## General Chemistry

Electrochemistry practice problems include questions on balancing redox reactions in acidic and basic solutions, calculating the cell potential, Eo/E at standard conditions using the table of standard reduction potentials, calculating the free energy change, ΔG based on the cell potential, calculating the cell potential under nonstandard conditions using the Nernst equation, calculating the equilibrium constant of the reaction based on the cell potential.

#### Practice

1.

Balance the following redox reactions occurring in acidic aqueous solution:

a) Al(s) + Fe2+(aq) → Al3+(aq) + Fe(s)

b) SO32(aq) + MnO4(aq) → SO42(aq) + Mn2+(aq)

c) Cr2O72-(aq) + C2O22-(aq) → Cr3+(aq) + CO2(g)

d) PbO2(s) + Mn2+(aq) + SO42-(aq) → PbSO4(s) + MnO4(aq)

e) MnO4(aq) + H2O2(aq) → Mn2+(aq) + O2(g)

a)

2Al(s) +  3Fe2+(aq) → 2Al3+(aq) + 3Fe(s)

b)

5SO32(aq) + 2MnO4(aq) + 6H+(aq) → 5SO42(aq) + 2Mn2+(aq) + 3H2O(l)

c)

3C2O22- + Cr2O72- + 14H+(aq) → 6CO2 + 2Cr3+ + 7H2O(l)

d)

2Mn2+ + 5PbO2 + 5SO42+ 4H+(aq) → 2MnO4+ 5PbSO4 + 2H2O(l)

e)

5H2O2(aq) + 2MnO4(aq) + 6H+(aq) → 5O2(g) + 2Mn2+(aq) + 8H2O(l)

Solution

a) Al(s) + Fe2+(aq) → Al3+(aq) + Fe(s)

1) Separate the half-reactions:

Al(s) → Al3+(aq) (oxidation)

Fe2+(aq) → Fe(s) (reduction)

2) Balance the elements. In this case, they are balanced, so proceed to the next step.

3) Balance the charges by adding electrons:

Al(s) → Al3+(aq) + 3e

Fe2+(aq) + 2e→ Fe(s)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

2Al(s) → 2Al3+(aq) + 6e

3Fe2+(aq) + 6e→ 3Fe(s)

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

2Al(s) → 2Al3+(aq) + 6e

3Fe2+(aq) + 6e→ 3Fe(s)
_____________________

2Al(s) +  3Fe2+(aq) + 6e→ 2Al3+(aq) + 6e+ 3Fe(s)

6) Cancel any species that appear in equal quantitates on both sides of the equation:

2Al(s) +  3Fe2+(aq) + 6e → 2Al3+(aq) + 6e + 3Fe(s)

2Al(s) +  3Fe2+(aq) → 2Al3+(aq) + 3Fe(s)

b) SO32(aq) + MnO4(aq) → SO42(aq) + Mn2+(aq)

1) Separate the half-reactions:

SO32(aq) → SO42(aq)

MnO4(aq) → Mn2+(aq)

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

H2O(l) + SO32(aq) → SO42(aq) + 2H+(aq)

MnO4(aq) + 8H+(aq) → Mn2+(aq) + 4H2O(l)

3) Balance the charges by adding electrons:

H2O(l) + SO32(aq) → SO42(aq) + 2H+(aq) + 2e

MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 2H2O(l)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

5H2O + 5SO32(aq) → 5SO42(aq) + 10H+(aq) + 10e

2MnO4(aq) + 16H+(aq) + 10e- → 2Mn2+(aq) + 4H2O(l)

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

5H2O + 5SO32(aq) → 5SO42(aq) + 10H+(aq) + 10e

2MnO4(aq) + 16H+(aq) + 10e → 2Mn2+(aq) + 8H2O(l)
______________________________________________

5H2O + 5SO32(aq) + 2MnO4(aq) + 16H+(aq) + 10e→ 5SO42(aq) + 10H+(aq) + 10e+ 2Mn2+(aq) + 8H2O(l)

6) Cancel any species that appear in equal quantitates on both sides of the equation:

5H2O(l) + 5SO32(aq) + 2MnO4(aq) + 16 6H+(aq) + 10e → 5SO42(aq) + 10H+(aq) + 10e + 2Mn2+(aq) + 8 3H2O(l)

5SO32(aq) + 2MnO4(aq) + 6H+(aq) → 5SO42(aq) + 2Mn2+(aq) + 3H2O(l)

We can rearrange the positions of species just to make the equation look a more conventional way:

2MnO4(aq) + 5SO32(aq) + 6H+(aq) → 2Mn2+(aq) + 5SO42(aq) + 3H2O(l)

c) Cr2O72- + C2O22- → Cr3+ + CO2

1) Separate the half-reactions:

Cr2O72- → Cr3+

C2O22- → CO2

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

Cr2O72- + 14H+(aq) → 2Cr3+ + 7H2O

C2O22-2CO2

3) Balance the charges by adding electrons. We need 6e on the left side of the first equation because the charges are +12 vs +6. The second equation needs two electrons on the right side to make the charges -2 on both sides:

Cr2O72- + 14H+(aq) + 6e → 2Cr3+ + 7H2O(l)

C2O22- → 2CO2 + 2e

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

Cr2O72- + 14H+(aq) + 6e → 2Cr3+ + 7H2O(l)

3C2O22- → 6CO2 + 6e

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

Cr2O72- + 14H+(aq) + 6e → 2Cr3+ + 7H2O(l)

3C2O22- → 6CO2 + 6e
______________________________________________

Cr2O72- + 14H+(aq) + 6e + 3C2O22- → 2Cr3+ + 7H2O + 6CO2 + 6e

6) Cancel any species that appear in equal quantitates on both sides of the equation:

Cr2O72- + 14H+(aq) + 6e + 3C2O22- → 2Cr3+ + 7H2O(l) + 6CO2 + 6e

3C2O22- + Cr2O72- + 14H+(aq) → 6CO2 + 2Cr3+ + 7H2O(l)

d) PbO2 + Mn2+ + SO42 → PbSO4 + MnO4

1) Separate the half-reactions:

Mn2+ → MnO4

PbO2 + SO42 → PbSO4

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

Mn2+ + 4H2O(l) → MnO4+ 8H+(aq)

PbO2 + SO42 + 4H+(aq) → PbSO4 + 2H2O(l)

3) Balance the charges by adding electrons.

Mn2+ + 4H2O(l) → MnO4+ 8H+(aq) + 5e

PbO2 + SO42 + 4H+(aq) + 2e→ PbSO4 + 2H2O(l)

4) Balance the number of electrons by multiplying the equation(s) with a whole number. We need to multiply the first half-reaction (oxidation half-reaction) by 2 and the reduction half-reaction by 5:

2Mn2+ + 8H2O → 2MnO4+ 16H+(aq) + 10e

5PbO2 + 5SO42 + 20H+(aq) + 10e→ 5PbSO4 + 10H2O(l)

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

2Mn2+ + 8H2O → 2MnO4+ 16H+(aq) + 10e

5PbO2 + 5SO42 + 20H+(aq) + 10e→ 5PbSO4 + 10H2O(l)
______________________________________________

2Mn2+ + 8H2O(l) + 5PbO2 + 5SO42 + 20 4H+(aq) + 10e → 2MnO4+ 16H+(aq) + 10e + 5PbSO4 + 10 2H2O(l)

2Mn2+ + 5PbO2 + 5SO42+ 4H+(aq) → 2MnO4+ 5PbSO4 + 2H2O(l)

e) MnO4(aq) + H2O2(aq) → Mn2+(aq) + O2(g)

1) Separate the half-reactions:

H2O2(aq) → O2(g)

MnO4(aq) → Mn2+(aq)

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

H2O2(aq) → O2(g) + 2H+(aq)

MnO4(aq) + 8H+(aq) + 5e→ Mn2+(aq) + 4H2O(l)

3) Balance the charges by adding electrons. The first reaction needs two electrons on the right, and for the second, we need 5 electrons on the left side:

H2O2(aq) → O2(g) + 2H+(aq) + 2e

MnO4(aq) + 8H+(aq) + 5e→ Mn2+(aq) + 4H2O(l)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

5H2O2(aq) → 5O2(g) + 10H+(aq) + 10e

2MnO4(aq) + 16H+(aq) + 10e→ 2Mn2+(aq) + 8H2O(l)

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

5H2O2(aq) → 5O2(g) + 10H+(aq) + 10e

2MnO4(aq) + 16H+(aq) + 10e→ 2Mn2+(aq) + 8H2O(l)
_____________________________________________

5H2O2(aq) + 2MnO4(aq) + 16 6H+(aq) + 10e → 5O2(g) + 10H+(aq) + 10e + 2Mn2+(aq) + 8H2O(l)

5H2O2(aq) + 2MnO4(aq) + 6H+(aq) → 5O2(g) + 2Mn2+(aq) + 8H2O(l)

2.

