How many faradays are required to reduce one moleof `MnO_(4)^(-)` to `Mn^(2+)`?

Options

A. `3`
B. `5`
C. `2`
D. `1`

Explanation

To determine how many Faradays are required to reduce one mole of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \), we can follow these steps:

### Step 1: Identify the oxidation states
In \( \text{MnO}_4^{-} \), the oxidation state of manganese (Mn) is +7. In \( \text{Mn}^{2+} \), the oxidation state of manganese is +2.

### Step 2: Calculate the change in oxidation state
The change in oxidation state can be calculated as follows:
\[ \text{Change in oxidation state} = \text{Final oxidation state} - \text{Initial oxidation state} = +2 - (+7) = -5 \]
This indicates that manganese is being reduced by 5 units.

### Step 3: Relate the change in oxidation state to Faradays
According to Faraday's laws of electrolysis, the number of Faradays required for a reduction reaction is equal to the absolute value of the change in oxidation state. Therefore, the number of Faradays required is:
\[ \text{Number of Faradays} = \left| -5 \right| = 5 \]

### Conclusion
Thus, to reduce one mole of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \), 5 Faradays are required.