# 5.E: Electrochemistry (Exercises)

## 17.1: Balancing Oxidation-Reduction Reactions

### Q17.1.1

If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?

### S17.1.1

In this example, we are given that a current of 2.5 Amps (A) is run through a circuit for 35 minutes, and we are asked to find the charge in Coulombs (C). We know that Amps is the charge (in Couloumbs) per second, which gives us this equation: $A =\frac{C }{s}$

To solve for Coulombs, we can rewrite the equation to:

$$C ={A}*{s}$$

When we plug in the values, we need to pay attention to units. Since we were given time in terms of minutes, we need to convert it to seconds. Subsequently, we get...

$$C={\frac {2.5 \ Coloumbs}{second}}*35 \ minutes*{\frac{60 \ seconds}{minute}}$$

Therefore...

$$C={5.25}*{10^3 \ Coulombs}$$

With correct significant figures, the answer should be 5.3 * 10^3 Coulombs of charge is moved through the circuit.

5.3 × 103 C

### Q17.1.2

For the scenario in the previous question, how many electrons moved through the circuit?

### S17.1.2

For the scenario in the previous question, how many electrons moved through the circuit?

If the total charge passing through the cell is equal to 5.3 × 103 C (from the previous question), then the number of electrons passing can be obtained by dividing the total charge (5.3x 103 C in this case) by the charge per electron (1.6 X 10-19 C). Charge per electron is always 1.6 x 10-19 C.

This is how you find the charge per electron:

$$\frac{96485\text{ Coulombs}}{\text{mole e}^-}\times\frac{1\text{ mole e}^-}{6.022\times10^{23}\text{ Electrons}} = 1.6\times10^{-19} C/$$e-

Therefore; number of electrons = (5.3 × 103 C)/(1.6 × 10-19 C/e-)

= 3.31 × 1022 number of electrons moved through the circuit.

### A17.1.2

3.31 × 1022 electrons

### Q17.1.3

For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

1. $$\ce{Fe^3+ + 3e- ⟶ Fe}$$
2. $$\ce{Cr ⟶ Cr^3+ + 3e-}$$
3. $$\ce{MnO4^2- ⟶ MnO4- + e-}$$
4. $$\ce{Li+ + e- ⟶ Li}$$

### S17.1.3

To determine if a reaction is being oxidized or reduced, you must determine the oxidation state of the elements on either side of the equation. If the oxidation state is increased, then the reaction was oxidized. If the oxidation state decreased, then the reaction was reduced. Alternatively, you can examine the number of electrons for the elements on either side of the reaction. If the number of electrons is reduced, then the reaction is an oxidation reaction. If the number of electrons is increased, then the reaction is a reduction reaction. We can remember this by the common chemistry mnemonic: OIL RIG. Oxidation is losing [electrons], Reduction is gaining [electrons. The electrons on either side of the half-reactions keep the charges balanced.

a) In this reaction, the oxidation states are very easy to find. The oxidation state of Fe3+ is (+3). The oxidation state of Fe is (0) because it is in elemental form. The oxidation state decreases from 3 to 0, so reduction is occurring.

b) This is similar to part A. Cr is in its elemental form, so its oxidation state is (0). Cr3+ is an ion, and its oxidation state is (+3). In this case, the oxidation state increases from 0 to 3, so oxidation is occurring.

c) For this reaction, you must find the oxidation states of Manganese in the permanganate compounds.

The first compound is MnO42- . You can set up a simple equation to find the oxidation state of Manganese.

$$X + 4(-2) = -2$$

X represents the oxidation state of Mn. It is added to 4(-2) because there are 4 oxygen molecules, and oxygen always has an oxidation state of (-2). The equation is equal to -2 because the overall compound has a charge of -2. Then, you solve for X. X is equal to 6, so the oxidation state of Manganese in MnO42- is (+6).

You repeat this process for MnO4- .

$$X + 4(-2) = -1$$

This time, X = +7, so the oxidation state of Mn in MnO4- is (+7).

The oxidation state for Manganese increased from 6 to 7, so oxidation occurred.

d) This problem is similar to the first 2. Li+ has an oxidation state of (+1), and Li has an oxidation state of (0). The oxidation state decreased from 1 to 0, so the reaction is going through reduction.

(a) reduction

(b) oxidation

(c) oxidation

(d) reduction

### Q17.1.4

For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.

1. $$\ce{Cl- ⟶ Cl2}$$
2. $$\ce{Mn^2+ ⟶ MnO2}$$
3. $$\ce{H2 ⟶ H+}$$
4. $$\ce{NO3- ⟶ NO}$$

### S17.1.4

oxidation definition: loss of electrons, increasing the oxidation number

reduction definition: gain of electrons, decreasing the oxidation number

-the sum of the oxidation numbers in a compound is zero

-the sum of the oxidation numbers in a polyatomic ion is equal to the charge

Solutions:

1. $$Cl^−⟶Cl_2$$

$$Cl^-$$ has an oxidation number of negative one, as the oxidation state of a lone ion is equal to its charge. Both of the chlorines in $$Cl_2$$ have a charge of zero. Therefore, there was an increase in the oxidation number from negative one to zero. An increase in oxidation number is oxidation.

2. $$Mn^+2⟶MnO_2$$

$$Mn^+2$$ has an oxidation number of plus two, aas the oxidation state of a lone ion is equal to its charge. Since each $$O$$ has an oxidation number of negative two, there is a total charge of negative four from both of the oxygens. Therefore, the $$Mn$$ has to have a plus four charge to balance out the negative four charge an keep the compound neutral. Since the $$Mn$$ increased in oxidation number from two to four, it is oxidation.

3. $$H_2⟶H^+$$

The hydrogens in the H2 have an oxidation number of zero. $$H^+$$ has an oxidation number of plus one. Therefore there was an increase in oxidation number, and it is oxidation.

4. $$NO_3^-⟶NO$$

Each of the oxygens has an oxidation number of negative two. Since the charge of $$NO_3^-$$ negative one, the charge of the nitrogen is plus five to balance the charges. The oxidation number of nitrogen in $$NO$$ is plus two. Therefore, the oxidation number of nitrogen decrease, and it is reduction.

### A17.1.4

(a) oxidation; (b) oxidation; (c) oxidation; (d) reduction

### Q17.1.5

Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half-reactions in an acidic solution.

1. $$\ce{Ca ⟶ Ca^2+ + 2e-,\: F2 + 2e- ⟶ 2F-}$$
2. $$\ce{Li ⟶ Li+ + e- ,\: Cl2 + 2e- ⟶ 2Cl-}$$
3. $$\ce{Fe ⟶ Fe^3+ + 3e- ,\: Br2 + 2e- ⟶ 2Br-}$$
4. $$\ce{Ag ⟶ Ag+ + e- ,\: MnO4- + 4H+ + 3e- ⟶ MnO2 + 2H2O}$$

### S17.1.5

We are given the oxidation and reduction half reactions for a reaction. We can add the half reactions together to get the overall reaction, but we must make sure that electrons are not involved in the final reaction equation. Therefore, we must make sure that the amount of electrons on the right side of the oxidation equation equal the amount of electrons on the left side of the reduction equation. This ensures that the electrons will cancel out in the final equation.

What is the purpose behind balancing these half reactions? To balance or confirm that a chemical reaction is a redox reaction we typically split the reaction into its corresponding half reactions and use them as a guide; oxidation indicates a loss of an electron, and reduction indicates gaining an electron-- in terms of oxidation states, or numbers, oxidation is an increase ( for ex. +1 to +2) and reduction is a decrease ( +2 to +1). If a reaction has both, that is one element is increasing in its oxidation state, and another is being reduced, then you can confirm that the reaction being studied is a redox reaction.

More on balancing redox reactions: here

Problem 1

Oxidation half reaction: $${Ca} \rightarrow {Ca^{2+}}+{2e}^{−}$$

Reduction half reaction: $${F}_{2}+{2e}^{−} \rightarrow {2F}^{−}$$

The electron number in each half reaction is equal, so the electrons cancel and we get our final equation.

${F_2}+{Ca} \rightarrow {2F^−}+{Ca^{2+}}$

Problem 2

Oxidation half reaction: $$2({Li} \rightarrow {Li^+}+{e^−})$$

$${2Li} \rightarrow {2Li^+}+{2e^−}$$

Reduction half reaction: $${Cl_2}+{2e^−} \rightarrow {2Cl^−}$$

We multiply the oxidation half reaction by 2 so that both half reactions involve 2 electrons. We then add the half reactions and cancel the electrons to get our overall reaction.

${Cl_2}+{2Li} \rightarrow {2Li^+}+{2Cl^−}$

Problem 3

Oxidation half reaction: $$2({Fe} \rightarrow {Fe^{3+}}+{3e^−})$$

$${2Fe} \rightarrow {2Fe^{3+}}+{6e^−}$$

Reduction half reaction: $$3({Br_2}+{2e^−} \rightarrow {2Br^−})$$

$${3Br_2}+{6e^−} \rightarrow {6Br^−}$$

We multiply the oxidation half reaction by 2 and the reduction half reaction by 3 so that both half reactions involve 6 electrons. We then add the half reactions and cancel the electrons to get our overall reaction.

${3Br_2}+{2Fe} \rightarrow {2Fe^{3+}}+{6Br^−}$

Problem 4

Oxidation half reaction: $$3({Ag} \rightarrow {Ag^+}+{e^−})$$

$${3Ag} \rightarrow {3Ag^+}+{3e^−})$$

Reduction half reaction: $${MnO^{−}_{4}}+{4H^+}+{3e^−} \rightarrow {MnO_2}+{2H_2O}$$

We multiply the oxidation half reaction by 3 so that both half reactions involve 3 electrons. We then add the half reactions and cancel the electrons to get our overall reaction.

${MnO^{−}_{4}}+{4H^+}+{3Ag} \rightarrow {3Ag^+}+{MnO_2}+{2H_2O}$

### A17.1.5

1. $$\ce{F2 + Ca ⟶ 2F- + Ca^2+}$$;
2. $$\ce{Cl2 + 2Li ⟶ 2Li+ + 2Cl-}$$;
3. $$\ce{3Br2 + 2Fe ⟶ 2Fe^3+ + 6Br-}$$;
4. $$\ce{MnO4 + 4H+ + 3Ag ⟶ 3Ag+ + MnO2 + 2H2O}$$

### Q17.1.5B

Balance the following in acidic solution:

1. $$\ce{H2O2 + Sn^2+ ⟶ H2O + Sn^4+}$$
2. $$\ce{PbO2 + Hg ⟶ Hg2^2+ + Pb^2+}$$
3. $$\ce{Al + Cr2O7^2- ⟶ Al^3+ + Cr^3+}$$

### S17.1.5B

a) i) Write the half reactions

$$Sn^{2+} \rightarrow Sn^{4+}$$ $$H_2O_2 \rightarrow H_2O$$

ii) Balance oxygen with H2O

$$Sn^{2+} \rightarrow Sn^{4+}$$ $$H_2O_2 \rightarrow H_2O + H_2O$$

iii) Balance hydrogen with H+

$$Sn^{2+} \rightarrow Sn^{4+}$$ $$H_2O_2 + {2H^+} \rightarrow H_2O + H_2O$$

iv) Balance charge with e-

$$Sn^{2+} \rightarrow Sn^{4+} + {2e^-}$$ $$H_2O_2 + 2H^+ +{2e^-} \rightarrow H_2O + H_2O$$

v) Combine half reactions

$$Sn^{2+} + H_2O_2 + 2H^+ +2e^- \rightarrow Sn^{4+} + 2e^- + H_2O + H_2O$$

vi) Simplify the reaction

$$Sn^{2+} + H_2O_2 + 2H^+ \rightarrow Sn^{4+} + 2H_2O$$

vii) Check if number of atoms and charges balance.

The charges on both sides equal 4+, and the number of atoms are equal on both sides.

b) i) Write the half reactions

$$Hg \rightarrow Hg_2^{2+}$$ $$PbO_2 \rightarrow Pb^{2+}$$

ii) Balance elements other the O and H

$$2Hg \rightarrow Hg_2^{2+}$$ $$PbO_2 \rightarrow Pb^{2+}$$

iii) Balance oxygen with H2O

$$2Hg \rightarrow Hg_2^{2+}$$ $$PbO_2 \rightarrow Pb^{2+}+2H_2O$$

iv) Balance hydrogen with H+

$$2Hg \rightarrow Hg_2^{2+}$$ $$PbO_2+4H^+ \rightarrow Pb^{2+}+2H_2O$$

v) Balance charge with e-

$$2Hg \rightarrow Hg_2^{2+}+2e^-$$ $$2e^-+PbO_2+4H^+ \rightarrow Pb^{2+}+2H_2O$$

vi) Combine half reactions

$$2Hg + 2e^- + PbO_2 +4H^+\rightarrow Hg_2^{2+} +2e^- + Pb^{2+}+2H_2O$$

vii) Simplify the reaction

$$2Hg + PbO_2 +4H^+\rightarrow Hg_2^{2+} + Pb^{2+}+2H_2O$$

viii) Check if number of atoms and charges balance.

The charges on both sides equal 4+, and the number of atoms are equal on both sides.

c) i) Write the half reactions

$$Al \rightarrow Al^{3+}$$ $$Cr_2O_7^{2-} \rightarrow Cr^{3+}$$

ii) Balance elements other the O and H

$$Al \rightarrow Al^{3+}$$ $$Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$$

iii) Balance oxygen with H2O

$$Al \rightarrow Al^{3+}$$ $$Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O$$

iv) Balance hydrogen with H+

$$Al \rightarrow Al^{3+}$$ $$14H^++Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O$$

v) Balance charge with e-

$$Al \rightarrow Al^{3+}+3e^-$$ $$6e^-+14H^++Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O$$

vi) Balance e-

$$2Al \rightarrow 2Al^{3+}+6e^-$$ $$6e^-+14H^++Cr_2O_7^{2-} \rightarrow 2Cr^{3+}+ 7H_2O$$

vii) Combine half reactions

$$6e^-+14H^++Cr_2O_7^{2-}+2Al \rightarrow 2Cr^{3+}+ 7H_2O+2Al^{3+}+ 6e^-$$

viii) Reduce

$$14H^++Cr_2O_7^{2-}+2Al \rightarrow 2Cr^{3+}+ 7H_2O+2Al^{3+}$$

ix) Check if number of atoms and charges balance.

The charges on both sides equal 12+, and the number of atoms are equal on both sides.

### A17.1.5B

a)

$$Sn^{2+} + H_2O_2 + 2H^+ \rightarrow Sn^{4+} + 2H_2O$$

b)

$$2Hg + PbO_2 +4H^+\rightarrow Hg_2^{2+} + Pb^{2+}+2H_2O$$

c)

$$14H^++Cr_2O_7^{2-}+2Al \rightarrow 2Cr^{3+}+ 7H_2O+2Al^{3+}$$

### Q17.1.6

Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

### S17.1.6

Oxidation: The net loss of electrons (the specific element increases in oxidation number, becoming more positive) ie. Zn(s) → Zn2++ 2e (0 charge becomes +2 charge, gain of electrons)

Reduction: The net gain of electrons (the specific element decreases in oxidation number, becoming more negative) ie. Cu2+ + 2e → Cu(s) (+2 charge becomes 0 charge, loss of electrons)

Oxidizing agent: The compound or element in a reaction that oxidizes the other compounds it bonds with, making them lose electrons and become more positive. The oxidizing agent in turn becomes reduced by the reducing agent.

