The charge required for the oxidation of one mole `Mn_3O_4` into `MnO_4^2-` in presence of alkaline medium is

Options

A. `5xx96500 C`
B. `96500 C`
C. `10xx96500 C`
D. `2xx96500 C`

Answer

To determine the charge required for the oxidation of one mole of `Mn3O4` into `MnO4^2-` in an alkaline medium, we can follow these steps:

### Step 1: Determine the oxidation states of manganese in both compounds.

1. **For `Mn3O4`:**
- Let the oxidation state of manganese be \( x \).
- The formula can be expressed as:
\[ 3x + 4(-2) = 0 \]
- Solving for \( x \):
\[ 3x - 8 = 0 \implies 3x = 8 \implies x = \frac{8}{3} \]

2. **For `MnO4^2-`:**
- Let the oxidation state of manganese be \( y \).
- The formula can be expressed as:
\[ y + 4(-2) = -2 \]
- Solving for \( y \):
\[ y - 8 = -2 \implies y = 6 \]

### Step 2: Calculate the change in oxidation state.

- The change in oxidation state for one manganese atom is:
\[ \Delta = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3} \]

### Step 3: Calculate the total change in charge for three manganese atoms.

- Since there are 3 manganese atoms in `Mn3O4`, the total change in charge is:
\[ \text{Total change} = 3 \times \frac{10}{3} = 10 \]

### Step 4: Relate the change in charge to Faraday's constant.

- According to Faraday's law, the charge required for a change of 1 mole of electrons is 1 Faraday (F), which is approximately 96500 Coulombs.
- Therefore, for a total change of 10, the charge required is:
\[ \text{Charge} = 10 \times F = 10 \times 96500 \, \text{C} \]

### Final Result:

The charge required for the oxidation of one mole of `Mn3O4` into `MnO4^2-` in the presence of an alkaline medium is:
\[ 10 \times 96500 \, \text{C} = 965000 \, \text{C} \]