Extra Credit 3

Q17.1.3

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

1. Fe³⁺ + 3e⁻⟶ Fe
2. Cr⟶Cr³⁺ + 3e⁻
3. MnO₄²⁻ ⟶ MnO₄^- + e^-
4. Li⁺ + e^- ⟶Li

S17.1.3

Both oxidation and reduction reactions involve the transfer of electrons between reactants to form products. Where oxidation is the loss of electron(s) while reduction is the gain of electron(s); an easy way to remember this is to use the acronym: OIL (Oxidation Is Loss) and RIG (Reduction Is Gain).

In these problems, the electrons are explicitly shown, which makes it easier to classify whether it is an oxidation or a reduction reaction.

a. Fe³⁺ + 3e^-⟶ Fe

For this problem, the electrons are on the left side of the arrow, denoting that they are being added to Fe³⁺ to create Fe.

Since the definition of oxidation is the loss of electrons and the definition of reduction is the gain of electrons, then it can be concluded that since the electrons are being added to the reactant side of the equation, the species is therefore gaining electrons.

To verify that this is correct, by looking at the charge of the species and how it changes from reactant to product can also help classify a reaction. Since Fe starts out with a +3 charge, and electrons have a negative charge, by adding 3 electrons, the resulting Fe species should be neutral, meaning that the electron was gained.

Therefore, this is a reduction reaction.

b. Cr ⟶ Cr³⁺ + 3e^-

For this problem, the electrons are on the right side of the arrow, denoting that they are being taken from Cr to create the positively-charged Cr³⁺.

Since the definition of oxidation is the loss of electrons and the definition of reduction is the gain of electrons, then it can be concluded that since the electrons are lost from the reactant side of the equation, the species is therefore losing electrons; it is being oxidized to a more positively-charged state.

To verify that this is correct, by looking at the charge of the species and how it changes from reactant to product can also help classify a reaction.

Since Cr starts out with a neutral charge, and electrons have a negative charge, by taking away 3 of its electrons, the resulting Cr species should be more positive (+3 charge here) because by losing some electrons that means that there will be more protons left, causing it to now carry a positive charge of +3.

Therefore, this is an oxidation reaction.

c. MnO₄^2- ⟶ MnO₄^- + e^-

For this problem, the electrons are on the right side of the arrow, denoting that they are being taken from MnO₄^2- to create the more positively-charged MnO₄^- .

Since the definition of oxidation is the loss of electrons and the definition of reduction is the gain of electrons, then it can be concluded that since the electrons are lost from the reactant side of the equation, the species is therefore losing electrons; it is being oxidized to a more positively-charged state.

To verify that this is correct, by looking at the charge of the species and how it changes from reactant to product can also help classify a reaction.

Since MnO₄^2- starts out with a -2 charge, and electrons have a negative charge, by taking away one of its electrons, the resulting MnO₄^- species should be more positive (-1 charge here, which is indeed more positive than -2) because by losing some electrons that means that there will be more protons left and less electrons, causing it to now carry a more positive charge of -1.

Therefore, this is an oxidation reaction.

d. Li⁺ + e^- ⟶ Li

For this problem, the electrons are on the left side of the arrow, denoting that they are being added to Li⁺ to create Li.

Since the definition of oxidation is the loss of electrons and the definition of reduction is the gain of electrons, then it can be concluded that since the electrons are being added to the reactant side of the equation, the species is therefore gaining electrons.

To verify that this is correct, by looking at the charge of the species and how it changes from reactant to product can also help classify a reaction.

Since Li starts out with a +1 charge, and electrons have a negative charge, by adding 1 electron, the resulting Li species should be neutral, meaning that the electron was gained.

Therefore, this is a reduction reaction.