Reducing agent: The compound or element in a reaction that reduces the other compounds it bonds with, making them gain electrons and become more negative. The reducing agent in turn it gets oxidized by the oxidizing agent.

a. Sn2+ is the reducing agent because it undergoes oxidation (Sn2+ becomes Sn4+ losing 2 electrons and becoming more positive)

b.H2O2 is oxidizing agent because it undergoes reduction (H2O2 becomes H2O gaining 2 electrons and becoming more negative)

c. Hg is reducing agent because it undergoes oxidation (Hg becomes Hg22+ losing 2 electrons and becoming more positive)

d.PbO2 is oxidizing agent because it undergoes reduction (PbO2 becomes Pb2+ gaining two electrons and becoming more negative)

e. Al is reducing agent because it undergoes oxidation (Al becomes Al3+ losing 3 electrons and becoming more positive)

f.Cr2O72− is oxidizing agent because undergoes reduction (Cr2O72− becomes Cr3+ gaining 3 electrons becoming more negative)

### A17.1.6

• oxidized: (a) Sn2+; (b) Hg; (c) Al
• reduced: (a) H2O2; (b) PbO2; (c) $$Cr_2O_7^{2−}$$
• oxidizing agent: (a) H2O2; (b) PbO2; (c) $$Cr_2O_7^{2−}$$
• reducing agent: (a) Sn2+; (b) Hg; (c) Al

### Q17.1.7

Balance the following in basic solution:

1. $$\ce{SO3^2-}(aq)+\ce{Cu(OH)2}(s)⟶\ce{SO4^2-}(aq)+\ce{Cu(OH)}(s)$$
2. $$\ce{O2}(g)+\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)$$
3. $$\ce{NO3-}(aq)+\ce{H2}(g)⟶\ce{NO}(g)$$
4. $$\ce{Al}(s)+\ce{CrO4^2-}(aq)⟶\ce{Al(OH)3}(s)+\ce{Cr(OH)4-}(aq)$$

### S17.1.7

1) $$\ce{SO3^2-}(aq)+\ce{Cu(OH)2}(s)⟶\ce{SO4^2-}(aq)+\ce{Cu(OH)}(s)$$

Step 1: Separate into two half reactions; reduction and oxidation. Reduction reactions gain electrons while oxidation reactions lose an electron.

$$\ce{SO3^2-}(aq)⟶\ce{SO4^2-}(aq)$$ (oxidation half reaction) This is the oxidation reaction because S(sulfur) goes from a +4 oxidation state to a +6 oxidation state, thus losing 2 electrons in the process.

$$\ce{Cu(OH)2}(s)⟶\ce{Cu(OH)}(s)$$ (reduction half reaction) This is the reduction reaction because Cu goes from a +2 oxidation state to a +1 oxidation state, thus gaining 1 electron in the process.

Step 2: Balance the two reactions.

To balance the oxidation reaction we must use water to balance the oxygens:

$$\ce{H2O}(l)+\ce{SO3^2-}(aq)⟶\ce{SO4^2-}(aq)$$

To balance the reduction reaction we must add OH^- to the right side:

$$\ce{Cu(OH)2}(s)⟶\ce{Cu(OH)}(s)+\ce{OH^-}(aq)$$

Step 3: Balance the hydrogen with protons(H+):

This is only necessary for the oxidation reaction:

$$\ce{H2O}(l)+\ce{SO3^2-}(aq)⟶\ce{SO4^2-}(aq)+{2H^+}(aq)$$

Step 4: Balance the charges with electrons:

Since Sulfur oxidizes from +4 to +6 we must add 2e- to the right side of the oxidation reaction:

$$\ce{H2O}(l)+\ce{SO3^2-}(aq)⟶\ce{SO4^2-}(aq)+{2H^+}(aq)+{2e-}$$

For the reduction reaction we only add 1e- because Cu reduces from 2+ to 1+:

$$\ce{Cu(OH)2}(s)+{1e-}⟶\ce{Cu(OH)}(s)+\ce{OH^-}(aq)$$

Step 5: Scale the reactions so they have the same number of electrons:

We must multiply the reduction reaction by 2 so the it has 2e- like the oxidation reaction:

$$\ce{2Cu(OH)2}(s)+{2e-}⟶\ce{2Cu(OH)}(s)+\ce{2OH^-}(aq)$$

Step 6:Add reactions and cancel out the electrons:

$$\ce{H2O}(aq)+\ce{SO3^2-}(aq)+\ce{2Cu(OH)2}(s)⟶\ce{SO4^2-}(aq)+{2H^+}(aq)+\ce{2Cu(OH)}(s)+\ce{2OH^-}(aq)$$

Step 7: Add OH- to balance H+. There are 2 net protons in this equation, so add 2 OH- ions to each side.

$$\ce{H2O}(aq)+\ce{SO3^2-}(aq)+\ce{2Cu(OH)2}(s)+\ce{2OH^-}(aq)⟶\ce{SO4^2-}(aq)+{2H^+}(aq)+\ce{2Cu(OH)}(s)+\ce{4OH^-}(aq)$$

Step 8:Combine OH- ions and H+ ions that are present on the same side to form water.

$$\ce{H2O}(aq)+\ce{SO3^2-}(aq)+\ce{2Cu(OH)2}(s)+\ce{2OH^-}(aq)⟶\ce{SO4^2-}(aq)+\ce{2Cu(OH)}(s)+\ce{2OH^-}(aq)+\ce{2H2O}(aq)$$

Cancel and Combine terms:

$$\ce{SO3^2-}(aq)+\ce{2Cu(OH)2}(s)⟶\ce{SO4^2-}(aq)+\ce{2Cu(OH)}(s)+\ce{H2O}(aq)$$

2)$$\ce{O2}(g)+\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)$$

Step 1:Separate into two half reactions; reduction and oxidation. Reduction reactions gain electrons while oxidation reactions lose an electron.

$$\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)$$ This is the oxidation half reaction because Mn goes from an oxidation state of +2 to an oxidation state of +4, thus losing two electrons in the process.

$$\ce{O2}(g)⟶\ce{MnO2}(s)$$ This is the reduction half reaction because Oxygen goes from an oxidation state of 0 to -4, thus gaining four electrons in the process.

Step 2: Balance the two half reactions:

In order to balance the oxidation reaction we must add 2 H+ to the right side to balance out the left side:

$$\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)+{2H^+}(aq)$$

To balance the reduction reaction all we need to do is add a Mn to the left side:

$$\ce{O2}(g)+{Mn}(s)⟶\ce{MnO2}(s)$$

Step 3: Balance the charges of the half reactions with electrons:

For the oxidation reaction, we need to add 2 electrons to the right side because Mn is oxidized from 2+ to 4+:

$$\ce{Mn(OH)2}(s)⟶\ce{MnO2}(s)+{2H^+}(aq)+{2e^-}$$

For the reduction half reaction, we must add 4 electrons to the left side because Oxygen is reduced from 0 to -4:

$$\ce{O2}(g)+{Mn}(s)+{4e^-}⟶\ce{MnO2}(s)$$

Step 4: Scale the reactions so that they both have the same number of electrons.

In order to balance the number of electrons for both half reactions, we must multiply the oxidation reaction by 2 so that it has 4 electrons just like the reduction half reaction. We leave the reduction half reaction how it is.

$$\ce{2Mn(OH)2}(s)⟶\ce{2MnO2}(s)+{4H^+}(aq)+{4e^-}$$

$$\ce{O2}(g)+{Mn}(s)+{4e^-}⟶\ce{MnO2}(s)$$

Step 5: Add both have reactions and cancel out the electrons:

$$\ce{2Mn(OH)2}(s)+\ce{O2}(g)+{Mn}(s)⟶\ce{3MnO2}(s)+{4H^+}(aq)$$

Step 6:Add OH- to balance H+. There are 4 net protons in this equation, so add 4 OH- ions to each side.

$$\ce{2Mn(OH)2}(s)+\ce{O2}(g)+{Mn}(s)+{4OH^-}(aq)⟶\ce{3MnO2}(s)+{4H^+}(aq)+{4OH^-}(aq)$$

Step 7:Combine OH- ions and H+ ions that are present on the same side to form water.

$$\ce{2Mn(OH)2}(s)+\ce{O2}(g)+{Mn}(s)+{4OH^-}(aq)⟶\ce{3MnO2}(s)+\ce{4H2O}(aq)$$ This is your answer.

3)$$\ce{NO3-}(aq)+\ce{H2}(g)⟶\ce{NO}(g)$$

Step 1: Separate into two half reactions; reduction and oxidation. Reduction reactions gain electrons while oxidation reactions lose an electron.

$$\ce{H2}(g)⟶\ce{2H^+}(aq)$$ This is the oxidation reaction because hydrogen loses 2 electrons in the process. We must balance the H2 on the reactants side with 2H+ on the product side.

$$\ce{NO3^-}(aq)⟶\ce{NO}(g)$$ This is the reduction half reaction because Nitrogen goes from an oxidation state of +5 to +2, gaining 3 electrons in the process.

Step 2: Balance the two half reactions:

The oxidation half reactions is already balanced from the previous step.

In order to balance the reduction half reaction we need to add 3 water molecules to the product side in order to balance the oxygen on the reactant side. Since we add 3 water molecules, the reactant side is now missing 6 H atoms. So we add 6H+ to the left side. The reduction half reaction now looks like this:

$$\ce{NO3^-}(aq)+\ce{6H^+}(aq)⟶\ce{NO}(g)+\ce{H2O}(l)$$

Step 3: Balance the charge of both reactions using electrons:

For the oxidation reaction must add 2 electrons to the right side because Hydrogen loses 2 electrons in the process.

$$\ce{H2}(g)⟶\ce{2H^+}(aq)+\ce{2e^-}$$

For the reduction half reaction we must add 5 electrons tot he left side because nitrogen gains 5 electrons in the process.

$$\ce{NO3^-}(aq)+\ce{6H^+}(aq)+\ce{5e^-}⟶\ce{NO}(g)+\ce{H2O}(l)$$

Step 4: Scale the two half reactions in order to cancel out the electrons:

In order to do this we must find a common factor of 2 electrons and 5 electrons. We find that it is 10. So we multiply the oxidation half reaction by a factor of 5 and the reduction half reaction by a factor of 2. We get the following:

$$\ce{5H2}(g)⟶\ce{10H^+}(aq)+\ce{10e^-}$$

$$\ce{2NO3^-}(aq)+\ce{12H^+}(aq)+\ce{10e^-}⟶\ce{2NO}(g)+\ce{2H2O}(l)$$

Step 5: Add both half reactions and cancel out the electrons:

$$\ce{5H2}(g)+\ce{2NO3^-}(aq)+\ce{12H^+}(aq)⟶\ce{2NO}(g)+\ce{10H^+}(aq)+\ce{6H2O}(l)$$

Step 6: Now cancel out the H+ on both sides and you have your final answer as follows:

$$\ce{5H2}(g)+\ce{2NO3^-}(aq)+\ce{2H^+}(aq)⟶\ce{2NO}(g)+\ce{6H2O}(l)$$

4)$$\ce{Al}(s)+\ce{CrO4^2-}(aq)⟶\ce{Al(OH)3}(s)+\ce{Cr(OH)4-}(aq)$$

Step 1: Separate into two half reactions; reduction and oxidation. Reduction reactions gain electrons while oxidation reactions lose an electron.

$$\ce{Al}(s)⟶\ce{Al(OH)3}(s)$$ This is the oxidation half reaction because Al goes from an oxidation state of 0 to +3.

$$\ce{CrO4^2-}(aq)⟶\ce{Cr(OH)4^-}(aq)$$ This is the reduction hafl reaction because Cr goes form and oxidation state of +7 to +5.

Step 2: Balance the half reactions.

In order to balance the oxidation half reaction we must add 3 OH- molecules to the reactants.

$$\ce{Al}(s)+\ce{3OH^-}(aq)⟶\ce{Al(OH)3}(s)$$

In order to balance the reduction half reaction we must balance the Hydrogen on the left side by adding 4 water molecules. By doing that we now have an uneven number of oxygen and hydrogen molecules on the right side, so we must add 4 OH-.

$$\ce{CrO4^2-}(aq)+\ce{4H2O}(l)⟶\ce{Cr(OH)4^-}(aq)+\ce{4OH^-}(aq)$$

Step 3: Add electrons to both half reactions to balance the charges.

$$\ce{Al}(s)+\ce{3OH^-}(aq)⟶\ce{Al(OH)3}(s)+\ce{3e^-}$$

$$\ce{CrO4^2-}(aq)+\ce{4H2O}(l)+\ce{2e^-}⟶\ce{Cr(OH)4^-}(aq)+\ce{4OH^-}(aq)$$

Step 4: Multiply by a common factor to cancel the electrons.

In order to cancel out the electrons we must find a common multiple. The common multiple is 6. So we multiply the oxidation half reaction by a factor of 2 and the reduction half reaction by a factor of 3.

$$\ce{2Al}(s)+\ce{6OH^-}(aq)⟶\ce{2Al(OH)3}(s)+\ce{6e^-}$$

$$\ce{3CrO4^2-}(aq)+\ce{12H2O}(l)+\ce{6e^-}⟶\ce{3Cr(OH)4^-}(aq)+\ce{12OH^-}(aq)$$

Step 5: Add both half reactions and cancel out electrons:

$$\ce{2Al}(s)+\ce{6OH^-}(aq)+\ce{3CrO4^2-}(aq)+\ce{12H2O}(l)⟶\ce{2Al(OH)3}(s)+\ce{3Cr(OH)4^-}(aq)+\ce{12OH^-}(aq)$$

Step 6: Cancel out common terms. The following is the final answer:

$$\ce{2Al}(s)+\ce{3CrO4^2-}(aq)+\ce{12H2O}(l)⟶\ce{2Al(OH)3}(s)+\ce{3Cr(OH)4^-}(aq)+\ce{6OH^-}(aq)$$

### A17.1.7

a.$\ce{SO3^2-}(aq)+\ce{2Cu(OH)2}(s)⟶\ce{SO4^2-}(aq)+\ce{2Cu(OH)}(s)+\ce{H2O}(aq)$

b.$\ce{2Mn(OH)2}(s)+\ce{O2}(g)+{Mn}(s)+{4OH^-}(aq)⟶\ce{3MnO2}(s)+\ce{4H2O}(aq)$

c.$\ce{5H2}(g)+\ce{2NO3^-}(aq)+\ce{2H^+}(aq)⟶\ce{2NO}(g)+\ce{6H2O}(l)$

d. $\ce{2Al}(s)+\ce{3CrO4^2-}(aq)+\ce{12H2O}(l)⟶\ce{2Al(OH)3}(s)+\ce{3Cr(OH)4^-}(aq)+\ce{6OH^-}(aq)$

### Q17.1.8

Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.

