In acidic solution, `KMnO_(4)`is reduced to

Options

A. `Mn^(2+)`
B. `MnO_(2)`
C. `Mn_(2)O_(3).
D. `MnO_(4)^(2-)`

Answer

To determine the product of the reduction of potassium permanganate (KMnO₄) in an acidic solution, we can follow these steps:

### Step 1: Identify the oxidation state of manganese in KMnO₄
In KMnO₄, the oxidation state of manganese (Mn) can be calculated as follows:
- Potassium (K) has an oxidation state of +1.
- Each oxygen (O) has an oxidation state of -2, and there are four oxygen atoms, contributing a total of -8.
- Let the oxidation state of Mn be x. The overall charge of the compound is neutral (0).

The equation can be set up as:
\[ x + 1 + (-8) = 0 \]
\[ x - 7 = 0 \]
\[ x = +7 \]

### Step 2: Determine the reduction process in acidic medium
In acidic conditions, KMnO₄ acts as a strong oxidizing agent. The Mn in KMnO₄ (with an oxidation state of +7) is reduced to a lower oxidation state.

### Step 3: Identify the final oxidation state of manganese in acidic medium
In acidic medium, the manganese is typically reduced to Mn²⁺, where the oxidation state of Mn is +2.

### Step 4: Write the half-reaction for the reduction
The half-reaction for the reduction of KMnO₄ in acidic medium can be written as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

### Conclusion
Thus, in acidic solution, KMnO₄ is reduced to Mn²⁺.

### Final Answer
The correct answer is **Mn²⁺**.

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