In the reaction
`2K_(3)[Fe(CN)_(6)] + 2KOH + H_(2)O_(2)rarr`
`2K_(4)[Fe(CN)_(6)] + 2H_(2)O + O_(2)`
`H_(2)O_(2)` acts as

Options

A. Reducing agent
B. Oxidising agent
C. Knocking agent
D. Bleaching agent

Answer

To determine the role of \( H_2O_2 \) in the given reaction, we can analyze the changes in oxidation states of the elements involved.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is:
\[ 2K_3[Fe(CN)_6] + 2KOH + H_2O_2 \rightarrow 2K_4[Fe(CN)_6] + 2H_2O + O_2 \]

2. **Determine the Oxidation States**:
- For \( K_3[Fe(CN)_6] \):
- Potassium (\( K \)) has an oxidation state of +1.
- The complex ion \( [Fe(CN)_6]^{3-} \) has an overall charge of -3 (since \( 3 \times +1 = +3 \)).
- The oxidation state of \( Fe \) in \( [Fe(CN)_6]^{3-} \) can be calculated as follows:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- Therefore, the oxidation state of \( Fe \) is +3.

- For \( K_4[Fe(CN)_6] \):
- Potassium (\( K \)) has an oxidation state of +1.
- The complex ion \( [Fe(CN)_6]^{4-} \) has an overall charge of -4 (since \( 4 \times +1 = +4 \)).
- The oxidation state of \( Fe \) in \( [Fe(CN)_6]^{4-} \):
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
- Therefore, the oxidation state of \( Fe \) is +2.

3. **Analyze the Change in Oxidation States**:
- The oxidation state of \( Fe \) changes from +3 to +2, indicating a reduction (gain of electrons).
- The oxidation state of oxygen in \( H_2O_2 \) is -1. In the products, \( O_2 \) has an oxidation state of 0, indicating that \( H_2O_2 \) is oxidized (loss of electrons).

4. **Identify the Role of \( H_2O_2 \)**:
- Since \( H_2O_2 \) is oxidized (its oxidation state increases from -1 to 0), it acts as a reducing agent because it donates electrons to reduce \( Fe^{3+} \) to \( Fe^{2+} \).

5. **Conclusion**:
- Therefore, in the given reaction, \( H_2O_2 \) acts as a **reducing agent**.

### Final Answer:
\( H_2O_2 \) acts as a reducing agent in the reaction.

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