Q19.1.1

Write the electron configurations for each of the following elements:

1. Sc
2. Ti
3. Cr
4. Fe
5. Ru

S19.1.1

The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and sub-shells. The electron configuration of each element is unique to its position on the periodic table where the energy level is determined by the period and the number of electrons is given by the atomic number of the element. There are four different types of orbitals (s, p, d, and f) which have different shapes and each orbital can hold a maximum of 2 electrons, but the p, d and f orbitals have different sub-levels, meaning that they are able to hold more electrons.

The periodic table is broken up into groups which we can use to determine orbitals and thus, write electron configurations:

Group 1 & 2: S orbital

Group 13 - 18: P orbital

Group 3 - 12: D orbital

Lanthanide & Actinides: F orbital

Each orbital (s, p, d, f) has a maximum number of electrons it can hold. An easy way to remember the electron maximum of each is to look at the periodic table and count the number of periods in each collection of groups.

Group 1 & 2: 2 (2 electrons total = 1 orbital x max of 2 electrons = 2 electrons)

Group 13 - 18: 6 (6 electrons total = 3 orbitals x 2 electrons max = 6 electrons)

Group 3 - 12: 10 (10 electrons total = 5 orbitals x 2 electrons max = 10 electrons)

Lanthanide & Actinides: 14 (14 electrons total = 7 orbitals x 2 electrons max = 14 electrons)

Electron fills the orbitals in a specific pattern that affects the order in which the long-hand versions are written:

Electron filling pattern: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f

An easier and faster way to write electron configurations is to use noble gas configurations as short-cuts. We are able to do this because the electron configurations of the noble gases always have all filled orbitals.

He: 1s²2s²

Ne: 1s²2s²2p⁶

Ar: 1s²2s²2p⁶3s²3p⁶

Kr: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

Xe: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶

Rn: 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶

The most common noble gas configuration used is Ar. When you want to use the noble gas configuration short-cut, you place the noble gas's symbol inside of brackets:

[Ar]

and then write it preceding the rest of the configuration, which is solely the orbitals the proceed after that of the noble gas.

a. Sc

Let's start off by identifying where Scandium sits on the periodic table: row 4, group 3. This identification is the critical basis we need to write its electron configuration.

By looking at Scandium's atomic number, 21, it gives us both the number of protons and the number of electrons. At the end of writing its electron configuration, the electrons should add up to 21.

At row 4, group 3 Sc, is a transition metal; meaning that its electron configuration will include the D orbital.

Now, we can begin to assign the 21 electrons of Sc to orbitals. As you assign electrons to their orbitals, you move right across the periodic table.

Its first 2 electrons are in the 1s orbital which is denoted as

1s²

where the "1" preceding the s denotes the fact that it is of row one, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 21-2=19 more electrons to assign.

Its next 2 electrons are in the 2s orbital which is denoted as

2s²

where the "2" preceding the s indicates that it is of row two, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 19-2=17 more electrons to assign.

Its next 6 electrons are in the 2p orbital which is denoted as

2p⁶

where the "2" preceding the p indicates that it is of row two, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 17-6=11 more electrons to assign.

Its next 2 electrons are in the 3s orbital which is denoted as

3s²

where the "3" preceding the s indicates that it is of row three, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 11-2=9 more electrons to assign.

Its next 6 electrons are in the 3p orbital which is denoted as

3p⁶

where the "3" preceding the p indicates that it is of row three, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 9-6=3 more electrons to assign.

Its next 2 electrons are in the 4s orbital which is denoted as

4s²

where the "4" preceding the s indicates that it is of row four, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 3-2=1 more electron to assign.

Its last electron would be alone in the 3 d orbital which is denoted as

3d¹

where the "3" preceding the d indicates that, even though it is technically of row 4, by disregarding the first row of H and He, this is the third row and it has an exponent of 1 because there is only 1 electron to be placed in the d orbital. Now we have assigned all of the electrons to the appropriate orbitals and sub-orbitals, so that the final, entire electron configuration is written as:

1s²2s²2p⁶3s²3p⁶4s²3d¹

This is the long-hand version of its electron configuration.

