2}$

$\sqrt{exp(-8.68)} > [H^+]$

${[H^+]} < 4.56 \times 10^{-5}$

We now can find an inequality for the pH:

$pH=−\log{[H^+]}$

$−\log{[H^+]} > −\log{4.56 \times 10^{-5}}$

$pH > 4.34$

A pH greater than 4.34 is needed for the reaction to take place at 25°C. This means the reaction favors basic conditions. We can check this answer by plugging in pH values greater than 4.34 to the Nernst equation. If our answer is right, the Nernst equation will yield a positive cell potential.

As a student, it is always in your best interest to check your answers; let us take a hypothetical pH value of 6 and see if the above answer is correct.

-0.257-(0.0592/2)log((10-6)2 = 0.0982. A positive E value indicated that the reaction is spontaneous.

### Q20.5.21

The reduction of Mn(VII) to Mn(s) by H2(g) proceeds in five steps that can be readily followed by changes in the color of the solution. Here is the redox chemistry:

1. MnO4−(aq) + e → MnO42−(aq); E° = +0.56 V (purple → dark green)
2. MnO42−(aq) + 2e + 4H+(aq) → MnO2(s); E° = +2.26 V (dark green → dark brown solid)
3. MnO2(s) + e + 4H+(aq) → Mn3+(aq); E° = +0.95 V (dark brown solid → red-violet)
4. Mn3+(aq) + e → Mn2+(aq); E° = +1.51 V (red-violet → pale pink)
5. Mn2+(aq) + 2e → Mn(s); E° = −1.18 V (pale pink → colorless)
6. Is the reduction of MnO4 to Mn3+(aq) by H2(g) spontaneous under standard conditions? What is E°cell?
7. Is the reduction of Mn3+(aq) to Mn(s) by H2(g) spontaneous under standard conditions? What is E°cell?

### S20.5.21

Ecell= Ecathode-Eanode

If E°cell is positive, the reaction will be spontaneous because it implies that there is a release of Gibb's free energy, or a -∆G.

For the reduction of MnO4-

cell=0.56V+2.26V+0.95V=+3.77V

Because E°cell is positive, the reduction is spontaneous.

For the reduction of Mn3+

cell=1.51V-1.18V=+.33V

Both reactions are spontaneous.

### Q20.5.22

Mn(III) can disproportionate (both oxidize and reduce itself) by means of the following half-reactions:

Mn3+(aq) + e → Mn2+(aq) E°=1.51 V
Mn3+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + e E°=0.95 V
1. What is E° for the disproportionation reaction?
2. Is disproportionation more or less thermodynamically favored at low pH than at pH 7.0? Explain your answer.
3. How could you prevent the disproportionation reaction from occurring?

### S20.5.22

1. First, we must know our equation for the E°. It is as follows: E°Cell = E°Cathode - E°Anode. We know that reduction occurs at the cathode and oxidation occurs at the anode. In the following equations, we must decide which one is the reduction half-reaction and which one is the oxidation half-reaction.

Mn3+(aq) + e→ Mn2+(aq) In this equation Mn3+ is gaining an electron therefore it is being reduced therefore this is the reduction half-reaction which occurs at the cathode.

Mn3+(aq) + 2H2O(l) → MnO2(s) + 4H+(aq) + eThe Oxidation Number of Mn in the MnO2 molecule is +4. The Oxidation Number of Mn went from +3 to +4 therefore it is being oxidized. Therefore this is the oxidation half-reaction that occurs at the anode.

Now, we take the corresponding E° values and plug them into E°Cell = E°Cathode - E°Anode in their corresponding spots. Therefore, our equation is as follows:

Cell = 1.51V-0.95V=0.56V. This is the E° value for disproportionation reaction.

2. A low pH will have more H+ ions than a pH of 7 becuase pH is a negative logarithm. This will increase Q in the Nernst equation (E=Eo−(0.0592V/n)logQ) thus the second term will increase. This second term is being subtracted from the E° term, therefore the E term will decrease. Becuase this is happening in the oxidation half-reaction, EAnode will decrease, therefore ECell will increase. A more positive ECell corresponds to a more negative ΔG. A more negative ΔG corresponds to a more thermodynamically favored process. Therefore, at a lower pH, the disproportionation will be more thermodynamically favored.

3. One way to prevent disproportionation from occurring is by removing water from the reaction.

### Q20.5.23

For the reduction of oxygen to water, E° = 1.23 V. What is the potential for this half-reaction at pH 7.00? What is the potential in a 0.85 M solution of NaOH?

### S20.5.23

In order to determine the potential at pH 7.00, it would be helpful to write the balanced half reaction.

Step 1. Balanced reaction

$$\\\mathrm{O_2}(g)\longrightarrow\mathrm{H_2O}(l)$$

Add $$\mathrm{H_2O}$$ to the right side in order to balance the oxygen.

$$\\\mathrm{O_2}(g)\longrightarrow\mathrm{H_2O}(l)+\mathrm{H_2O}(l)$$

Then add $$\mathrm{H^+}$$ to the left side to balance hydrogen

$$\\\mathrm{4H^+}(aq)+\mathrm{O_2}(g)\longrightarrow\mathrm{H_2O}(l)+\mathrm{H_2O}(l)$$

The total charge of the left side is +4 ( $$\mathrm{O_2}$$ has a 0 charge, there are 4 protons with a total of +4 charge) , while the charge of the right side is 0 (water has a neutral charge). To balance the charges, add 4 electrons to the left side.

$$\\\mathrm{4e^-}+\mathrm{4H^+}(aq)+\mathrm{O_2}(g)\longrightarrow\mathrm{2H_2O}(l)$$

From this equation we can see the reactants and the products which would help us with calculating the reaction quotient, $$\mathit{Q}$$, which we will plug in the Nernst equation.

Step 2. Find $$\mathit{Q}$$.

$$\\\mathit{Q}=\mathrm{\dfrac{1}{P_{O_2}[H^+]^4}}$$

In this equation, the numerator (product) is has an activity of 1 because water is in its liquid state. The denominator is the activity of $$\mathrm{O_2}$$ (given by the pressure of oxygen because it is a gas) and the activity (concentration) of the protons. Note: the concentration of the proton is raised to the power of 4 because of its coefficient.

$$\\\mathit{Q}=\mathrm{\dfrac{1}{1\,atm\,\,[1\times10^{-7}\, M]^4}}$$

Here, we plugged in values. Assuming, it is in standard conditions, the pressure of oxygen is 1 atm. Given that the pH is 7.00, we can calculate the $$\mathrm{H^+}$$ concentration using $$\mathrm{[H^+]} = \mathrm{10^{-pH}}$$.

Simply the equation to get:

$$\\\mathit{Q}=\mathrm{1\times10^{28}}$$

Now we plug in $$Q$$, $$n$$ (number of electrons), and the given $$E^o$$ into the Nernst equation

$$E = E^o - \dfrac{0.0592\, V}{n} \log Q$$

$$E = 1.23\,V - \dfrac{0.0592\, V}{4} \log (1\times10^{28})$$

Simplify and get:

$$E = 1.23\,V- \dfrac{0.0592\, V}{4} (28)$$

$$E = 1.23\,V - 0.414\,V$$

$$E = 1.23\,V - 0.414\,V$$

$$E = 0.816\,V$$

The potential for this half reaction is 0.816 V at a pH of 7.00

In a 0.85 M NaOH solution:

In order to solve this problem, we must figure out the pH of the solution. Given that the concentration of NaOH is 0.85 M, and NaOH is a strong base, the hydroxide ion concentration is also 0.85 M. Since we want the pH, we need to convert the concentration into pOH first.