### S17.1.8

• A species is oxidized if its oxidation number increases because it loses an electron (i.e becomes more positive/ less negative).
• Example: $$Zn^{+} \rightarrow Zn^{2+}$$
• A species is reduced if its oxidation number decreases because it gains an electron (i.e becomes more negative/ less positive).
• Example: $$Zn^{2+} \rightarrow Zn^{+}$$
• The oxidizing agent is the species which accepts the electron (i.e the reduced species).
• The reducing agent is the species which gives away the electron (i.e the oxidized species).
1. $$SO_{3}^{2-}(aq)+Cu(OH)_{2}(s) \rightarrow SO_{4}^{2-}(aq)+Cu(OH)(s)$$
• $$SO_{3}^{2-}(aq) \rightarrow SO_{4}^{2-}(aq)$$
• Oxidation State: $$(+2) \rightarrow (+4)$$
• Since it becomes more positive, this species is oxidized
• Since it gave up electrons to become more positive, it is a reducing agent
• $$Cu(OH)_{2}(s) \rightarrow Cu(OH)(s)$$
• Oxidation State: $$(+1) \rightarrow (0)$$
• Since it becomes less positive, this species is reduced
• Since it gains electrons to become less positive, it is an oxidizing agent.
2. $$O_{2}(g)+Mn(OH)_{2}(s) \rightarrow MnO_{2}(s)$$
• Oxidation States: $$(O_{2}^{+0})+(Mn^{2+} O^{-2}H^{+1}) \rightarrow Mn^{4+}O^{4-}$$
• O2: $$(0) \rightarrow (-4)$$; reduced, oxidizing agent
• Mn: $$(+2) \rightarrow (+4)$$; oxidized, reducing agent
• Note: OH becomes water and it is not part of the reaction
3. $$NO_{3}^{-}(aq)+H_{2}(g) \rightarrow NO(g)$$
• $$NO_{3}^{-}(aq) \rightarrow NO(g)$$
• Oxidation States: $$N^{+5} O_{3}^{2-} \rightarrow N^{+2}O^{2-}$$
• N: $$(+5) \rightarrow (+2)$$; reduced, oxidizing agent
• O: $$(-6) \rightarrow (-2)$$; oxidized, reducing agent
4. $$Al(s)+CrO_{4}^{2-}(aq) \rightarrow Al(OH)_{3}(s)+Cr(OH)_{4}^{-}$$
• $$Al(s) \rightarrow Al(OH)_{3}(s)$$
• Oxidation State: $$(0) \rightarrow (+3)$$
• Oxidized therefore it is a reducing agent
• $$CrO_{4}^{2-}(aq) \rightarrow Cr(OH)_{4}^{-}$$
• Oxidation State: $$(+6) \rightarrow (+3)$$
• Reduced, therefore it is an oxidizing agent

### A17.1.8

Oxidized = reducing agent: (a) $$\ce{SO3^2-}$$; (b) Mn(OH)2; (c) H2; (d) Al; reduced = oxidizing agent: (a) Cu(OH)2; (b) O2; (c) $$\ce{NO3-}$$; (d) $$\ce{CrO4^2-}$$

### Q17.1.9

1. Why is it not possible for hydroxide ion (OH−) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?
2. Why is it not possible for hydrogen ion (H+) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?

### A17.1.9

(a) In an acidic solution, [H+] > $$1.0x10^{-7}M$$ > [OH-]. The hydroxide ion cannot appear as a reactant because its concentration will be very close to zero. If it were produced, it would instantly react with the excess H+ ions to produce H2O. Thus, hydroxide ion should not appear as a reactant or product in acidic solution.

(b) In basic solution, [OH] > $$1.0x10^{-7}M$$ > [H+]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not appear as a reactant or product in basic solution.

### Q17.1.10

Why must the charge balance in oxidation-reduction reactions?

### A17.1.10

In redox reactions, the oxidation half-reactions lose electrons, while the reduction half-reactions gain electrons. The electrons lost by the oxidation half-reactions should all be gained by the reduction half-reactions. The number of electrons and the electric charge in the overall reactions do not change. Therefore, the charge in redox reactions must be balanced.

## 17.2: Galvanic Cells

### Q17.2.1

Write the following balanced reactions using cell notation. Use platinum as an inert electrode, if needed.

1. $$\ce{Mg}(s)+\ce{Ni^2+}(aq)⟶\ce{Mg^2+}(aq)+\ce{Ni}(s)$$
2. $$\ce{2Ag+}(aq)+\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)$$
3. $$\ce{Mn}(s)+\ce{Sn(NO3)2}(aq)⟶\ce{Mn(NO3)2}(aq)+\ce{Au}(s)$$
4. $$\ce{3CuNO3}(aq)+\ce{Au(NO3)3}(aq)⟶\ce{3Cu(NO3)2}(aq)+\ce{Au}(s)$$

### S17.2.1

Write the following balanced reactions using cell notation:

a. $$Mg(s) + Ni^{2+}(aq)\rightarrow Mg^{2+}(aq)+Ni(s)$$

Solution: First identify the oxidation and reduction half reactions:

$oxidation: Mg(s)\rightarrow Mg^{2+}(aq) + 2e^-$

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

Mg loses 2 electrons during the reaction to become Mg2+ which means that it is oxidized and found at the anode. Ni2+ gains 2 electrons and is reduced to Ni and found at the cathode. In cell notation, the anode is on the left side and the cathode is on the right. The cathode and anode are the two half cells that are separated by a double line that represents the salt bridge. The solid metal represents the electrode of the reaction and is found at the very end of either end. Within each half cell, reactants are listed first, then products. Each chemical species of a different phase in the reaction is separated by a single vertical line.

$Mg(s)|Mg^{2+}(aq)||Ni^{2+}(aq)|Ni(s)$

b. $$2Ag^+(aq)+Cu(s)\rightarrow Cu^{2+}(aq)+2Ag(s)$$

Solution: First identify the oxidation and reduction half reactions:

$oxidation: Cu(s)\rightarrow Cu^{2+}(aq) + 2e^-$

$reduction: 2Ag^+(aq) + 2e^-\rightarrow 2Ag(s)$

2Ag+ gains 2x1 electrons during the reaction to become 2Ag which means that it is reduced and found at the cathode. Cu loses 2 electrons and is oxidized to Cu2+ and found at the anode. The anode is always at the left of the notation and the cathode is at the right. The cathode and anode are the two half cells that are separated by a double line that represents the salt bridge. The solid metal represents the electrode of the reaction and is found at the very end of either end. Within each half cell, reactants are listed first, then products. Each chemical species in the reaction is separated by a single vertical line.

$Cu(s)|Cu^{2+}(aq)||Ag^+(aq)|Ag(s)$

c. $$Mn(s) +Sn(NO_3)_2(aq)\rightarrow Mn(NO_3)_2(aq)+Sn(s)$$

Solution: First identify the oxidation and reduction half reactions:

$oxdation: Mn(s)\rightarrow Mn^{2+}(aq) + 2e^-$

$reduction: Sn^{2+}(aq) + 2e^-\rightarrow Sn(s)$

In the compound Sn(NO3)2, Sn has a charge of 2+. Sn2+ gains 2 electrons during the reaction to become Sn(s) which means that it is reduced and found at the cathode. Mn loses 2 electrons and is oxidized to Mn2+ and found at the anode. The anode is always at the left of the notation and the cathode is at the right. The cathode and anode are the two half cells that are separated by a double line that represents the salt bridge. The solid metal represents the electrode of the reaction and is found at the very end of either end. Within each half cell, reactants are listed first, then products. Each chemical species in the reaction is separated by a single vertical line.

$Mn(s)|Mn^{2+}(aq)||Sn^{2+}(aq)|Sn(s)$

d. $$3CuNO_3(aq)+Au(NO_3)_3(aq)\rightarrow 3Cu(NO_3)_2(aq)+Au(s)$$

Solution: First identify the oxidation and reduction half reactions:

$oxidation: 3Cu^+(aq)\rightarrow 3Cu^{2+}(aq) + 3e^-$

$reduction: Au^{3+}(aq) + 3e^-\rightarrow Au(s)$

In the compound Au(NO3)3, Au has a charge of 3+. Au3+ gains 3 electrons during the reaction to become Au(s) which means that it is reduced and found at the cathode. The copper in CuNO3 has a charge of +1. Cu loses 1 electron to have a charge of +2 and forms the complex Cu(NO3)2. This is an oxidation half reaction and is found at the anode. The anode is always at the left of the notation and the cathode is at the right. The cathode and anode are the two half cells that are separated by a double line that represents the salt bridge. The solid metal represents the electrode of the reaction and is found at the very end of either end. Within each half cell, reactants are listed first, then products. Each chemical species in the reaction is separated by a single vertical line. Because copper is in ion form, a platinum electrode will be used at the anode. Since Cu+ and Cu2+ are both aqueous, they are separated by a comma instead of a vertical line.

$Pt(s)|Cu^+(aq), Cu^{2+}(aq)||Au^{3+}(aq)|Au(s)$

### A17.2.1

(a) $$\ce{Mg}(s)│\ce{Mg^2+}(aq)║\ce{Ni+}(aq)│\ce{Ni}(s)$$; (b) $$\ce{Cu}(s)│\ce{Cu^2+}(aq)║\ce{Ag+}(aq)│\ce{Ag}(s)$$; (c) $$\ce{Mn}(s)│\ce{Mn^2+}(aq)║\ce{Sn^2+}(aq)│\ce{Sn}(s)$$; (d) $$\ce{Pt}(s)│\ce{Cu+}(aq),\: \ce{Cu^2+}(aq)║\ce{Au^3+}(aq)│\ce{Au}(s)$$

### Q17.2.2

Given the following cell notations, determine the species oxidized, species reduced, and the oxidizing agent and reducing agent, without writing the balanced reactions.

1. $$\ce{Mg}(s)│\ce{Mg^2+}(aq)║\ce{Cu^2+}(aq)│\ce{Cu}(s)$$
2. $$\ce{Ni}(s)│\ce{Ni^2+}(aq)║\ce{Ag+}(aq)│\ce{Ag}(s)$$

### S17.2.2

The anode half-cell comes first in the cell notation (the species involved in the oxidation half-reaction, to the left of the double vertical lines), followed by the cathode (species involved in the reduction half-reaction, to the right of the double vertical lines). With the given half reactions, the reactants are described first, followed by the products. So as one reads the cell notation, their eyes move in the direction of electron flow. Its also important to note that spectator ions are not included in the cell notation. The double vertical line represents the salt bridge connecting the two different solutions. Each single vertical line represents a phase boundary, separating species of different phases. The phase of each chemical species is shown in paranthesis, and if the electrolytes are not in standard conditions then concentrations/pressure are included in parantheses along with the phase notation. If no concentration/pressure is noted then the reaction is assumed to be taking place under standard conditions (1.00 M/ 1.00 atm and 298K). The oxidizing agent is the species that accepts electrons. Thus, the oxidizing agent is also the species that is reduced. The reducing agent is the specie that donates electrons. Thus, it is also the specie that is oxidized.

1. From the cell notation, we can tell that Mg(s) and Mg2+(aq) are involved in the oxidation half-reaction since they are written to the left of the double vertical lines meaning

$\text{ Mg}(\mathit{ s}) \rightarrow \text{ Mg}^\text{ 2+}(\mathit{ aq})+\text{ 2 e}^-$

and Cu(s) and Cu2+(aq) are involved in the reduction half-reaction since they are written to the right of the double vertical lines meaning

$\text{ Cu}^\text{ 2+}(\mathit{ aq})+\text{ 2 e}^- \rightarrow \text{ Cu}(\mathit{ s})$

Mg2+(aq) is the specie that donates electrons so it is the reducing agent and the specie that is oxidized. Cu2+(aq) is the specie that accepts electrons so it is the oxidizing agent and the species that is reduced.

1. From the cell notation, we can tell that Ni(s) and Ni2+(aq) are involved in the oxidation half-reaction since they are written to the left of the double vertical lines meaning

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

and Ag(s) and Ag+(aq) are involved in the reduction half-reaction since they are written to the right of the double vertical lines meaning

$\text{ Ag}^\text{ +}(\mathit{ aq})+\text{ e}^- \rightarrow \text{ Ag}(\mathit{ s})$

Ni2+(aq) is the specie that donates electrons so it is the reducing agent and the specie that is oxidized. Ag+(aq) is the specie that accepts electrons so it is the oxidizing agent and the species that is reduced.

### A17.2.2

a) Mg2+(aq) = oxidized ; Cu2+(aq) = reduced ; Mg2+(aq)= reducing agent ; Cu2+(aq)= oxidizing agent

b) Ni2+(aq) = oxidized ; Ag+(aq) = reduced ; Ni2+(aq) = reducing agent ; Ag+(aq) = oxidizing agent

### Q17.2.3

For the cell notations in the previous problem, write the corresponding balanced reactions.

### S17.2.3

[Mg(s)+Cu^{_{2+}}(aq)\rightarrow Mg^{_{2+}}(aq)+Cu(s)\]

Solution:

We are asked to write the corresponding balanced reactions based on the cell notation.

First, we need to identify and sperate the half reactions. Note that 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:$$Mg(s)\rightarrow Mg^{_{2+}}(aq)$$

Reduction half reaction: $$Cu^{_{2+}}(aq)\rightarrow Cu(s)$$

Second, we need to balance each half reactions.

Each half equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:

1. Balance elements in the equation other than O and H.
2. Balance the oxygen atoms by adding the appropriate number of water (H2O) molecules to the opposite side of the equation.
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.
4. Add up the charges on each side. Make them equal by adding enough electrons (e-) to the more positive side.
5. * If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH- must be added to turn the remaining H+ into water molecules.

In this problem, as no O or H atoms are involved, we simply need to balance all elements, than adding e- to equal the changes.

Oxidation half reaction: $$Mg(s)\rightarrow Mg^{_{2+}}(aq)+2e^{_{-}}$$

Reduction half reaction:$$Cu^{_{2+}}(aq)+2e^{_{-}}\rightarrow Cu(s)$$

Finally, we adding two half equations together to from the final balanced reaction.

Half reactions are added together by this order:

1. Making sure the numbers of e- in each half reactions are equal. If they are not equal, they must be multiplied by the lowest common multiple to be made the same.
2. Adding to half equation together. Canceling out the electrons and common terms to form one balanced equation.

In this problem, numbers of e- in each half reactions are equal. Thus:

Balanced Equation : $$Mg(s)+Cu^{_{2+}}(aq)\rightarrow Mg^{_{2+}}(aq)+Cu(s)$$

Check the equation to make sure it is balanced.

$Ni(s)+2Ag^{_{+}}(aq)\rightarrow Ni^{_{2+}}(aq)+2Ag(s)$

Solution:

We are asked to write the corresponding balanced reactions based on the cell notation.

First, we need to identify and sperate the half reactions. Note that 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: $$Mg(s)\rightarrow Mg^{_{2+}}(aq)$$

Reduction half reaction: $$Ag^{_{+}}(aq)\rightarrow Ag(s)$$

Second, we need to balance each half reactions.

In this problem, as no O or H atoms are involved, we simply need to balance all elements, than adding e- to equal the changes.

Oxidation half reaction: $$Ni(s)\rightarrow Ni^{_{2+}}(aq)+2e^{_{-}}$$

Reduction half reaction: $$Ag^{_{+}}(aq)+e^{_{-}}\rightarrow Ag(s)$$

Finally, we adding two half equations together to from the final balanced reaction.

Half reactions are added together by this order:

1. Making sure the numbers of e- in each half reactions are equal. If they are not equal, they must be multiplied by the lowest common multiple to be made the same.
2. Adding to half equation together. Canceling out the electrons and common terms to form one balanced equation.

In this problem, numbers of e- in each half reactions are not equal. To make it equal, the reduction half reaction need to be multiply by two. Then adding two equations together. Thus:

Oxidation half reaction: $$Ni(s)\rightarrow Ni^{_{2+}}(aq)+2e^{_{-}}$$

Reduction half reaction: $$2Ag^{_{+}}(aq)+2e^{_{-}}\rightarrow 2Ag(s)$$

Balanced Equation : $$Ni(s)+2Ag^{_{+}}(aq)\rightarrow Ni^{_{2+}}(aq)+2Ag(s)$$

### A17.2.3

(a) $$\ce{Mg}(s)+\ce{Cu^2+}(aq)⟶\ce{Mg^2+}(aq)+\ce{Cu}(s)$$; (b) $$\ce{2Ag+}(aq)+\ce{Ni}(s)⟶\ce{Ni^2+}(aq)+\ce{2Ag}(s)$$

### Q17.2.4

Balance the following reactions and write the reactions using cell notation. Ignore any inert electrodes, as they are never part of the half-reactions.

