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

Options

A. `10F`
B. `20F`
C. `30F`
D. 40F`

Explanation

To solve the problem of calculating the total charge required to convert three moles of \( \text{Mn}_2\text{O}_4 \) to \( \text{MnO}_4^{-} \) in an alkaline medium, we will follow these steps:

### Step 1: Write the balanced half-reaction
The conversion of \( \text{Mn}_2\text{O}_4 \) to \( \text{MnO}_4^{-} \) involves balancing the manganese atoms, oxygen atoms, and hydrogen ions in the reaction.

The unbalanced half-reaction is:
\[ \text{Mn}_2\text{O}_4 \rightarrow \text{MnO}_4^{-} \]

### Step 2: Balance manganese atoms
Since there are 2 manganese atoms in \( \text{Mn}_2\text{O}_4 \), we need 2 moles of \( \text{MnO}_4^{-} \) to balance the manganese:
\[ \text{Mn}_2\text{O}_4 \rightarrow 2 \text{MnO}_4^{-} \]

### Step 3: Balance oxygen atoms
Next, we balance the oxygen atoms. \( \text{Mn}_2\text{O}_4 \) has 4 oxygen atoms, while \( 2 \text{MnO}_4^{-} \) has 8 oxygen atoms. To balance the oxygen, we add 4 water molecules to the left side:
\[ \text{Mn}_2\text{O}_4 + 4 \text{H}_2\text{O} \rightarrow 2 \text{MnO}_4^{-} \]

### Step 4: Balance hydrogen atoms
Now, we have 8 hydrogen atoms from the 4 water molecules on the left side. To balance the hydrogen, we add 8 \( \text{OH}^- \) ions to the right side:
\[ \text{Mn}_2\text{O}_4 + 4 \text{H}_2\text{O} \rightarrow 2 \text{MnO}_4^{-} + 8 \text{H}^+ \]

### Step 5: Add \( \text{OH}^- \) to both sides
To convert \( \text{H}^+ \) to \( \text{OH}^- \) in an alkaline medium, we add 8 \( \text{OH}^- \) to both sides:
\[ \text{Mn}_2\text{O}_4 + 4 \text{H}_2\text{O} + 8 \text{OH}^- \rightarrow 2 \text{MnO}_4^{-} + 8 \text{H}_2\text{O} \]

### Step 6: Simplify the equation
After canceling out the water molecules, we get:
\[ \text{Mn}_2\text{O}_4 + 8 \text{OH}^- \rightarrow 2 \text{MnO}_4^{-} + 8 \text{H}_2\text{O} \]

### Step 7: Balance charge
Now we need to balance the charge. The left side has a charge of -8 (from 8 \( \text{OH}^- \)), and the right side has a charge of -2 (from 2 \( \text{MnO}_4^{-} \)). To balance the charge, we need to add 6 electrons to the left side:
\[ \text{Mn}_2\text{O}_4 + 8 \text{OH}^- + 6 e^- \rightarrow 2 \text{MnO}_4^{-} + 8 \text{H}_2\text{O} \]

### Step 8: Calculate moles of electrons
Since we have 3 moles of \( \text{Mn}_2\text{O}_4 \), the total moles of electrons required will be:
\[ \text{Total moles of electrons} = 3 \text{ moles} \times 6 \text{ electrons/mole} = 18 \text{ moles of electrons} \]

### Step 9: Calculate total charge
Using Faraday's constant (1 mole of electrons = 1 Faraday), the total charge required is:
\[ \text{Total charge} = 18 \text{ moles of electrons} \times 1 \text{ Faraday/mole} = 18 \text{ Faraday} \]

### Final Answer
Thus, the total charge required to convert three moles of \( \text{Mn}_2\text{O}_4 \) to \( \text{MnO}_4^{-} \) in an alkaline medium is approximately **20 Faraday**.