Balance the following redox reactions occurring in basic aqueous solution:

a) Mn2+(aq) + ClO3(aq) → MnO2(s) + ClO2(g)

b) MnO4(aq) + Br(aq) → MnO2 + BrO3(aq)

c) MnO4(aq)  + CN(aq) → CNO(aq)  + MnO2(s)

d) H2O2(aq) + ClO2(aq) → ClO2(aq) + O2(g)

e) MnO4(aq) + Fe(OH)2(s) → MnO2(s) + Fe(OH)3(s)

a)

Mn2+ +  ClO3 → MnO2 + ClO2

b)

Br(aq) + 2MnO4(aq)  → BrO3(aq) + 2MnO2 + 2OH(aq)

c)

3CN(aq) + 2MnO4(aq) + H2O(l) → 3CNO(aq) + 2MnO2(s) + 2OH(aq)

d)

H2O2(aq) + 2ClO2(aq) + 2OH(aq) → O2(g) + 2ClO2(aq) + 2H2O(l)

e)

MnO4(aq) + 3Fe(OH)2(s) + 2H2O(l) → MnO2(s) + 3Fe(OH)3(s) + OH(aq)

Solution

a) Mn2+ + ClO3 → MnO2 + ClO2

1) Separate the half-reactions:

Mn2+ → MnO2

ClO3 → ClO2

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

Mn2+ + 2H2O → MnO2 + 4H+(aq)

ClO3 + 2H+(aq) → ClO2 + H2O(l)

3) Balance the charges by adding electrons.

Mn2+ + 2H2O → MnO2 + 4H+(aq) + 2e

ClO3 + 2H+(aq) + e→ ClO2 + H2O(l)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

Mn2+ + 2H2O → MnO2 + 4H+(aq) + 2e

2ClO3 + 4H+(aq) + 2e→ 2ClO2 + 2H2O(l)

5) Once everything is balanced, add the half-reactions together and cancel any species that appear in equal quantitates on both sides of the equation.

Mn2+ + 2H2O → MnO2 + 4H+(aq) + 2e

2ClO3 + 4H+(aq) + 2e→ 2ClO2 + 2H2O(l)
_____________________________________________

Mn2+ + 2H2O + 2ClO3 + 4H+ + 2e → MnO2 + 4H+ + 2e + 2ClO2 + 2H2O

Mn2+ +  ClO3 → MnO2 + ClO2

We don’t need to add any OH ions on any side of the equation, since there are no H+ ions. Therefore, for this reaction, the balanced equation in basic and acidic solutions is identical.

b) MnO4(aq) + Br(aq) → MnO2 + BrO3(aq)

1) Separate the half-reactions:

Br(aq) → BrO3(aq)

MnO4(aq)  → MnO2

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

Br(aq) + 3H2O(l) → BrO3(aq) + 6H+(aq)

MnO4(aq)  + 4H+(aq) → MnO2 + 2H2O(l)

3) Balance the charges by adding electrons.

Br(aq) + 3H2O(l) → BrO3(aq) + 6H+(aq) + 6e

MnO4(aq)  + 4H+(aq) + 3e→ MnO2 + 2H2O(l)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

Br(aq) + 3H2O(l) → BrO3(aq) + 6H+(aq) + 6e

2MnO4(aq)  + 8H+(aq) + 6e→ 2MnO2 + 4H2O(l)

5) Once everything is balanced, add the half-reactions together and cancel any species that appear in equal quantitates on both sides of the equation.

Br(aq) + 3H2O(l) → BrO3(aq) + 6H+(aq) + 6e

2MnO4(aq)  + 8H+(aq) + 6e→ 2MnO2 + 4H2O(l)
_____________________________________________

Br(aq) + 3H2O + 2MnO4(aq)  + 8 2H+ + 6e→ BrO3(aq) + 6H+ + 6e + 2MnO2 + 4H2O(l)

Br(aq) + 2MnO4(aq)  + 2H+(aq) → BrO3(aq) + 2MnO2 + H2O(l)

6) This is the balanced red-ox reaction in acidic media, so to obtain the one in a basic solution, we need to add 2OH on both sides of the equation:

Br(aq) + 2MnO4(aq)  + 2H+(aq) + 2OH(aq) → BrO3(aq) + 2MnO2 + H2O(l) + 2OH(aq)

The two H+ and OH ions are written as two water molecules, and one of them is canceled with the one on the right side:

Br(aq) + 2MnO4(aq)  + 2H2O(l) → BrO3(aq) + 2MnO2 + H2O(l) + 2OH(aq)

Br(aq) + 2MnO4(aq)  → BrO3(aq) + 2MnO2 + 2OH(aq)

c) MnO4(aq)  + CN(aq) → CNO-(aq)  + MnO2(s)

1) Separate the half-reactions:

CN(aq) → CNO(aq)

MnO4(aq)  → MnO2(s)

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

CN(aq) + H2O(l) → CNO(aq) + 2H+(aq)

MnO4(aq)  + 4H+(aq) → MnO2(s) + 2H2O(l)

3) Balance the charges by adding electrons.

CN(aq) + H2O(l) → CNO(aq) + 2H+(aq) + 2e

MnO4(aq)  + 4H+(aq) + 3e→ MnO2(s) + 2H2O(l)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

3CN(aq) + 3H2O(l) → 3CNO(aq) + 6H+(aq) + 6e

2MnO4(aq)  + 8H+(aq) + 6e→ 2MnO2(s) + 4H2O(l)

5) Once everything is balanced, add the half-reactions together and cancel any species that appear in equal quantitates on both sides of the equation.

3CN(aq) + 3H2O(l) → 3CNO(aq) + 6H+ + 6e

2MnO4(aq)  + 8H+(aq) + 6e→ 2MnO2(s) + 4H2O(l)
_______________________________________

3CN(aq) + 3H2O(l) + 2MnO4(aq)  + 8 2H+(aq) + 6e → 3CNO(aq) + 6H+ + 6e + 2MnO2(s) + 4H2O(l)

3CN(aq) + 2MnO4(aq) + 2H+(aq) → 3CNO(aq) + 2MnO2(s) + H2O(l)

6) This is the balanced red-ox reaction in acidic media, so to obtain the one in a basic solution, we need to add 2OH on both sides of the equation:

3CN(aq) + 2MnO4(aq) + 2H+(aq) + 2OH(aq) → 3CNO(aq) + 2MnO2(s) + H2O(l) + 2OH(aq)

The two H+ and OH ions are written as two water molecules, and one of them is canceled with the one on the right side:

3CN(aq) + 2MnO4(aq) + 2H2O(l) → 3CNO(aq) + 2MnO2(s) + H2O + 2OH(aq)

3CN(aq) + 2MnO4(aq) + H2O(l) → 3CNO(aq) + 2MnO2(s) + 2OH(aq)

d) H2O2(aq) + ClO2(aq) → ClO2(aq) + O2(g)

1) Separate the half-reactions:

H2O2(aq) → O2(g)

ClO2(aq) → ClO2(aq)

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

H2O2(aq) → O2(g) + 2H+(aq)

ClO2(aq) → ClO2(aq)

3) Balance the charges by adding electrons.