So for Sc, its short-hand version of its electron configuration would therefore be:

[Ar] 4s²3d¹

b. Ti

Start off by identifying where Titanium sits on the periodic table: row 4, group 4, meaning it has 22 electrons total. Titanium is one element to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 2 electrons remaining, so therefore the final orbital will be denoted as:

3d²

If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Titanium will be:

1s²2s²2p⁶3s²3p⁶4s²3d²

So for Ti, its short-hand version of its electron configuration would therefore be:

[Ar] 4s²3d²

c. Cr

Start off by identifying where Chromium sits on the periodic table: row 4, group 6, that means it has a total of 24 electrons. But first, Cr, along with Mo, Nb, Ru, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 24 electrons that its configuration would look like:

1s²2s²2p⁶3s²3p⁶4s²3d⁴

which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that there are 4 electrons in its 3 d orbital. The most stable configurations are half-filled (d⁵) and full orbitals (d¹⁰), so the elements with electrons resulting in ending with the d⁴ or d⁹ are so unstable that we write its stable form instead, where an electron from the preceding s orbital will be moved to fill the d orbital, resulting in a stable orbital.

If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Chromium will be:

1s²2s²2p⁶3s²3p⁶4s¹3d⁵

So for Cr, its short-hand version of its electron configuration would therefore be:

[Ar] 4s¹3d⁵

d. Fe

Start off by identifying where Iron sits on the periodic table: row 4, group 8, meaning it has 26 electrons total. This is 5 elements to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 6 electrons remaining, so therefore the final orbital will be denoted as:

3d⁶

If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Iron will be:

1s²2s²2p⁶3s²3p⁶4s²3d⁶

So for Fe, its short-hand version of its electron configuration would therefore be:

[Ar] 4s²3d⁶

e. Ru

Start off by identifying where Ruthenium sits on the periodic table: row 5, group 8, that means it has a total of 44 electrons. But first, as stated earlier, Ru, along with Cr, Mo, Nb, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 44 electrons that its configuration would look like:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s² 4d⁶

which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that, even though there are 4 paired electrons, there are also 4 electrons unpaired. This results in a very unstable configuration, so to restore stability, we have to use a configuration that has the most paired electrons, which would be to take an electron from the s orbital and place it in the d orbital to create:

5s¹4d⁷

If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Ru will be:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹4d⁷

So for Cr, its short-hand version of its electron configuration would therefore be:

[Kr] 5s¹4d⁷

Q19.2.3

Give the coordination number for each metal ion in the following compounds:

1. [Co(CO₃)₃]³⁻ (note that CO₃²⁻ is bidentate in this complex)
2. [Cu(NH₃)₄]²⁺
3. [Co(NH₃)₄Br₂]₂(SO₄)₃
4. [Pt(NH₃)₄][PtCl₄]
5. [Cr(en)₃](NO₃)₃
6. [Pd(NH₃)₂Br₂] (square planar)
7. K₃[Cu(Cl)₅]
8. [Zn(NH₃)₂Cl₂]

S19.2.3

The coordination number is the total number of points of attachment to the central element which can vary from 2 to 16, but is usually 4 if tetrahedral or square planar structure and 6 if octahedral structure. (Tetrahedral and octahedral structures are the two most common structures in a coordinate complex.) It is quite simple to determine the coordination number for a metal ion in a complex, you must simply find the number the points of attachment to the metal ion.

a. [Co(CO₃)₃]³⁻ (note that CO₃²⁻ is bidentate in this complex)

Since CO₃²⁻ is bidentate in this complex, it has 2 attachment points to the metal ion. So, since there are three of CO₃²⁻, the coordination number would be

3 (CO₃²⁻ ions) x 2 (attachment points from bidentate CO₃²⁻) = 6 total attachment points

therefore, the coordination number is 6

b. [Cu(NH₃)₄]²⁺

This complex has 4 of NH₃ attached to the metal ion, Cu, so the coordination number would be

4 (NH₃) x 1 (attachment point per NH₃) = 4 total attachment points

therefore, the coordination number is 4

c. [Co(NH₃)₄Br₂]₂(SO₄)₃

This complex is more complicated, but the coordination number can be found using the same method. The compound outside of the brackets ([X]), which is (SO₄)₃ in this case, is not attached to the metal ion, so it is not relevant for the coordination number.