To find the pOH, use $$\mathrm{pOH=-log[OH^-]}$$

$$\mathrm{pOH=-log[0.85]}$$

$$\mathrm{pOH=0.07058}$$

Now that we have pOH, we can determine pH by subtracting pOH from 14 (pOH+pH=14).

$$\mathrm{pH=14-0.0758}$$

$$\mathrm{pH=13.929}$$

Now, find the $$H^+$$ concentration using the pH (use $$\mathrm{[H^+]} = \mathrm{10^{-pH}}$$ ).

$$\mathrm{[H^+]} = \mathrm{10^{-13.929}}$$

$$\mathrm{[H^+]} = 1.1766\times10^{-14}$$

We can now plug in this value to find $$Q$$.

$$\\\mathit{Q}=\mathrm{\dfrac{1}{1\,atm\,\,[1.1765\times10^{-14}\,M]^4}}$$

$$\\\mathit{Q}=5.1999\times10^{55}$$

Now plug this into the Nernst equation

$$E = 1.23\,V - \dfrac{0.0592\, V}{4} \log (5.1999\times10^{55})$$

$$E = 1.23\,V - \dfrac{0.0592\, V}{4} (55.716)$$

$$E = 1.23\,V - 0.8246\,V$$

$$E = 0.405\,V$$

In 0.85 M of NaOH, the cell potential is 0.41 V.

yes; E° = 0.41 V

### Q20.5.24

The biological molecule abbreviated as NADH (reduced nicotinamide adenine dinucleotide) can be formed by reduction of NAD+ (nicotinamide adenine dinucleotide) via the half-reaction NAD+ + H+ + 2e → NADH; E° = −0.32 V.

1. Would NADH be able to reduce acetate to pyruvate?
2. Would NADH be able to reduce pyruvate to lactate?
3. What potential is needed to convert acetate to lactate?
acetate + CO2 + 2H+ +2e → pyruvate +H2O E° = −0.70 V
pyruvate + 2H+ + 2e → lactate E° = −0.185 V

### S20.5.24

1. In order to start the problem, you need to flip the reaction for NADH. In doing so, the signs for E° switches and becomes positive indicating a spontaneous reaction.

$$NADH \to (NAD^{+}+H^{+}+2e^{−} ; E° = 0.32 V$$

The next step to take is to add the two E° (both the NADH and acetate) to check if they will create a spontaneous reaction.

$$0.32 V (NADH) + -0.70 V (acetate) = -0.38 V$$

Since the reaction's cell potential is negative, it indicates that the reaction is not spontaneous. Therefore, NADH will not be able to reduce acetate to pyruvate.

2. The same applies to the second part except this time, the E° of the NADH and pyruvate are added together.

$$0.32 V (NADH) + -0.185 V (acetate) = 0.135 V$$

Since this yields a positive value, that means the reaction is spontaneous. Therefore, NADH will be able to reduce pyruvate to lactate.

3. What potential is needed to convert acetate to lactate?

First, find E°acetate→lactate by subtracting the larger reduction potential by the smaller reduction potential:

acetate→lactate= E°lactate-E°acetate= -0.185 - (-0.70)= 0.52 V

a) No

B) Yes

C) 0.52 V

### Q20.5.25

Given the following biologically relevant half-reactions, will FAD (flavin adenine dinucleotide), a molecule used to transfer electrons whose reduced form is FADH2, be an effective oxidant for the conversion of acetaldehyde to acetate at pH 4.00?

acetate + 2H+ +2e → acetaldehyde + H2O E° = −0.58 V

### S20.5.25

(Oxidation) Acetate + 2H+ + 2e- —► Acetaldehyde + H2O E° = −0.58 V

(Reduction) FAD + 2H+ + 2e- —► FADH2 E° = −0.18 V

Using the $$\Delta E° = E°_{cathode} - E°_{anode}$$ formula, we can calculate the E°cell. Then we can plug in the E°cell value into the $$\Delta G°= -nFE°_{cell}$$ formula to determine the spontaneity of the mechanism. If the reaction is spontaneous, then FAD would be an effective oxidant. Because the equation full equation does not depend on Hydrogen ion concentration, the voltage remains the same at acidic concentrations.

$$\Delta E° = E°_{cathode} - E°_{anode} = -0.18V - (-0.58V)$$

= + 0.40V

$$\Delta G°= -nFE°_{cell}$$

ΔG° = -(2 mols of electrons transferred) (96,485 C/mol) (+0.40 V)

= -77188 J

Since ΔG° for this reaction is a negative value, this reaction is spontaneous. Therefore, FAD is an effective oxidant for the conversion of Acetaldehyde to Acetate at pH 4.00.

Yes.

### Q20.5.26

Ideally, any half-reaction with E° > 1.23 V will oxidize water as a result of the following half-reaction:

$O_2(g) + 4H^+(aq) + 4e^− → 2H_2O(l)$

1. Will $$FeO_4^{2−}$$ oxidize water if the half-reaction for the reduction of Fe(VI) → Fe(III) is FeO42−(aq) + 8H+(aq) + 3e → Fe3+(aq) + 4H2O; E° = 1.9 V?
2. What is the highest pH at which this reaction will proceed spontaneously if [Fe3+] = [FeO42−] = 1.0 M and $$P_\mathrm{O_2}$$= 1.0 atm?

### S20.5.26

1. Since the FeO42- reaction, with a Eo of 1.9 V, satisfies the requirement that E° > 1.23 V, FeO42- will oxidize water.

2. The first step of determining this is to write the full balanced equation. It should first be noted that the O2 equation needs to be an oxidation reaction, whereas right now it shows reduction. Turning it around, it looks like this: 2H2O(l) →O2(g) + 4H+(aq) + 4e. However, we cannot combine the equations yet because the electrons are not balanced; O2 loses 4 electrons, while FeO42- only gains three. Thus we must multiply the O2 equation by three and the FeO42- equation by four.

4FeO42−(aq) + 32H+(aq) + 12e + 6H2O(l)→ 4Fe3+(aq) + 16H2O(l) + 3O2(g) + 12H+(aq) + 12 e-

The electrons will cancel, but so will everything else in red; some simple subtraction gives us the final full equation:

4FeO42−(aq) + 20H+(aq) +→ 4Fe3+(aq) + 10H2O(l) + 3O2(g)

For determining the highest pH we will use the following equation: $$E_{cell} = E_{cell}^o - \dfrac{.0592 \, V}{n} \ln Q$$

We need to determine Eocell, Q, and n. Eocell is determined by subtracting the cathode Eo from the anode Eo. The cathode is where the reduction reaction happens, which in this case is the FeO4- reaction, and the anode is where oxidation happens, which is the O2 reaction. So Eocell = 1.9-1.23 = .67 V.

n is the number of electrons transferred; looking back at when we balanced our equation, this equals 12.