1. $$\ce{Al}(s)+\ce{Zr^4+}(aq)⟶\ce{Al^3+}(aq)+\ce{Zr}(s)$$
2. $$\ce{Ag+}(aq)+\ce{NO}(g)⟶\ce{Ag}(s)+\ce{NO3-}(aq) \hspace{20px} \textrm{(acidic solution)}$$
3. $$\ce{SiO3^2-}(aq)+\ce{Mg}(s)⟶\ce{Si}(s)+\ce{Mg(OH)2}(s) \hspace{20px} \textrm{(basic solution)}$$
4. $$\ce{ClO3-}(aq)+\ce{MnO2}(s)⟶\ce{Cl-}(aq)+\ce{MnO4-}(aq) \hspace{20px} \textrm{(basic solution)}$$

### S17.2.4

1: Al(s)+Zr4+(aq) ⟶ Al3+(aq)+Zr(s)

Half Reactions

Al(s) ⟶ Al3+(aq) (oxidation)

Zr4+(aq) ⟶ Zr(s) (reduction)

Step 1 Elements other than O and H are already balanced

Step 2 No oxygen atoms so no need to add water molecules

Step 3 No water molecules added so no need to add H+ ions

Step 4 Al(s) ⟶ Al3+(aq)+3e-

Zr4+(aq)+4e- ⟶ Zr(s)

Step 5 4(Al(s) ⟶ Al3+(aq)+3e-)

3(Zr4+(aq)+4e- ⟶ Zr(s))

Step 6 4Al(s)+3Zr4+(aq)+12e- ⟶ 4Al3+(aq)+12e-+3Zr(s)

4Al(s)+3Zr4+(aq) ⟶ 4Al3+(aq)+3Zr(s)

Cell Notation

Al half-equation(oxidation) will be written down first and the the Zr half equation(reduction) will come after the salt bridge. Do not forget the concentrations but since they are not given you do not need to include them.

Al(s)|Al3+||Zr4+|Zr(s)

2:Ag+(aq)+NO(g)⟶ Ag(s)+NO3-(aq)(acidic solution)

Half Reactions

Ag+(aq) ⟶ Ag(s) (reduction)

NO(g) ⟶ NO3-(aq) (oxidation)

Step 1 Elements other than O and H are already balanced

Step 2 No oxygen atoms so no need to add water molecules for first half-equation

NO(g)+2H2O ⟶ NO3-(aq)

Step 3 No water molecules added in the first half-reaction so no need to add H+ ions in first half-equation

NO(g)+2H2O ⟶ NO3-(aq)+4H+

Step 4 Ag+(aq)+e-⟶ Ag(s)

NO(g)+2H2O ⟶ NO3-(aq)+4H++3e-

Step 5 3(Ag+(aq)+e-⟶ Ag(s))

NO(g)+2H2O ⟶ NO3-(aq)+4H++3e-

Step 6 3Ag+(aq)+3e-+NO(g)+2H2O ⟶ 3Ag(s)+NO3-(aq)+4H++3e-

3Ag+(aq)+NO(g)+2H2O⟶ 3Ag(s)+NO3-(aq)+4H+

Cell Notation

NO half-equation(oxidation) will be written down first and the the Ag half equation(reduction) will come after the salt bridge. You Begin with Pt because half-equation does not have a solid.

Pt(s)|NO(g)|NO3-||Ag+|Ag(s)

3:SiO2−3(aq)+Mg(s)⟶ Si(s)+Mg(OH)2(s)(basic solution)

Half Reactions

SiO2−3(aq) ⟶ Si(s) (reduction)

Mg(s) ⟶ Mg(OH)2(s) (oxidation)

Step 1 Elements other than O and H are already balanced

Step 2 SiO2−3(aq) ⟶ Si(s)+2H2O

Mg(s)+2H2O⟶ Mg(OH)2(s)

Step 3 SiO2−3(aq)+4H+ ⟶ Si(s)+2H2O

Mg(s)+2H2O⟶ Mg(OH)2(s)+2H+

Step 4 SiO2−3(aq)+4H++e-⟶ Si(s)+2H2O

Mg(s)+2H2O⟶ Mg(OH)2(s)+2H++2e-

Step 5 2(SiO2−3(aq)+4H++e-⟶ Si(s)+2H2O)

Mg(s)+2H2O⟶ Mg(OH)2(s)+2H++2e-

Step 6 2SiO2−3(aq)+8H++2e-+Mg(s)+2H2O⟶ 2Si(s)+4H2O+Mg(OH)2(s)+2H++2e-

2SiO2−3(aq)+6H++Mg(s)⟶ 2Si(s)+2H2O+Mg(OH)2(s)

*Basic Solution so must add 6OH- to both sides since we have 6H+ left on one side

2SiO2−3(aq)+6H++Mg(s)+6OH-⟶ 2Si(s)+2H2O+Mg(OH)2(s)+6OH-

2SiO2−3(aq)+6H2O+Mg(s)⟶ 2Si(s)+2H2O+Mg(OH)2(s)+6OH-

2SiO2−3(aq)+4H2O+Mg(s)⟶ 2Si(s)+Mg(OH)2(s)+6OH-

Cell Notation

Mg half-equation(oxidation) will be written down first and the the Si half equation(reduction) will come after the salt bridge.

Mg(s),Mg(OH)2(s)|| SiO2−3|Si(s)

4:ClO3(aq)+MnO2(s)⟶ Cl(aq)+MnO4(aq)(basic solution)

Half Reactions

ClO3(aq)⟶ Cl(aq) (reduction)

MnO2(s) ⟶ MnO4(aq) (oxidation)

Step 1 Elements other than O and H are already balanced

Step 2 ClO3(aq)⟶ Cl(aq) +3H2O

MnO2(s) +2H2O⟶ MnO4(aq)

Step 3 ClO3(aq)+6H+⟶ Cl(aq) +3H2O

MnO2(s) +2H2O⟶ MnO4(aq) +4H+

Step 4 ClO3(aq)+6H++6e-⟶ Cl(aq) +3H2O

MnO2(s) +2H2O⟶ MnO4(aq) +4H++3e-

Step 5 ClO3(aq)+6H++6e-⟶ Cl(aq) +3H2O

2(MnO2(s) +2H2O⟶ MnO4(aq) +4H++3e-)

Step 6 ClO3(aq)+6H++6e-+2MnO2(s) +4H2O⟶ Cl(aq) +3H2O+2MnO4(aq) +8H++6e-

ClO3(aq)+2MnO2(s) +H2O⟶ Cl(aq) +2MnO4(aq) +2H+

*Basic Solution so must add 2OH- to both sides since we have 2H+

ClO3(aq)+2MnO2(s) +H2O+2OH-⟶ Cl(aq) +2MnO4(aq) +2H++2OH-

ClO3(aq)+2MnO2(s) +H2O+2OH-⟶ Cl(aq) +2MnO4(aq) +2H2O

ClO3(aq)+2MnO2(s) +2OH-⟶ Cl(aq) +2MnO4(aq) +H2O

Cell Notation

Mn half-equation(oxidation) will be written down first and the the Cl half equation(reduction) will come after the salt bridge. You end with Pt because half-equation does not have a solid.

MnO2(s)|MnO4||ClO3(1.0M),Cl|Pt(s)

### A17.2.4

a)4Al(s)+3Zr4+(aq) ⟶ 4Al3+(aq)+3Zr(s)

Al(s)|Al3+||Zr4+|Zr(s)

b)3Ag+(aq)+NO(g)+2H2O⟶ 3Ag(s)+NO3-(aq)+4H+

Pt(s)|NO(g)|NO3-||Ag+|Ag(s)

c)2SiO2−3(aq)+4H2O+Mg(s)⟶ 2Si(s)+Mg(OH)2(s)+6OH-

Mg(s),Mg(OH)2(s)|| SiO2−3|Si(s)

d)ClO3(aq)+2MnO2(s) +2OH-⟶ Cl(aq) +2MnO4(aq) +H2O

MnO2(s)|MnO4||ClO3(1.0M),Cl|Pt(s)

### Q17.2.5

Identify the species oxidized, species reduced, and the oxidizing agent and reducing agent for all the reactions in the previous problem.

### S17.2.5

Note: To find the species that is is being oxidized, notice that it is the species that is known as the reducing agent because it is the species that is losing electrons.

In equation 1. Al(s) undergoes the reaction and transforms from a 0 charge to a positive 3 charge. Because it lost electrons it is being oxidized thus it is considered the reducing agent .

In equation 2. NO(g) undergoes the reaction and transforms from a 0 charge to a positive 3 charge. Because it lost electrons it is being oxidized thus it is considered the reducing agent .

In equation 3. Mg(s) undergoes the reaction and transforms from a positive 0 charge to a positive 2 charge. Because it lost electrons it is being oxidized thus it is considered the reducing agent .

In equation 4. MnO2(s) because the Mn ion undergoes the reaction and transforms from a 4 charge to a positive 7 charge. Because it lost electrons it is being oxidized thus it is considered the reducing agent .

Note: To find the species that is being reduced, notice that it is the species that is known as the oxidizing agent because it is the species that is gaining electrons.

In equation 1. Zr4+ undergoes the reaction and changes from a positive 4 charge to a Zr with charge 0.

In equation 2. Ag+ undergoes the reaction and changes from a positive 1 charge to a Ag with charge 0.

In equation 3. SiO23- because the Si ion undergoes the reaction and changes from a positive 1 charge to a Si with charge 0.

In equation 4. ClO3- because the Cl ion undergoes the reaction and changes from a positive 5 charge to a Zr with charge -1.

### A17.2.5

Species oxidized = reducing agent: (a) Al(s); (b) NO(g); (c) Mg(s); and (d) MnO2(s); Species reduced = oxidizing agent: (a) Zr4+(aq); (b) Ag+(aq); (c) $$\ce{SiO3^2-}(aq)$$; and (d) $$\ce{ClO3-}(aq)$$

### Q17.2.6

From the information provided, use cell notation to describe the following systems:

1. In one half-cell, a solution of Pt(NO3)2 forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO3)2 solution with all solute concentrations 1 M.
2. The cathode consists of a gold electrode in a 0.55 M Au(NO3)3 solution and the anode is a magnesium electrode in 0.75 M Mg(NO3)2 solution.
3. One half-cell consists of a silver electrode in a 1 M AgNO3 solution, and in the other half-cell, a copper electrode in 1 M Cu(NO3)2 is oxidized.

### S17.2.6

1. a) In order to start this problem, you need to write the half-reactions of each to determine which species is being oxidized and which species is being reduced. After the half reactions are balanced, you can then look at the reduction potentials of the reactions to see which reactant is being reduced and which is oxidized. The compound with the higher reduction potential will be the compound that is reduced in the reaction.

For platinum: $Pt^{2+}(aq)\rightarrow Pt(s)$

The $$NO_{3}^{-}$$ helps us determine the charge of Pt. Since there is 2 $$NO_{3}^{-}$$ions, Pt needs to be 2+ to balance it out. We now need to add the necessary electrons in order for both the reactants side and the products side to be equal in charge. Since the reactant side is +2 and the product side is 0, we need to add 2 electrons to the reactant side, giving the balanced half reaction.

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

We now need to see if Pt is getting reduced or oxidized. Since Pt goes from a +2 charge to a 0 charge, it is ultimately being reduced.

Now we move onto the next half-reaction.

For Copper: $$Cu(s)\rightarrow Cu^{2+}(aq)$$

Again, we use $$NO_{3}^{-}$$ to help us determine the charge of Cu. We now need to balance it out with electrons. This time, the product side is 2+ and the reactant side is 0. We need to add two electrons to the product side, giving the balanced half reaction.

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

Because the two electrons cancel out, the overall reaction is $Cu(s)+Pt^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pt(s)$

Writing in cell notation, we go from anode to cathode. The electrodes are on the edges and the change in phase is separated by one line. The double line symbolizes the salt bridge. Add the phases and the molarity as well. We don't need to add the multiplication factors because E° doesn't depend on it.

So the cell notation for this is

$Cu(s)|Cu^{2+}(aq, 1 M)||Pt^{2+}(aq, 1 M)|Pt(s)$

b) This time, the anode and cathode have been identified already. Mg is the anode, so it's being oxidized.

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

We add two electrons to the product side since its +2 and the reactant side is 0.

Au is the cathode, so it is being reduced.

$Au^{3+}(aq)+3e^{-}\rightarrow Au(s)$

We add three electrons to the reactant side since its +3 and the product side is 0.

Because the electrons don’t cancel out(3/2 ratio), we need to multiply the half reaction by 3 for the anode and multiply the half reaction by 2 for the cathode. We multiply by these numbers because those are the least common multiples.

The overall reaction is

$3Mg(s)+2Au^{3+}(aq)\rightarrow 2Au(s)+3Mg^{2+}(aq)$

Writing in cell notation, we go from anode to cathode. The electrodes are on the edges and the change in phase is separated by one line. The double line symbolizes the salt bridge. Add the phases and the molarity as well. We don't need to add the multiplication factors because E doesn't depend on it. E is not in standard conditions(25 C, 1 atm, 1 M) since Mg and Au are not 1 M.

So the cell notation for this is

$Mg(s)|Mg^{2+}(aq, 0.75M)||Au^{3+}(aq, 0.55M)|Au(s)$

c) Copper is already stated as being oxidized so it is the anode.

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

We add two electrons to the product side since its +2 and the reactant side is 0.

For Ag, it is the cathode and is therefore being reduced.

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

We add one electron to the reactant side since its +1 and the product side is 0.

Since its a 2/1 ratio, we multiply the half reaction of Ag by 2 so that the electrons cancel out.

The overall reaction becomes

$2Ag^{+}(aq)+Cu(s) \rightarrow Cu^{2+}(aq)+ 2Ag(s)$

Writing in cell notation, we go from anode to cathode. The electrodes are on the edges and the change in phase is separated by one line. The double line symbolizes the salt bridge. Add the phases and the molarity as well. We don't need to add the multiplicative factors because E° doesn't depend on it.

So the cell notation for this is

$Cu(s)| Cu^{2+}(aq, 1 M)||Ag^{+}(aq, 1 M)|Ag(s)$

### A17.2.6

a. Pt(S) | Pt+2 (aq, 1.0 M) || Cu+2 (aq, 1.0 M) | Cu(s)

b. Au(s) | Au+3 (aq, 0.55 M) || Mg+2 (aq, 0.75 M) | Mg(s)

c. Ag(s) | Ag+ (aq, 1.0 M) || Cu+2 (aq, 1.0 M) | Cu(s)

### Q17.2.7

Why is a salt bridge necessary in galvanic cells like the one below?

In this standard galvanic cell, the half-cells are separated; electrons can flow through an external wire and become available to do electrical work.

### A17.2.7

In galvanic cells like the one pictured above, the salt bridge is necessary to maintain a balance of charges within the internal circuit, as electrons are continuously moving from one half cell to the other. Specifically, in the cell, electrons flow from the anode to the cathode. The oxidation reaction that occurs at the anode generates both electrons and positively charged ions. The electrons move through the wire toward the cathode, leaving an unbalanced positive charge in this vessel (anode). Therefore, in order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell. A similar (but reversed) situation is found in the cathodic cell, where the migration of positively charged ions from the salt bridge are moving into this half cell in order to counteract the influx of electrons. In the absence of a salt bridge, opposing charges will continue to build up in each half cell and the reaction will be brought to a halt due to the charge imbalance. In essence, with a salt bridge, which is usually composed of an inert electrolyte such as sodium nitrate (which is pictured above), each half-cell remains electrically neutral and a current can continue to flow through the circuit.

### Q17.2.8

For each of the following reactions, list the substance reduced, the substance oxidized, the reducing agent, and the oxidizing agent.

1. 6H+ + MnO4- + 5SO32- → 5SO42- + 2Mn2+ + 3H2O
2. 8H+ + Cr2O72- + 6HI → 2Cr3+ + 312 + 7H2O
3. 3Cl2 + 6OH- → CIO3- + 5Cl- + 3H2O

### S17.2.8

When a substance is reduced, it's oxidation number is lowered as the substance is gaining electrons. The reduced substance gains electrons from the oxidized substance which loses its electrons. When a substance is oxidized, it's oxidation number increases as the substance is losing electrons.The substance being oxidized is the reducing agent as it reduces another species in the equation. The substance being reduced is the oxidizing agent because its accepting electrons allowing the other substance to become oxidized.