H2O2(aq) → O2(g) + 2H+(aq) + 2e

ClO2(aq) + e → ClO2(aq)

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

H2O2(aq) → O2(g) + 2H+(aq) + 2e

2ClO2(aq) + 2e → 2ClO2(aq)

5) Once everything is balanced, add the half-reactions together and cancel any species that appear in equal quantitates on both sides of the equation.

H2O2(aq) → O2(g) + 2H+ + 2e

2ClO2(aq) + 2e → 2ClO2(aq)
_______________________________________

H2O2(aq) + 2ClO2(aq) + 2e → O2(g) + 2H+ + 2e + 2ClO2(aq)

H2O2(aq) + 2ClO2(aq) → O2(g) + 2H+ + 2ClO2(aq)

6) This is the balanced red-ox reaction in acidic media, so to obtain the one in a basic solution, we need to add 2OH on both sides of the equation:

H2O2(aq) + 2ClO2(aq) + 2OH(aq) → O2(g) + 2H+ + 2ClO2(aq) + 2OH(aq)

The two H+ and OH ions are written as two water molecules:

H2O2(aq) + 2ClO2(aq) + 2OH(aq) → O2(g) + 2ClO2(aq) + 2H2O(l)

e) MnO4(aq) + Fe(OH)2(s) → MnO2(s) + Fe(OH)3(s)

1) Separate the half-reactions:

MnO4(aq) → MnO2(s)

Fe(OH)2(s) → Fe(OH)3(s)

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

MnO4(aq) + 4H+(aq) → MnO2(s) + 2H2O(l)

Fe(OH)2(s) + H2O(l) → Fe(OH)3(s) + H+(aq)

3) Balance the charges by adding electrons.

MnO4(aq) + 4H+(aq) + 3e→ MnO2(s) + 2H2O(l)

Fe(OH)2(s) + H2O(l) → Fe(OH)3(s) + H+(aq) + e

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

MnO4(aq) + 4H+(aq) + 3e→ MnO2(s) + 2H2O(l)

3Fe(OH)2(s) + 3H2O(l) → 3Fe(OH)3(s) + 3H+(aq) + 3e

5) Once everything is balanced, add the half-reactions together and cancel any species that appear in equal quantitates on both sides of the equation.

MnO4(aq) + 4H+(aq) + 3e→ MnO2(s) + 2H2O(l)

3Fe(OH)2(s) + 3H2O(l) → 3Fe(OH)3(s) + 3H+(aq) + 3e

_______________________________________

MnO4(aq) + 4H+(aq) + 3e + 3Fe(OH)2(s) + 3H2O(l) → MnO2(s) + 2H2O(l) + 3Fe(OH)3(s) + 3H+ + 3e

MnO4(aq) + H+(aq) + 3Fe(OH)2(s) + H2O(l) → MnO2(s) + 3Fe(OH)3(s)

6) This is the balanced red-ox reaction in acidic media, so to obtain the one in a basic solution, we need one OH on both sides of the equation:

MnO4(aq) + H+(aq) + OH(aq) + 3Fe(OH)2(s) + H2O(l) → MnO2(s) + 3Fe(OH)3(s) + OH(aq)

The H+ and OH ions are written as a water molecule:

MnO4(aq) + 3Fe(OH)2(s) + 2H2O(l) → MnO2(s) + 3Fe(OH)3(s) + OH(aq)

3.

For each redox reaction, sketch a voltaic cell, label the anode and cathode, and indicate the half-reaction that occurs at each electrode. Show the direction of electron flow and the species present in each solution.

a) Cu2+(aq) + 2e → Cu(s)

Sn(s) → Sn2+(aq) + 2e

b) Cd2+(aq) + 2e → Cd(s)

Mn(s) → Mn2+(aq) + 2e

a)

b)

4.

Given the values of standard reduction potentials, determine the redox reaction that will occur in a galvanic cell. Sketch a voltaic cell, label the anode and cathode, and indicate the half-reaction that occurs at each electrode. Show the direction of electron flow and the species present in each solution.

Pb2+(aq) + 2e → Pb(s) Eo = -0.13 V

Ag+(aq) + e → Ag(s) Eo = 0.80 V

Solution

In a galvanic cell, the Eo must be greater than 0. Therefore, the only possibility is to reverse the half-reaction of the Pb2+.

Pb(s) → Pb2+(aq) + 2e  Eo = +0.13 V

The overall balanced equation for the redox reaction would be:

2Ag+(aq) + Pb(s) → 2Ag(s) + Pb2+(aq)

Eo = +0.13 V + 0.80 V = 0.93 V

Agis the oxidizing agent, and it is being reduced, while Pb(s) is the reducing agent and it is being oxidized. This corresponds to what we know from the table of standard reduction potentials: The species standing higher in the table are a stronger oxidizing agent

5.

Calculate the E°cell for each balanced redox reaction and determine if the reaction is spontaneous as written.

a) Ni(s) + Mg2+(aq) → Ni2+(aq) + Mg(s)

b) 2Cl(aq) + Pb2+ (aq) → Cl2(l) + Pb(s)

c) Cd(s) + Pb2+(aq) → Cd2+(aq) + Pb(s)

d) Fe(s) + 3Ag+(aq) → Fe3+(aq) + 3Ag(s)

e) Ca2+(aq) + Fe(s) → Ca(s) + Fe2+(aq)

a)

Nonspontaneous

b)

Nonspontaneous

c)

Spontaneous

d)

Spontaneous

e)

Nonspontaneous

Solution

a) Separate the half-reactions and look up the Eo values from a table of standard reduction potentials. When reversing the reaction, change the sign for Eo.

Ni(s) → Ni2+(aq) + 2eEo = +0.23 V

Mg2+(aq) + 2e→ Mg(s) Eo = -2.37 V

The overall cell potential is the sum of the potentials for the half-reactions.

Eo = +0.23 V + (-2.37 V) = -2.14 V

The cell potential is negative, therefore, the reaction is nonspontaneous.

An alternative approach is to write the reactions without changing the sign of Eo, and then use the formula for the total cell potential when the Eo of the anode is subtracted from the Eo of the anode.

In this case, the cathode would work on the reaction Mg2+(aq) + 2e→ Mg(s) since it is a reduction reaction, and the anode is the oxidation of Ni: Ni(s) → Ni2+(aq) + 2e.

The cell potential is equal to:

Eo = EocathodeEoanode = -2.37 V – (-0.23 V) = -2.14 V

Both approaches are correct, however, I found my students like the method of changing the sign of Eo when the reaction is reversed more since, in this case, we only need to add the potentials without worrying about what is being subtracted from what.

Therefore, we will follow this strategy from now on.

b) 2Cl(aq) + Pb2+ (aq) → Cl2(l) + Pb(s)

Separate the half-reactions and look up the Eo values from a table of standard reduction potentials. When reversing the reaction, change the sign for Eo.

2Cl(aq) → Cl2(l) + 2eEo = 1.36 V

Pb2+ (aq) + 2e→ Pb(s) Eo = -0.13 V

The overall cell potential is the sum of the potentials for the half-reactions.

Eo = -1.36 V + (-0.13 V) = -1.49 V

The cell potential is negative, therefore, the reaction is nonspontaneous.

c) Cd(s) + Pb2+(aq) → Cd2+(aq) + Pb(s)

Separate the half-reactions and look up the Eo values from a table of standard reduction potentials. When reversing the reaction, change the sign for Eo.