Q is the reaction quotient, equal to the concentration of the products over the concentration of the reactants with their coefficients as exponents. Working with the numbers given us ([Fe3+] = [FeO42−] = 1.0 M and PO2= 1.0 atm), we can create this equation: $$Q = \dfrac{[\ce{O_2}]^3[\ce{Fe^{3+}}]^4}{[\ce{H^+}]^{20}[\ce{FeO_4^{2-}}]^4}$$ $$Q = \dfrac{[1 M]^3[1 M]^4}{[\ce{H^+}]^{20}[\ce{1 atm}]^4} = \dfrac{1}{[\ce{H^+}]^{20}}$$

The entire point of this problem is to solve for [H+] when Ecell=0, determining what the lowest concentration of H+ can be with the reaction still being spontaneous, Ecell being positive. So we can plug in the values that we have and solve for [H+]: $$0 \lt .67 - \dfrac{.0592}{12} \ln \dfrac{1}{[\ce{H^+}]^{20}}$$ $$-.67 \lt - \dfrac{.0592}{12} \ln \dfrac{1}{[\ce{H^+}]^{20}}$$ $$-.67\left(\dfrac{12}{.0592}\right) \lt -\ln \dfrac{1}{[\ce{H^+}]^{20}}$$

Using a few logarithm laws, we can manipulate the natural log to make solving for H+ easier: $$-.67\left(\dfrac{12}{.0592}\right) \lt \ln {[\ce{H^+}]^{20}}$$ $$-135.811 \lt 20\ln {[\ce{H^+}]}$$ $$e^\left({\dfrac{-135.811}{20}}\right) \lt [\ce{H^+}]$$ $$[\ce{H^+}] \gt .0011243$$

This equation proves, then, that the pH needs to be lower than -log(.0011243)=2.95 for this reaction to be spontaneous under the given conditions.

a) Yes

b) 2.95

### Q20.5.27

Under acidic conditions, ideally any half-reaction with E° > 1.23 V will oxidize water via the reaction

$2H_2O(l) → O_2(g) + 4H^+(aq) + 4e^−.$

• Will aqueous acidic KMnO4 evolve oxygen with the formation of MnO2?
• At pH 14.00, what is E° for the oxidation of water by aqueous KMnO4 (1 M) with the formation of MnO2?
• At pH 14.00, will water be oxidized if you are trying to form MnO2 from MnO42− via the reaction 2MnO42−(aq) + 2H2O(l) → 2MnO2(s) + O2(g) + 4OH(aq)?

### S20.5.27

Begin by writing the two half reactions

O2(g) + 4H+(aq) + 4e → 2H2O(l)

MnO4- + 4H+ + 3e--->MnO2 + 2H2O(l)

Combine the two equations.

-We want the MnO4 to be the reactant

2H2O(l) --> O2(g) + 4H+(aq) + 4e Flipped Around

MnO4- + 4H+ + 3e--->MnO2 + 2H2O

-Sum the two equations, making sure to multiply the equations such that the number of electrons are the same

3(2H2O(l) --> O2(g) + 4H+(aq) + 4e)

2(MnO4- + 4H+ + 3e--->MnO2 + 2H2O)

6H2O + 4MnO2 + 16H+ --> 3O2 + 12H+ + 4MnO2 + 8H2O

-Simplify

4MnO2 + 4H+ --> 3O2 + 4MnO2 + 2H2O

• Because oxygen is a product of this reaction, acidic KMnO4 will evolve oxygen.

2:

Begin by writing the two half reactions

MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l) 1.68V

O2 + 4H+ +4e- --> 2H2O 1.23V

Determine which will be oxidized and which will be reduced

-Use the standard cell potential, E°cell = E°(cathode) - E°(anode) , to determine which is cathode and which is anode

-A spontaneous reaction would have an E°cell >0

1.68V - (-1.23V) = 2.91V

1.23V - (1.68V) = -0.45V

-Therefore MnO4- will be the cathode, the species that will be reduced and O2 will be the anode, the species that is oxidized.

Combine the two equations so that MnO4- is reduced and O2 is oxidized

MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l) 1.68V

O2 + 4H+ +4e- --> 2H2O 1.23V

-Flip the O2 equation

MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l) 1.68V

2H2O --> O2 + 4H+ +4e- -1.23V

-Manipulate both equations so that they have the same number of electrons

4(MnO4- + 4H+ + 3e---> MnO2 + 2H2O(l)) 1.68V

3(2H2O --> O2 + 4H+ +4e-) -1.23V

-Combine them and simplify

4MnO4- + 4H+ --> 4MnO2 + 2H2O + 3O2

Calculate the E°cell potential

-Use the standard cell potential, E°cell = E°(cathode) - E°(anode).

E°cell = 1.68V - (-1.23V) = 2.91V

Use the Nernst Equation to calculate Ecell @ pH 14

Q = product concentration over reactant concentration

$E =\frac{8.3145*298}{12*96485.33}*ln\frac{1}{10^{-14}}$

E = 2.63V

• Because E is positive, the reaction is spontaneous at pH 14.

3:

Begin by writing the three half reactions and potentials at standard.

MnO4- + 4H+ + 3e--->MnO2 + 2H2O 1.68V

MnO42- --> MnO4- + e- -0.56V

2H2O --> O2 + 4H+ +4e- -1.23V

Sum the first two equations without eliminating the electrons

MnO42- --> 4H+ + 2e- --> MnO2 + 2H2O 2.252V

Sum the third equation

MnO42- --> 4H+ + 2e- --> MnO2 + 2H2O 2.252V

2H2O --> O2 + 4H+ +4e- -1.23V

2MnO42- + 4H+ --> O2 + 2MnO2 + 2H2O 3.482V

Compensate for basic reaction

2MnO42- + 4H+ --> O2 + 2MnO2 + 2H2O 3.482V

-Add OH- to both sides of the equation and simplify

2MnO42- + 2H2O --> O2 + 2MnO2 + 4OH- 3.482V

Use Nernst equation to determine E @ pH 14

-Where E is the final cell potential, E° is the cell potential at standard conditions, R is the ideal gas constant, T is the temperature in K, n is the number of electrons exchanged, F is Faraday's constant and Q is concentrations in:

2MnO42- + 4H+ --> O2 + 2MnO2 + 2H2O 3.482V

$E =\3.482 - frac{8.3145*298}{4*96485.33}*ln\frac{1}{(10^-14)}$

E = 3.482

• Therefore water will be oxidized because in the equation, it is already shown to donate its electrons, and the reaction is spontaneous at pH 14.

a) Yes

b) 2.91 V

c) Yes

### Q20.5.28

Complexing agents can bind to metals and result in the net stabilization of the complexed species. What is the net thermodynamic stabilization energy that results from using CN as a complexing agent for Mn3+/Mn2+?

Mn3+(aq) + e → Mn2+(aq) E° = 1.51 V
Mn(CN)63−(aq) + e → Mn(CN)64− E° = −0.24 V

### A20.5.28

Since the process is spontaneous we know that $$\text{E}^°_\text{cell}$$ must be positive. For this to be true we use the equation

$\text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode}$

and substitute the given standard reduction potentials into the appropriate places.

$\text{E}^°_\text{cell} = 1.51\text{ V} - (-0.24)\text{ V} =1.75 \text{ V}$

We can find the net thermodynamic stabilization energy since it is synonymous with the change in free energy ΔG of the system by the equation

$\text{ΔG} = -\text{nF}\text{E}^°_\text{cell}$

Where "n" is the number of moles electrons transferred in the overall reaction and "F" is Faraday's constant "F" is 96486 J/(V $$\times$$ mol of e-).

$\text{ΔG} = -(1\text{ mol}\text{ of}\text{ e}^-)\times\frac{96486\text{ J}}{\text{V}\times\text{mol}\text{ of}\text{ e}^-}\times1.75 \text{ V}$

$\text{ΔG} = -168850.5\text{ J} \approx -1.69\times10^5\text{ J}$

The net thermodynamic stabilization energy is about -1.69 $$\times$$ 105 J or -169 kJ.

