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)
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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) SO₃^2–(aq) + MnO₄^–(aq) → SO₄^2–(aq) + Mn²⁺(aq)

1) Separate the half-reactions:

SO₃^2–(aq) → SO₄^2–(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) + SO₃^2–(aq) → SO₄^2–(aq) + 2H⁺(aq)

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

3) Balance the charges by adding electrons:

H₂O(l) + SO₃^2–(aq) → SO₄^2–(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 + 5SO₃^2–(aq) → 5SO₄^2–(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 + 5SO₃^2–(aq) → 5SO₄^2–(aq) + 10H⁺(aq) + 10e^–

2MnO₄^–(aq) + 16H⁺(aq) + 10e^– → 2Mn²⁺(aq) + 8H₂O(l)
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5H₂O + 5SO₃^2–(aq) + 2MnO₄^–(aq) + 16H⁺(aq) + 10e^– → 5SO₄^2–(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) + 5SO₃^2–(aq) + 2MnO₄^–(aq) + 16 6H⁺(aq) + 10e^– → 5SO₄^2–(aq) + 10H⁺(aq) + 10e^– + 2Mn²⁺(aq) + 8 3H₂O(l)

 5SO₃^2–(aq) + 2MnO₄^–(aq) + 6H⁺(aq) → 5SO₄^2–(aq) + 2Mn²⁺(aq) + 3H₂O(l)

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

2MnO₄^–(aq) + 5SO₃^2–(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5SO₄^2–(aq) + 3H₂O(l)

c) Cr₂O₇^2- + C₂O₂^2- → Cr³⁺ + CO₂

1) Separate the half-reactions:

Cr₂O₇^2- → Cr³⁺

C₂O₂^2- → CO₂

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

Cr₂O₇^2- + 14H⁺(aq) → 2Cr³⁺ + 7H₂O

C₂O₂^2- → 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:

Cr₂O₇^2- + 14H⁺(aq) + 6e^– → 2Cr³⁺ + 7H₂O(l)

C₂O₂^2- → 2CO₂ + 2e^–

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

Cr₂O₇^2- + 14H⁺(aq) + 6e^– → 2Cr³⁺ + 7H₂O(l)

3C₂O₂^2- → 6CO₂ + 6e^–

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

Cr₂O₇^2- + 14H⁺(aq) + 6e^– → 2Cr³⁺ + 7H₂O(l)

3C₂O₂^2- → 6CO₂ + 6e^–
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Cr₂O₇^2- + 14H⁺(aq) + 6e^– + 3C₂O₂^2- → 2Cr³⁺ + 7H₂O + 6CO₂ + 6e^–

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

Cr₂O₇^2- + 14H⁺(aq) + 6e^– + 3C₂O₂^2- → 2Cr³⁺ + 7H₂O(l) + 6CO₂ + 6e^–

3C₂O₂^2- + Cr₂O₇^2- + 14H⁺(aq) → 6CO₂ + 2Cr³⁺ + 7H₂O(l)

d) PbO₂ + Mn²⁺ + SO₄^2– → PbSO₄ + MnO₄^–

1) Separate the half-reactions:

Mn²⁺ → MnO₄^–

PbO₂ + SO₄^2– → 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 + SO₄^2– + 4H⁺(aq) → PbSO₄ + 2H₂O(l)

 3) Balance the charges by adding electrons.

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

PbO₂ + SO₄^2– + 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 + 5SO₄^2– + 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 + 5SO₄^2– + 20H⁺(aq) + 10e^– → 5PbSO₄ + 10H₂O(l)
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2Mn²⁺ + 8H₂O(l) + 5PbO₂ + 5SO₄^2– + 20 4H⁺(aq) + 10e^– → 2MnO₄^– + 16H⁺(aq) + 10e^– + 5PbSO₄ + 10 2H₂O(l)

 2Mn²⁺ + 5PbO₂ + 5SO₄^2– + 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)
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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)

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)
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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)
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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)
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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)
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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^–

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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)

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

Pb(s) → Pb²⁺(aq) + 2e^–  E^o = +0.13 V

The overall balanced equation for the redox reaction would be:

2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺(aq)

E^o = +0.13 V + 0.80 V = 0.93 V

Ag⁺ is 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

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

Ni(s) → Ni²⁺(aq) + 2e^– E^o = +0.23 V

Mg²⁺(aq) + 2e^– → Mg(s) E^o = -2.37 V

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

E^o = +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 E^o, and then use the formula for the total cell potential when the E^o of the anode is subtracted from the E^o 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:

E^o = E^o_cathode – E^o_anode = -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 E^o 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 E^o values from a table of standard reduction potentials. When reversing the reaction, change the sign for E^o.