Cd(s) → Cd2+(aq) + 2eEo = +0.40 V

Pb2+ (aq) + 2e→ Pb(s) Eo = -0.13 V

The overall cell potential is the sum of the potentials for the half-reactions.

Eo = 0.40 V + (-0.13 V) = 0.27 V

The cell potential is positive, therefore, the reaction is spontaneous.

d) Fe(s) + 3Ag+(aq) → Fe3+(aq) + 3Ag(s)

Separate the half-reactions and look up the Eo values from a table of standard reduction potentials. When reversing the reaction, change the sign for Eo.

Fe(s) → Fe3+(aq) + 3eEo = +0.036 V

Ag+ (aq) + e→ Ag(s) Eo = +0.80 V

The overall cell potential is the sum of the potentials for the half-reactions.

Eo = +0.036 V + 0.80 V = 0.836 V

The cell potential is positive, therefore, the reaction is spontaneous.

Notice that we did not balance the number of electrons and it is simply because multiplying a half-reaction with a coefficient does not change the potential of that reaction.

e) Ca2+(aq) + Fe(s) → Ca(s) + Fe2+(aq)

Separate the half-reactions and look up the Eo values from a table of standard reduction potentials. When reversing the reaction, change the sign for Eo.

Fe(s) → Fe2+(aq) + 2eEo = +0.45 V

Ca2+(aq) + 2e→ Ca(s) Eo = -2.76 V

The overall cell potential is the sum of the potentials for the half-reactions.

Eo = +0.45 V + (-2.76 V) = -2.31 V

The cell potential is negative, therefore, the reaction is nonspontaneous.

6.

Balance the following redox reaction occurring in acidic media, and using a table for standard electrode potentials, determine the Eocell at 25 oC.

Cd(s) + Cr2O72−(aq) → Cd2+(aq) + Cr3+(aq)

Eocell = +1.73 V

Solution

1) Separate the half-reactions:

Cr2O72- → Cr3+

Cd → Cd2+

2) Balance the elements. Add H2O to balance the O atoms, and H+ to balance the H atoms.

Cr2O72- + 14H+(aq) → 2Cr3+ + 7H2O

Cd → Cd2+

3) Balance the charges by adding electrons. We need 6e on the left side of the first equation because the charges are +12 vs +6. The second equation needs two electrons on the right side to make the charges 0 on both sides:

Cr2O72- + 14H+(aq) + 6e → 2Cr3+ + 7H2O(l)

Cd → Cd2+ + 2e

4) Balance the number of electrons by multiplying the equation(s) with a whole number.

Cr2O72- + 14H+(aq) + 6e → 2Cr3+ + 7H2O(l)

3Cd → 3Cd2+ + 6e

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

Cr2O72- + 14H+(aq) + 6e → 2Cr3+ + 7H2O(l)

3Cd → 3Cd2+ + 6e
______________________________________________

Cr2O72- + 14H+(aq) + 6e + 3Cd  → 2Cr3+ + 7H2O(l) + 3Cd2+ + 6e

6) Cancel any species that appear in equal quantitates on both sides of the equation:

Cr2O72- + 14H+(aq) + 6e + 3Cd  → 2Cr3+ + 7H2O(l) + 3Cd2+ + 6e

Cr2O72- + 14H+(aq) + 3Cd  → 2Cr3+ + 7H2O(l) + 3Cd2+

Now that we have the overall reaction, we need to separate it into the oxidation and reduction half-reactions to look up their potentials in the table:

Oxidation:   3Cd  → 3Cd2+ + 6e–     Eo = +0.40 V

Reduction:  Cr2O72- + 14H+(aq) + + 6e → 2Cr3+ + 7H2O(l)    Eo = +1.33 V

Eocell = 0.40 V + +1.33 V = +1.73 V

The cell potential is positive, and therefore, the reaction is spontaneous.

7.

Determine the balanced net reaction from the following two half-reactions and calculate the Eocell at 25 oC.

NO(g) + 2H2O(l) ⟶ NO3(aq) + 4H+(aq) + 3e E° = -0.96 V

MnO2(s) + 4H+(aq) + 2e ⟶ Mn2+(aq) + 2H2O(l) E° = 1.21 V

E°cell = 0.25 V

Solution

The first reaction is an oxidation reaction as the nitrogen goes from +2 to +5 oxidation state, while the second is a reduction reaction where the manganese goes from +4 to +2. Therefore, there is no need to reverse any of the reactions. We only need to balance the number of electrons by multiplying the first equation by 2 and the second equation by 3.

2NO(g) + 4H2O(l) ⟶ 2NO3(aq) + 8H+(aq) + 6e E° = -0.96 V

3MnO2(s) + 12H+(aq) + 6e ⟶ 3Mn2+(aq) + 6H2O(l) E° = 1.21 V

Notice that the reaction potentials do not change with coefficients. The net equation would be:

2NO(g) + 4H2O(l) ⟶ 2NO3(aq) + 8H+(aq) + 6e E° = 0.96 V

3MnO2(s) + 12H+(aq) + 6e ⟶ 3Mn2+(aq) + 6H2O(l) E° = 1.21 V

___________________________________________________

2NO(g) + 4H2O(l) + 3MnO2(s) + 12 4H+(aq) + 6e ⟶ 2NO3(aq) + 8H+(aq) + 6e + 3Mn2+(aq) + 6 2H2O(l)

2NO(g) + 3MnO2(s) + 4H+(aq) ⟶ 2NO3(aq) + 3Mn2+(aq) + 2H2O(l)

E°cell = 1.21 V – 0.96 V = 0.25 V

8.

Using a table for standard electrode potentials, find a metal that can be used to reduce Al3+ ions but not Na+ ions.

Mg

Solution

The strength of reducing power increasing as we go down the table of standard reduction potentials. So, metals that are below Al3+ can reduce it to Al, and since we don’t want to reduce the Na+ ions, the metal should be above Na. The only metal that is below Al3+ and above Na+ is Mg.

You can also test this by writing the half-reactions. The Eo must be positive in order for the reaction to occur:

We switch the sign of Efor the reversed reaction:

Al3+(aq) + 2e– → Al(s)    E= -1.66 V

Mg(s) → Mg2+(aq) + 2e–  E+2.37 V

Eocell = 2.37 – 1.66 = 0.71 V

9.

Using a table for standard electrode potentials, determine if Fe can be reduced by Sn2+ ions.

Fe can reduce Sn2+ ions.

Solution

The strength of reducing power increasing as we go down the table of standard reduction potentials. So, metals that are below Sn2+ can reduce it to Sn, and because Fe is below Sn, it can reduce Sn2+ ions.

You can also test this by writing the half-reactions. The Eo must be positive in order for the reaction to occur:

We switch the sign of Efor the reversed reaction:

Sn2+(aq) + 2e– → Sn(s)    E= -0.14 V

Fe(s) → Fe2+(aq) + 2e–  E= +0.45 V

Eocell = 0.45 – 0.14 = 0.31 V

Notice that the iron cannot be oxidized to Fe3because the Eocell  won’t be positive in that case.

10.

Using a table for standard electrode potentials, determine if H2(g) is capable of reducing Cd2+(aq).

Hcannot reduce Cd2+.

Solution

Hcannot reduce Cd2+ because it stands above in the table of reduction potentials.

You can also test this by writing the half-reactions. The Eo must be positive in order for the reaction to occur:

We switch the sign of Efor the reversed reaction:

Cd2+(aq) + 2e– → Cd(s)   E= -0.14 V

H2(g) → 2H+(aq) + 2e E° = 0 V

Eocell = -0.14 + 0 = -0.14 V

11.