1. Mn is the substance reduced. MnO4- is the oxidizing agent. S is the substance oxidized. SO32- is the reducing agent.

2. Cr is the substance reduced. Cr2O72- is the oxidizing agent. I is being oxidized. HI is the reducing agent.

3. Cl is the substance reduced. Cl2 (gas) is the oxidizing agent. Cl is also the substance oxidized (Cl2-->ClO3-). Cl2 is the reducing agent.

### Q17.2.9

• An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.
• An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode part of the anode or cathode? Explain.

### A17.2.9

A cathode in an electrochemical cell is where reduction occurs. This means that the electrode is accepting electrons. For example, in a zinc-copper galvanic cell, copper is accepting electrons from zinc; thus, copper is gaining mass at the cathode. An anode in an electrochemical cell is where oxidation occurs. This means the electrode is donating electrons. When zinc donates its electrons to copper in a zinc-copper galvanic cell, zinc ions would be displaced into the solution and thus zinc loses mass.

Active electrodes participate in the oxidation-reduction reaction. Since metals form cations, the electrode would lose mass if metal atoms in the electrode were to oxidize and go into solution. Oxidation occurs at the anode.

### Q17.2.10

The mass of three different metal electrodes, each from a different galvanic cell, were determined before and after the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes. The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, given the label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass. Make an educated guess as to which electrodes were active and which were inert electrodes, and which were anode(s) and which were the cathode(s).

### S17.2.10

When a galvanic cell circuit is closed, the electrons flow from the anode to the cathode. As a reaction progresses, the anode loses mass since it is oxidized (loses electrons) into metal cations that go into solution, while the cathode gains mass as the cations in solution are reduced (gains electrons) and deposited on the cathode. If the electrode doesn't change in mass, it must be inert since it's only serving as a medium for electron transfer.

Based on that, electrode A is the cathode since it increased in mass; electrode B is the inert cathode since it did not change in mass; and electrode C is the anode since it lost mass.

## 17.3: Standard Reduction Potentials

### Q17.3.1

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

1. $$\ce{Mg}(s)+\ce{Ni^2+}(aq)⟶\ce{Mg^2+}(aq)+\ce{Ni}(s)$$
2. $$\ce{2Ag+}(aq)+\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)$$
3. $$\ce{Mn}(s)+\ce{Sn(NO3)2}(aq)⟶\ce{Mn(NO3)2}(aq)+\ce{Sn}(s)$$
4. $$\ce{3Fe(NO3)2}(aq)+\ce{Au(NO3)3}(aq)⟶\ce{3Fe(NO3)3}(aq)+\ce{Au}(s)$$

### S17.3.1

The standard cell potential (E°cell) is the voltage/potential difference produced from the oxidation (happens at the anode) and reduction (happens at the cathode) half reactions in the galvanic cell. The first tool that we need to calculate for E°cell is the Standard Reduction Potentials, where negative potentials (top) are more likely to oxidize than the positive potentials (bottom) which are more likely to reduce. The second tool that we need is the formula, E°cell= E°(cathode)-E°(anode). From the result, we can determine the spontaneity of the reaction by looking at the sign of the E°cell, if E°cell is positive then the reaction is spontaneous and non-spontaneous if E°cell is negative.

1. Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)

Step 1: Write the cell half reactions and their respective cell potentials:

$Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)$ with $$E^o=-2.356\, V$$

$Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)$ with $$E^o=-0.25\, V$$

Step 2: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:

Anode (oxidation half reaction): $Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)$ with $$E^o=-2.356\, V$$

Cathode (reduction half reaction): $Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)$ with $$E^o=-0.25\, V$$

Step 3: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (-0.25 V)-(-2.356 V) =+2.106 V$

You can also determine the standard cell potential by adding the half reactions together. This allows us to add the potential of the anode and cathode together. Remember that the charge (sign) of the potential changes when you flip the reaction.

Example:

Anode (oxidation half reaction): $Mg(s)\rightarrow Mg^{2+}(aq)+2e^{-}$ with $$E^o=2.356\, V$$

Cathode (reduction half reaction): $Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)$ with $$E^o=-0.25\, V$$

$2.356 + (-0.25) = +2.106 V$

This reaction is spontaneous because the cell potential (E°) is positive, +2.106 V.

2. 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)

Step 1: Write the cell half reactions and their respective cell potentials:

$Ag^{+}(aq)+e^{-}\rightarrow Ag(s)$ with $$E^o=+0.800\, V$$

$Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)$ with $$E^o=+0.340\, V$$

Step 2: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:

**One important thing to remember is that when we multiply a coefficient to balance out the electrons between the half reactions, the cell potential of each reaction doesn't change.**

Anode (oxidation half reaction) : $Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)$ with $$E^o=+0.340\, V$$

Cathode (reduction half reaction) : $2(Ag^{+}(aq)+e^{-}\rightarrow Ag(s))$ with $$E^o=+0.800\, V$$

Step 3: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+0.800 V)-(+0.340 V) =+0.460 V$

This reaction is spontaneous because the cell potential (E°) is positive, +0.460V.

3. Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)

Step 1: Rewrite the overall reaction in the net ionic equation form (because it separates in solution) and cancel out the spectator ion, in this case is NO3-, and balance the equation if needed :

$Mn(s)+Sn^{2+}(aq)+2NO^{-}_3(aq)\rightarrow Mn^{2+}(aq)+2NO^{-}_3(aq)+Sn(s)$

Which simplify into: $Mn(s)+Sn^{2+}(aq)\rightarrow Mn^{2+}(aq)+Sn(s)$ where the spectator ion, NO3-, is taken out from both sides.

Step 2: Write the cell half reactions and their respective cell potentials:

$Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)$ with $$E^o=-1.19\, V$$

$Sn^{2+}(aq)+2e^{-}\rightarrow Sn(s)$ with $$E^o=-0.137\, V$$

Step 3: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:

Anode (oxidation half reaction) : $Mn^{2+}(aq)+2e^{-}\rightarrow Mn(s)$ with $$E^o=-1.19\, V$$

Cathode (reduction half reaction) : $Sn^{2+}(aq)+2e^{-}\rightarrow Sn(s)$ with $$E^o=-0.137\, V$$

Step 4: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (-0.137 V)-(-1.19 V) =+1.053 V$

This reaction is spontaneous because the cell potential (E°) is positive, +1.053 V.

4. 3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)

Step 1: Rewrite the overall reaction in the net ionic equation form (because it separates in solution) and cancel out the spectator ion, in this case is NO3-, and balance the equation if needed:

$3Fe^{2+}(aq)+6NO^{-}_3(aq)+Au^{3+}(aq)+3NO^{-}_3(aq)\rightarrow 3Fe^{3+}(aq)+9NO^{-}_3+Au(s)$

Which simplify into: $3Fe^{2+}(aq)+Au^{3+}(aq)\rightarrow 3Fe^{3+}(aq)+Au(s)$

Step 2: Write the cell half reactions and their respective cell potentials:

$Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)$ with $$E^o=+0.771\, V$$

$Au^{3+}(aq)+3e^{-}\rightarrow Au(s)$ with $$E^o=+1.52\, V$$

Step 3: Using the Standard Reduction Potentials table, determine where each half reaction happens (anode or cathode) and balance the electrons between the half reactions:

**One important thing to remember is that when we multiply a coefficient to balance out the electrons between the half reactions, the cell potential of each reaction doesn't change.**

Anode (oxidation half reaction) : $3(Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq))$ with $$E^o=+0.771\, V$$

Cathode (reduction half reaction) : $Au^{3+}(aq)+3e^{-}\rightarrow Au(s)$ with $$E^o=+1.52\, V$$

Step 4: Calculate the standard cell potential, E°cell, and determine the spontaneity of the reaction:

$E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = (+1.52 V)-(+0.771 V) =+0.749V$

This reaction is spontaneous because the cell potential (E°) is positive, +0.749 V.

### A17.3.1

(a) +2.106 V (spontaneous); (b) +0.460 V (spontaneous); (c) +1.053 V (spontaneous); (d) +0.749 V (spontaneous)

### Q17.3.2

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

1. $$\ce{Mn}(s)+\ce{Ni^2+}(aq)⟶\ce{Mn^2+}(aq)+\ce{Ni}(s)$$
2. $$\ce{3Cu^2+}(aq)+\ce{2Al}(s)⟶\ce{2Al^3+}(aq)+\ce{2Cu}(s)$$
3. $$\ce{Na}(s)+\ce{LiNO3}(aq)⟶\ce{NaNO3}(aq)+\ce{Li}(s)$$
4. $$\ce{Ca(NO3)2}(aq)+\ce{Ba}(s)⟶\ce{Ba(NO3)2}(aq)+\ce{Ca}(s)$$

### S17.3.2

Since we are looking for the standard cell potential, we will be using following equation:

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$

To find out the $$\mathrm{E^{\circ}_{cathode}}$$ and the $$\mathrm{E^{\circ}_{anode}}$$, we will need to refer to a Table of Standard Reduction Potentials like Table P2.

As we know, spontaneity is determined by Gibbs Free Energy $$\mathrm{(\Delta{G})}$$ in the following manner:

• $$\mathrm{-\, \Delta{G}}$$ means the reaction is spontaneous
• $$\mathrm{+\, \Delta{G}}$$ means that the reaction is non-spontaneous

To determine the spontaneity of a redox equation, we refer the following equation:

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

By calculating the standard cell potential, we can predict the spontaneity of the reaction without actually doing to the calculation in the following manner:

• $$\mathrm{+\, E^{\circ}_{cell}}$$ means that the reaction will be spontaneous (because the positive cell potential will keep the negative sign in the equation)
• $$\mathrm{-\, E^{\circ}_{cell}}$$ means that the reaction will be non-spontaneous (because the negative cell potential cancels out the negative sign in the equation)
1. We are given the following chemical equation: $$\mathrm{Mn(s) + Ni^{2+}(aq) \rightarrow Mn^{2+}(aq) + Ni(s)}$$

First, let's write out the half-reactions (based on the reduction potential table):

$$\mathrm{Mn^{2+} + 2e^{-} \rightarrow Mn(s)}$$

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

Then, we look at the Table of Standard Reduction Potential values to get the $$\mathrm {E^{\circ}_{cell}}$$ of each half-reaction and determine the cathode and anode.

Half-Reaction $$\mathrm{\, E^{\circ}\, (V)}$$
$$\mathrm{Mn^{2+} + 2e^{-} \rightarrow Mn(s)}$$ –1.17 V Cathode
$$\mathrm{Ni^{2+}(aq) + 2e^{-} \rightarrow Ni(s)}$$ –0.257 V Anode

Now, we can use $$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$ to figure out our standard cell potential and determine its spoantaneity

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{-0.257\, V} -\; \mathrm{(-1.17\, V)}$$

$$\mathrm {E^{\circ}_{cell}} = \mathbf{+0.913\, V}$$

Since $$\mathrm {E^{\circ}_{cell}}$$ is positive, this reaction is spontaneous.

1. We are given the following chemical equation: $$\mathrm{3\, Cu^{2+}(aq) + 2\, Al(s) \rightarrow 2\, Al^{3+}(aq) + 2\, Cu(s)}$$

First, let's write out the half-reactions (based on the reduction potential table):

$$\mathrm{Cu^{2+}(aq) + e^{-} \rightarrow Cu^{+}(aq)}$$

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

Then, we look at the Table of Standard Reduction Potential values to get the $$\mathrm {E^{\circ}_{cell}}$$ of each half-reaction and determine the cathode and anode.

Half-Reaction $$\mathrm{\, E^{\circ}\, (V)}$$
$$\mathrm{Cu^{2+}(aq) + e^{-} \rightarrow Cu^{+}(aq)}$$ +0.3419 V Cathode
$$\mathrm{Al^{3+}(aq) + 3e^{-} \rightarrow Al(s)}$$ –1.676 V Anode

Now, we can use $$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$ to figure out our standard cell potential and determine its spoantaneity

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{+0.3419\, V} -\; \mathrm{(-1.676\, V)}$$

$$\mathrm {E^{\circ}_{cell}} = \mathbf{+2.018\, V}$$

Since $$\mathrm {E^{\circ}_{cell}}$$ is positive, this reaction is spontaneous.

1. We are given the following chemical equation: $$\mathrm{Na(s) + LiNO_3(aq) \rightarrow NaNO_3(aq) + Li(s)}$$

First, let's write out the half-reactions (based on the reduction potential table):

$$\mathrm{Li^{+} + e^{-} \rightarrow Li(s)}$$

$$\mathrm{Na^{+}(aq) + e^{-} \rightarrow Na(s)}$$

Then, we look at the Table of Standard Reduction Potential values to get the $$\mathrm {E^{\circ}_{cell}}$$ of each half-reaction and determine the cathode and anode.

Half-Reaction $$\mathrm{\, E^{\circ}\, (V)}$$
$$\mathrm{Li^{+} + e^{-} \rightarrow Li(s)}$$ –3.040 V Cathode
$$\mathrm{Na^{+}(aq) + e^{-} \rightarrow Na(s)}$$ –2.713 V Anode

Now, we can use $$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$ to figure out our standard cell potential and determine its spoantaneity

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{–3.040\, V} -\; \mathrm{(–2.713\, V)}$$

$$\mathrm {E^{\circ}_{cell}} = \mathbf{-0.327\, V}$$

Since $$\mathrm {E^{\circ}_{cell}}$$ is negative, this reaction is non-spontaneous.

1. We are given the following chemical equation: $$\mathrm{Ca(NO_3)_2(aq) + Ba(s) \rightarrow Ba(NO_3)_2(aq) + Ca(s)}$$

First, let's write out the half-reactions (based on the reduction potential table):

$$\mathrm{Ca^{2+} + 2e^{-} \rightarrow Ca(s)}$$

$$\mathrm{Ba^{2+} + 2e^{-} \rightarrow Ba(s)}$$

Then, we look at the Table of Standard Reduction Potential values to get the $$\mathrm {E^{\circ}_{cell}}$$ of each half-reaction and determine the cathode and anode.

Half-Reaction $$\mathrm{\, E^{\circ}\, (V)}$$
$$\mathrm{Ca^{2+} + 2e^{-} \rightarrow Ca(s)}$$ –2.84 V Cathode
$$\mathrm{Ba^{2+} + 2e^{-} \rightarrow Ba(s)}$$ –2.91 V Anode

Now, we can use $$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$ to figure out our standard cell potential and determine its spoantaneity

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{E^{\circ}_{cathode}} -\; \mathrm{E^{\circ}_{anode}}$$

$$\mathrm {E^{\circ}_{cell}} =\, \mathrm{–2.84\, V} -\; \mathrm{(–2.91\, V)}$$

$$\mathrm {E^{\circ}_{cell}} = \mathbf{+0.07\, V}$$

Since $$\mathrm {E^{\circ}_{cell}}$$ is positive, this reaction is spontaneous.

### A17.3.2

a. .913 (Positive; spontaneous)

b. 2.018

(Positive; spontaneous)

c.

=-.327 (negative; non-spontaneous)

d.

=+.07

(Positive; spontaneous)

### Q17.3.3

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions?

$\ce{Cu}(s)│\ce{Cu^2+}(aq)║\ce{Au^3+}(aq)│\ce{Au}(s)$

### S17.3.3

The components on the left side of the shorthand notation refers to the reaction going on in anode, while the double lines signify the salt bridge, and the right side refers to the reaction going on in the cathode. The solid Cu(s) and Au(s) are the electrodes in the galvanic cell. The single line between each of these elements separates phases, going from solid, to liquids, gases, and finally aqueous.

Remember that the anode releases electrons and is therefore where the oxidation reaction occurs, while the cathode gains electrons and is where the reduction reaction occurs.

Since the Cu(s) and Cu2+ are present on the left side, this will be at our anode and resulting oxidation reaction.

