Given the following standard reduction potentials: PbO2 + 4H⁺ + 2e⁻ - Pb²+ + 2H2O, E° = 1.455 V MnO4⁻ + 8H⁺ + 5e⁻ - Mn²+ + 4H2O, E° = 1.51 V Ce⁴+ + e⁻ - Ce³+, E° = 1.61 V H2O2 + 2H⁺ + 2e⁻ - 2H2O, E° = 1.77 V Which of the following statements is correct under standard conditions?

Question

Given the following standard reduction potentials: PbO2 + 4H⁺ + 2e⁻ -> Pb²+ + 2H2O, E° = 1.455 V MnO4⁻ + 8H⁺ + 5e⁻ -> Mn²+ + 4H2O, E° = 1.51 V Ce⁴+ + e⁻ -> Ce³+, E° = 1.61 V H2O2 + 2H⁺ + 2e⁻ -> 2H2O, E° = 1.77 V Which of the following statements is correct under standard conditions?

Accepted Answer

Ce⁴+ will oxidise Pb²+ to PbO2. Ce⁴+ (E° = 1.61 V) is a stronger oxidant than PbO2 (E° = 1.455 V), so Ce⁴+ can oxidise Pb²+ to PbO2. MnO4⁻ (E° = 1.51 V) can also oxidise Pb²+ to PbO2 since 1.51 > 1.455. H2O2 (1.77 V) vs MnO4⁻ (1.51 V): H2O2 would reduce MnO4⁻ to Mn²+, not oxidise Mn²+. PbO2 (1.455 V) cannot oxidise Mn²+ (requires an agent with E° > 1.51 V).

Solution

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