The equation
`3MnO_4^(2-)+4H^(+)to2MnO_4^(-)+MnO_2+2H_2O` represents

Options

A. reduction
B. disproportionation
C. oxidation in acidic medium
D. reduction in acidic medium.

Explanation

To analyze the given chemical equation and determine its classification, we will follow these steps:

### Step 1: Write down the equation
The equation given is:
\[ 3 \text{MnO}_4^{2-} + 4 \text{H}^+ \rightarrow 2 \text{MnO}_4^{-} + \text{MnO}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Assign oxidation states
We need to assign oxidation states to manganese (Mn) in each compound:
- In \(\text{MnO}_4^{2-}\):
- Let the oxidation state of Mn be \(x\).
- The oxidation state of O is \(-2\).
- The equation becomes:
\[ x + 4(-2) = -2 \]
\[ x - 8 = -2 \]
\[ x = +6 \]

- In \(\text{MnO}_4^{-}\):
- Let the oxidation state of Mn be \(y\).
- The equation becomes:
\[ y + 4(-2) = -1 \]
\[ y - 8 = -1 \]
\[ y = +7 \]

- In \(\text{MnO}_2\):
- Let the oxidation state of Mn be \(z\).
- The equation becomes:
\[ z + 2(-2) = 0 \]
\[ z - 4 = 0 \]
\[ z = +4 \]

### Step 3: Identify changes in oxidation states
- Mn in \(\text{MnO}_4^{2-}\) changes from +6 to +7 in \(\text{MnO}_4^{-}\) (oxidation).
- Mn in \(\text{MnO}_4^{2-}\) changes from +6 to +4 in \(\text{MnO}_2\) (reduction).

### Step 4: Determine the type of reaction
Since the same species (Mn) is undergoing both oxidation (increase in oxidation state) and reduction (decrease in oxidation state), this reaction is classified as a **disproportionation reaction**.

### Conclusion
The reaction represented by the equation:
\[ 3 \text{MnO}_4^{2-} + 4 \text{H}^+ \rightarrow 2 \text{MnO}_4^{-} + \text{MnO}_2 + 2 \text{H}_2\text{O} \]
is a **disproportionation reaction**.

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