### Q20.5.29

You have constructed a cell with zinc and lead amalgam electrodes described by the cell diagram Zn(Hg)(s)∣Zn(NO3)2(aq)∥Pb(NO3)2(aq)∣Pb(Hg)(s). If you vary the concentration of Zn(NO3)2 and measure the potential at different concentrations, you obtain the following data:

Zn(NO3)2 (M) Ecell (V)
0.0005 0.7398
0.002 0.7221
0.01 0.7014
1. Write the half-reactions that occur in this cell.
2. What is the overall redox reaction?
3. What is E°cell? What is ΔG° for the overall reaction?
4. What is the equilibrium constant for this redox reaction?

### S20.5.29

1.) Based on the cell diagram, the left side of the cell notation is the oxidation half reaction, and the right side of the cell notation is the reduction half reaction. Thus:

Oxidation Half Reaction: $$Zn(s)+2NO_3^{_{-}}(aq)\rightarrow Zn(NO_3)_2(aq)+2e^{_{-}}$$

Reduction Half Reaction: $$Pb(NO_3)_2(aq) + 2e^{_{-}}\rightarrow Pb(s)+2NO_3^{_{-}}(aq)$$

2.) To get the overall redox reaction, we need to balance both half equations, than add them together, cancelling other common elements. Detailed procedure can be found at S17.2.3

Overall redox reaction: $$Zn(s)+(Pb(NO_3)_2(aq)\rightarrow Zn(NO_3)_2(aq)+Pb(s)$$

3.)

$$E°cell=E°cathode−E°anode$$

Substitute the value of standard reduction potential of E°cathode (reduction half reaction) and the value of standard reduction potential of E°anode (oxidation half reaction) base on the Standard Reduction Potential graph.( See S20.3.16)

$E°cell=-0.13V - (-0.79V) = 0.63V$

ΔG°=-nFE°. n is the number of electron transferred in overall reaction. F, the Faraday constant, equals to 96485 C/mole. E° is the cell potential.

In this problem, n=2, and E°cell =0.63V.

$ΔG°=-nFE°= - 2 mole* 96485 C/mole * 0.63 V = -121571 J=-1.2*10^{_{5}}J.$

4.) ΔG°=-RTIn(k). R, the idea gas constant, equals to 8.31453 J/(mole*K). T is the temperature in Kelvin. K is the equilibrium constant.

In this problem, ΔG°=-1.2*105J. As T is not given, we assumed at the reaction is under standard condition, in which T= 298K.

$ΔG°=RTInk$

$-1.2*10^{_{5}}J = -8.31453*298* \natural(k)$

$k = 1.1*10^{_{21}}$

### Q20.5.30

Hydrogen gas reduces Ni2+ according to the following reaction: Ni2+(aq) + H2(g) → Ni(s) + 2H+(aq); E°cell = −0.25 V; ΔH = 54 kJ/mol.

1. What is K for this redox reaction?
2. Is this reaction likely to occur?
3. What conditions can be changed to increase the likelihood that the reaction will occur as written?
4. Is the reaction more likely to occur at higher or lower pH?

### S20.5.30

1. Step 1 Split into its half-equations

Ni2+(aq)+2e-→ Ni(s) (reduction)

H2(g)→2H+(aq)+2e- (oxidation)

Step 2 Find Δ Go using Δ Go=-nFEocell

n=2 mol e-

F=96,485 C/mol

Δ Go=-(2 mol e-)(96,485 kJ/V·mol)(-0.25V)

Δ Go=48242.5 J/mol

Step 3 Find K using Δ Go=-RTlnK

R= 8.3145 J /mol·K

T= temperature not given so we use room temp, 298 K

48242.5 J/mol=-(8.3145 J /mol·K)(298K)lnK

48242.5 J/mol=-2477.721 J/mol (lnK)

-19.47=lnK

e-19.47=K

K=3.5 X 10-9

2. Because we found out that Δ Go was positive, we know that the reaction is non-spontaneous. Therefore we can say the reaction is unlikely to occur.

3. Given that ΔH = 54 kJ/mol is positive we know the reaction is endothermic, meaning that heat is on the reactants side. If we were to increase the temperature it would result in the reaction favoring the products since raising temperature would add more reactants and by Le Chatelier's Principle we know that increasing reactants will push a reaction toward the product's side.

4. This reaction will more likely occur in a low PH because that would mean we have a lower H+ concentration. Since H+ is a product in this reaction, lowering its concentration would result in the reaction moving forward because again we look at Le Chatelier's Principle which tells us that when we reduce the amount of products in a reaction, the reaction will move forward to favor those products and in our case make the reaction more likely to happen.

### Q20.5.31

The silver–>silver bromide electrode has a standard potential of 0.07133 V. What is Ksp of AgBr?

### S20.5.31

First, we must find the standard reduction potentials of each side of the reaction. We are given that this reaction is done with a Silver to Silver Bromide electrode. We can find these following equations in a standard reduction potential chart:

$$Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$$ $$E^{o}=0.8 V$$

$$AgBr(s) + e^{-} \rightarrow Ag(s) + Br^{-}(aq)$$ $$E^{o}= 0.07133 V$$

We also need to know how $$AgBr$$ dissolves and divides into its ions:

$AgBr(s) \rightarrow Ag^{+}(aq) + Br^{-}(aq)$

$K_{sp}= [Ag^{+}][Br^{-}]$

We want to manipulate our two equations to get it in the form of our dissolving equation. First, we must flip our first equation so that we have the Ag+ ion on the correct side. As we flip our standard reduction equation, we must also flip our E° value.

$$Ag(s) \rightarrow Ag^{+}(aq) + e^{-}$$ $$E^{o}=-0.8 V$$

To find our E°cell value, we must add our two E° values according to the following equation.

$E^{o}_{cell}= E^{o}_{cathode} + E^{o}_{anode}$

$E^{o}_{cell}= 0.01733V + (-0.8V)$

$E^{o}_{cell}= -0.7287 V$
Now that we have our Eºcell value, we can use the Nernst Equation to find our K value, which will resemble Ksp in this case.

$E = E^{o}cell - \frac{0.0591}{n}logK$

We can plug in our $$E^{o}$$ value into the equation as well as our number of electrons. In this case it is one electron.

$0 = -0.7287 - \frac{0.0591}{1}logK$

We can add -0.7287 to the other side.
$0.7287 = -0.0591logK$

We divide 0.0591 onto the other side.
$\frac{0.7287}{-0.0591} = log K$

Solve the left side of the equation.
$-12.33 = log K$

Raise each side to the power of 10 to cancel out the log.
$10^{-12.33} = K$

Solve.

$K_{sp}= 4.7 * 10^{-13}$

## 20.6: Cell EMF Under Nonstandard Conditions

Problems folded into 20.5. Must tease out

## 20.7: Batteries and Fuel Cells

### Q20.7.1

What advantage is there to using an alkaline battery rather than a Leclanché dry cell?

### S20.7.1

Before analyzing the advantage of these two types of batteries, we must first understand the three different types of batteries:

1. Primary batteries. These are the common types of batteries that come to mind when people think of batteries. As a primary cell is used, chemical reactions in the battery use up the chemicals that generate the power, which is the reason that these batteries are not rechargeable. These batteries make up 90% of the market.

