`H_(2)O_(2)` reduces `MnO_(4)^(-)` ion to

Options

A. `Mn^(+)`
B. `Mn^(2+)`
C. `Mn^(3+)`
D. `Mn^(-)`

Explanation

To solve the problem of determining what the `MnO4^(-)` ion is reduced to when it reacts with `H2O2`, we can follow these steps:

### Step 1: Write the unbalanced chemical equation
The first step is to write the unbalanced equation for the reaction between `MnO4^(-)` and `H2O2`. The initial reaction can be represented as:
\[ \text{MnO}_4^{-} + \text{H}_2\text{O}_2 \rightarrow \text{Mn}^{2+} + \text{H}_2\text{O} + \text{O}_2 \]

### Step 2: Identify oxidation states
Next, we need to identify the oxidation states of manganese in `MnO4^(-)` and in the product `Mn^{2+}`.
- In `MnO4^(-)`, manganese has an oxidation state of +7.
- In `Mn^{2+}`, manganese has an oxidation state of +2.

### Step 3: Determine the change in oxidation state
Now, we can determine the change in oxidation state for manganese:
- The change from +7 to +2 indicates that manganese is being reduced. The reduction involves a decrease in oxidation state by 5.

### Step 4: Balance the reaction
To balance the reaction, we need to ensure that both mass and charge are balanced. The balanced equation for the reaction can be written as:
\[ 2 \text{MnO}_4^{-} + 5 \text{H}_2\text{O}_2 + 6 \text{H}^{+} \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + 5 \text{O}_2 \]

### Step 5: Conclusion
From the balanced equation, we can see that `MnO4^(-)` is reduced to `Mn^{2+}` in the presence of `H2O2`. Therefore, the final answer is:
\[ \text{MnO}_4^{-} \text{ is reduced to } \text{Mn}^{2+} \]