((4 (NH₃) + 2 (Br)) x 1 (attachment point per NH₃ or Br₂)) = 6 total attachment points

therefore, the coordination number is 6

d. [Pt(NH₃)₄][PtCl₄]

This complex, [Pt(NH₃)₄], has 4 of NH₃ attached to the metal ion, Pt, so there are

4 (NH₃) x 1 (attachment point per NH₃) = 4 attachment points

But there is another complex that needs to be accounted for. This complex, [PtCl₄], has 4 of Cl attached to the metal ion, Pt, so there are

4 (Cl) x 1 (attachment point per Cl) = 4 attachment points

So for the whole complex, [Pt(NH₃)₄][PtCl₄], the coordination number would be a sum of its attachment points:

4 attachment points + 4 attachment points = 8 total attachment points

therefore, the coordination number is 8

e. [Cr(en)₃](NO₃)₃

Since en is a bidentate in this complex, that means that it has 2 attachment points to the metal ion. The compound outside of the brackets ([X]) is not attached to the metal ion, so it is not relevant for the coordination number. So, since there are three of en, the coordination number would be

3 (en) x 2 (attachment points from bidentate en) = 6 total attachment points

therefore, the coordination number is 6

f. [Pd(NH₃)₂Br₂] (square planar)

This complex has 2 of NH₃ and 2 of Br attached to the metal ion, Pd, so the coordination number would be

(2 (NH3) + 2 (Br)) x 1 (attachment point per NH₃ or Br₂) = 4 total attachment points

therefore, the coordination number is 4

g. K₃[Cu(Cl)₅]

This complex has 5 of Cl attached to the metal ion, Cu, so the coordination number would be

5 (Cl) x 1 (attachment point per Cl) = 5 total attachment points

therefore, the coordination number is 5

h. [Zn(NH₃)₂Cl₂]

This complex has 2 of NH₃ and 2 of Cl attached to the metal ion, Zn, so the coordination number would be

(2 (NH3) + 2 (Cl)) x 1 (attachment point per NH₃ or Cl) = 4 total attachment points

therefore, the coordination number is 4

Q12.3.15

Nitrogen(II) oxide reacts with chlorine according to the equation:

2 NO(g) + Cl₂(g) ⟶ 2 NOCl (g)

The following initial rates of reaction have been observed for certain reactant concentrations:

What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl₂? What is the rate constant? What are the orders with respect to each reactant?

S12.3.16

Here, I labeled each value which makes it easier to work through.

The rate equation to describe the rate's dependence on the concentrations of NO and Cl₂ is

rate=k[NO]^m[Cl₂]ⁿ

where: "m" is the reaction order with respect to NO

"n" is the reaction order with respect to Cl₂

"k" is the rate constant

All of these values have to be determined experimentally, so we will be using the data provided.

The most straightforward method to determine reaction orders is the mathematical approach:

To determine "m", which is the reaction order with respect to [NO], we must first choose the data where [Cl₂] is held constant.

So we will be using the data: 1a, 2a, and 1c, 2c.

Create a ratio:

\[\dfrac{rate 2c}{rate 1c} = \dfrac{4.56}{1.14} = \dfrac{(1^m)}{(0.5^m)}\]

Solve for m.

4=2^m

m=2, which is the reaction order with respect to [NO].

To determine "n", which is the reaction order with respect to [Cl₂], we do the same thing as we did above.

We will be using the data: 2b, 3b, and 2c, 3c.