2. Secondary batteries. This is an electrical battery that can be charged, discharged into a load, and recharged many times. It is composed of one or more electrochemical cells and are typically more expensive than primary batteries. An example is a lead-acid storage battery in cars.

3. Fuel cells. These are not "true" batteries because they are not self-contained but they are nonetheless added as a third category.

Now that we know that both are primary batteries, we narrow our comparison to within this first category:

a. Leclanché cells are just another word for the cells used in the common dry-cell, zinc-carbon battery. This cell is made up of an outer zinc container which is the "anode," while the cathode is a central carbon rod.

• cathode (reduction):

$2MnO_{2(s)} + 2NH^+_{4(aq)} + 2e^− \rightarrow Mn_2O_{3(s)} + 2NH_{3(aq)} + H_2O_{(l)} \label{Eq1}$

• anode (oxidation):

$Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^− \label{Eq2}$

The Zn(s) oxidized to form Zn2+ ions at the anode and these ions react with NH3 formed at the cathode and Cl- ions present in solution.

• overall:

$2MnO_{2(s)} + 2NH_4Cl_{(aq)} + Zn_{(s)} \rightarrow Mn_2O_{3(s)} + Zn(NH_3)_2Cl_{2(s)} + H_2O_{(l)} \label{Eq3}$

The dry cell is inexpensive to manufacture but it is not efficient in producing electrical energy because only the relatively small fraction of the MnO2 that is near the cathode is reduced and only a small fraction of the zinc cathode is consumed as the cell discharges. Dry cells leak more easily than an alkaline cell. In addition, dry cells have a limited shelf life because the Zn(s) anode reacts spontaneously with NH4Cl in the electrolyte, and dry cells cannot work properly in all conditions for example in cold temperatures the efficiency of the cell decereases immensly.

Diagram of a Leclanché cell:

b. Alkaline batteries

The alkaline battery is just a Leclanché cell that can operate under alkaline (basic) conditions. The half-reactions that occur in an alkaline battery are:

• cathode (reduction)

$2MnO_{2(s)} + H_2O_{(l)} + 2e^− \rightarrow Mn_2O_{3(s)} + 2OH^−_{(aq)} \label{Eq4}$

• anode (oxidation):

$Zn_{(s)} + 2OH^−_{(aq)} \rightarrow ZnO_{(s)} + H_2O_{(l)} + 2e^− \label{Eq5}$

• overall:

$Zn_{(s)} + 2MnO_{2(s)} \rightarrow ZnO_{(s)} + Mn_2O_{3(s)} \label{Eq6}$

This battery works better at low temperatures compared to the Leclanché cell which is why some prefer it even though it might be heavier than other batteries. The alkaline battery is also better for the environment because it doesn't cause as much destruction when thrown away. In addition, the alkaline cell has twice the amount of energy density as the Leclanché, allowing it to produce the same amount of voltage while having a longer shelf life.

Diagram of the alkaline cell:

### A20.7.1

An alkaline battery is a Leclanché cell that can operate under alkaline conditions. The Alkaline cell has twice the energy density of a Leclanché cell, which allows it to produce the same voltage while having a longer shelf life and better performance. Although the Leclanché cell is inexpensive to manufacture, it corrodes easier as opposed to the Alkaline cell, which, unused, could last up to seven years and does not run out of power easily. The Alkaline cell also works well at low temperature compared to the Leclanché cell. The Alkaline battery may be heavier than other batteries but it is better for the environment and does not post serious health complications.

### Q20.7.2

Why does the density of the fluid in lead–acid batteries drop when the battery is discharged?

### S20.7.2

In a lead-acid battery, two types of lead are acted upon electrochemically by an electrolytic solution. When a lead-acid battery is discharged, the electrolyte divides into H2 and SO4.

The mechanics of a lead-acid battery relies on the following redox reaction:

${Pb}(s)+{PbO_2}(s)+{2H_2SO_4}(aq)→{2PbSO_4}(s)+{2H_2O}(l)$

Two lead plates, one coated with lead dioxide, act as the electrodes in the battery along with sulfuric acid and water. As the battery is discharged, the lead dioxide-coated cathode facilitates the reduction reaction with the ionized sulfuric acid and hydrogen ions to produce solid lead sulfate and water, whilst the other lead anode facilitates the oxidation reaction with ionized sulfuric acid to produce hydrogen ions. The density of 98% sulfuric acid (1.84 g/cm3) is higher than that of water (1.00 g/cm3), so as the battery is discharged removing sulfuric acid and producing water, the density of the fluid in these batteries will begin to decrease and approach 1.00 g/cm3.

1.00 g/cm3

### Q20.7.3

What type of battery would you use for each application and why?

1. powering an electric motor scooter
2. a backup battery for a smartphone
3. powering an iPod

### S20.7.3

1. For an electric motor scooter, you would want a battery that you can recharge so you that you can use up the electricity in the scooter and recharge it for further use. Thus, you would want a secondary battery for this purpose. An example that we've learnt for this purpose is a Lead-Acid (storage) battery which can also be found in cars but it can also be used for an electric motor scooter as well. Secondary batteries can be recharged easillly unlike primary batteries and is therefore cost and energy efficient. Moreover it is not as harmful to the environment as primary cells.
2. Assuming this is an external battery (would be hard to replace the battery yourself...), you would want to have a primary battery that cannot be recharged as this would have a single use. However, there are three primary batteries that could be used:
• Lechanche dry cell (alkali battery)
• although this dry cell may be inexpensive to manufacture, it only produces about 1.55V and is not efficient in producing electrical energy as only a small amount is produced. Moreover, there is a limited shelf life where the dry cell may corrode and have its contents leak out. You would want a battery that is able to be stored until needed so the dry cell does not fulfill this category.
• Lithium iodine battery
• this cell type is the best for this purpose as these batteries can be long-lived and are reliable. Thus, they are the most common type of battery where you would not want to replace the battery. They are also portable yet powerful and can therefore be used to power such devices.
3. For the use of an iPod, you would want a rechargeable battery as the battery would be drained after each use, however, you would want to recharge it. Thus, having a secondary battery would be the most useful for this purpose. Examples that we have learnt could be the Nickel-Cadmium (NiCad) battery, Nickel-metal hydride battery (NiMH), and Lead-Acid (Lead storage) battery. Howeover, the Lead-Acid (lead storage) battery is most commonly used in cars and is far too large for a small ipod so the NiCad and NiMH would be preferred for thus purpose.

### Q20.7.3

2. lithium–iodine battery
3. NiCad, NiMH, or lithium ion battery (rechargeable)

### Q20.7.4

Why are galvanic cells used as batteries and fuel cells? What is the difference between a battery and a fuel cell? What is the advantage to using highly concentrated or solid reactants in a battery?

### S20.7.4

Galvanic cells are used to harness the energy (to create a flow of electrons) of spontaneous redox reactions. In order to do so, we have to separate the chemicals of the half-reactions, the anode (where oxidation happens) and the cathode (where reduction happens) which is joined by an apparatus, salt bridge, that allows the electrons to flow from the anode to the cathode (Figure 20.7.4).

This is why galvanic cells are used as batteries and fuel cells: the spontaneous redox reaction supply energy which is used to perform work.

A battery cell and a fuel cell are both galvanic cells, however, a battery cell has all the reactants necessary to produce electricity. On the other hand, a fuel cell requires constant external supply of one or more reactants to generate electricity. It is not used as a storage device of electrical energy.

The advantage to using highly concentrated or solid reactants in a battery is that the concentration of the reactants and products don't change greatly as the battery is discharged. As a result during the discharge process, output voltage pretty much stays constant.