2Cl^–(aq) → Cl₂(l) + 2e^– E^o = –1.36 V

Pb²⁺ (aq) + 2e^– → Pb(s) E^o = -0.13 V

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

E^o = -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 E^o values from a table of standard reduction potentials. When reversing the reaction, change the sign for E^o.

Cd(s) → Cd²⁺(aq) + 2e^– E^o = +0.40 V

Pb²⁺ (aq) + 2e^– → Pb(s) E^o = -0.13 V

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

E^o = 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 E^o values from a table of standard reduction potentials. When reversing the reaction, change the sign for E^o.

Fe(s) → Fe³⁺(aq) + 3e^– E^o = +0.036 V

Ag⁺ (aq) + e^– → Ag(s) E^o = +0.80 V

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

E^o = +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 E^o values from a table of standard reduction potentials. When reversing the reaction, change the sign for E^o.

Fe(s) → Fe²⁺(aq) + 2e^– E^o = +0.45 V

Ca²⁺(aq) + 2e^– → Ca(s) E^o = -2.76 V

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

E^o = +0.45 V + (-2.76 V) = -2.31 V

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

1) Separate the half-reactions:

Cr₂O₇^2- → Cr³⁺

Cd → Cd²⁺

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

Cr₂O₇^2- + 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:

Cr₂O₇^2- + 14H⁺(aq) + 6e^– → 2Cr³⁺ + 7H₂O(l)

Cd → Cd²⁺ + 2e^–

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

Cr₂O₇^2- + 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.

Cr₂O₇^2- + 14H⁺(aq) + 6e^– → 2Cr³⁺ + 7H₂O(l)

3Cd → 3Cd²⁺ + 6e^–
______________________________________________

Cr₂O₇^2- + 14H⁺(aq) + 6e^– + 3Cd  → 2Cr³⁺ + 7H₂O(l) + 3Cd²⁺ + 6e^–

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

Cr₂O₇^2- + 14H⁺(aq) + 6e^– + 3Cd  → 2Cr³⁺ + 7H₂O(l) + 3Cd²⁺ + 6e^–

Cr₂O₇^2- + 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^–     E^o = +0.40 V

Reduction:  Cr₂O₇^2- + 14H⁺(aq) + + 6e^– → 2Cr³⁺ + 7H₂O(l)    E^o = +1.33 V

E^o_cell = 0.40 V + +1.33 V = +1.73 V

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

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

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 E^o must be positive in order for the reaction to occur:

We switch the sign of E^o for the reversed reaction:

Al³⁺(aq) + 2e^– → Al(s)    E^o = -1.66 V

Mg(s) → Mg²⁺(aq) + 2e^–  E^o = +2.37 V

E^o_cell = 2.37 – 1.66 = 0.71 V

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 E^o must be positive in order for the reaction to occur:

We switch the sign of E^o for the reversed reaction:

Sn²⁺(aq) + 2e^– → Sn(s)    E^o = -0.14 V

Fe(s) → Fe²⁺(aq) + 2e^–  E^o = +0.45 V

E^o_cell = 0.45 – 0.14 = 0.31 V

Notice that the iron cannot be oxidized to Fe³⁺ because the E^o_cell  won’t be positive in that case.

H₂ cannot reduce Cd²⁺ because it stands above in the table of reduction potentials.

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

We switch the sign of E^o for the reversed reaction:

Cd²⁺(aq) + 2e^– → Cd(s)   E^o = -0.14 V

H₂(g) → 2H⁺(aq) + 2e^–  E° = 0 V

E^o_cell = -0.14 + 0 = -0.14 V

ΔG° is correlated to E^o_cell with the following formula:

ΔG° = –nFE^o_cell

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^–  E^o = +0.76 V

Pb²⁺(aq) + 2e^– → Pb(s) E^o = -0.13 V

E^o_cell = +0.76 V + (-0.13 V) = 0.63 V

Once again, what we did here is looked up the values of E^o and changed the sign of E^o for the reaction that is reversed. In this case, it is the oxidation of Zn. The E^o (Zn²⁺ → Zn(s)) is -0.76 V and because we have the reverse reaction, we changed the sign and simply added the E^o 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 E^o are not changed for reversed reactions, and instead, a minus sign is introduced in the formula of E^o_cell:

E^o_cell = E^o_cathode – E^o_anode

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

E^o_cathode, then is equal to -0.13 V, and the cell potential would be:

