`HOCl+H_2O_2rarrH_3O^++Cl^(-)+O_2`
`I_2+H_2O_2+2OH^(-)rarr2I^(-)+2H_2O+O_2`
Choose the correct option

Options

A. `H_2O_2` acts as rfucing and oxidising agent respectively in equation (A) and (B)
B. `H_2O_2` acts as oxidising agent in equation (A) and (B)
C. `H_2O_2` acts as reducing agent in equation (A) and (B)
D. `H_2O_2` act as oxidizing and reducing agent rspctively in equation (A) and (B)

Answer

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

### Step 1: Analyze the first reaction
The first reaction is:
\[ \text{HOCl} + \text{H}_2\text{O}_2 \rightarrow \text{H}_3\text{O}^+ + \text{Cl}^- + \text{O}_2 \]

- In this reaction, we need to determine the oxidation states of the elements involved.
- The oxidation state of Cl in HOCl is +1, and in Cl⁻ it is -1.
- The oxidation state of O in H₂O₂ is -1, and in O₂ it is 0.

From this, we can see that:
- Cl is reduced from +1 to -1 (gaining electrons).
- O in H₂O₂ is oxidized from -1 to 0 (losing electrons).

Thus, H₂O₂ acts as a reducing agent in this reaction.

### Step 2: Analyze the second reaction
The second reaction is:
\[ \text{I}_2 + \text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow 2\text{I}^- + 2\text{H}_2\text{O} + \text{O}_2 \]

- Again, we determine the oxidation states.
- The oxidation state of I in I₂ is 0, and in I⁻ it is -1.
- The oxidation state of O in H₂O₂ is -1, and in O₂ it is 0.

From this, we can see that:
- I is reduced from 0 to -1 (gaining electrons).
- O in H₂O₂ is oxidized from -1 to 0 (losing electrons).

Thus, H₂O₂ also acts as a reducing agent in this reaction.

### Step 3: Conclusion
In both reactions, H₂O₂ is acting as a reducing agent because it causes the oxidation of another species (Cl in the first reaction and I in the second reaction) while being oxidized itself.

### Final Answer
The correct option is that H₂O₂ is a reducing agent in both cases.

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