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

To solve the problem, we need to determine the charge required for the reduction of 1 mole of MnO₄⁻ to MnO₂.

1. Understanding the Reduction Process:
The reduction of MnO₄⁻ to MnO₂ involves a change in the oxidation state of manganese (Mn). We start by identifying the oxidation states:

• In MnO₄⁻, the oxidation state of Mn is +7 (since each oxygen is -2, and the total charge is -1).
• In MnO₂, the oxidation state of Mn is +4 (since each oxygen is -2, and the compound is neutral).

2. Calculating the Change in Oxidation State:
The change in the oxidation state of Mn is:

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

3. Determining the Total Charge:
For 1 mole of MnO₄⁻, the number of electrons transferred is 3 moles. Since 1 Faraday (F) corresponds to the charge of 1 mole of electrons, the total charge required is:

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

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