Cell Potential, Free Energy, and the Equilibrium Constant

Use tabulated electrode potentials to calculate ΔG°rxn for each reaction at 25 °C.

a) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

b) Sn2+(aq) + Mg(s) → Sn(s) + Mg2+(aq)

c) Br2(l) + 2I(aq) → 2Br(aq) + I2(s)

d) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

e) O2(g) + 4H+(aq) + Cu(s) → Cu2+(aq) + 2H2O(l)

f) MnO2(s) + 4H+(aq) + Cd(s) → Mn2+(aq) + 2H2O(l ) + Cd2+(aq)

a)

ΔG° = -120 kJ

b)

ΔG° = -430 kJ

c)

ΔG° = -110 kJ

d)

ΔG° = -1.16 x 103 kJ

e)

ΔG° = -340 kJ

f)

ΔG° = -310 kJ

Solution

ΔG° is correlated to Eocell with the following formula:

ΔG° = –nFEocell

Where n is the moles of electrons and F is the Faraday’s constant which is equal to 96,485 C. Therefore, the plan is to first calculate the cell potential and then use it to determine the ΔG°.

a) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

Separate the half-reactions:

Zn(s) → Zn2+(aq) + 2e–  Eo = +0.76 V

Pb2+(aq) + 2e→ Pb(s) Eo = -0.13 V

Eocell = +0.76 V + (-0.13 V) = 0.63 V

Once again, what we did here is looked up the values of Eo and changed the sign of Eo for the reaction that is reversed. In this case, it is the oxidation of Zn. The Eo (Zn2+ → Zn(s)) is -0.76 V and because we have the reverse reaction, we changed the sign and simply added the Eo values to find the cell potential. This method seemed to be less confusing to my students, and that is how will do in these practice problems.

Alternatively, and this is what you will see in most textbooks, the signs of Eo are not changed for reversed reactions, and instead, a minus sign is introduced in the formula of Eocell:

Eocell = Eocathode Eoanode

In this example, Zn(s) → Zn2+(aq) + 2e–  is the oxidation, and therefore, it occurs on the anode. The Eoanode = -0.76 V as per the table of standard reduction potentials.

Eocathode, then is equal to -0.13 V, and the cell potential would be:

Eocell = Eocathode Eoanode = -0.13 V – (-0.76 V) = 0.63 V

Now that we have the value of Eocell, we can use it in the formula for ΔG°:

ΔG° = –nFEocell

$\Delta G^\circ {\rm{ }} = {\rm{ }} – nF{E^o}_{cell}\; = \, – 2\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,{\rm{0}}{\rm{.63}}\,{\rm{V}}\,{\rm{ = }}\, – {\rm{2}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{0}}{\rm{.63}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 1.2\,{\rm{ \times }}\,{10^5}\,{\rm{J}}\,{\rm{ = }}\, – 1.2\,{\rm{ \times }}\,{10^2}\,{\rm{kJ}}$

b) Sn2+(aq) + Mg(s) → Sn(s) + Mg2+(aq)

1) Separate the half-reactions, and switch the sign of Eo for the reversed reaction:

Sn2+(aq) + 2e→ Sn(s)    Eo = -0.14 V

Mg(s) → Mg2+(aq) + 2e–  Eo = +2.37 V

2) Calculate the Eocell by adding the potentials of the two reactions:

Eocell = -0.14 V + 2.37 V = 2.23 V

3) Use the value of Eocell to determine the ΔG°:

ΔG° = –nFEocell

$\Delta G^\circ {\rm{ }} = {\rm{ }} – nF{E^o}_{cell}\; = \, – 2\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,{\rm{2}}{\rm{.23}}\,{\rm{V}}\,{\rm{ = }}\, – {\rm{2}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{2}}{\rm{.23}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 4.30\,{\rm{ \times }}\,{10^5}\,{\rm{J}}\,{\rm{ = }}\, – 430\,{\rm{kJ}}$

c) Br2(l) + 2I(aq) → 2Br(aq) + I2(s)

1) Separate the half-reactions, and switch the sign of Eo for the reversed reaction:

Br2(l) + 2e → 2Br(aq) Eo = +1.09 V

2I(aq) → I2(s) + 2e–  Eo = 0.54 V

2) Calculate the Eocell by adding the potentials of the two reactions:

Eocell = 1.09 V + (-0.54 V) = 0.55 V

3) Use the value of Eocell to determine the ΔG°:

ΔG° = –nFEocell

$\Delta G^\circ {\rm{ }} = {\rm{ }} – nF{E^o}_{cell}\; = \, – 2\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,{\rm{0}}{\rm{.55}}\,{\rm{V}}\,{\rm{ = }}\, – {\rm{2}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{0}}{\rm{.55}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 1.1\,{\rm{ \times }}\,{10^5}\,{\rm{J}}\,{\rm{ = }}\, – 110\,{\rm{kJ}}$

d) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

1) Separate the half-reactions, and switch the sign of Eo for the reversed reaction:

Al(s) → Al3+(aq) + 3e Eo = +1.66 V

Cu2+(aq) + 2e–  → Cu(s) Eo = +0.34 V

Even though adding coefficients does not change the potential of a reaction, it is important to do it when you need to calculate the ΔG° because the formula includes the number of moles of electrons. To balance the number of electrons in both equations, we multiply the first by two, and the second by three, and the number of electrons will be six:

2Al(s) → 2Al3+(aq) + 6e Eo = +1.66 V

3Cu2+(aq) + 6e–  → 3Cu(s) Eo = +0.34 V

2) Calculate the Eocell by adding the potentials of the two reactions:

Eocell = 1.66 V + 0.34 V = 2.00 V

3) Use the value of Eocell to determine the ΔG°:

ΔG° = –nFEocell

$\Delta G^\circ {\rm{ }} = {\rm{ }} – nF{E^o}_{cell}\; = \, – 6\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,{\rm{2}}{\rm{.00}}\,{\rm{V}}\,{\rm{ = }}\, – {\rm{6}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{2}}{\rm{.00}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 1.16\;{\rm{ \times }}\,{10^6}\,{\rm{J}}\,{\rm{ = }}\, – 1.16\;{\rm{ \times }}\,{10^3}\,{\rm{kJ}}$

e) O2(g) + 4H+(aq) + Cu(s) → Cu2+(aq) + 2H2O(l)

1) Separate the half-reactions, and switch the sign of Eo for the reversed reaction:

O2(g) + 4H+(aq) + 4e→ 2H2O(lEo = +1.23 V

Cu(s) → Cu2+(aq) + 2e–  Eo = 0.34 V

Balance the number of electrons:

O2(g) + 4H+(aq) + 4e→ 2H2O(lEo = +1.23 V

2Cu(s) → 2Cu2+(aq) + 4e–  Eo = -0.34 V

2) Calculate the Eocell by adding the potentials of the two reactions:

Eocell = 1.23 V + (-0.34 V) = 0.89 V

3) Use the value of Eocell to determine the ΔG°:

ΔG° = –nFEocell

$\Delta G^\circ {\rm{ }} = {\rm{ }} – nF{E^o}_{cell}\; = \, – 4\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,0.89\,{\rm{V}}\,{\rm{ = }}\, – {\rm{4}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{0}}{\rm{.89}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 3.4\,{\rm{ \times }}\,{10^5}\,{\rm{J}}\,{\rm{ = }}\, – 340\,{\rm{kJ}}$

f) MnO2(s) + 4H+(aq) + Cd(s) → Mn2+(aq) + 2H2O(l) + Cd2+(aq)

1) Separate the half-reactions, and switch the sign of Eo for the reversed reaction:

MnO2(s) + 4H+(aq) + 2e–  → Mn2+(aq) + 2H2O(lEo = +1.21 V

Cd(s) → Cd2+(aq) + 2e–  Eo = +0.40 V

2) Calculate the Eocell by adding the potentials of the two reactions:

Eocell = 1.21 V + (0.40 V) = 1.61 V

3) Use the value of Eocell to determine the ΔG°:

ΔG° = –nFEocell

$\Delta G^\circ {\rm{ }} = {\rm{ }} – nF{E^o}_{cell}\; = \, – 2\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,1.61\,{\rm{V}}\,{\rm{ = }}\, – {\rm{2}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{1}}{\rm{.61}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 3.1\,{\rm{ \times }}\,{10^5}\,{\rm{J}}\,{\rm{ = }}\, – 310\,{\rm{kJ}}$

Notice that all the reactions had a positive Eo indicating that the reactions are spontaneous and can be used to run a galvanic cell. Importantly, when Eo is positive, the ΔG° is negative which is also an indicator of a spontaneous process.