Create a ratio:

\[\dfrac{rate 3c}{rate 2c} = \dfrac{9.12}{4.56} = \dfrac{(1^n)}{(0.5^n)}\]

Solve for n.

2=2ⁿ

n=1, which is the reaction order with respect to [Cl₂].

Now, to determine "k", we simply plug the values into the original equation

rate=k[NO]^m[Cl₂]ⁿ

where we can any row of data values, I will chose 1a, 1b, and 1c.

1.14(mol/L/h)=k(0.50mol/L)²(0.50mol/L)¹

k= 9.12 L²mol^-2h^-1

Q12.6.7

Write the rate equation for each of the following elementary reactions:

a. O₃ ⟶ O₂ + O

b. O₃ + Cl ⟶ O₂ + ClO

c. ClO + O ⟶ Cl + O₂

d. O₃ + NO ⟶ NO₂ + O₂

e. NO₂ + O ⟶ NO + O₂

S12.6.7

Rate equations are dependent on the reactants and not the products, so to write a rate equation, include the rate constant "k" multiplied to the concentration of the reactants multiplied by each other.

a. O₃ ⟶ O₂ + O

To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".

Rate = k[O₃]

b. O₃ + Cl ⟶ O₂ + ClO

To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".

Rate = k[O₃][Cl]

c. ClO + O ⟶ Cl + O₂

To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".

Rate = k[ClO][O]

d. O₃ + NO ⟶ NO₂ + O₂

To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".

Rate = k[O₃][NO]

e. NO₂ + O ⟶ NO + O₂

To write this reaction's rate equation, only focus on the reactant(s) and its/their concentration and multiplying that by a rate constant, "k".

Rate = k[NO₂][O]

Q21.4.19

Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of

S21.4.19

The half life is the period of time required for the concentration of a reactant to decrease to one-half its initial value. For radioactive decay, we look at the half life equation for first order reactions (radioactive decays such as the one mentioned above are always first order), where a given set of reaction conditions is a constant and the half life is independent of the concentration of the reactants.

So we can manipulate the equation:

\[ln\dfrac{[A]_0}{[A]}\] = kt

with [A]=.5[A]₀ since half life is half of the reactant left to get the new equation:

t_1/2 = \[\dfrac{0.693}{k}\]

Now, to solve the problem and find the rate constant, 'k", we simply plug in the values.

6 hours = \[\dfrac{0.693}{k}\]

which yields

0.12 h^-1 as the rate constant, "k".

Q20.3.7

Copper(II) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO₄ solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of ZnCl₂?

S20.3.7

A piece of zinc metal is placed in a beaker of aqueous CuSO₄ solution, the blue color fades with time and the zinc strip begins to erode, and a black solid forms around the zinc strip because of the face that the solid zinc is reacting with the CuSO₄ to form solid copper and aqueous zinc (2+) ion. The fact that the blue color is fading means that the copper is becoming solid and is no longer a compound with SO₄. The black solid that is forming around the zinc strip is the copper that is forming.

The half reactions for this are:

Zn (s) → Zn²⁺ (aq)

CuSO₄ (aq) → Cu (s)

Here, the changes that are occurring are phase changes. The solid zinc is becoming aqueous zinc (2+) ion and the CuSO₄ is becoming a solid copper.

If a piece of copper metal is placed in a colorless aqueous solution of ZnCl₂, then the solid copper will eventually become CuCl₂ solution that will be a pale blue color and the zinc would become a solid.

Q20.5.18

In acidic solution, permanganate (MnO₄⁻) oxidizes Cl⁻ to chlorine gas, and MnO₄⁻ is reduced to Mn²⁺(aq).

a. Write the balanced chemical equation for this reaction.

b. Determine E°_cell.

c. Calculate the equilibrium constant.

S20.5.18

a. To write the balanced equation of this redox reaction you have to understand what a redox reaction is.

An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species.

Oxidation is the loss of electrons while reduction is gain of electrons.