Since the Au(s) and Au3+ are present on the right side, this will be at our cathode and resulting reduction reaction.

Step 1: Write out the individual half reactions

$Cu^{2+} \rightarrow Cu(s)$

$Au^{3+} \rightarrow Au(s)$

Step 2: Balance charge of half reactions by adding e- half reactions

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

$Au^{3+}+3e^- \rightarrow Au(s)$

Step 3: Determine which element is the cathode and which element is the anode by using the standard reduction potential table found here. Whichever half reaction has a higher standard reduction potential, it will make a better oxidizing agent (cathode). The half-reaction that will be the cathode, its reactants will be used as the reactants in the overall reaction, and the half-reaction that is the anode, its products will be used as the reactants in the overall reaction.

Anode: $Cu^{2+}+2e^- \rightarrow Cu(s)\: \:\: \: E^{\circ}_{cell}=0.3419$

Cathode: $Au^{3+}+3e^- \rightarrow Au(s)\: \:\: \: E^{\circ}_{cell}=1.52$

Step 4: To combine the two half-reactions, multiply the half-reactions so that the electrons equal to a common multiple.

$3(Cu^{2+}+2e^- \rightarrow Cu(s))$

$2(Au^{3+}+3e^- \rightarrow Au(s))$

$3Cu^{2+}+6e^- \rightarrow 3Cu(s)$

$2Au^{3+}+6e^- \rightarrow 2Au(s)$

Step 5: Flip the appropriate reactions depending on whether it is the cathode or anode, and cancel the electrons.

$3Cu(s) \rightarrow 3Cu^{2+}$

$2Au^{3+} \rightarrow 2Au(s)$

Solution:

$3Cu(s) + 2Au^{3+} (aq) \rightarrow 3Cu^{2+} +2Au(s)$

Oxidation Half Reaction

• We can see that $$E^{\circ}_{anode}$$ = 0.3419 V for $$Cu^{2+}+2e^- \rightarrow Cu(s)$$

Reduction Half Reaction

• We can see that $$E^{\circ}_{cathode}$$ = 1.52 V for $$Au^{3+}+3e^- \rightarrow Au(s)$$

Step 1: Substitute reduction potential values into the equation, $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$

$$E^{\circ}_{cell}$$ = 1.52V - 0.3419V = +1.18V > 0

This positive value of this cell potential indicates that the reaction is spontaneous in the forward direction.

### A17.3.3

$$\ce{3Cu}(s)+\ce{2Au^3+}(aq)⟶\ce{3Cu^2+}(aq)+\ce{2Au}(s)$$; +1.18 V; spontaneous

### Q17.3.4

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

### S17.3.4

The fact that the reaction occurs at 25 °C and both of the electrodes have a value of 1M, this reaction is occurring under standard conditions.

Step 1: split the reaction into the silver nitrate solution and zinc nitrate. Due to the fact that Nitrate is in both metals, it is a spectator ion and due to this will not appear in the overall reaction.

(In Ag half reaction I just put e^- because she put e-)

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

Step 2: to determine if the reaction is spontaneous you must determine whether or not Ecell is negative or positive. A negative Ecell signifies a nonspontaneous reaction whereas a positive Ecell means that the reaction is spontaneous.

Determine the standard reduction potentials of each reaction by referring to an SRP table.

The value for Ag+(aq) + e- →Ag(s) is 0.7996V and Zn2+(aq) + 2e-→Zn(s) has a value of -0.7618 V

Step 3: The equation for Ecell is Ecell=cathode-anode which could also be written as Ecell=reduction-oxidation

In this case The electrons for the reactions are both on the reactant side of the equation. A trick to discern which one is the anode and the cathode is the anode is always the smaller value, in this case that is Zn2+(aq) + 2e-→Zn(s) -0.7618 V

Step 4: Substitute in the values and solve

Ecell= 0.7996-(-0.7618)

Ecell= 1.56 V Since it is positive the reaction is spontaneous.Q17.3.4

According to the standard potential formula, the answer will be 1.56 (positive which means it is spontaneous in standard conditions). E~1.56

### Q17.3.5

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

### S17.3.5

To begin solving this problem, first separate the overall reaction into separate half reactions using the appropriate chemical species mentioned in the problem. You know that Cadmium is in solid metal form, and it is oxidized to its +2 charge ion. You also know that Aluminum Nitrate's aluminum species is reduced and becomes solid Aluminum. Al in Al(NO3)3 is Al3+. Once you figure out which species is oxidized and reduced using their respective oxidation numbers, you can write out the half reactions. Therefore, you obtain the following half reactions:

A - oxidation: Cd(s) --> Cd2+(aq) + 2e-

B - reduction: Al3+(aq) + 3e- --> Al(s)

Next, multiply each half reaction by the correct coefficient to cancel out the electrons. This yields:

A(3) - oxidation: 3Cd(s) --> 3Cd2+(aq) + 6e-

B(2) - reduction: 2Al3+(aq) + 6e- --> 2Al(s)

Add the two balanced half reactions to obtain an overall balanced equation:

3Cd(s) + 2Al3+(aq) --> 3Cd2+(aq) + 2Al(s)

To determine if this reaction is spontaneous at standard conditions, you have to determine the Ecell (Ecathode - Eanode). You know that the cadmium half reaction is the oxidation reaction, so it is the anode. The aluminum reaction is the reduction reaction, so it is the cathode. Using the Standard Reduction Potentials table, you can determine the Ecell of each half reaction.

A - oxidation: Cd(s) --> Cd2+(aq) + 2e- has a standard reduction potential of -0.40V.

B - reduction: Al3+(aq) + 3e- --> Al(s) has a standard reduction potential of -1.66V.

-1.66V - (-0.40V) = -1.26V.

Since Ecell is negative for this cell, the reaction is NON-SPONTANEOUS at standard conditions.

### A17.3.5

$$\ce{3Cd}(s)+\ce{2Al^3+}(aq)⟶\ce{3Cd^2+}(aq)+\ce{2Al}(s)$$; −1.259 V; nonspontaneous

### Q17.3.6

Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br2(l) is the same as for Br2(aq).

### S17.3.6

Start by writing out reduction equations for each side of the cell diagram, the left side of the cell diagram is always the anode and the right side is always the cathode.

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

Since the left side is always the anode, this side is also the oxidation half-reaction meaning electrons are lost. The reason there are only two electrons is because hydrogen goes from a 0 oxidation state to two +1 states needing two negative charges to balance each side.

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

Since the right side of the cell diagram is the cathode it is the reduction half-reaction meaning electrons are gained. The reason there are only two electrons is because bromine goes from a 0 oxidation state to two -1 states needing two negative charges to balance each side.

OVERALL EQUATION: H2(g) + Br2(aq) 2Br-(aq) + 2H+(aq)*

*the electrons cancel out, two on the reactants side and two on the products side.

Now we will write out each equation as a reduction half-reaction in order to find its standard cell potential according to the redox tower.

$H^{+}(aq) + 2e^{-}\rightarrow H^{2}(g)$ Eocell of H+/H2 = 0.0 V

$Br_{2}(g) + 2e ^{-} \rightarrow 2Br^{-}(aq)$ Eocell of Br2/Br- = 1.09 V

Refer to the Table of Standard Reduction Potentials to determine the cell potential of each half reaction and use the equation below to calculate the standard cell potential (Eocell )

Eocell = Eocathode - Eoanode

Eocell of H+/H2 = 0.0 V

Eocell of Br2/Br- = 1.09 V

Now if we refer to the equation Eocell = Eocathode - Eoanode we know the left half of cell diagram is the anode half(oxidation) and the right side is the cathode half (reduction), so plug in the standard values:

Eocell = 1.09V + 0

Eocell = 1.09V

Now to determine spontaneity of the corresponding system, we must use the equations below along with the standard cell potential:

∆G = -nFE°cell

n being the number of electrons, F being Farradays constant (96458 Coloumbs) and E being the standard cell potential.

∆G = -(2)(96458 C)(1.09 V)

When ∆G>0 it is non-spontaneous

When ∆G<0 the reaction is spontaneous.

Therefore, when ∆G= -210,278 J or -210.3 kJ the reaction is spontaneous

### A17.3.6

1.09 V;spontaneous

## 17.4: The Nernst Equation

### Q17.4.1

For the standard cell potentials given here, determine the ΔG° for the cell in kJ.

1. 0.000 V, n = 2
2. +0.434 V, n = 2
3. −2.439 V, n = 1

### S17.4.1

In this question we must find standard gibbs free energy (ΔG°) when being given voltage and number of electrons transferred. We can solve these problems using an equation relating the two values:

$\Delta G^{\circ}-n\cdot F\cdot E^{\circ}$

For each problem we are given E° (cell potential), n (number of moles of electrons transferred), and F (faraday's constant = 96.485 kJ/V*mol). Thus, we can plug in our known values into the equation to solve for ΔG°.

1. First begin by listing the given values:

E°= 0.000 V, n = 2, F = 96.485 Kj/V*mol

Now plug into the known equation:

$\Delta G^{\circ}=-(2 mol)\cdot (96.485 \frac{kj}{mol\cdot V})\cdot (0.00V)$

Since our E° is 0, it is easy to see that the product of our given values will also be 0. Thus, we know that our ΔG° will also be 0.00 kJ

2. Once again, we begin by listing the given values.

E° = 0.434 V, n = 2, F = 96.485 kJ/V*mol

We can plug into the known equation, making sure that the units of every value will cancel out to leave us with kJ:

$\Delta G^{\circ}=-(2 mol)\cdot (96.485 \frac{kj}{mol\cdot V})\cdot (0.0434V)$

We can see that the mol of electrons will cancel with the mols from faraday's constant, and the volts from our cell potential will

also cancel out with the volts from faraday's constant, leaving us with kJ in the final answer

ΔG° = -83.75 kJ

3. Once again, we can solve the same way, but this time our E° is negative:

E°= -2.439 V, n = 1, F = 96.485 Kj/V*mol

$\Delta G^{\circ}=-(1 mol)\cdot (96.485 \frac{kj}{mol\cdot V})\cdot (-2.439V)$

ΔG° = 235.33 kJ

### A17.4.1

(a) 0 kJ/mol; (b) −83.7 kJ/mol; (c) +235.3 kJ/mol

### Q17.4.2

For the ΔG° values given here, determine the standard cell potential for the cell.

1. 12 kJ/mol, n = 3
2. −45 kJ/mol, n = 1

### S17.4.2

a. We are given ΔG° (Gibb's Free Energy under standard conditions), and the number of moles. We are being asked to find Standard Cell Potential, and so we can use the following equation:

$\Delta&space;G^&space;\circ&space;=&space;-nFE^\circ$

Solving for E°,

$E^&space;\circ&space;=&space;\frac{\Delta&space;G^\circ}{-nF}$.

We know that ΔG°=12kJ/mol, n=3, and F is Faraday's constant, which is 96,485 C/mol.

$E^&space;\circ&space;=&space;\frac{12&space;\frac{kJ}{mol}}{-3*96,485&space;\frac{C}{mol}}$

However, cell potential is expressed in terms of volts, and this equation gives an answer in terms of kJ/C. We must convert from kJ to J, which gives units of J/C. Because 1 volt = 1J/C, our answer will be in the proper units.

$E^&space;\circ&space;=&space;\frac{12&space;\frac{kJ}{mol}}{-3*96,485&space;\frac{C}{mol}}*\frac{1,000&space;J}{1&space;kJ}*\frac{1V*C}{1J}=-0.0415V$

This answer makes sense because we started with a positive ΔG° value and got a negative E° value, which both indicate that the reaction is not spontaneous. E° and ΔG° values for a reaction should always have opposite signs.

E°<0 not spontaneous

E°>0 spontaneous

b. Solving this problem the same way,

$E^&space;\circ&space;=&space;\frac{\Delta&space;G^\circ}{-nF}$.

$E^&space;\circ&space;=&space;\frac{-45&space;\frac{kJ}{mol}}{-1*96,485&space;\frac{C}{mol}}*\frac{1,000&space;J}{1&space;kJ}*\frac{1V*C}{1J}=0.466V$

Again, this solution makes sense because the reaction had a negative ΔG° value, indicating that it is spontaneous, and the positive E° value confirms this.

a) -0.0415 V

b) 0.466 V

### Q17.4.3

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

1. $$\ce{Hg}(l)+\ce{S^2-}(aq,\: 0.10\:M)+\ce{2Ag+}(aq,\: 0.25\:M)⟶\ce{2Ag}(s)+\ce{HgS}(s)$$
2. The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.
3. The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br2(l) is the same as that of Br2(aq).

### S17.4.3

1. Hg(l)+S2−(aq,0.10M)+2Ag+(aq,0.25M)⟶2Ag(s)+HgS(s)

To find the Standard Cell Potential:

a) Eo=EoCathode-EoAnode

$\text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode})$

b)You use The formula listed above. For that, you will need to look at the Standard Cell Potentials chart. Remember that within electrochemical reactions, oxidation occurs in the Anode and reduction occurs in the Cathode. In which for this case indicates that the oxidation half-reaction is:

Hg(l)+H2(g)⟶HgS(s)+2H++2e- Eo=-0.72V

$\text{Hg}(\mathit{l})+\text{H}_2(\mathit{g}) \rightarrow \text{HgS}(\mathit{s})+\text{2H}^{+}+\text{2e}^-$

$\text{E}^o(\mathit{Anode})=-0.72V$

The reduction half reaction is:

c)Ag+(aq)+e- ⟶Ag(s) (This is a reduction because of gain of electrons)

$\text{Ag}^{+}(\mathit{aq})+\text{e}^- \rightarrow \text{Ag}(\mathit{s})$

EoCathode=0.7994

$\text{E}^o(\mathit{Cathode})=0.7994V$

Plugging the two cell potentials into the equation:

Eo=0.7994-(-0.72)

$\text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode})$

$\text{E}^o= (0.7994) - (-0.72)$

d)Eo=1.5V

$\text{E}^o= 1.5V$

Because the Standard Cell Potential is positive, the reaction is spontaneous!

To Find the Cell potential

Because the given chemical reaction is not under standard conditions, to find the cell potential of the reaction we must use the Nernst equation!

$${E_{cell}}={E^o}-{\frac{0.0592}{n}}{(logQ)}$$

n=number of electrons, in this case 2
Q=The equilibrium constant that we find by using the concentrations given to us in the formula; Q= (0.10)(0.25)2 =0.00625

Then we plug it all into the formula:

$${E_{cell}}={1.51}-{\frac{0.0592}{2}}{(log(0.00625))}$$

$${E_{cell}}={1.43V}$$

It is Spontaneous since it the Ecell is positive.

2.The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.

First, write the corresponding half-reactions:

Al3+(aq) + 3 e => Al(s) Eo=-1.662

$\text{Al}^{3+}(\mathit{aq})+\text{3e}^- \rightarrow \text{Al}(\mathit{s}) \; \text{E}^o = -1.662V$

Ni2+(aq) + 2 e => Ni(s) Eo=-0.257

$\text{Ni}^{2+}(\mathit{aq}) + \text{2e}^- \rightarrow \text{Ni}(\mathit{s}) \; \text{E}^o = -0.257V$

Since Nickel is the most positive in terms of Standard Potential Values is the reduction, which will be in the cathode and Aluminum in the Anode, oxidation.

Looking at the given cell potential of the half-reactions we can identify which reaction would take place at the Cathode and Anode. Because the Nickel has a more positive Standard Cell Potential value we can infer that it would most likely be reduced, while the more negative Standard Cell Potential from Aluminum would indicate it would be oxidized. Thus, reduction of Nickel occurs at the cathode and the oxidation of Aluminum occurs at the Anode.

To find the Standard Cell Potential:

a) Eo=EoCathode-EoAnode

$\text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode})$

Eo=-0.25-(-1.66)=1.41V

$\text{E}^o= (-0.257) - (-1.662)$

$\text{E}^o = 1.405V$

Because the Standard Cell Potential is positive, the reaction is spontaneous!

