The following reactions show the `H_(2)O_(2)` behaviour in A and B reactions as
(1) `PbS_((s))+4H_(2)O_(2(aq))rarrPbSO_(4(s))+4H_(2)O_((l))`
(2) `HOCl+H_(2)O_(2)rarr H_(3)O^(+)+Cl^(-)+O_(2)`

Options

A. oxidising in acidic medium and reducing in basic medium
B. reducing in acidic medium and oxidising in basic medium
C. oxidising in acidic medium and reducing in acidic medium
D. reducing in acidic medium and oxidising in acidic medium

Answer

To analyze the behavior of hydrogen peroxide (H₂O₂) in the given reactions, we need to determine its role in each reaction by examining the oxidation states of the elements involved.

### Step 1: Analyze Reaction A
The first reaction is:
\[ \text{PbS}_{(s)} + 4 \text{H}_2\text{O}_2_{(aq)} \rightarrow \text{PbSO}_4_{(s)} + 4 \text{H}_2\text{O}_{(l)} \]

1. **Identify Oxidation States:**
- For PbS: Lead (Pb) is in the +2 oxidation state and Sulfur (S) is in the -2 oxidation state.
- For PbSO₄: Lead (Pb) is in the +2 oxidation state and Sulfur (S) is in the +6 oxidation state.
- For H₂O₂: The oxidation state of oxygen in H₂O₂ is -1.

2. **Determine Changes in Oxidation States:**
- Sulfur changes from -2 (in PbS) to +6 (in PbSO₄), indicating that sulfur is oxidized.
- Oxygen changes from -1 (in H₂O₂) to -2 (in H₂O), indicating that oxygen is reduced.

3. **Conclusion for Reaction A:**
- H₂O₂ acts as an **oxidizing agent** because it oxidizes sulfur from -2 to +6 while itself being reduced from -1 to -2.

### Step 2: Analyze Reaction B
The second reaction is:
\[ \text{HOCl} + \text{H}_2\text{O}_2 \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^- + \text{O}_2 \]

1. **Identify Oxidation States:**
- For HOCl: Chlorine (Cl) is in the +1 oxidation state.
- For Cl⁻: Chlorine (Cl) is in the -1 oxidation state.
- For O₂: The oxidation state of oxygen is 0.

2. **Determine Changes in Oxidation States:**
- Chlorine changes from +1 (in HOCl) to -1 (in Cl⁻), indicating that chlorine is reduced.
- Oxygen changes from -1 (in H₂O₂) to 0 (in O₂), indicating that oxygen is oxidized.

3. **Conclusion for Reaction B:**
- H₂O₂ acts as a **reducing agent** because it reduces chlorine from +1 to -1 while itself being oxidized from -1 to 0.

### Final Conclusion
- In reaction A, H₂O₂ acts as an **oxidizing agent** in acidic medium.
- In reaction B, H₂O₂ acts as a **reducing agent** in acidic medium.