12.

Calculate the equilibrium constant for each redox reaction in problem 11.

a) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

b) Sn2+(aq) + Mg(s) → Sn(s) + Mg2+(aq)

c) Br2(l) + 2I(aq) → 2Br(aq) + I2(s)

d) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

e) O2(g) + 4H+(aq) + Cu(s) → Cu2+(aq) + 2H2O(l)

f) MnO2(s) + 4H+(aq) + Cd(s) → Mn2+(aq) + 2H2O(l ) + Cd2+(aq)

a)

K = 1.91 x 1021

b)

K = 2.33 x 1075

c)

K = 3.88 x 1018

d)

K = 6.07 x 10202

e)

K = 1.44 x 1060

f)

K = 2.59 x 1054

Solution

There are two equations we can use to calculate the equilibrium constant. One is linked through the Eocell, and the other, through the ΔG° which is also calculated through the Eocell. Yes, we can calculate the ΔG° using the standard free energies of formations, however, this is not related to electrochemistry and we won’t focus on that here.

${E^o}_{{\rm{cell}}}\, = \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,K$

$K\; = \;{e^{\frac{{nE}}{{0.0257}}}}$

For reaction (a), Eocell = 0.63 V, therefore, K would be:

$K\; = \;{e^{\frac{{2\, \cdot \,0.63}}{{0.0257}}}}\, = \,{e^{49.0}}\, = \;1.91\, \times \,{10^{21}}$

The second approach:

The equilibrium constant is correlated with the standard free energy change (ΔG°) by this formula:

ΔG° = –RT ln K

Therefore, the equilibrium constant would be equal to:

$\ln \,K\; = \;\frac{{ – \Delta G^\circ }}{{RT}}$

$K\; = \;{e^{\frac{{ – \Delta G^\circ }}{{RT}}}}$

a) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

We determined that ΔG° = -120 x 103 J, however, this is after rounding off the significant figures. When calculating the K, you should not round off the value of ΔG° but rather use it as it is and round off the value of K since being in the exponent, it changes the value of K significantly:

$K\; = \;{e^{\frac{{{\rm{ – }}\left( {{\rm{ – 121,571}}{\rm{.1}}\,{\rm{J}}} \right)}}{{{\rm{8}}{\rm{.314}}\,{\rm{J/K}} \cdot {\rm{mol}}\, \cdot {\rm{298}}\;{\rm{K}}}}}}\, = \,{e^{49.0}}\; = \;1.90\, \times \,{10^{21}}\,$

And now, compare to what we’d get when using the rounded value of ΔG°:

$K\; = \;{e^{\frac{{{\rm{ – }}\left( {{\rm{ – 120,000}}\,{\rm{J}}} \right)}}{{{\rm{8}}{\rm{.314}}\,{\rm{J/K}} \cdot {\rm{mol}}\, \cdot {\rm{298}}\;{\rm{K}}}}}}\, = \,{e^{48.4}}\; = \;1.08\, \times \,{10^{21}}\,$

For the next problems, we will use the first approach to calculate the K.

b) Sn2+(aq) + Mg(s) → Sn(s) + Mg2+(aq)

For reaction (b), Eocell = 2.23 V, therefore, K would be:

$K\; = \;{e^{\frac{{2\, \cdot \,2.23}}{{0.0257}}}}\, = \;2.33\, \times \,{10^{75}}$

c) Br2(l) + 2I(aq) → 2Br(aq) + I2(s)

For reaction (c), Eocell = 0.55 V, therefore, K would be:

$K\; = \;{e^{\frac{{2\, \cdot \,0.55}}{{0.0257}}}}\, = \;3.88\, \times \,{10^{18}}$

d) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

For reaction (d), Eocell = 2.00 V, and the number of electron moles is 6, therefore, K would be:

$K\; = \;{e^{\frac{{6\, \cdot \,2.00}}{{0.0257}}}}\, = \;6.07\, \times \,{10^{202}}$

e) O2(g) + 4H+(aq) + Cu(s) → Cu2+(aq) + 2H2O(l)

For reaction (e), Eocell = 0.89 V, and the number of electron moles is 4, therefore, K would be:

$K\; = \;{e^{\frac{{4\, \cdot \,0.89}}{{0.0257}}}}\, = \;1.44\, \times \,{10^{60}}$

f) MnO2(s) + 4H+(aq) + Cd(s) → Mn2+(aq) + 2H2O(l ) + Cd2+(aq)

For reaction (f), Eocell = 1.61 V, therefore, K would be:

$K\; = \;{e^{\frac{{2\, \cdot \,1.61}}{{0.0257}}}}\, = \;2.59\, \times \,{10^{54}}$

13.

Use an appendix for the free energies of formation to calculate the standard cell potential for the oxidation reaction of ethylene, C2H4 by permanganate ion, MnO4.

5C2H4(g) + 12MnO4(aq) + 36H+(aq) → 10CO2(g) + 12Mn2+(aq) + 28H2O(l)

The standard free energy of formation for Mn2+ ions is -228 kJ/mol.

1.43 V

Solution

The plan is to calculate the ΔG° rxn and use in the expression ΔG° = –nFEocell to determine the Eocell.

To calculate 𝚫G°rxn, we need to subtract the standard free energies of formations of the reactants from the standard free energies of formations of products, both multiplied by their stoichiometric coefficients:

ΔG°rxn = ΣnpΔGof (products) – ΣnrΔGo(reactants)

Where np and nr are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard free energies of formations of elements or their molecules in standard states are equal to zero.

5C2H4(g) + 12MnO4(aq) + 36H+(aq) → 10CO2(g) + 12Mn2+(aq) + 28H2O(l)

ΔG°rxn = [10 ΔGo CO2(g) + 12 ΔGo Mn2+(aq) + 28 ΔGo H2O(l)] – [5 ΔGo C2H4(g) + 12 ΔGo MnO4(aq) + 36 ΔGo H+(aq)]

ΔG°rxn = [10 (-394) + 12 (-228) + 28 x (-237)] – [5 x 68 + 12 (-449) + 36 x 0]

ΔG°rxn = [-3940 -2736 – 6636] – [340 – 5388 + 0] = -8264 kJ = -8.264 x 106 J

From the expression ΔG° = –nFEocell, we can write that:

${E^o}_{{\rm{cell}}}\, = \, – \frac{{\Delta G{^\circ _{rxn}}\;}}{{nF}}$

So, at this point, we need to obtain the number of electrons by separating the half-reactions. First separate, then balance the half-reaction equations, and add electrons when needed as we did for balancing redox reactions:

5C2H4(g) + 12MnO4(aq) + 36H+(aq) → 10CO2(g) + 12Mn2+(aq) + 28H2O(l)
____________________________________________________________

12MnO4(aq) + 36H+(aq) → 12Mn2+(aq) + 28H2O(l)

5C2H4(g) → 10CO2(g)

Balance the atoms and charges:

12MnO4(aq) + 96H+(aq) + 60e → 12Mn2+(aq) + 48H2O(l)

5C2H4(g) + 20H2O(l) → 10CO2(g) + 60H+(aq) + 60e

There are 60 electrons in each half-reaction, therefore, n = 60. To match the units, we will write the value of ΔG° in C·V.

${E^o}_{{\rm{cell}}}\, = \, – \frac{{ – 8.264\; \times \,{{10}^6}\,{\rm{C}}\,{\rm{ \times }}\;{\rm{V\;}}}}{{60\, \times 96,485\,{\rm{C}}}} = \,1.43\,{\rm{V}}$

14.

