The charge required for the reduction of 1 mol of \( MnO_4^- \) to \( MnO_2 \) is
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- 1 F
- 5 F
- 3 F
- 6 F
The Correct Option is C
Solution and Explanation
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 $\boxed{3 \text{ F}}$.
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