`MnO_(4)^(2-)` on reduction in acidic medium froms

Options

A. `MnO_(4)`
B. `Mn^(+ +)`
C. `MnO_(4)^(2-)`
D. `Mn`

Answer

To solve the question regarding the reduction of \( \text{MnO}_4^{2-} \) in acidic medium, we will follow these steps:

### Step 1: Identify the species involved
We start with the manganate ion \( \text{MnO}_4^{2-} \). In acidic medium, we need to consider how this ion behaves during reduction.

### Step 2: Write the half-reaction for the reduction
In acidic medium, the manganate ion \( \text{MnO}_4^{2-} \) can be reduced to manganese(II) ion \( \text{Mn}^{2+} \). The half-reaction can be written as follows:

\[ \text{MnO}_4^{2-} + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

### Step 3: Balance the half-reaction
In this half-reaction, we need to ensure that both mass and charge are balanced. The left side has a charge of \( 2- + 8+ = 6+ \) and the right side has a charge of \( 2+ \). The addition of 5 electrons on the left side balances the charge.

### Step 4: Conclusion
Thus, when \( \text{MnO}_4^{2-} \) is reduced in acidic medium, it forms \( \text{Mn}^{2+} \).

### Final Answer
The final product of the reduction of \( \text{MnO}_4^{2-} \) in acidic medium is \( \text{Mn}^{2+} \).

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