Total charge required to convert three moles of `Mn_(3)O_(4)` to `MnO_(4)^(-2)` in present of alkaline medium

Options

A. (a) 5F
B. (b) 10F
C. (c) 20F
D. (d) None of these

Answer

To solve the problem of calculating the total charge required to convert three moles of `Mn_(3)O_(4)` to `MnO_(4)^(-2)` in an alkaline medium, we can follow these steps:

### Step 1: Determine the oxidation states
1. **Find the oxidation state of Mn in `Mn_(3)O_(4)`**:
- Let the oxidation state of Mn be \( x \).
- The formula for the compound is \( 3x + 4(-2) = 0 \) (since the compound is neutral).
- This simplifies to \( 3x - 8 = 0 \), thus \( 3x = 8 \) and \( x = \frac{8}{3} \).

2. **Find the oxidation state of Mn in `MnO_(4)^(-2)`**:
- Let the oxidation state of Mn be \( y \).
- The formula is \( y + 4(-2) = -2 \).
- This simplifies to \( y - 8 = -2 \), thus \( y = 6 \).

### Step 2: Calculate the change in oxidation state
- The oxidation state changes from \( \frac{8}{3} \) to \( 6 \).
- The change in oxidation state for one mole of Mn is:
\[ \Delta = \left( \frac{8}{3} - 6 \right) = \left( \frac{8}{3} - \frac{18}{3} \right) = -\frac{10}{3} \]
- Since there are 3 moles of `Mn_(3)O_(4)`, the total change in oxidation state for 3 moles is:
\[ 3 \times \left( -\frac{10}{3} \right) = -10 \]

### Step 3: Calculate the number of electrons required
- Each mole of `Mn_(3)O_(4)` requires 10 electrons to convert to `MnO_(4)^(-2)`.
- Therefore, for 3 moles:
\[ 3 \times 10 = 30 \text{ moles of electrons} \]

### Step 4: Calculate the total charge
- According to Faraday's law, 1 mole of electrons corresponds to a charge of approximately \( 96500 \) coulombs (1 Faraday).
- Thus, for 30 moles of electrons, the total charge \( Q \) is:
\[ Q = 30 \times 96500 = 30F \]

### Final Answer
The total charge required to convert three moles of `Mn_(3)O_(4)` to `MnO_(4)^(-2)` in an alkaline medium is \( 30F \).

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