Sample problems redox and electrochemical reactions

1. Question: The half-reaction for the reduction of MnO₄⁻ to Mn²⁺ in acidic solution is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. What mass of MnO₄⁻ can be reduced by 1 mol of electrons? Solution: 5 moles of electrons reduce 1 mole of MnO₄⁻. Thus, 1 mole of electrons reduces 1/5 mole of MnO₄⁻, or 1/5 x 158.04 g/mol = 31.61 g.

2. Question: The half-reaction for the oxidation of Fe²⁺ to Fe³⁺ in acidic solution is Fe²⁺ → Fe³⁺ + e⁻. How many grams of Fe²⁺ can be oxidized by 3 mol of electrons? Solution: 1 mole of electrons oxidizes 1 mole of Fe²⁺. Thus, 3 moles of electrons oxidize 3 moles of Fe²⁺, or 3 x 55.85 g/mol = 167.55 g.
3. Question: The reduction of Cu²⁺ to Cu (Cu²⁺ + 2e⁻ → Cu) has an E° of 0.34 V, and the oxidation of Zn to Zn²⁺ (Zn → Zn²⁺ + 2e⁻) has an E° of -0.76 V. What is the standard cell potential? Solution: E°cell = E°cathode – E°anode = 0.34 V – (-0.76 V) = 1.1 V
4. Question: The reduction of Cl₂ to Cl⁻ (Cl₂ + 2e⁻ → 2Cl⁻) has an E° of 1.36 V. What is the standard potential for the reverse (oxidation) reaction? Solution: The standard potential for the reverse reaction is the negative of the standard potential for the forward reaction, so E° = -1.36 V.
5. Question: A cell has an anode half-reaction of Zn → Zn²⁺ + 2e⁻ (E° = -0.76 V) and a cathode half-reaction of Ag⁺ + e⁻ → Ag (E° = 0.80 V). What is the cell potential? Solution: E°cell = E°cathode – E°anode = 0.80 V – (-0.76 V) = 1.56 V
6. Question: What is the standard cell potential for a cell with an anode half-reaction of Fe²⁺ → Fe³⁺ + e⁻ (E° = -0.77 V) and a cathode half-reaction of Ag⁺ + e⁻ → Ag (E° = 0.80 V)? Solution: E°cell = E°cathode – E°anode = 0.80 V – (-0.77 V) = 1.57 V
7. Question: How many grams of Cr are needed to react with an excess of Ag⁺ according to the reaction 2Cr + 3Ag⁺ → 2Cr³⁺ + 3Ag, if 107.9 g of Ag are produced? Solution: The reaction ratio is 2:3. Thus, 107.9 g Ag is 107.9 g / 107.87 g/mol = 1 mol. This requires 1 x 2/3 mol of Cr, or 2/3 x 52.00 g/mol = 34.67 g.
8. Question: The reaction of Fe with Cu²⁺ according to the equation Fe + Cu²⁺ → Fe²⁺ + Cu has an E° of 0.78 V. How much work can be obtained from this reaction if 55.85 g of Fe are consumed? Solution: This amount of Fe is 1 mol, and 1 mol of electrons are transferred in the reaction. The work is nFE, where n = 1, F = 96485 C/mol, and E = 0.78 V. So, work = 1 x 96485 C x 0.78 J/C = 75258 J.
9. Question: The reaction of Zn with H⁺ according to the equation Zn + 2H⁺ → Zn²⁺ + H₂ has an E° of -0.76 V. How much work can be obtained from this reaction if 65.38 g of Zn are consumed? Solution: This amount of Zn is 1 mol, and 2 mol of electrons are transferred in the reaction. The work is nFE, where n = 2, F = 96485 C/mol, and E = -0.76 V. So, work = 2 x 96485 C x -0.76 J/C = -146377 J.
10. Question: The reduction of MnO₂ to Mn²⁺ (MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O) has an E° of 1.23 V. What is the standard potential for the reverse (oxidation) reaction? Solution: The standard potential for the reverse reaction is the negative of the standard potential for the forward reaction, so E° = -1.23 V.