To Find the Cell potential

a) $${E_{cell}}={E^o}-{\frac{0.0592}{n}}{(logQ)}$$

n=number of electrons, in this case 6
Q=The equilibrium constant that we find by using the concentrations given to us in the formula; Q= (0.25)3(0.015)2 =3.5156E-6

$Q= \frac{(0.015)^2}{(0.25)^3}$

$Q= 0.0144$

All this was from the balanced formula of the reaction, which indicates 6 electrons and the nickel with a coefficient of 3 and Al with a coefficient of 2.

Then we plug it all into the formula:

$${E_{cell}}={1.41}-{\frac{0.0592}{6}}{(log(3.516E-6)}$$

$${E_{cell}}={1.405}-{\frac{0.0592}{6}}{(log(.0144)}$$

Ecell=1.46V

$\text{E}(\mathit{Cell}) = 1.423V$

Spontaneous since the Standard Cell is positive.

3. The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br2(l) is the same as that of Br2(aq)

a) First write the formula:

Al3+(aq) + 3 e => Al(s) Eo=-1.662V

$\text{Al}^{3+}(\mathit{aq})+\text{3e}^- \rightarrow \text{Al}(\mathit{s}) \; \text{E}^o = -1.662V$

Br2(l) + 2 e => 2 Br(aq) Eo=+1.066 1.0873

$\text{Br}_2(\mathit{l})+\text{2e}^- \rightarrow \text{2Br}^-(\mathit{aq}) \; \text{E}^o = 1.0873V$

To find the Standard Cell Potential:

a) Eo=EoCathode-EoAnode

$\text{E}^o= \text{E}^o(\mathit{Cathode}) - \text{E}^o(\mathit{Anode})$

Eo=-1.66-1.066=-2.726V

$\text{E}^o= (-1.662) - (1.0873)$

$\text{E}^o= -2.749V$

Because the Cell Potential is negative the reaction is non-spontaneous!

To Find the Cell potential

$${E_{cell}}={E^o}-{\frac{0.0592}{n}}{(logQ)}$$

n=number of electrons, in this case 6
Q=The equilibrium constant that we find by using the concentrations given to us in the formula; Q= (0.11)3(0.023)2 =7.04E-7

$Q= \frac{(0.11)^3}{(0.023)^2}$

$Q= 2.52$

All this was from the balanced formula of the reaction, which indicates 6 electrons and the Br with a coefficient of 3 and Al with a coefficient of 2.

Then we plug it all into the formula:

Ecell=-2.726-$$frac{0.0592}{6})(log(7.04E-7$$

$${E_{cell}}={2.726}-{\frac{0.0592}{6}}{(log(7.04E-7)}$$

$${E_{cell}}={-2.749}-{\frac{0.0592}{6}}{(log(2.52)}$$

Ecell=-2.665V

$\text{E}(\mathit{Cell}) = -2.753V$

Non Spontaneous since Standard Cell potential was negative

### A17.4.3

1. standard cell potential: 1.50 V, spontaneous; cell potential under stated conditions: 1.43 V, spontaneous;
2. standard cell potential: 1.405 V, spontaneous; cell potential under stated conditions: 1.423 V, spontaneous;
3. standard cell potential: −2.749 V, nonspontaneous; cell potential under stated conditions: −2.757 V, nonspontaneous

### Q17.4.4

Determine ΔG and ΔG° for each of the reactions in the previous problem.

### S17.4.4

1. $$Hg (l) + S_{2}^{-} (aq,0.10 M) + 2Ag (aq, 0.25 M)\rightarrow 2Ag (s) + HgS (s)$$

Solution:

GIVEN:

Standard Cell Potential: $$1.50 V$$ -- spontaneous

Cell Potential Under Conditions: $$1.43 V$$ -- spontaneous

UNKNOWN:

$$ΔG$$ : ?

$$ΔG°$$: ?

Equations to use:

$$ΔG = -n\times F\times E_{cell}$$

$$ΔG° = -n\times F\times E°_{cell}$$

First, solve for $$ΔG$$

a. Determine the equation to use

$$ΔG = -n\times F\times E_{cell}$$

b. Identify our known values

$$n$$ = ?

Step 1:

Write out the half cells

1) $$HgS + 2e^{-}\rightarrow Hg + S_{2}^{-}$$

2) $$Ag^{-} + e^{-}\rightarrow Ag$$

Step 2:

Balance and determine the number of electrons transferred

equation 1 × (1)

equation 2 × (2)

total of 2 electrons transferred

$$F = 96 485 C mol^{-1}$$

$$E_{cell} = 1.43 V$$

c. Plug it into the equation and solve

$$ΔG = -2 e^{-}\times 96 485 C mol^{-}\times 1.43 V$$

$$ΔG = -2.76\times 10^{-5} J K^{-1} mol^{-1}$$

Then, solve for $$ΔG°$$

a. Determine the equation to use

$$ΔG° = -n\times F\times E°cell$$

b. Identify our known values

$$n = 2 e^{-}$$

$$F = 96 485 C mol^{-1}$$

$$E°_{cell} = \(1.50 V$$

c. Plug it into the equation and solve

$$ΔG° = -2 e^{-}\times 96 485 C mol ^{-1}\times 1.50 V$$

$$ΔG° = -2.89\times 10^{-5} J K^{-1} mol^{-1}$$

2. The galvanic cell made from a half-cell consisting of an aluminum electrode in $$0.015 M$$ aluminum nitrate solution and a half-cell consisting of a nickel electrode in $$0.25 M$$ nickel(II) nitrate solution.

$$2Al(s) + 3Ni^{+2}(aq, 0.25 M) \rightarrow 3Ni(s) + 2Al^{+3}(aq, 0.015 M)$$

Solution:

GIVEN:

Standard Cell Potential: $$1.405 V$$ -- spontaneous

Cell Potential Under Conditions: $$1.423 V$$ -- spontaneous

UNKNOWN:

$$ΔG$$ : ?

$$ΔG°$$: ?

Equations to use:

$$ΔG = -n\times F\times E_{cell}$$

$$ΔG° = -n\times F\times E°_{cell}$$

First, solve for $$ΔG$$

a. Determine the equation to use

$$ΔG = -n\times F\times E_{cell}$$

b. Identify our known values

$$n$$ = ?

Step 1:

Write out the half cells

1) $$Al (s) \rightarrow Al^{+3} (aq, 0.015 M) + 3e^{-}$$

2) $$Ni^{2+} (aq, 0.25 M) + 2e^{-}\rightarrow Ni (s)$$

Step 2:

Balance the charges of the equations

Equation 1 × (2)

Equation 2 × (3)

= $$6 e^{-}$$ transferred

$$F = 96 485 C mol^{-}$$

$$Ecell = 1.423 V$$

c. Plug it into the equation and solve

$$ΔG = -6 e^{-}\times 96 485 C mol^{-1}\times 1.423 V$$

$$ΔG = -8.24\times 10^{5} J K^{-1} mol^{-1}$$

Then, solve for $$ΔG°$$

a. Determine the equation to use

$$ΔG° = -n\times F\times E°cell$$

b. Identify our known values

$$n = 6 e^{-}$$

$$F = 96 485 C mol^{-1}$$

$$E°cell = 1.405 V$$

c. Plug it into the equation and solve

$$ΔG° = -6 e^{-}\times 96 485 C mol^{-1}\times 1.405 V$$

$$ΔG° = -8.13\times 10^{5} J K^{-1} mol^{-1}$$

3. The cell made of a half-cell in which $$1.0 M$$ aqueous bromine is oxidized to $$0.11 M$$ bromide ion and a half-cell in which aluminum ion at $$0.023 M$$ is reduced to aluminum metal. Assume the standard reduction potential for $$Br_{2} (l)$$ is the same as that of $$Br_{2} (aq)$$.

⇒ $$3Br_{2}(aq, 1.0 M) + 2Al^{3+}(aq, 0.023 M) → 2Al(s) + 6Br^{-}(aq, 0.11 M)$$

Solution:

GIVEN:

Standard Cell Potential: $$−2.749 V$$, non-spontaneous

Cell Potential Under Conditions: $$−2.757 V$$, non-spontaneous

UNKNOWN:

$$ΔG$$ : ?

$$ΔG°$$: ?

Equations to use:

$$ΔG = -n\times F\times E_{cell}$$

$$ΔG° = -n\times F\times E°_{cell}$$

First, solve for $$ΔG$$

a. Determine the equation to use

$$ΔG = -n\times F\times E_{cell}$$

b. Identify our known values

$$n$$ = ?

Step 1:

Write out the half cells

1) $$Br_{2} (aq, 1.0M) \rightarrow 2Br^{-} (aq, 0.11M) + 2e^{-}$$

2) $$Al^{3+} (aq, 0.023M) + 3e^{-} \rightarrow Al (s)$$

Balance the charges of the equations

Equation 1 × (3)

Equation 2 × (2)

= $$6 e^{-} transferred$$

$$F = 96 485 C mol^{-1}$$

$$E_{cell} = -2.757 V$$

c. Plug it into the equation and solve

$$ΔG = -6 e^{-}\times 96 485 C mol^{-} \times -2.757 V$$

$$ΔG = 1.59 \times 10^{6} J K^{-1} mol^{-1}$$

Then, solve for $$ΔG°$$

a. Determine the equation to use

$$ΔG° = -n\times F\times E°_{cell}$$

b. Identify our known values

$$n = 6 electrons$$

$$F = 96 485 C mol^{-1}$$

$$E°_{cell} = -2.749 V$$

c. Plug it into the equation and solve

$$ΔG° = -6 e^{-} \times 96 485 C mol^{-1} \times -2.749 V$$

$$ΔG° = 1.59 \times 10^{6} J K^{-1} mol^{-1}$$

### Q17.4.5

Use the data in Table P1 to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.

1. $$\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)$$
2. $$\ce{CdS}(s)⇌\ce{Cd^2+}(aq)+\ce{S^2-}(aq) \hspace{40px} \textrm{at 377 K}$$
3. $$\ce{Hg^2+}(aq)+\ce{4Br-}(aq)⇌\ce{[HgBr4]^2-}(aq)$$
4. $$\ce{H2O}(l)⇌\ce{H+}(aq)+\ce{OH-}(aq) \hspace{40px} \textrm{at 25 °C}$$

### S17.4.5

Use the data in Table P1 to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.

1. $$\ce{AgCl}(s)⇌\ce{Ag+}(aq)+\ce{Cl-}(aq)$$
2. $$\ce{CdS}(s)⇌\ce{Cd^2+}(aq)+\ce{S^2-}(aq) \hspace{40px} \textrm{at 377 K}$$
3. $$\ce{Hg^2+}(aq)+\ce{4Br-}(aq)⇌\ce{[HgBr4]^2-}(aq)$$
4. $$\ce{H2O}(l)⇌\ce{H+}(aq)+\ce{OH-}(aq) \hspace{40px} \textrm{at 25 °C}$$

Needed equation: $E^{_{cell}^{\circ }}= \frac{RT}{nF}lnK$

Variables:

• R is a constant that is equal to $$8.3145 J\cdot mol^{-1}\cdot K^{-1}$$
• T is the temperature in Kelvin (this is stated in the problem)
• F is Faraday's Constant which is equal to $$96485 J\cdot V^{-1}\cdot mol^{-1}$$
• n is the number of electrons transferred between oxidants and reductants. We will use the redox half reactions to find n.
• $$E^\circ_{\ce{cell}}$$ (the standard potential of the reaction) is given by the equation $$E_{cell}^{\circ}=E_{cath}^{\circ}-E_{anode}^{\circ}$$. The standard potential of the cathode (the reduction half reaction) and the anode (the oxidation half reaction) can be found in Table P1.
• K is the equilibrium constant that we are solving for.

Solve:

a.) First find the redox half reactions where chlorine is being reduced and silver is being oxidized. We know this because Ag goes from an oxidation number of 0 to +1 meaning it lost an electron (oxidation) so Cl must have been reduced in order to accept the electron. The following half reactions and their standard redox potentials were found from Table P1.

Cathode/Reduction half reaction: $$AgCl(s)+e^{-}\rightarrow Ag(s)+Cl^{-}$$ $$E_{cath}^{\circ}=.2223$$

Anode/Oxidation half reaction: $$Ag(s)\rightarrow Ag^{+}(aq)+e^{-}$$ $$E_{anode}^{\circ}=.7996$$

From this we can find $$E_{cell}^{\circ}=.2223-.7996=-.5773$$ and that n=1 since 1 electron is being transferred (If you had an unequal number of electrons between the equations, you'd scale the equations as you would if you were balancing and then would use the coefficient of electrons in the properly scaled half reactions). Also T=298.15K as stated in the problem. We now have $$E_{cell}^{\circ}$$, T, n, F (constant), and R(constant). By pluggin in our variables we get $$k=1.7\times 10^{-10}$$.

Following this same method for b, c, and d you get

b.)

• T=377
• anode/oxidation: $$Cd^{2+}+2e^{-}\rightarrow Cd(s))$$ E°anode=-0.4030
• cathode/reduction: $$S(s)+2e^{-}\rightarrow S^{2-}$$ E°cath=-0.4070
• cell=-.4070+.4030=-.004
• n=2 (2 electrons being transferred)

$$k=2.6\times 10^{-21}$$

c.)

• T=298.15K
• anode/oxidation: $$Br_{2}+2e^{-}\rightarrow 2Br^{-}$$ E°anode=1.087
• cathode/reduction: $$2Hg^{2+}+2e^{-}\rightarrow Hg_{2}^{2+}$$ E°cath=.911
• cell=.911-1.087=-.176
• n=2

$$k=1.1\times 10^{-6}$$

d.)

• T(K)=T(°C)+273.15=25+273.15=298.15K
• anode/oxidation: $$2H^{+}+2e^{-}\rightarrow H_{2}(g)$$ E°anode=0
• cathode/reduction: $$H_{2}O+e^{-}\rightarrow \frac{1}{2}H_{2}(g)+OH^{-}$$ E°cath=-.828
• E°cell=-.828
• n=2 (even though one of the equations has 1 electron, you'd have to scale it up to 2 in order for the equations to balance)

$$k=1.0\times 10^{-26}$$

## 17.5: Batteries and Fuel Cells

### Q17.5.1

What are the desirable qualities of an electric battery?

### S17.5.1

The purpose of a battery is to store potential energy in its chemical form until it is needed where it is then converted to electric energy. Desirable qualities of an electric battery include low cost of materials, low toxicity and ease of disposal, and a high capacity. Furthermore, they should be lightweight, resistant to heat and humidity, and avoid leaks when the battery is used according to instruction.

### Q17.5.2

List some things that are typically considered when selecting a battery for a new application.

### A17.5.2

Because batteries can be used in many different things it is important to look at how each part of the battery will affect the new application.

You will want to look at:

• Cost (you don't want your product to be too expensive to produce)
• Materials (what materials are best suited for oxidation/reduction in the conditions needed for the application)

• The type of battery:

• Energy requirements, how long does the battery need to last and at what output
• Mass (important for functionality and ability to transport)
• Toxicity of materials and how to dispose of dead batteries (because batteries use many toxic materials, it is important to consider how to dispose of them and if it is possible to dispose of them safely) Toxic materials include acid, lead, nickel, cadmium, lithium, and mercury.
• Access to materials (the type of material is important but the availability is also important depending on whether or notthe product is going to be mass produced)

### Q17.5.3

Consider a battery made from one half-cell that consists of a copper electrode in 1 M CuSO4 solution and another half-cell that consists of a lead electrode in 1 M $$Pb(NO_3)_2$$ solution.

1. What are the reactions at the anode, cathode, and the overall reaction?
2. What is the standard cell potential for the battery?
3. Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.
4. Suppose sulfuric acid is added to the half-cell with the lead electrode and some $$PbSO_4(s)$$ forms. Would the cell potential increase, decrease, or remain the same?

