Question

The charge required for the reduction of 1 mol of \( MnO_4^- \) to \( MnO_2 \) is

• A. 1 F
• B. 5 F
• C. 3 F
• D. 6 F

Hint:

The number of Faradays required for a redox reaction corresponds to the total number of electrons involved in the oxidation or reduction of the species.

Answer 1

The Correct Option is C

The objective is to determine the charge needed to reduce 1 mole of MnO₄⁻ to MnO₂.

1. Reduction Process Analysis:
The reduction of MnO₄⁻ to MnO₂ entails a change in the manganese (Mn) oxidation state. The oxidation states are established as follows:

• In MnO₄⁻, Mn has an oxidation state of +7 (given oxygen's -2 state and a net charge of -1).
• In MnO₂, Mn has an oxidation state of +4 (given oxygen's -2 state and a neutral compound).

2. Oxidation State Change Calculation:
The variation in Mn's oxidation state is:

$ +7 \text{ (initial)} - +4 \text{ (final)} = +3 $
This indicates that each Mn atom gains 3 electrons during reduction.

3. Total Charge Determination:
For 1 mole of MnO₄⁻, 3 moles of electrons are transferred. As 1 Faraday (F) represents the charge of 1 mole of electrons, the total required charge is:

$ 3 \text{ moles of electrons} \times 1 \text{ F/mole} = 3 \text{ F} $

4. Conclusion:
The charge required for the reduction of 1 mole of MnO₄⁻ to MnO₂ is $\boxed{3 \text{ F}}$.