In which of the following reactions, `H_(2)O_(2)` is acting as a reducing agent?

Options

A. `SO_(2)+H_(2)O_(2) to H_(2)SO_(4)`
B. `2KI+H_(2)O_(2) to 2KOH+I_(2)`
C. `PbS+4H_(2)O_(2) to PbSO_(4)+4H_(2)O`
D. `Ag_(2)O+H_(2)O_(2) to 2Ag+H_(2)O+O_(2)`

Answer

To determine in which of the given reactions hydrogen peroxide (H₂O₂) acts as a reducing agent, we need to analyze the oxidation states of the elements involved in each reaction. A reducing agent is a substance that donates electrons to another substance, causing the other substance to be reduced while itself gets oxidized.

Let's analyze each option step by step:

### Step 1: Analyze the first reaction
**Reaction:** SO₂ + H₂O₂ → H₂SO₄

- **Oxidation states:**
- In SO₂: Sulfur (S) is +4, Oxygen (O) is -2.
- In H₂O₂: Oxygen (O) is -1.
- In H₂SO₄: Sulfur (S) is +6, Oxygen (O) is -2.

- **Change in oxidation state:**
- Sulfur goes from +4 (in SO₂) to +6 (in H₂SO₄), indicating oxidation.
- H₂O₂ is not reducing anything here; it is acting as an oxidizing agent.

### Step 2: Analyze the second reaction
**Reaction:** 2KI + H₂O₂ → 2KOH + I₂

- **Oxidation states:**
- In KI: Potassium (K) is +1, Iodine (I) is -1.
- In H₂O₂: Oxygen (O) is -1.
- In KOH: Potassium (K) is +1, Oxygen (O) is -2, Hydrogen (H) is +1.
- In I₂: Iodine (I) is 0.

- **Change in oxidation state:**
- Iodine goes from -1 (in KI) to 0 (in I₂), indicating oxidation.
- H₂O₂ is not reducing anything here; it is acting as an oxidizing agent.

### Step 3: Analyze the third reaction
**Reaction:** PbS + 4H₂O₂ → PbSO₄ + 4H₂O

- **Oxidation states:**
- In PbS: Lead (Pb) is +2, Sulfur (S) is -2.
- In H₂O₂: Oxygen (O) is -1.
- In PbSO₄: Lead (Pb) is +2, Sulfur (S) is +6, Oxygen (O) is -2.

- **Change in oxidation state:**
- Sulfur goes from -2 (in PbS) to +6 (in PbSO₄), indicating oxidation.
- H₂O₂ is not reducing anything here; it is acting as an oxidizing agent.

### Step 4: Analyze the fourth reaction
**Reaction:** Ag₂O + H₂O₂ → 2Ag + H₂O + O₂

- **Oxidation states:**
- In Ag₂O: Silver (Ag) is +1, Oxygen (O) is -2.
- In H₂O₂: Oxygen (O) is -1.
- In Ag: Silver (Ag) is 0.
- In H₂O: Hydrogen (H) is +1, Oxygen (O) is -2.
- In O₂: Oxygen (O) is 0.

- **Change in oxidation state:**
- Silver goes from +1 (in Ag₂O) to 0 (in Ag), indicating reduction.
- Oxygen goes from -1 (in H₂O₂) to 0 (in O₂), indicating oxidation.
- Here, H₂O₂ is reducing silver and itself is getting oxidized.

### Conclusion
In the fourth reaction, H₂O₂ is acting as a reducing agent because it reduces silver (Ag) from +1 to 0 while itself gets oxidized from -1 to 0.

### Final Answer
**H₂O₂ acts as a reducing agent in the reaction:** Ag₂O + H₂O₂ → 2Ag + H₂O + O₂.

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