In aqueous alkaline solution two electron reduction of `HO_(2)^(-)` gives

Options

A. `HO^(-)`
B. `H_(2)O`
C. `O_(2)`
D. `O_(2)^(-)`

Explanation

To solve the problem of the two-electron reduction of `HO2^(-)` in an aqueous alkaline solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the species and its oxidation state**:
The species `HO2^(-)` (hydroperoxide ion) contains two oxygen atoms. In this ion, the oxidation state of each oxygen atom is -1.

2. **Determine the reduction process**:
In a reduction process, the species gains electrons. Since we are looking for a two-electron reduction, `HO2^(-)` will gain 2 electrons.

3. **Write the half-reaction for the reduction**:
The half-reaction for the reduction of `HO2^(-)` can be written as:
\[ HO2^{-} + 2e^{-} \rightarrow ? \]

4. **Balance the oxygen atoms**:
In alkaline solution, we typically balance oxygen by adding water (H2O) to the side that needs oxygen. Since we have 2 oxygen atoms in `HO2^(-)`, we will need to add water to balance the reaction. For every water added, we will also need to add hydroxide ions (OH^-) to maintain the balance of the alkaline medium.

5. **Add water and hydroxide ions**:
We can add 2 water molecules to the left side:
\[ HO2^{-} + 2e^{-} + 2H2O \rightarrow ? \]
This addition gives us 4 oxygen atoms on the left side (2 from `HO2^-` and 2 from `2H2O`).

6. **Form hydroxide ions**:
Now, we will form hydroxide ions (OH^-) from the water. Since we added 2 water molecules, we will have to add 4 hydroxide ions to the right side to balance the oxygen:
\[ HO2^{-} + 2e^{-} + 2H2O \rightarrow 4OH^{-} \]

7. **Final balanced equation**:
The final balanced equation for the reduction of `HO2^(-)` in an alkaline solution is:
\[ HO2^{-} + 2e^{-} + 2H2O \rightarrow 2OH^{-} \]

### Conclusion:
The product of the two-electron reduction of `HO2^(-)` in an aqueous alkaline solution is `OH^-`.