The charge required for the reduction of 1 mole of `MnO_(4)^(-)` to `MnO_(2)` is :

Options

A. 1F
B. 3F
C. 5F
D. 7F

Explanation

To determine the charge required for the reduction of 1 mole of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \), we can follow these steps:

### Step 1: Determine the Oxidation States
First, we need to find the oxidation states of manganese in both \( \text{MnO}_4^{-} \) and \( \text{MnO}_2 \).

- In \( \text{MnO}_4^{-} \):
- Let the oxidation state of Mn be \( x \).
- The oxidation state of oxygen is -2.
- The equation becomes:
\[ x + 4(-2) = -1 \]
\[ x - 8 = -1 \implies x = +7 \]
- Thus, the oxidation state of Mn in \( \text{MnO}_4^{-} \) is +7.

- In \( \text{MnO}_2 \):
- Again, let the oxidation state of Mn be \( y \).
- The equation becomes:
\[ y + 2(-2) = 0 \]
\[ y - 4 = 0 \implies y = +4 \]
- Thus, the oxidation state of Mn in \( \text{MnO}_2 \) is +4.

### Step 2: Calculate the Change in Oxidation State
Now, we calculate the change in oxidation state during the reduction:
\[ \text{Change in oxidation state} = +7 \text{ (from MnO}_4^{-}) - +4 \text{ (from MnO}_2) = +3 \]
This indicates that 3 electrons are required for the reduction of 1 mole of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \).

### Step 3: Determine the Charge Required
According to Faraday's laws of electrolysis, 1 mole of electrons corresponds to 1 Faraday of charge (approximately 96500 coulombs). Therefore, for 3 moles of electrons, the charge required can be calculated as:
\[ \text{Charge} = 3 \text{ moles of electrons} \times 1 \text{ Faraday} = 3 \text{ Faraday} \]
Thus, the total charge required for the reduction of 1 mole of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \) is 3 Faraday.

### Final Answer
The charge required for the reduction of 1 mole of \( \text{MnO}_4^{-} \) to \( \text{MnO}_2 \) is **3 Faraday**.

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