`KMnO_(4)`gets reduced to

Options

A. `K_(2)MnO_(4)` in neutral medium
B. `MnO_(2)` in acidic medium
C. `Mn^(2+)` in alkaline medium
D. `MnO_(2)` in neutral medium

Explanation

To determine what potassium permanganate (`KMnO4`) gets reduced to, we need to consider the reduction reaction that occurs in an acidic medium. Here’s a step-by-step breakdown of the process:

### Step 1: Identify the species involved
Potassium permanganate is represented as `KMnO4`. In solution, it dissociates into `K^+` and `MnO4^-` ions. The `MnO4^-` ion is the species that undergoes reduction.

**Hint:** Remember that `KMnO4` dissociates into potassium ions and permanganate ions in solution.

### Step 2: Understand the reduction process
In an acidic medium (presence of `H^+` ions), the `MnO4^-` ion is reduced. The reduction half-reaction for permanganate in acidic solution can be written as follows:

\[ MnO4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H2O \]

This equation shows that `MnO4^-` gains 5 electrons (`5e^-`) and is reduced to `Mn^{2+}`.

**Hint:** Look for the number of electrons involved in the reduction process and the products formed.

### Step 3: Write the final reduced form
From the reduction half-reaction, we can see that the `MnO4^-` ion is reduced to `Mn^{2+}` in the presence of hydrogen ions and water. Therefore, the final reduced form of `KMnO4` in acidic medium is `Mn^{2+}`.

**Hint:** Focus on the product of the reduction reaction to identify the final reduced species.

### Conclusion
Thus, when `KMnO4` gets reduced in an acidic medium, it is converted to `Mn^{2+}`.

**Final Answer:** `KMnO4` gets reduced to `Mn^{2+}`.