Total charge required for the oxidation of two moles `Mn_3O_4 ` into `MnO_4^(2-)` in presence of alkaline medium is

Options

A. 5 F
B. 10 F
C. 20 F
D. None of these

Answer

To solve the problem of calculating the total charge required for the oxidation of two moles of \( \text{Mn}_3\text{O}_4 \) into \( \text{MnO}_4^{2-} \) in an alkaline medium, we can follow these steps:

### Step 1: Determine the oxidation states of manganese in \( \text{Mn}_3\text{O}_4 \) and \( \text{MnO}_4^{2-} \)

1. For \( \text{Mn}_3\text{O}_4 \):
- Let the oxidation state of manganese be \( x \).
- The equation for the oxidation states is:
\[ 3x + 4(-2) = 0 \]
- Simplifying gives:
\[ 3x - 8 = 0 \implies 3x = 8 \implies x = \frac{8}{3} \]
- Thus, the oxidation state of manganese in \( \text{Mn}_3\text{O}_4 \) is \( +\frac{8}{3} \).

2. For \( \text{MnO}_4^{2-} \):
- The oxidation state of manganese is \( y \).
- The equation for the oxidation states is:
\[ y + 4(-2) = -2 \]
- Simplifying gives:
\[ y - 8 = -2 \implies y = 6 \]
- Thus, the oxidation state of manganese in \( \text{MnO}_4^{2-} \) is \( +6 \).

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

- The change in oxidation state for one manganese atom is:
\[ \Delta = \left( +6 - \frac{8}{3} \right) = \left( \frac{18}{3} - \frac{8}{3} \right) = \frac{10}{3} \]
- Since there are 3 manganese atoms in \( \text{Mn}_3\text{O}_4 \):
\[ \text{Total change} = 3 \times \frac{10}{3} = 10 \]

### Step 3: Determine the number of electrons transferred

- The total number of electrons lost during the oxidation of \( \text{Mn}_3\text{O}_4 \) to \( \text{MnO}_4^{2-} \) is 10 electrons.

### Step 4: Calculate the total charge required

- The charge (in coulombs) can be calculated using Faraday's law, where 1 mole of electrons corresponds to 96500 coulombs (1 Faraday).
- The total charge required for the oxidation of 2 moles of \( \text{Mn}_3\text{O}_4 \) is:
\[ \text{Charge} = n \times F = 10 \text{ electrons} \times 2 \text{ moles} \times 96500 \text{ C/mol} = 20 \times 96500 \text{ C} = 1930000 \text{ C} \]
- In terms of Faraday, this is:
\[ \text{Charge} = 20 \text{ Faraday} \]

### Final Answer

The total charge required for the oxidation of two moles of \( \text{Mn}_3\text{O}_4 \) into \( \text{MnO}_4^{2-} \) in an alkaline medium is **20 Faraday**.

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