### Q20.7.5

This reaction is characteristic of a lead storage battery:

Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

If you have a battery with an electrolyte that has a density of 1.15 g/cm3 and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?

### S20.7.5

Let's assume that there is one liter of the electrolyte solution. Using the density, convert this volume into mass (use conversion factors $$(\frac{1cm^3}{1mL})$$ and $$(\frac{1000mL}{1L})$$).

$$(1 L\ soln.)(\frac{1.15g}{1cm^3})(\frac{1cm^3}{1mL})(\frac{1000mL}{1L})=1150g\ soln.$$

The solution contains 30.0% sulfuric acid by mass.

$$(1150g\ soln.)(0.300)=345g\ H_{2}SO_{4}$$

Find the moles of sulfuric acid in solution (molar mass $$H_{2}SO_{4}=98.079g/mol$$).

$$(345g\ H_{2}SO_{4})(\frac{1mol\ H_{2}SO_{4}}{98.079g\ H_{2}SO_{4}})=3.52mol\ H_{2}SO_{4}$$

Since we assumed there was only one liter of electrolyte solution, the molarity is $$[H_{2}SO_{4}] = 3.52M$$.
Looking at the reaction $$Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)→2PbSO_{4}(s)+2H_{2}O(l)$$, $$H_{2}SO_{4}$$ is the only species in solution, so the reaction quotient equation is

$$Q=\frac{[products]}{[reactants]}=\frac{1}{[H_{2}SO_{4}]^2}=\frac{1}{(3.52M)^2}=0.0807$$

Using the Nearst equation, if we plug in $$lnQ=ln(0.0807)=-2.52$$,

$$E=E^{\circ}-\frac{RT}{nF}lnQ$$
$$E=E^{\circ}-\frac{RT}{nF}(-2.52)=E^{\circ}+2.52(\frac{RT}{nF})$$

$$\frac{RT}{nF}$$ will always be positive since it's a ratio of positive constants, temperature, and moles of electrons. Therefore, $$E>E^{\circ}$$.The potential is greater than the standard cell because the increment of product (when calculated, $$H_2SO_4$$ is 3.52M, which is higher than the standard condition 1M), according to Le Chatelier's principle, will cause the reaction to drive forward. This will cause the reaction to become more spontaneous, which is seen as the E cell potential is greater than the standard E cell.

### A20.7.5

1. [H2SO4] = 3.52 M; E > E°

## 20.8: Corrosion

### Q20.8.1

Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?

### S20.8.1

You would expect a bent nail to corrode more rapidly than a straight nail because corrosion tends to begin at an area in which the metal is under stress, for example where the nail is bent. Also, when you bend a nail, the sacrificial plating on the outside of the nail may be cracked when bent, therefore speeding up the oxidation of the metal and the corrosion process.

It is also possible that bending the nail would increase the surface area of the nail by stretching it. Creating more surface area on which oxidation can take place.

### Q20.8.2

What does it mean when a metal is described as being coated with a sacrificial layer? Is this different from galvanic protection?

### S20.8.2

Corrosion is the unwanted oxidation of a material. This is a natural process which converts a refined metal to a more chemically stable form. Both sacrificial layers and galvanic protection prevent corrosion. One can prevent corrosion by adding a protective layer. A layer of zinc can be added to steel to prevent corrosion. This is considered an example of galvanic protection. A sacrificial layer is essentially the same thing. When a metal is coated with a sacrificial layer, it simply means that the sacrificial layer is more likely to be oxidized, and thus will protect whatever it is covering.

### Q20.8.3

Why is it important for automobile manufacturers to apply paint to the metal surface of a car? Why is this process particularly important for vehicles in northern climates, where salt is used on icy roads?

### S20.8.3

The paint that is applied to the metal surface of a car is important because it prevents corrosion by preventing oxygen and water from coming into direct contact with the car's metal surface. Corrosion is a naturally occurring process in which metals are oxidized in the presence of moisture or air. In northern climates, the salt used on the icy roads lowers the freezing point of water, and so the roads won't be as icy and skidding is less likely to happen. Because of this, paint is even more necessary in these climates because salt is an electrolyte that increases water's conductivity, facilitating the flow of electric current between cathodic and anodic sites. So without the paint, the rate of corrosion of the metal on the car's surface would increase. The paint therefore forms a protective layer on top of the metal, keeping the oxygen and water from directly touching the metal and causing its corrosion.

## 20.9: Electrolysis

### Q20.9.1

Why might an electrochemical reaction that is thermodynamically favored require an overvoltage to occur?

### S20.9.1

Overvoltage occurs when there is a requirement for additional voltage to be added to an electrolytic cell in order for the rate of reaction to occur as it would in an ideal setting. Although a reaction may already be thermodynamically favored prior to the additional voltage, there a few reasons that it may still be necessary. First, although a reaction may be thermodynamically favored and thus exothermic, it may be advancing at a less than optimal rate due to high activation energy. Overvoltage can aid in lowering the activation energy, and thus increasing the rate, all while the reaction remains thermodynamically favored. Additionally, in some reactions in which gases are produced, the flow of electrons is slowed by the gases being formed. Once again, overvoltage can alleviate the slower rate even though the reaction was already favorable.

### Q20.9.2

How could you use an electrolytic cell to make quantitative comparisons of the strengths of various oxidants and reductants?

### S20.9.2

If we set one molecule (in our case we used the reduction of hydrogen) as a baseline zero we can begin to compare the corresponding voltage.

In example:

$Zn(s) + 2 H+(aq) \rightarrow Zn2+(aq) + H2(g)$

The reactions are:

$2H^{+}(aq) + 2e^{-} \rightarrow H_2(g)$

$Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}$

The voltmeter should read around +.76V meaning we must apply .76 V to the system. This gives the baseline value of Zn oxidation of .76V and its reduction would be the opposite -.76V.

If we have a voltmeter we can quantitatively track how many volts are given off by the electrolysis or how many volts are needed to push the reaction forward. Each volt value is used as quantitative comparison between various elements. If we continue to compare values in the electrolysis with hydrogen then we will have a quantitative scale comparing the results. And that is what we currently have and its called a Table of Standard Reduction Potentials which compares all the values to hydrogen and hydrogen is the baseline of 0.0V.

### S20.9.2

If we know the stoichiometry of an electrolysis or electrochemical reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction, or in other words, we can calculate the quantity of material that is oxidized or reduced at an electrode. We can determine strength of oxidants and reductants by calculating the number of moles of electrons transferred when a known current is passed through a cell. We can also use the stoichiometry of the reaction and the total charge transferred to calculate the amount of product formed or the amount of metal deposited in an electroplating process.

### Q20.9.3

Why are mixtures of molten salts, rather than a pure salt, generally used during electrolysis?

### S20.9.3

Mixtures of molten salts are used over pure salts during electrolysis because they have good electric conductivity, a higher heat capacity, better thermal conductivity and can act as a solvent. The pure salts have higher melting points alone, but when they are mixed, their combined melting point (known as the eutectic point) is lower. This helps the reaction move forward, making electrolysis easier.

### Q20.9.4

Two solutions, one containing Fe(NO3)2·6H2O and the other containing the same molar concentration of Fe(NO3)3·6H2O, were electrolyzed under identical conditions. Which solution produced the most metal? Justify your answer.

### S20.9.4

1. To find which solution produces the most metal, you must first split the two compounds into their respective half reactions that form once they are electrolyzed.

