The reaction `MnO_(4)^(-)+e^(-)hArrMnO_(4)^(2)` takes place in

Options

A. a basic medium
B. acidic medium
C. neutral medium
D. both acidic and basic medium

Answer

To determine the medium in which the reaction \( \text{MnO}_4^{-} + e^{-} \rightarrow \text{MnO}_4^{2-} \) takes place, we can analyze the behavior of the permanganate ion (\( \text{MnO}_4^{-} \)) in different pH conditions.

### Step-by-Step Solution:

1. **Identify the species involved**: The reaction involves the permanganate ion (\( \text{MnO}_4^{-} \)) and an electron (\( e^{-} \)).

2. **Understand the behavior of \( \text{MnO}_4^{-} \)**:
- In **acidic medium**, \( \text{MnO}_4^{-} \) is reduced to \( \text{Mn}^{2+} \).
- In **neutral medium**, \( \text{MnO}_4^{-} \) is reduced to \( \text{MnO}_2 \).
- In **basic medium**, \( \text{MnO}_4^{-} \) can be reduced to \( \text{MnO}_4^{2-} \).

3. **Determine the correct medium**: Since the question specifically asks about the reduction of \( \text{MnO}_4^{-} \) to \( \text{MnO}_4^{2-} \), we focus on the basic medium where this reduction occurs.

4. **Conclusion**: The reaction \( \text{MnO}_4^{-} + e^{-} \rightarrow \text{MnO}_4^{2-} \) takes place in a **basic medium**.

### Final Answer:
The reaction \( \text{MnO}_4^{-} + e^{-} \rightarrow \text{MnO}_4^{2-} \) takes place in **basic medium**.