In which of the following reactions `H_(2)O_(2)` acts as a reducing agent?
1. `H_(2)O_(2)+2H^(+)+2e^(-)to2H_(2)O`
2. `H_(2)O_(2)-2e^(-)toO_(2)+2H^(+)`
3. `H_(2)O_(2)+2e^(-)to2OH^(-)`
4. `H_(2)O_(2)+2OH^(-)-2e^(-)toO_(2)+2H_(2)O`

Options

A. 1,2
B. 3,4
C. 1,3
D. 2,4

Answer

To determine in which reactions \( H_2O_2 \) (hydrogen peroxide) acts as a reducing agent, we need to analyze the four given reactions based on the changes in oxidation states of the elements involved.

### Step-by-Step Solution

1. **Understanding Reducing Agents**: A reducing agent is a substance that donates electrons in a chemical reaction, leading to the reduction of another species. In the process, the reducing agent itself gets oxidized.

2. **Analyzing Each Reaction**:
- **Reaction 1**:
\[ H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O \]
- Here, \( H_2O_2 \) is converting to \( H_2O \). The oxidation state of oxygen in \( H_2O_2 \) is -1, and in \( H_2O \), it is -2. Since the oxidation state decreases, \( H_2O_2 \) is gaining electrons (reduction) and acts as an oxidizing agent.

- **Reaction 2**:
\[ H_2O_2 - 2e^- \rightarrow O_2 + 2H^+ \]
- In this reaction, \( H_2O_2 \) is converting to \( O_2 \). The oxidation state of oxygen in \( H_2O_2 \) is -1, and in \( O_2 \), it is 0. Since the oxidation state increases, \( H_2O_2 \) is losing electrons (oxidation) and acts as a reducing agent.

- **Reaction 3**:
\[ H_2O_2 + 2e^- \rightarrow 2OH^- \]
- In this case, \( H_2O_2 \) is converting to \( OH^- \). The oxidation state of oxygen in \( H_2O_2 \) is -1, and in \( OH^- \), it remains -2. Since the oxidation state is decreasing, \( H_2O_2 \) is gaining electrons (reduction) and acts as an oxidizing agent.

- **Reaction 4**:
\[ H_2O_2 + 2OH^- - 2e^- \rightarrow O_2 + 2H_2O \]
- Here, \( H_2O_2 \) is again converting to \( O_2 \). The oxidation state of oxygen in \( H_2O_2 \) is -1, and in \( O_2 \), it is 0. Since the oxidation state increases, \( H_2O_2 \) is losing electrons (oxidation) and acts as a reducing agent.

3. **Conclusion**:
- From the analysis, \( H_2O_2 \) acts as a reducing agent in **Reaction 2** and **Reaction 4**.

### Final Answer
Thus, the correct answer is **D: 2 and 4**.