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 determining the quantity of electricity in Faraday required to reduce 0.5 moles of MnO4^- to Mn^2+ in acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reduction of permanganate ion (MnO4^-) to manganese ion (Mn^2+) in acidic medium can be represented as:
\[ \text{MnO}_4^- + \text{8H}^+ + \text{5e}^- \rightarrow \text{Mn}^{2+} + \text{4H}_2\text{O} \]
This equation shows that 1 mole of MnO4^- is reduced to 1 mole of Mn^2+ and involves the transfer of 5 electrons.

2. **Calculate the Moles of Electrons for 0.5 Moles**:
Since the reaction shows that 1 mole of MnO4^- requires 5 moles of electrons, for 0.5 moles of MnO4^-, the number of moles of electrons required can be calculated as:
\[ \text{Moles of electrons} = 0.5 \text{ moles MnO}_4^- \times 5 \text{ moles e}^- = 2.5 \text{ moles e}^- \]

3. **Convert Moles of Electrons to Faraday**:
One mole of electrons corresponds to one Faraday (approximately 96500 coulombs). Therefore, the quantity of electricity in Faraday required for 2.5 moles of electrons is:
\[ \text{Faraday required} = 2.5 \text{ moles e}^- \times 1 \text{ Faraday/mole e}^- = 2.5 \text{ Faraday} \]

4. **Conclusion**:
The quantity of electricity required to reduce 0.5 moles of MnO4^- to Mn^2+ in acidic medium is **2.5 Faraday**.

### Final Answer:
The quantity of electricity required is **2.5 Faraday**.

---