E^o_cell = E^o_cathode – E^o_anode = -0.13 V – (-0.76 V) = 0.63 V

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

ΔG° = –nFE^o_cell

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

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

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

Sn²⁺(aq) + 2e^– → Sn(s)    E^o = -0.14 V

Mg(s) → Mg²⁺(aq) + 2e^–  E^o = +2.37 V

2) Calculate the E^o_cell by adding the potentials of the two reactions:

E^o_cell = -0.14 V + 2.37 V = 2.23 V

3) Use the value of E^o_cell to determine the ΔG°:

ΔG° = –nFE^o_cell

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

c) Br₂(l) + 2I^–(aq) → 2Br^–(aq) + I₂(s)

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

Br₂(l) + 2e^– → 2Br^–(aq) E^o = +1.09 V

2I^–(aq) → I₂(s) + 2e^–  E^o = –0.54 V

2) Calculate the E^o_cell by adding the potentials of the two reactions:

E^o_cell = 1.09 V + (-0.54 V) = 0.55 V

3) Use the value of E^o_cell to determine the ΔG°:

ΔG° = –nFE^o_cell

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

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

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

Al(s) → Al³⁺(aq) + 3e^– E^o = +1.66 V

Cu²⁺(aq) + 2e^–  → Cu(s) E^o = +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^– E^o = +1.66 V

3Cu²⁺(aq) + 6e^–  → 3Cu(s) E^o = +0.34 V

2) Calculate the E^o_cell by adding the potentials of the two reactions:

E^o_cell = 1.66 V + 0.34 V = 2.00 V

3) Use the value of E^o_cell to determine the ΔG°:

ΔG° = –nFE^o_cell

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

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

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

O₂(g) + 4H⁺(aq) + 4e^– → 2H₂O(l)  E^o = +1.23 V

Cu(s) → Cu²⁺(aq) + 2e^–  E^o = –0.34 V

Balance the number of electrons:

O₂(g) + 4H⁺(aq) + 4e^– → 2H₂O(l)  E^o = +1.23 V

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

2) Calculate the E^o_cell by adding the potentials of the two reactions:

E^o_cell = 1.23 V + (-0.34 V) = 0.89 V

3) Use the value of E^o_cell to determine the ΔG°:

ΔG° = –nFE^o_cell

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

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

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

MnO₂(s) + 4H⁺(aq) + 2e^–  → Mn²⁺(aq) + 2H₂O(l)  E^o = +1.21 V

Cd(s) → Cd²⁺(aq) + 2e^–  E^o = +0.40 V

2) Calculate the E^o_cell by adding the potentials of the two reactions:

E^o_cell = 1.21 V + (0.40 V) = 1.61 V

3) Use the value of E^o_cell to determine the ΔG°:

ΔG° = –nFE^o_cell

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

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

There are two equations we can use to calculate the equilibrium constant. One is linked through the E^o_cell, and the other, through the ΔG° which is also calculated through the E^o_cell. 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), E^o_cell = 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), E^o_cell = 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), E^o_cell = 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), E^o_cell = 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), E^o_cell = 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), E^o_cell = 1.61 V, therefore, K would be:

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

The plan is to calculate the ΔG° _rxn and use in the expression ΔG° = –nFE^o_cell to determine the E^o_cell.

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 = Σn_pΔG^o_f (products) – Σn_rΔG^o_f (reactants)

Where n_p and n_r 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 ΔG^o CO₂(g) + 12 ΔG^o Mn²⁺(aq) + 28 ΔG^o H₂O(l)] – [5 ΔG^o C₂H₄(g) + 12 ΔG^o MnO₄^–(aq) + 36 ΔG^o 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° = –nFE^o_cell, 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)
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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.

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

There are two parts to this practice problem; one is to determine if E_cell is larger or smaller than E°_cell, and the second is to actually calculate the E_cell 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 E_cell 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

E_cell = -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 E_cell is smaller than E°_cell.

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

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

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

Because Q = 2.5, the difference of E_cell and E°_cell is not large. However, it is important to remember the pattern: if Q > 1, E_cell < E°_cell, and if Q < 1, E_cell > 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, E_cell should be larger than E°_cell.

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

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

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 E_cell to determine the ΔG:

ΔG = –nFE_cell

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

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 E^o by separating the half-reactions to find the potential of each electrode in the table:

2MnO4^–(aq) + 16H⁺(aq) + 10e^– → 2Mn²⁺(aq) + 8H₂O(l) E^o = +1.51 V

10Br^–(aq) → 5Br₂(l) + 10e^– E^o = -1.09 V

E^o_cell = 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:

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

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 E_cell:

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

E_cell > E^o 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.

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

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.

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

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

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

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.

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

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

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