The Nernst Equation

A voltaic cell runs the following redox reaction:

Fe2+(aq) + Mg(s) → Mg2+(aq) + Fe(s)

Determine whether Ecell is larger or smaller than E°cell and determine the Ecell for the following concentrations:

a) when [Fe2+] = 1.0 M and [Mg2+] = 2.5 M

b) when [Fe2+] = 3.0 M and [Mg2+] = 1.0 M

a)

1.908 V

b)

1.934 V

Solution

There are two parts to this practice problem; one is to determine if Ecell is larger or smaller than E°cell, and the second is to actually calculate the Ecell at the given concentrations.

On the bases of both is the Nernst equation:

$E\; = \,{E^o}\; – \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,Q$

Remember, V is just a unit here, and if it confuses you, you can write the equation as:

$E\; = \,{E^o}\; – \,\frac{{0.0257}}{n}\,\ln \,Q$

Where Q is the reaction quotient, and it defines, based on the concentrations of products and reactants, whether Ecell is larger or smaller than E°cell.

However, before moving on to that part, we want to make sure that this is a galvanic cell i.e. the reaction is spontaneous.

Fe2+(aq) + 2e→ Fe(s)  E° = -0.45 V

Mg(s) → Mg2+(aq) + 2e–  E° = +2.37 V

Ecell = -0.45 V + 2.37 V = 1.92 V

The cell potential is positive so, at standard conditions, it is a spontaneous reaction. This means any changes in concentrations that increase the tendency of the forward reaction to occur will increase the cell potential.

Consequently, if changing the concentrations decreases the tendency of the forward reaction to occur, it

will decrease the cell potential and increases the tendency of the reverse reaction to occur.

So, the next part is to use the Le Chatelier’s principle and determine how each of the concentration set will affect the tendency of the reaction and therefore, the cell potential.

a) when [Fe2+] = 1.0 M and [Mg2+] = 2.5 M

Fe2+(aq) + Mg(s) → Mg2+(aq) + Fe(s)

The concentration of the product, Mg2+ is higher than what it is at standard conditions (1 M), therefore, according to the Le Chatelier’s principle, the reaction will tend to occur in the reverse direction. This also means that Ecell is smaller than E°cell.

Let’s also plug the numbers into the Nernst equation and determine the Ecell:

$E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{[F}}{{\rm{e}}^{{\rm{2 + }}}}{\rm{]}}}}$

$E\; = \,1.92\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[2}}{\rm{.5]}}}}{{{\rm{[1}}{\rm{.0]}}}}\; = \;1.908\,{\rm{V}}$

Because Q = 2.5, the difference of Ecell and E°cell is not large. However, it is important to remember the pattern: if Q > 1, Ecell < E°cell, and if Q < 1, Ecell > E°cell.

Review the principle of reaction quotient if this is still confusing.

b) when [Fe2+] = 3.0 M and [Mg2+] = 1.0 M

There is more reactant in the system than there is at standard conditions, so the reaction will tend to shift forward and consume some of it. Therefore, Ecell should be larger than E°cell.

$E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{[F}}{{\rm{e}}^{{\rm{2 + }}}}{\rm{]}}}}$

$E\; = \,1.92\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[1]}}}}{{{\rm{[3}}{\rm{.0]}}}}\; = \;1.934\,{\rm{V}}$

15.

Calculate E°, E, and ΔG for the following cell reaction:

Mn(s) + Pb2+(aq) → Mn2+(aq) + Pb(s)

[Mn2+] = 0.164 M, [Pb2+] = 0.038

E°cell  = 1.05 V

Ecell  = 1.03 V

ΔG = -199 kJ

Solution

To calculate the E°, we separate the half-reactions, and look up the standard potentials:

Mn(s) → Mn2+(aq) + 2e  E° = +1.18 V

Pb2+(aq) + 2e→ Pb(s)  E° = -0.13 V

E°cell  = +1.18 V – 0.13 V = 1.05 V

To calculate the E, we use the Nernst equation:

$E\; = \,{E^o}\; – \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,Q$

$E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{[P}}{{\rm{b}}^{{\rm{2 + }}}}{\rm{]}}}}$

$E\; = \,1.05\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[0}}{\rm{.164]}}}}{{{\rm{[0}}{\rm{.038]}}}}\; = \,1.031\,{\rm{V}}$

And now, we use the value of Ecell to determine the ΔG:

ΔG = –nFEcell

$\Delta G{\rm{ }} = {\rm{ }} – nF{E_{cell}}\; = \, – 2\,\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}\,{\rm{ \times }}\,\frac{{{\rm{96,485}}\,{\rm{C}}}}{{\cancel{{{\rm{mol }}{{\rm{e}}^{\rm{ – }}}}}}}\;{\rm{ \times }}\,{\rm{1}}{\rm{.031V}}\,{\rm{ = }}\, – {\rm{2}}\;{\rm{ \times }}\,{\rm{96,485}}\,\cancel{{\rm{C}}}\,{\rm{ \times }}\,{\rm{1}}{\rm{.031}}\frac{{\rm{J}}}{{\cancel{{\rm{C}}}}}\,{\rm{ = }}\, – 1.99\,{\rm{ \times }}\,{10^5}\,{\rm{J}}\,{\rm{ = }}\, – 199\,{\rm{kJ}}$

16.

A galvanic cell operates on the following redox reaction:

2MnO4(aq) + 10Br(aq) + 16H+(aq) → 2Mn2+(aq) + 5Br2(l) + 8H2O(l)

Calculate the cell potential at 25 oC, given the concentrations of the aqueous components are: [MnO4] = 0.0250 M, [Br] = 0.0150 M, [Mn2+] = 0.380 M, and [H+] = 0.64 M.

1.37 V

Solution

To calculate the cell potential, we again use the Nernst equation because the concentrations are not 1 M.

$E\; = \,{E^o}\; – \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,Q$

We need to first calculate the Eo by separating the half-reactions to find the potential of each electrode in the table:

2MnO4(aq) + 16H+(aq) + 10e → 2Mn2+(aq) + 8H2O(l) Eo = +1.51 V

10Br(aq) → 5Br2(l) + 10e Eo = -1.09 V

Eocell = 1.51 – 1.09 = 0.42 V

The quotient is equal to:

$Q\; = \,\frac{{{{{\rm{[M}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}^{\rm{2}}}}}{{{{{\rm{[Mn}}{{\rm{O}}_{\rm{4}}}^{\rm{ – }}{\rm{]}}}^{\rm{2}}}{{{\rm{[B}}{{\rm{r}}^{\rm{ – }}}{\rm{]}}}^{{\rm{10}}}}{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}}^{{\rm{16}}}}}}$

$Q\; = \,\frac{{{{\left( {{\rm{0}}{\rm{.380}}} \right)}^{\rm{2}}}}}{{{{\left( {{\rm{0}}{\rm{.0250}}} \right)}^{\rm{2}}}{{\left( {{\rm{0}}{\rm{.0150}}} \right)}^{{\rm{10}}}}{{\left( {{\rm{0}}{\rm{.64}}} \right)}^{{\rm{16}}}}}}\; = \;5.06\; \times \;{10^{23}}$

There are 10 electrons involved in the reaction, therefore, for the Nernst equation, we have:

$E\; = \,1.51\,{\rm{V}}\; – \,\frac{{0.0257\;{\rm{V}}}}{{10}}\,\ln \,{\rm{5}}{\rm{.06}}\, \times \,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\, = \,1.37\,{\rm{V}}$

17.

