case-I : `I_2 + H_2O_2 to O_2 + 2I^- +2H^+`
case-II: `2H^+H_2O_2+ 2OCl^- to Cl_2 + O_2 +2H_2O`

Options

A. In case-I `H_2O_2` act as oxidising agent and in case-II `H_2O_2` act as reducing agent.
B. In both case `H_2O_2` act as reducing agent.
C. In both case `H_2O_2` act as oxidising agent.
D. In case-I `H_2O_2` act as reducing agent and in case-II `H_2O_2` act as oxidising agent.

Answer

To solve the given problem, we need to analyze the two cases provided and determine the role of hydrogen peroxide (H₂O₂) in each reaction.

### Step-by-Step Solution

**Case I: Reaction Analysis**
1. **Write the reaction**:
\[ I_2 + H_2O_2 \rightarrow O_2 + 2I^- + 2H^+ \]

2. **Identify oxidation states**:
- In \(I_2\), iodine (I) has an oxidation state of 0.
- In \(I^-\), iodine has an oxidation state of -1.
- In \(H_2O_2\), the oxidation state of oxygen is -1 (for each O).
- In \(O_2\), oxygen has an oxidation state of 0.

3. **Determine changes in oxidation states**:
- Iodine is reduced from 0 to -1 (gaining electrons).
- Oxygen in H₂O₂ is oxidized from -1 to 0 (losing electrons).

4. **Conclusion for Case I**:
- Since H₂O₂ causes iodine to be reduced (it gains electrons), H₂O₂ acts as an **oxidizing agent** in this case.

**Case II: Reaction Analysis**
1. **Write the reaction**:
\[ 2H^+ + H_2O_2 + 2OCl^- \rightarrow Cl_2 + O_2 + 2H_2O \]

2. **Identify oxidation states**:
- In \(OCl^-\), chlorine (Cl) has an oxidation state of +1.
- In \(Cl_2\), chlorine has an oxidation state of 0.
- In \(H_2O_2\), the oxidation state of oxygen is -1 (for each O).
- In \(O_2\), oxygen has an oxidation state of 0.

3. **Determine changes in oxidation states**:
- Chlorine is reduced from +1 in \(OCl^-\) to 0 in \(Cl_2\) (gaining electrons).
- Oxygen in H₂O₂ is oxidized from -1 to 0 (losing electrons).

4. **Conclusion for Case II**:
- Since H₂O₂ causes chlorine to be reduced (it gains electrons), H₂O₂ acts as an **oxidizing agent** in this case as well.

### Final Conclusion
- In both cases, H₂O₂ is acting as an **oxidizing agent** because it causes the reduction of iodine in Case I and the reduction of chlorine in Case II.

### Answer
The correct answer is that H₂O₂ is acting as an oxidizing agent in both cases.

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