Hydrogen peroxide is a powerful oxidising agent, both in the acidic and alkaline medium.
In acidic medium: `H_(2)O_(2)+2H^(o+)+2e^(ɵ)to2H_(2)O`
In alkaline medium: `H_(2)O_(2)+2e^(ɵ)to2overset(Theta)(O)H`
Hydrogen peroxide acts as a reducing agent towards powerful oxidising agents.
In acidic medium: `H_(2)O_(2)to2H^(o+)+O_(2)+2e^(ɵ)` In alkaline medium, however, its reducing nature is more effective.
`H_(2)O_(2)to2H^(o+)+O_(2)+2e^(ɵ)`
In which of the following reactions, `H_2O_2` act as a reducing agent?

Options

A. `2KI+H_(2)O_(2)to2KOH+I_(2)`
B. `KNO_(2)+H_(2)O_(2)toKNO_(3)+H_(2)O`
C. `Na_(2)SO_(3)+H_(2)O_(2)toNa_(2)SO_(4)+H_(2)O`
D. `PbO_(2)+H_(2)O_(2)toPbO+H_(2)O+O_(2)`

Explanation

To determine in which of the given reactions hydrogen peroxide (H₂O₂) acts as a reducing agent, we need to analyze each reaction and identify the changes in oxidation states of the involved species. A reducing agent is a substance that donates electrons and gets oxidized in the process.

### Step-by-Step Solution:

1. **Identify the Reactions**: We have four reactions to analyze:
- Reaction 1: 2KI + H₂O₂ → 2KOH + I₂
- Reaction 2: KNO₂ + H₂O₂ → KNO₃ + H₂O
- Reaction 3: Na₂SO₃ + H₂O₂ → Na₂SO₄ + H₂O
- Reaction 4: PbO₂ + H₂O₂ → PbO + H₂O + O₂

2. **Analyze Reaction 1**:
- Oxidation state of I in KI is -1.
- Oxidation state of I in I₂ is 0.
- Since iodine is oxidized (from -1 to 0), KI is acting as a reducing agent, and H₂O₂ is acting as an oxidizing agent.

3. **Analyze Reaction 2**:
- Oxidation state of N in KNO₂ is +3.
- Oxidation state of N in KNO₃ is +5.
- Here, KNO₂ is oxidized (from +3 to +5), meaning KNO₂ is the reducing agent, and H₂O₂ is the oxidizing agent.

4. **Analyze Reaction 3**:
- Oxidation state of S in Na₂SO₃ is +4.
- Oxidation state of S in Na₂SO₄ is +6.
- Sulfur is oxidized (from +4 to +6), indicating Na₂SO₃ is the reducing agent, and H₂O₂ is the oxidizing agent.

5. **Analyze Reaction 4**:
- Oxidation state of Pb in PbO₂ is +4.
- Oxidation state of Pb in PbO is +2.
- Lead is reduced (from +4 to +2), indicating that PbO₂ is the oxidizing agent. Since H₂O₂ is also involved, we need to check its role. In this case, H₂O₂ is being oxidized to O₂, meaning it acts as a reducing agent.

6. **Conclusion**:
- In the first three reactions, H₂O₂ acts as an oxidizing agent.
- In the fourth reaction, H₂O₂ acts as a reducing agent.

Therefore, the answer is that **H₂O₂ acts as a reducing agent in Reaction 4**.