A voltaic cell consists of a Mg/Mg2+ half-cell and a Sn/Sn2+ half-cell at 25 °C. The initial concentrations of Sn2+ and Mg2+ are 1.80 M and 0.250 M, respectively.

a) Calculate the initial cell potential.

b) What is the cell potential when the concentration of Sn2+ dropped to 0.650 M?

c) Determine the concentrations of Sn2+ and Mg2+ ions when the cell potential falls to 2.15 V.

a)

E = 2.26 V

b)

E = 2.22 V

c)

[Mg2+] = 2.04595 M

[Sn2+] = 0.00405 M

Solution

a) To calculate the initial cell potential, we determine the E°, by separating the half-reactions such E° > 0. Because Sn/Sn2+ has a higher potential, it will be the oxidizing agent, and therefore, taking the electrons from the Mg2+ ion:

Mg(s) → Mg2+(aq) + 2e  E° = +2.37 V

Sn2+(aq) + 2e→ Sn(s)  E° = -0.14 V

The overall reaction is the sum of the two half-reactions:

Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s)

E°cell  = +2.37 V – 0.14 V = 2.23 V

Now, this is the standard cell potential which is when the concentrations are 1 M. However, we have [Sn2+] = 1.80 M, and [Mg2+] = 0.25 M, and therefore, we use the Nernst equation to calculate the Ecell:

$E\; = \,2.23\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[0}}{\rm{.25]}}}}{{{\rm{[1}}{\rm{.80]}}}}\; = \,2.26\,{\rm{V}}$

Ecell > Eo and this is what we expect when Q < 1.

b) To calculate the E, we again use the Nernst equation.

$E\; = \,{E^o}\; – \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,Q$

$E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{[S}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}}$

We are given the concentration of Sn2+ ions (0.650 M), so to use the Nernst equation, we also need the [Mg2+]. Because of the 1:1 stochiometric ratio, [Mg2+] increases by as much as [Sn2+] decreases. The initial concentration of Sn2+ is 1.80 M 1 molar, so if [Sn2+] drops to 0.650 M, 1.15 M (1.8 – 0.65) of it must be reacted and this is by how much [Mg2+] increases. Therefore, [Mg2+] = 0.250 + 1.15 = 1.4 M.

$E\; = \,2.23\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[1}}{\rm{.40]}}}}{{{\rm{[0}}{\rm{.65]}}}}\; = \;2.22\,{\rm{V}}$

c) Here, we need to determine the concentrations of the ions given the values of E° = 2.23 V, and E = 2.15 V.

The total concentration of the ions is 1.80 M + 0.250 M = 2.05 M, and it will stay constant throughout the reaction only that [Sn2+] will be decreasing, and [Mg2+] increase by the same amount. So, we can assign [Mg2+] = x M, and [Sn2+] = (2.05 -x) M and use the Nernst equation to determine the x:

$E\; = \,2.23\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{\rm{x}}}{{{\rm{2}}{\rm{.05}}\,{\rm{ – }}\,{\rm{x}}}}\; = \;2.15\,{\rm{V}}$

$E\; = \,2.23\,{\rm{V}}\; – \,2.15\,{\rm{V}}\; = \;\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{\rm{x}}}{{{\rm{2}}{\rm{.05}}\,{\rm{ – }}\,{\rm{x}}}}$

$\frac{{2\,\left( {2.23\,{\rm{V}}\; – \,2.15\,{\rm{V}}} \right)\,}}{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}\; = \,{\rm{ln}}\,\frac{{\rm{x}}}{{{\rm{2}}{\rm{.05}}\,{\rm{ – }}\,{\rm{x}}}}$

$6.2257\; = \,{\rm{ln}}\,\frac{{\rm{x}}}{{{\rm{2}}{\rm{.05}}\,{\rm{ – }}\,{\rm{x}}}}$

${e^{6.2257}}\; = \,\frac{{\rm{x}}}{{{\rm{2}}{\rm{.05}}\,{\rm{ – }}\,{\rm{x}}}}\; = \,505.6$

Solving for x, we find that x = 2.04595, and this is the [Mg2+].

The concentration of Sn2+ ions would then be: 2.05 – 2.04595 = 0.00405 M

Note: You can also use the notation [Mg2+] instead of x if that helps keeping track of things easier:

$E\; = \,2.23\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{[M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{]}}}}{{{\rm{2}}{\rm{.05}}\,{\rm{ – }}\,{\rm{[M}}{{\rm{g}}^{{\rm{2 + }}}}{\rm{]}}}}\; = \;2.15\,{\rm{V}}$

Alternatively, we could assign the concentration of Sn2+ as (1.80 – x) M, and [Mg2+] = (0.25 + x) M for the time when E drops to 2.15, and the Nernst equation would look like this:

$E\; = \,2.23\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,\frac{{{\rm{0}}{\rm{.25}}\,{\rm{ + }}\,{\rm{x}}}}{{{\rm{1}}{\rm{.80}}\,{\rm{ – }}\,{\rm{x}}}}\; = \;2.15\,{\rm{V}}$

Solving for x, we find that x = 1.79595. Therefore,

[Mg2+] = (0.25 + 1.79595) = 2.0459 M

[Sn2+] = (1.80 – x) = 0.00405 M

If the fraction in the ln term is not what you want to see at all, just write Q and use the unknowns once the value of Q is calculated.

$E\; = \,2.23\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,Q\; = \;2.15\,{\rm{V}}$

18.

A voltaic cell consists of a Zn/Zn2+ half-cell and a H2/H+ half-cell at 25 °C. Calculate the initial cell potential if [Zn2+] = 0.0250 M, [H+]  = 2.80 M, P(H2) = 0.45 atm.

E = 0.84 V

Solution

The first step is to obtain a correct net reaction such that the cell potential is positive. Zn is below H2 in the reduction potential table, therefore, it will reduce the H+ ions to hydrogen gas.

Zn(s) → Zn2+(aq) + 2e  E° = +0.76 V

2H+(aq) + 2e→ H2(g)  E° = 0 V

The overall reaction is the sum of the two half-reactions:

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

E°cell  = +0.76 V + 0 V = 0.76 V

Next, we calculate the quotient to use it in the Nernst equation.

$Q = \frac{{{P_{{{\rm{H}}_{\rm{2}}}}}\, \times \,{\rm{[Z}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}}{{{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}}^{\rm{2}}}}}\; = \,\frac{{0.45\, \times \,0.0250}}{{{{2.80}^2}}}\; = \,0.001435$

$E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,Q$

$E\; = \,0.76\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,{\rm{0}}{\rm{.001435}}\,{\rm{ = }}\,{\rm{0}}{\rm{.84 V}}$

19.

Determine the emf of a cell consisting of a Sn2+/Sn half-cell and a Pt/H+/H2 half-cell if [Sn2+] = 0.12 M, [H+] = 0.0650 M, and P(H2) = 1.25 atm?

E = 0.0941 V

Solution

The first step is to obtain a correct net reaction such that the cell potential is positive. Sn is below H2 in the reduction potential table, therefore, it will reduce the H+ ions to hydrogen gas.

Sn(s) → Sn2+(aq) + 2e  E° = +0.14 V

2H+(aq) + 2e→ H2(g)  E° = 0 V

The overall reaction is the sum of the two half-reactions:

Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)

E°cell  = +0.14 V + 0 V = 0.14 V

Next, we calculate the quotient to use it in the Nernst equation.

$Q = \frac{{{P_{{{\rm{H}}_{\rm{2}}}}}\, \times \,{\rm{[S}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{]}}}}{{{{{\rm{[}}{{\rm{H}}^{\rm{ + }}}{\rm{]}}}^{\rm{2}}}}}\; = \,\frac{{1.25\, \times \,0.120}}{{{{0.0650}^2}}}\; = \,35.5$

$E\; = \,{E^o}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,Q$

$E\; = \,0.14\,{\rm{V}}\; – \,\frac{{{\rm{0}}{\rm{.0257}}\;{\rm{V}}}}{{\rm{2}}}\,{\rm{ln}}\,{\rm{35}}{\rm{.5}}\,{\rm{ = }}\,{\rm{0}}{\rm{.0941 V}}$