For Fe(NO3)2⋅6H2O, the resulting half reaction would be: $$Fe^{2+}(aq)+2e^{-} \rightarrow Fe(s)$$

For Fe(NO3)3⋅6H2O, the resulting half reaction would be: $$Fe^{3+}(aq)+3e^{-} \rightarrow Fe(s)$$

2. The fact that both compounds were electrolyzed under identical conditions indicates that the same amount of electrons were transferred. This means that:

If a certain amount of electrons were transferred during the electrolysis of Fe(NO3)2·6H2O, then that specific amount of electrons divided by 2 will provide you with the moles of iron metal produced. The same is true for Fe(NO3)3·6H2O. If a certain amount of electrons were transferred during the electrolysis of Fe(NO3)3·6H2O, then that amount of electrons divided by 3 will provide you with the moles of iron metal produced. Thus, Fe(NO3)2·6H2O will produce the most metal.

### Q20.9.5

The electrolysis of molten salts is frequently used in industry to obtain pure metals. How many grams of metal are deposited from these salts for each mole of electrons?

1. AlCl3
2. MgCl2
3. FeCl3

### S20.9.5

1.) The problem asks for how much metal is plated. So we are looking at what happens when solid Al is formed (we do not consider chlorine because it is not a solid; but we do use to to find the charge on the Al ion as 3+). So we are looking at what happens when aluminum is plated which is given by this reaction $Al^{3+}(aq)+3{e^-}→Al(s)$ All we are doing here is converting moles to grams per electron via how many moles are given in the balanced half reaction. So here for every one mole of aluminum plated, there are three moles of electrons. We can then use the molar mass of aluminum to find grams of metal: Our work follows: $\frac{1mol \ Al}{3 mol \ e{^-}} \ \frac{26.981539g \ Al}{1mol \ Al}= \ 8.99g \ per \ mol \ e{^-}$

2.) We use the same concept here but the reaction this time is: $Mg^{2+}(aq)+2{e^-}→Mg(s).$ And we have two electrons per mole Magnesium. SO we do:

$\frac{1mol \ Mg}{2 mol \ e{^-}} \ \frac{24.305g \ Mg}{1mol \ Mg}= \ 12.2g \ per \ mol \ e{^-}$

3.) For this problem we only have three moles of electrons per iron: $Fe^{3+}(aq)+3{e^-}→Fe(s).$ So, we follow the procedure for one but for iron this time: $\frac{1mol \ Fe}{3 mol \ e{^-}} \ \frac{55.845g \ Fe}{1mol \ Fe}= \ 18.6g \ per \ mol \ e{^-}$

a)

$8.99g \ per \ mol \ e{^-}$

b)

$12.2g \ per \ mol \ e{^-}$

c)

$18.6g \ per \ mol \ e{^-}$

### Q20.9.6

Electrolysis is the most direct way of recovering a metal from its ores. However, the Na+(aq)/Na(s), Mg2+(aq)/Mg(s), and Al3+(aq)/Al(s) couples all have standard electrode potentials (E°) more negative than the reduction potential of water at pH 7.0 (−0.42 V), indicating that these metals can never be obtained by electrolysis of aqueous solutions of their salts. Why? What reaction would occur instead?

### S20.9.6

Electrolysis is the use of an electric current to stimulate a non-spontaneous reaction. Electrolysis can never occur with the stated conditions because the salts, Na+ and Cl-, are not better oxidizing agents than H2O. You can see this by viewing a table of standard reduction potentials. Sodium, Magnesium, and Aluminum have lower values than water. As a result, H2O will be reduced instead of the metals. The only way electrolysis can take place is when the elements are less reactive (better oxidizing agents) than water.

### Q20.9.7

What volume of chlorine gas at standard temperature and pressure is evolved when a solution of MgCl2 is electrolyzed using a current of 12.4 A for 1.0 h?

### S20.9.7

Reaction: $$MgCl_{2}+2e^{-}\rightarrow Mg+Cl_{2}$$

Equations: $$n_{e}=\frac{It}{F}$$ and PV=nRT

Variables:

• $$n_{e}$$ is moles of electrons
• I is the current which is given to be 12.4A
• t is the time in seconds which is $$1.0 hr\frac{3600 seconds}{1 hr}$$
• F is Faradays constant which equal 96485 J*mol-2*V-1
• Standard Pressure and Temperature: T=273K P=1 atm
• R is a constant which equals 0.0821 L*atm*mol-1*K-1

Solve: $$n_{e}$$=$$\frac{12.4A*3600s}{96485 J*V^{-1}*mol^{-1}}$$=0.463 moles of electrons

Now convert moles of electrons to moles of chlorine

$$0.463 moles of e^{-1}*\frac{1 mol Cl_{2}}{2 moles of e^{-1}}$$=0.231 moles of Cl2

Now use the ideal gas equation to solve for moles of chlorine.

$V=\frac{nRT}{P}=\frac{0.231*0.0821*273}{1}=5.2 L$

5.2 L

### Q20.9.8

What mass of copper metal is deposited if a 5.12 A current is passed through a Cu(NO3)2 solution for 1.5 h.

### S20.9.8

1. Write out the chemical equation you will need for this problem. In this case, since we are working with copper, the equation will be

$$Cu^{2+} + 2e^- \rightarrow Cu$$

2. We now know that for every 2 moles of electrons transferred, we get 1 mole of copper. We now need to plug our given values into the following equation:

$n_e = \frac{I×t}{F}$

Where ne is the total number of moles of electrons transferred, I is the current in amps, t is the time in seconds and F is Faraday's constant (96485 coulombs/mol of electrons)

$$(5.12 A) \times (1.5hrs) \times \left(\frac{3600s}{1hr}\right) \times \left(\frac{1 mol\,e^-}{96485 C}\right) = 0.2866 mol$$

3. Finally, we convert moles of electrons to moles of copper by dividing by 2 according to our half equation above, then convert moles of copper to grams of copper using stoiciometry:

$$(0.2866 mol) \times \left(\frac{1 mol Cu(NO_3)_2}{2 mol\,e^-}\right) \times \left(\frac{1molCu}{1molCu(NO_3)_2}\right) \times \left(\frac{63.546gCu}{1 mol Cu}\right) = 9.105g\,of \,Cu$$

### Q20.9.9

What mass of PbO2 is reduced when a current of 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery?

### S20.9.9

Because this is a battery there is a redox equation going on inside. By looking at the standard reduction potential table we get:

$Pb^2+ +2e^- \rightarrow Pb(s)$

We look at the lead part of the reaction because it is being reduced in PbO2.

To solve this problem we need to use this equation:

$\textrm{n}=\dfrac{\textrm{I x t}}{\textrm{F}}$

I= current (amps), t= time (seconds), F= Faraday’s constant (9.65x104 C/mole), n= # of moles of e-'s

An amp is a rate of Columbus per second.

This equation allows us to convert current to moles of electrons from there we can convert moles of electrons to mass.

By looking at the problem we see that I=5.0 A, T=2.0 h and F= 9.65x104 C/mole we are trying to find n.

We first have to convert time into seconds from hours we do this by converting hours to minutes then minutes into seconds:

$2\text{ hours}\times\frac{60\text { minutes}}{1\text{ hour}}\times\frac{60\text { seconds}}{1\text{ minute}} = 7200\text{ seconds}$

Next we "plug and chug" by plugging in all the known numbers into the equation:

$\text{ n}\ =\frac{5\text { A}\times 7200\text { seconds}}{96500 \text{ C/mole}}$

$\text{ n}\ = .373 \text{ moles of electrons}$

Because 2 electrons are transferred in the equation above we have to divide .373 by 2.

