Which of the following equation depicts reducing nature of `H_(2)O_(2)`?

Options

A. `2[Fe(CN)_(6)]^(4-)+ 2H^(+) + H_(2)O_(2) to 2[Fe(CN)_(6)]^(3-)+2H_(2)O`
B. `I_(2)+H_(2)O_(2)+ 2OH^(-) to 2I^(-) + 2H_(2)O +O_(2)`
C. `Mn^(2+) +H_(2)O_(2) to Mn^(4+) + 2OH^(-)`
D. `PbS + 4H_(2)O_(2) to PbSO_(4) + 4H_(2)O`

Explanation

To determine which equation depicts the reducing nature of \( H_2O_2 \), we need to analyze the oxidation states of the elements involved in each equation. The reducing agent is the one that donates electrons and gets oxidized itself, while the other species gets reduced.

### Step-by-Step Solution:

1. **Understand Reducing Nature**:
- A substance is considered a reducing agent if it reduces another substance and, in the process, gets oxidized itself. This means that the reducing agent loses electrons.

2. **Analyze Each Option**:
- We will analyze the oxidation states of the reactants and products in each equation to see if \( H_2O_2 \) is acting as a reducing agent.

3. **Option 1**:
- **Equation**: \( Fe^{2+} \rightarrow Fe^{3+} \)
- **Oxidation State Change**: Iron goes from +2 to +3, indicating it is getting oxidized. Since \( H_2O_2 \) is not reducing iron, this option is incorrect.

4. **Option 2**:
- **Equation**: \( I_2 \rightarrow 2I^{-} \)
- **Oxidation State Change**: Iodine goes from 0 to -1, indicating it is getting reduced. Meanwhile, \( H_2O_2 \) is oxidized from -1 to 0. This shows that \( H_2O_2 \) is reducing iodine, making this a correct option.

5. **Option 3**:
- **Equation**: \( Mn^{2+} \rightarrow Mn^{4+} \)
- **Oxidation State Change**: Manganese goes from +2 to +4, indicating it is getting oxidized. Thus, \( H_2O_2 \) is not reducing manganese, making this option incorrect.

6. **Option 4**:
- **Equation**: \( PbS \rightarrow PbSO_4 \)
- **Oxidation State Change**: Lead remains at +2 in both states. There is no change in oxidation state, indicating no reducing or oxidizing behavior. Thus, this option is also incorrect.

7. **Conclusion**:
- The only equation that shows the reducing nature of \( H_2O_2 \) is **Option 2**, where it reduces iodine while itself gets oxidized.

### Final Answer:
The equation that depicts the reducing nature of \( H_2O_2 \) is **Option 2**.