Given below are the two reactions of `H_(2)O_(2)`. Mark the correct statement which follows.
(i) `2KMnO_(4) + 3H_(2)SO_(4) + 5H_(2)O_(2) to K_(2)SO_(4) + 2MnSO_(4) + 8H_(2)O + 5O_(2)`
(ii) `2Cr(OH)_(3) + 4NaOH + 3H_(2)O_(2) to 2Na_(2)CrO_(4) + 8H_(2)O`

Options

A. (i) Show oxidising nature of `H_(2)O_(2)` and (ii) shows reducing nature of `H_(2)O_(2)`.
B. In (i) `H_(2)O_(2)` acts as a reducing agent and in (ii) it acts as an oxidising agent.
C. In both (i) and (ii), `H_(2)O_(2)` acts as an oxidising agent.
D. In both (i) and (ii), `H_(2)O_(2)` acts as a reducing agent.

Explanation

To solve the question regarding the reactions of hydrogen peroxide (H₂O₂), we need to analyze both reactions provided and determine the role of H₂O₂ in each case.

### Step-by-Step Solution:

1. **Analyze Reaction (i):**
- The reaction is:
\[ 2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2 \]
- Determine the oxidation state of manganese (Mn) in KMnO₄:
- Let the oxidation state of Mn be \( x \).
- The equation for the oxidation state is:
\[ x + 1 + 4(-2) = 0 \implies x - 8 + 1 = 0 \implies x = +7 \]
- In MnSO₄, the oxidation state of Mn is:
- \( +2 \) (since sulfur is \( +6 \) and oxygen is \( -2 \)).
- Thus, Mn is reduced from \( +7 \) to \( +2 \).

2. **Determine the role of H₂O₂ in Reaction (i):**
- H₂O₂ is being oxidized to O₂, indicating that it is losing electrons.
- Since H₂O₂ is causing Mn to be reduced, it is acting as a **reducing agent**.

3. **Analyze Reaction (ii):**
- The reaction is:
\[ 2Cr(OH)_3 + 4NaOH + 3H_2O_2 \rightarrow 2Na_2CrO_4 + 8H_2O \]
- Determine the oxidation state of chromium (Cr) in Cr(OH)₃:
- The oxidation state of Cr is \( +3 \) (since OH⁻ is \( -1 \) and there are three OH⁻ ions).
- In Na₂CrO₄, the oxidation state of Cr is:
- Let it be \( y \):
\[ 2 + y + 4(-2) = 0 \implies y - 8 + 2 = 0 \implies y = +6 \]
- Thus, Cr is oxidized from \( +3 \) to \( +6 \).

4. **Determine the role of H₂O₂ in Reaction (ii):**
- H₂O₂ is being reduced to H₂O, indicating that it is gaining electrons.
- Since H₂O₂ is causing Cr to be oxidized, it is acting as an **oxidizing agent**.

5. **Conclusion:**
- In Reaction (i), H₂O₂ acts as a reducing agent.
- In Reaction (ii), H₂O₂ acts as an oxidizing agent.

### Final Answer:
- The correct statement is that H₂O₂ is acting as a reducing agent in Reaction (i) and as an oxidizing agent in Reaction (ii).