### S17.5.3

a) Since the anode is the site where oxidation occurs while the cathode is where reduction occurs, the first thing we should determine is which of the central elements (Pb or Cu) is reduced and which is oxidized, which we can determine by referring to the table of standard reduction potentials. The higher the standard reduction potential, the more the element or compound wants to be reduced.

$Cu^{2+}(aq)+2e^-\to Cu(s) \qquad E^\circ=0.34V$

$Pb^{2+}(aq)+2e^-\to Pb(s) \qquad E^\circ=-0.13V$

So, as we can see, since $$Pb^{2+}$$ has a lower standard reduction potential, it will be oxidized while $$Cu^{2+}$$, which has a higher standard reduction potential, will be reduced.

So at the anode: $$Pb(s) \to Pb^{2+}(aq)+2e^- \qquad E^\circ=-0.13V$$ The original solution was incorrect (the anode had a reduction reaction).

While at the cathode: $$Cu^{2+}(aq)+2e^-\to Cu(s) \qquad E^\circ=0.34V\] With the overall reaction being: $$Cu^{2+}(aq) + Pb(s) \to Pb^{2+}(aq) + Cu(s)$$ The original solution was incorrect (the overall reaction included $$Pb{SO_4}$$ and $$CuSO_4$$, making the reaction unbalanced). b) The standard cell potential for the battery ($$E^\circ$$) is given by the equation: $$E^\circ = E^{\circ}_{cathode} - E^{\circ}_{anode}$$ $E^{\circ}_{cathode} = 0.34V$ $E^{\circ}_{anode} = -0.13V$ Note: the standard reduction value stays the same regardless if the reaction is reversed (i.e. $$Cu^{2+}(aq)+2e^-\to Cu(s)$$ to $$Cu(s) \to Cu^{2+}(aq)+2e^-$$) $0.34V - (-0.13V) = 0.47V = E^{\circ}_{cell}$ c) Since the $$E^{\circ}_{cell}$$ for this reaction is only 0.47V, it would take at least three of these cells (0.47V times 3) to replace a dry cell battery (since we need 1.0V - 1.5V). This $$E^{\circ}_{cell}$$ alone would not be able to replace a dry cell battery, but if multiple of these cells are put together then the answer is yes. d) In order to determine if the cell potential increases or decreases, we need to use the Nernst Equation and compare the cell potential under the standard and the non standard conditions. $E_{cell} = E^{\circ}_{cell} - \left ( \frac{RT}{nF} \right )lnQ$ By adding more SO4, the only factor that changes in this equation is Q, which, according to our overall reaction, is $Q=\frac{[Pb^{2+}]}{[Cu^{2+}]}$ So, at standard conditions, this ratio Q equals one. By adding more SO4 in the half cell with the lead electrode, more $$Pb{SO_4}$$(s) forms. More $$Pb{SO_4}$$(s) means more $$Pb^{2+}$$ is used up meaning less $$Pb{{(NO_3)}_2}$$(aq), therefore lowering Q under 1 under these conditions. As a result of Q being less of 1, the ln of any number lower than 1 is negative, so the Nernst equation in this case becomes $E_{cell} = E^{\circ}_{cell} + \left ( \frac{RT}{nF} \right )(positive \ number)$ As a result, the cell potential increases. ### Q17.5.4 Consider a battery with the overall reaction: $Cu(s)+2Ag^+(aq)⟶2Ag(s)+Cu^{2+}(aq)$ 1. What is the reaction at the anode and cathode? 2. A battery is “dead” when it has no cell potential. What is the value of $$Q$$ when this battery is dead? 3. If a particular dead battery was found to have $$[Cu^{2+}] = 0.11\, M$$, what was the concentration of silver ion? ### A17.5.4 (a) anode: $Cu(s)⟶Cu^{2+}(aq)+2e^-$ with $$E^o\, (anode)=0.34\, V$$ cathode: $2 \times (Ag^+(aq) + e^-⟶ Ag(s))$ with $$E^∘(cathode)=0.7996\, V$$ (b) $$3.5 \times 10^{15}$$ (c) $$5.6 \times 10^{−9}\, M$$ ### Q17.5.5 An inventor proposes using a SHE (standard hydrogen electrode) in a new battery for smartphones that also removes toxic carbon monoxide from the air: • Anode: $CO(g)+H_2O(l)⟶CO_2(g)+2H^+(aq)+2e^−$ with $$E^o (anode)=−0.53\, V$$ • Cathode: $2H^+(aq)+2e^−⟶H_2(g)$ with $$E^o (cathode)=0\, V$$ • Overall reaction: $CO(g)+H_2O(l)⟶CO_2(g)+H_2(g)$ with $$E^o_{cell}=+0.53\, V$$ Would this make a good battery for smartphones? Why or why not? ### A17.5.5 A standard hydrogen electrode would not be a good battery for smartphones. Before we can really answer the question, we have to know what a standard hydrogen electrode is and how does it function. 1. platinum electrode 2. hydrogen gas 3. solution of an acid 4. hydroseal for prevention of the oxygen interference 5. reservoir through which the second half-element of the galvanic cell should be attached. The connection can be direct, through a narrow tube to reduce mixing, or through a salt bridge, depending on the other electrode and solution. This creates an ionically conductive path to the working electrode of interest. This is a scheme of a standard hydrogen electrode. The redox reaction occurs at the platinized platinum electrode. The electrode is dipped in an solution with hydrogen ions meant to be reduced spontaneously to hydrogen gas. Though this electrode may sound useful, there are difficulties and problems that arise. The most common difficulty is the preparation of the platinized surface and in controlling the concentration of the reactants. Other limitations of a standard hydrogen electrodes are the following: 1. It is not convenient to assemble the apparatus. 2. It is difficult to maintain the pressure of hydrogen gas at exactly 1 bar without using sophisticated apparatus, especially in a closed system such as a battery. 3. Platinum is expensive 4. Even minor impurities impurities present in the hydrogen gas and acid could very easily poison the Pt plate and thus affect the equilibrium at the electrode. For impractical and humorous reasons of a standard hydrogen electrode are the following: 1. Hydrogen gas is extremely flammable. If your battery is damaged there is a chance that it will ignite the hydrogen gas being generated and cause your phone to explode and catch on fire. For these reasons, the standard hydrogen electrode is an impractical tool for battery making. ### Q17.5.6 Why do batteries go dead, but fuel cells do not? ### A17.5.6 Batteries and fuel cells are both "rechargable" forms of energy storage. While fuel cells are recharged by adding more reactants or reagents to the fuel cell, a battery is recharged by passing electricity through the battery. Unlike fuel cells however, batteries can suffer from continued recharging as dendrites will slowly form at the cathode. Once the dendrite from the cathode connects with the anode, the battery is dead. ### Q17.5.7 Explain what happens to battery voltage as a battery is used, in terms of the Nernst equation. ### Q17.5.8 Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures. ### A17.5.8 $$E_{cell}$$, as described in the Nernst equation, has a term that is directly proportional to temperature. At low temperatures, this term is decreased, resulting in a lower cell voltage provided by the battery to the device—the same effect as a battery running dead. ## 17.6: Corrosion ### Q17.6.1 Which member of each pair of metals is more likely to corrode (oxidize)? 1. Mg or Ca 2. Au or Hg 3. Fe or Zn 4. Ag or Pt ### Q17.6.2 Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is mostly iron, so use −0.447 V as the standard reduction potential for steel. Mg and Zn ### Q17.6.2 Aluminum $$(E^\circ_{\ce{Al^3+ /Al}} = \mathrm{−2.07\: V})$$ is more easily oxidized than iron $$(E^\circ_{\ce{Fe^3+ /Fe}} = \mathrm{−0.477\: V})$$, and yet when both are exposed to the environment, untreated aluminum has very good corrosion resistance while the corrosion resistance of untreated iron is poor. Explain this observation. ### Q17.6.3 If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon. ### A17.6.3 Both examples involve cathodic protection. The (sacrificial) anode is the metal that corrodes (oxidizes or reacts). In the case of iron (−0.447 V) and zinc (−0.7618 V), zinc has a more negative standard reduction potential and so serves as the anode. In the case of iron and copper (0.34 V), iron has the smaller standard reduction potential and so corrodes (serves as the anode). ### Q17.6.4 Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact? ### A17.6.4 Since B corrodes when with A, B oxidizes in comparison to A, or B is the donor of electrons (anode) and A is the cathode. When A and C are together, C is the cathode and A is the anode since C is the metal that corrodes. This means that A can donate its electrons to C, but accepts electrons from B. Hence, since donors are higher up (or have positive E values) on the standard reduction potentials chart, B would be highest since it donates to A. A would be next on the chart (or between metals B and C) since it donates electrons to C, causing itself (metal A) to oxidize/corrode. On the standard reduction potentials table, B would have the greatest reduction potential > A > C. When B and C come into contact, since B is higher up (with larger E value) than C, it would be the donor of electrons, or would lose electrons and become oxidized. This oxidation would lead to metal B becoming corroded, so metal B corrodes when in contact with metal C. ### Q17.6.5 Why would a sacrificial anode made of lithium metal be a bad choice despite its $$E^\circ_{\ce{Li+ /Li}} = \mathrm{−3.04\: V}$$, which appears to be able to protect all the other metals listed in the standard reduction potential table? ### A17.6.5 While the reduction potential of lithium would make it capable of protecting the other metals, this high potential is also indicative of how reactive lithium is; it would have a spontaneous reaction with most substances. This means that the lithium would react quickly with other substances, even those that would not oxidize the metal it is attempting to protect. Reactivity like this means the sacrificial anode would be depleted rapidly and need to be replaced frequently. (Optional additional reason: fire hazard in the presence of water.) ## 17.7: Electrolysis ### Q17.7.1 Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Table P1 are the same as those at each of the melting points. Assume the efficiency is 100%. 1. CaCl2 2. LiH 3. AlCl3 4. CrBr3 ### Q17.7.2 What mass of each product is produced in each of the electrolytic cells of Q17.7.1 if a total charge of 3.33 × 105 C passes through each cell? Assume the voltage is sufficient to perform the reduction. ### A17.7.2 1. $$\mathrm{mass\: Ca=69.1\: g\\ mass\:Cl_2=122\: g}$$ 2. $$\mathrm{mass\: Li=23.9\: g\\ mass\: H_2=3.48\: g}$$ 3. $$\mathrm{mass\: Al=31.0\: g\\ mass\:Cl_2=122\: g}$$ 4. $$\mathrm{mass\: Cr=59.8\: g\\ mass\:Br_2=276\: g}$$ ### Q17.7.3 How long would it take to reduce 1 mole of each of the following ions using the current indicated? Assume the voltage is sufficient to perform the reduction. 1. Al3+, 1.234 A 2. Ca2+, 22.2 A 3. Cr5+, 37.45 A 4. Au3+, 3.57 A ### Q17.7.4 A current of 2.345 A passes through the cell shown in Figure for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?) ### S17.7.4 The overall equation for this reaction is represented below: $2H_2O(l) \rightarrow 2H_2 (g)+ O_2 (g)$ where $${E^{\circ}}_{cell}=−1.229 V$$ The cathode or reduction half reaction: $4H^+ (aq) + 4e^- \rightarrow 2H_2( g)$ $${E^{\circ}}_{cathode}=0 V$$ The anode or oxidation half reaction: $2H_2O(l) \rightarrow O_2(g)+ 4H^+ (aq) + 4e^-$ $$E^{\circ}_{anode}=+1.229 V$$ As four electrons are present in both half reactions, we know there are four moles of electrons transferred. Using the current in Amps, the time interval the current is applied and the moles of electrons we found, we can find the amount of $$H_2$$ gas that formed. Assuming standard conditions (1 atm as given and 298 K), we can use three equations to find the volume of hydrogen gas formed: $Coulombs ={Amps}\cdot{seconds}$ $96,485 \ Coulombs = 1 \ mole \ of \ electrons$ $PV=nRT$ Where: P is the Pressure (in atm), V is the Volume (in Liters), n is the number of moles of $$H_2$$ gas, R is the universal gas constant [in (L * atm) / (mol * K)] and T is the temperature in Kelvin (K). $45 \,minute \cdot \frac{60\,seconds}{1\,minute} = 2,700\,seconds$ $moles\, e^{-}\,=\frac{2.345\,A \times 2,700\,seconds}{96,485\frac{Coulumbs}{mol\,e^{-}}}$ Leaving us with 0.065 moles e- transferred. According to this equation:$$2H^{+}+2e^{-} \to H_2\]

We have 2 moles of electrons for every mol of H2 so

$0.065\,mol\,e^{-} \cdot \frac{1\,mol\,H_{2}}{2\,mol\,e^{-}} = 0.0325\,mol\,H_{2}$

Since we are trying to find volume, we can use PV=nRT where P stands for pressure (in atm in this case), V stands for volume (in liters in this case), R is the gas constant, n is the number of moles (of H2 in this case), and T is for temperature (in Kelvin). So since we are trying to find the volume in liters (L):

Note: since the pressure is at 1 atm, that means we are at STP (Standard Temperature and Pressure)

$V=\frac{nRT}{P}$

$V=\frac{(0.0325\,mol\,H_{2}) \times (0.08206\ L*atm*mol^{-1}K^{-1}) \times (298K)}{1\,atm} V=0.795L$

(It is important to apply significant figures rules)

A17.7.4

0.79 L

### Q17.7.5

An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO3)2 solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm3. Assume the efficiency is 100%.

### S17.7.5

Here, we first need to determine how much zinc there is in order to use the density of zinc to figure out the volume of zinc and then the surface area. To find the amount of zinc, we must first understand that this question refers to the process of electroplating. Electroplating is a process in which an electrical current is run through a solution of dissolved ions, in turn reducing them, and then forming a thin layer of itself onto another metal. This is very important to know because without knowledge of this being a reduction reaction, it would be near impossible to find the solution. However, because we do know that this process is rooted in reduction, we can use the reduction equation of zinc to solve this problem.

$\text {Zn}^\text{ 2+} + 2\text {e}^- \rightarrow \text {Zn}$

From the reduction equation, we have to first solve for the amount of electrons. We first determine the total amount of energy like so (A or ampere-seconds are equivalent to one C)

$2.599\text{ A} \times \frac{60\text{ seconds}}{1\text{ min}} \times \frac{60\text{ minutes}}{1\text{ hour}} = 9356.4\text{ C}$

And then we can use the amount of charge carried by each electron to determine the how many electrons are running through the system.

$\frac {9356.4\text{ C}}{1.6 \times 10^{-19} \text {C} \text { per } \text {electron}} = 5.85 \times 10^{22} \text {electrons }$

From that, we convert the electrons into moles using Avogadro's constant so we can use the 2:1 stoichiometric ratio between the electrons and zinc to determine how many moles of zinc have been plated.

$\frac {5.85 \times 10^{22} \text {electrons }}{6.022 \times 10^{23}} = .097\text { moles} \text { of } \text {electrons}$

$\frac {.097 \text { moles} \text { of } \text {electrons}}{2} = .049\text { moles} \text { of } \text { Zn}$

Now we can find the grams of zinc there are in order to see how much volume it took up.

$.049\text{ mol Zn} \times \frac{65.39 \text { grams}}{ 1 \text { mol Zn}} \times \frac{ 1\text { cm}^3}{7.14 \text { grams}} = .44 \text { cm}^3$

Finally, we can use the amount of volume and use how thick the zinc was plated on to determine the total surface area of the zinc.

$0.01123\text{ mm} \times \frac{1 \text { cm}}{ 10 \text { mm}} = 0.001123 \text { cm}$

$\frac {0.44 \text { cm}^3}{0.001123 \text { cm }} = 392 \text { cm}^2$

A17.7.5

392 cm2