In acidic medium is converted to Mn2+. The quantity of electricity in faraday required to reduce `0.5` mole `MnO_(4)^(-)` to `Mn^(2+)` would be :

Options

A. `2.5`
B. 5
C. 1
D. 0.5

Explanation

To solve the problem of calculating the quantity of electricity in Faraday required to reduce 0.5 moles of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \) in acidic medium, we can follow these steps:

### Step 1: Determine the change in oxidation state of manganese
In \( \text{MnO}_4^{-} \), the oxidation state of manganese (Mn) can be calculated as follows:
- Let the oxidation state of Mn be \( x \).
- The equation for the oxidation state is:
\[ x + 4(-2) = -1 \]
Simplifying this gives:
\[ x - 8 = -1 \implies x = +7 \]
- In \( \text{Mn}^{2+} \), the oxidation state of Mn is \( +2 \).

### Step 2: Calculate the change in oxidation state
The change in oxidation state from \( +7 \) to \( +2 \) is:
\[ \text{Change} = +7 - (+2) = +5 \]

### Step 3: Determine the number of electrons involved
Since the change in oxidation state is \( +5 \) for one manganese atom, it means that each \( \text{MnO}_4^{-} \) ion requires 5 electrons to be reduced to \( \text{Mn}^{2+} \).

### Step 4: Calculate the total number of electrons for 0.5 moles of \( \text{MnO}_4^{-} \)
For 0.5 moles of \( \text{MnO}_4^{-} \):
\[ \text{Total electrons} = 0.5 \, \text{moles} \times 5 \, \text{electrons/mole} = 2.5 \, \text{electrons} \]

### Step 5: Convert the number of moles of electrons to Faraday
1 Faraday is defined as the charge of 1 mole of electrons, which is approximately \( 96500 \, \text{C} \). Therefore, the quantity of electricity in Faraday required to reduce 0.5 moles of \( \text{MnO}_4^{-} \) is:
\[ \text{Quantity of electricity} = 2.5 \, \text{Faraday} \]

### Final Answer
The quantity of electricity in Faraday required to reduce 0.5 moles of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \) is **2.5 Faraday**.

---