The quantity of electricity requried to reduce 0.05 mol of `MnO_(4)^(-)` to `Mn^(2+)` in acidic medium would be

Options

A. 0.01 F
B. 0.05 F
C. 0.15 F
D. 0.25 F

Answer

To solve the problem of calculating the quantity of electricity required to reduce 0.05 mol of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \) in acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Change in Oxidation State**:
- The oxidation state of manganese in \( \text{MnO}_4^{-} \) is +7.
- The oxidation state of manganese in \( \text{Mn}^{2+} \) is +2.
- Therefore, the change in oxidation state is from +7 to +2, which means that manganese is reduced by gaining 5 electrons.

2. **Determine the Moles of Electrons Required**:
- For the reduction of 1 mole of \( \text{MnO}_4^{-} \), 5 moles of electrons are required.
- Thus, for 0.05 moles of \( \text{MnO}_4^{-} \), the moles of electrons required can be calculated using the ratio:
\[ \text{Moles of electrons} = 0.05 \, \text{mol} \, \text{MnO}_4^{-} \times \frac{5 \, \text{mol electrons}}{1 \, \text{mol} \, \text{MnO}_4^{-}} = 0.25 \, \text{mol electrons} \]

3. **Convert Moles of Electrons to Faraday**:
- Since 1 mole of electrons corresponds to 1 Faraday, the quantity of electricity required for 0.25 moles of electrons is:
\[ \text{Quantity of electricity} = 0.25 \, \text{mol electrons} = 0.25 \, \text{Faraday} \]

4. **Final Answer**:
- Therefore, the quantity of electricity required to reduce 0.05 mol of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \) in acidic medium is **0.25 Faraday**.