$\dfrac{.373\text{ moles of electrons}}{2\text{ electrons transferred}} = .1865\text{ moles}$

Now that we have the moles of electrons we can convert from moles to grams.

We do this by multiplying the number of moles by the mass of PbO2.

$.1865\text{ moles}\times\dfrac{239.2\text{ grams}}{1\text{ mole}} = 44.61\text{ grams}$

44.61 grams of PbO2 is reduced when 5.0 A is withdrawn over a period of 2.0 h from a lead storage battery.

### A20.9.9

44.61 grams of PbO2 is reduced.

### Q20.9.10

Electrolysis of Cr3+(aq) produces Cr2+(aq). If you had 500 mL of a 0.15 M solution of Cr3+(aq), how long would it take to reduce the Cr3+ to Cr2+ using a 0.158 A current?

### S20.9.10

In order to find the time it would take to reduce Cr3+ to Cr2+ using a 0.158 A current, we can use the following equation:

$$n_{e}=\frac{It}{F}$$

Where t is the time in seconds, F is Faraday's constant, I is the current, and ne is the number of electrons transferred. We can find the number of electrons transferred by using the following equation

$$Cr^{3+} + e^{-} \to Cr^{2+}$$

and by finding out how many mols of $$Cr^{3+}$$ we started with by using dimensional analysis

$$M=\frac{moles}{Volume}$$

$$moles=MV$$

$$moles=0.15\frac{moles}{L}0.5L$$

$$0.075\,moles\,of\,Cr^{3+}$$

We see that the moles of Cr3+ and the moles of e- is a one to one ratio so we have

$$0.075\,moles\,of\,e^{-} = n_{e}$$

now that we have all of our data we plug them into the equation solving for time (t)

$$t=\frac{n_{e}F}{I}$$

$$t=\frac{0.075\,moles\,of\,e^{-} \cdot 96485 \frac{C}{mol\, e^{-}}}{0.158 A}$$

Since 1A = 1 Coulumb (C), all of the units cancel out, leaving us with

$$t=45799.8s$$

So it would take about 45800 seconds to reduce the Cr3+ to Cr2+ using a 0.158 A current

We can convert this time into hours to better understand the length of this process.

$$45799.8s \times \frac{1 \, s}{3600 \, hr} = 12.7 hr$$

### Q20.9.11

Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.

1. AgNO3
2. RbI

### S20.9.11

Solution (a)

When $$AgNO_3$$ is electrolyzed in an aqueous solution, because the bond between silver and nitrate is ionic, in a polar solution (water), it will dissociate. Therefore $AgNO_3 \rightarrow Ag^+ + NO_{3}^-$ Electrolysis occurs when a system has an $$E^{o}<0$$, since the reaction is non-spontaneous and therefore needs an external energy source to push the reaction in a non-spontaneous direction. Because the solution contains $$Ag^+$$ ions and $$Ag^+$$ cannot lose another electron due to a high ionization energy, it will gain electrons be reduced: $Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)} \;\;\; E° = 0.7996\; V$ Using the table of Standard Reduction Potentials, water, the aqueous solution has the reduction half reactions: $O_2{(g)} + 4H^+{(aq)} +4e^− \rightarrow 2H_2O{(l)} \;\;\; E° = 1.229\; V$ $H^+{(aq)} +OH^-{(aq)} \rightarrow H_2O{(l)} \;\;\; E° = 0\; V$ $$E°=E°_{cathode}-E°_{anode}$$ and since reduction occurs at the cathode, $$E°=\; 0.7996-E°_{anode}$$. In order for E°<0, $$E°_{anode}>\; 0.7996\; V$$. Only $$O_2{(g)} + 4H^+{(aq)} +4e^− \rightarrow 2H_2O{(l)}$$ has $$E°>\; 0.7996\; V$$. So oxidation (lose of electrons) occurs at the anode, reduction (gain of electrons) occurs at the cathode: $Cathode:\;\;\;\; Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)}$ $Anode:\;\;\;\; 2H_2O{(l)} \rightarrow O_2{(g)} + 4H^+{(aq)} +4e^−$ Therefore, Ag(s) is produced at the cathode, and $$O_2{(g)}$$ is produced at the anode.

Solution (b)

When $$RbI$$ is electrolyzed in an aqueous solution, because the bond between Rubidium and Iodine is ionic, in a polar solution (water), it will dissociate: $RbI \rightarrow Rb^+ + I^-$Electrolysis occurs when a system has an $$E^{o}<0$$, since the reaction is non-spontaneous and therefore needs an external energy source to push the reaction in a non-spontaneous direction. Because the solution contains $$I^-$$ ions and $$I^-$$ cannot gain another electron due to a low electronegativity, it will lose electrons and be oxidized: $2I^-{(aq)} \rightarrow I_2{(s)} + 2e^-$Using the table of Standard Reduction Potentials, the standard reduction potential for the reduction of $$I_2{(s)}$$ is $$E°=\; 0.5355\; V$$ and this is the $$E°_{anode}$$ because oxidation occurs at the anode. Water, the aqueous solution has a reduction reaction where water $$(H_2O{(l)})$$ is a reactant that is reduced: $H_2O{(l)} + 2e^- \rightarrow H_2{(g)} + 2OH^-{(aq)}\;\;\; E°=\; -0.828\; V$E°<0 for $$E°=E°_{cathode}-E°_{anode}$$ for electrolysis and since reduction occurs at the cathode $$E°_{cathode}=\; -0.828\; V$$, and since oxidation occurs at the anode $$E°_{anode}=\; 0.5355\; V$$: $Cathode:\;\;\;\; 2H_2O{(l)} + 2e^- \rightarrow H_2{(g)} + 2OH^-{(aq)}$ $Anode:\;\;\;\; 2I^-{(aq)} \rightarrow I_2{(s)} + 2e^-$ $E°=E°_{cathode}-E°_{anode}$ $E°=-0.828-0.5355\; V$ $E°=-1.364\; V\; so\; E°<0$Therefore, $$I_2{(g)}$$ is produced at the anode, and $$H_2{(g)}$$ is produced at the cathode.

### A20.9.11

1. cathode: Ag(s); anode: O2(g);
2. cathode: H2(g); anode: I2(s)

### Q20.9.12

Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.

1. MgBr2
2. Hg(CH3CO2)2
3. Al2(SO4)3

### S20.9.12

Recall where oxidation and reduction take place: oxidation occurs at the anode, and reduction occurs at the cathode. This question also pertains to the activity series in order to determine what is being reduced and what is being oxidized. Since these are all in aqueous solutions, they will involve the half reaction of water, which is: 2H2O(aq) + 2e- → H2(g) + 2OH- when is is being reduced, or 2H2O (aq) → O2(g) + 4H+(aq) + 4e- when it is being oxidized.

1. MgBr2

Anode: 2Br-(aq)→ Br2(l) + 2e-

Cathode: 2H2O(aq) + 2e- → H2(g) + 2OH-

2. Hg(CH3CO2)2

Anode: 2H2O (aq) → O2(g) + 4H+(aq) + 4e-

Cathode: Hg2+(aq) + 2e- → Hg(l)

3. Al2(SO4)3

Anode: 2H2O (aq) → O2(g) + 4H+(aq) + 4e-

Cathode: 2H2O(aq) + 2e- → H2(g) + 2OH-