From the information given, 2 half reactions can be determined:

MnO₄^- → Mn²⁺

Cl^- → Cl₂

For each half reaction, follow these steps to balance a redox reaction under acidic conditions:

1. Balance elements in the equation other than O and H.

MnO₄^- → Mn²⁺

2. Balance the oxygen atoms by adding the appropriate number of water (H₂O) molecules to the opposite side of the equation.

MnO₄^- → Mn²⁺ + 4H₂O

3. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H⁺ ions to the opposite side of the equation.

8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O

4. Add up the charges on each side. Make them equal by adding enough electrons (e^-) to the more positive side. (Rule of thumb: e^- and H⁺ are almost always on the same side.)

5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O

5. Balance elements in the equation other than O and H.

2Cl^- → Cl₂

6. Balance the oxygen atoms by adding the appropriate number of water (H₂O) molecules to the opposite side of the equation.

2Cl^- → Cl₂

(there are no other elements besides Cl here, so it can be left alone for this step)

7. Balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom) by adding H⁺ ions to the opposite side of the equation.

2Cl^- → Cl₂

(there are no other elements besides Cl here, so it can be left alone for this step)

8. Add up the charges on each side. Make them equal by adding enough electrons (e^-) to the more positive side. (Rule of thumb: e^- and H⁺ are almost always on the same side.)

2Cl^- → Cl₂ + 2e^-

Now, looking at both balanced half reactions:

9. The e^- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.

(5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O) x 2

(2Cl^- → Cl₂ + 2e^-) x 5

———————————————————————————

10e^- + 16H⁺ + 2MnO₄^- → 2Mn²⁺ + 8H₂O

10Cl^- → 5Cl₂ + 10e^-

10. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.

10e^- + 16H⁺ + 2MnO₄^- + 10Cl^- → 2Mn²⁺ + 8H₂O + 5Cl₂ + 10e^-

16H⁺ + 2MnO₄^- + 10Cl^- → 2Mn²⁺ + 8H₂O + 5Cl₂

MnO₄⁻+ 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O(l)

Cl₂(g) + 2e⁻ ⇌ 2Cl⁻

b. The standard cell potential (E^o_cell) is the difference of the potentials of the half reactions at 2 electrodes, which forms the voltage of that cell:

E^o_cell = E^o_cathode - E^o_anode

The cathode is where reduction happens and the anode is where oxidation happens, so

5e^- + 8H⁺ + MnO₄^- → Mn²⁺ + 4H₂O is where reduction is happening, so it is the cathode

2Cl^- → Cl₂ + 2e^- is where oxidation is happening, so it is the anode

The standard reduction potential values of the two half reactions are given in a table where:

E^o for MnO₄⁻+ 8H⁺ + 5e⁻ ⇌ Mn²⁺ + 4H₂O(l) = 1.51 V

E^o for Cl₂(g) + 2e⁻ ⇌ 2Cl⁻ = 1.358 V

———————————————————————————————

E^o_cell = 1.51V - 1.358V = 0.152 V

c. To calculate the equilibrium constant from the standard cell potential, you use the Nernst equation:

\[E^{0} _{cell} = \dfrac{RT}{nF} ln K\]

Here, the problem assumes standard state, so we know that R=(8.3145 J)/(mol*K), T=298 K, F=96485 J or 96485 J/(V*mol). Using this data we get (RT/F)=.025693 and substitute the calculated value of (RT/F) into RT/nF to get .025693/n. Now we get the new Nernst equation:

\[E^{0} _{cell} = \dfrac{0.025693}{n} ln K\]

where: "n" is the total number of electrons being transferred and "K" is the equilibrium constant.

Plugging in the values, we get:

The number of electrons transferred is found by looking at the balanced half reactions as shown above in part a, and so we see that 2 electrons are transferred.

\[0.152 V = \dfrac{0.025693}{10} ln K\]

K= 4.93E25

(For part c, the original solution was incorrect.)