STATEMENT-1 `H_(2)O_(2) to H_(2)O + (1)/(2)O_(2)` . This is an example of disproportionation reaction.
STATEMENT-2 `H_(2)O_(2)` can act as a oxidising as well as reducing agent .

Answer

To analyze the statements regarding the reaction of hydrogen peroxide (H₂O₂), we will break down the process step by step.

### Step 1: Understanding Disproportionation Reaction
A disproportionation reaction is a specific type of redox reaction in which a single substance is both oxidized and reduced, resulting in two different products.

### Step 2: Writing the Reaction
The reaction given is:
\[ \text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \frac{1}{2} \text{O}_2 \]

### Step 3: Assigning Oxidation States
Next, we need to determine the oxidation states of the elements involved in the reaction.

1. In H₂O₂ (hydrogen peroxide):
- The oxidation state of hydrogen (H) is +1.
- Let the oxidation state of oxygen (O) be \( x \).
- The equation for H₂O₂ becomes:
\[ 2(+1) + 2x = 0 \]
\[ 2 + 2x = 0 \]
\[ 2x = -2 \]
\[ x = -1 \]
- Therefore, the oxidation state of oxygen in H₂O₂ is -1.

2. In H₂O (water):
- The oxidation state of oxygen is -2.
- The equation for H₂O becomes:
\[ 2(+1) + x = 0 \]
\[ 2 + x = 0 \]
\[ x = -2 \]

3. In O₂ (elemental oxygen):
- The oxidation state of oxygen is 0.

### Step 4: Identifying Oxidation and Reduction
Now, we can identify the changes in oxidation states:
- Oxygen in H₂O₂ changes from -1 (in H₂O₂) to -2 (in H₂O) - this is a reduction.
- Oxygen in H₂O₂ changes from -1 (in H₂O₂) to 0 (in O₂) - this is an oxidation.

### Step 5: Conclusion on Disproportionation
Since H₂O₂ is both oxidized and reduced in this reaction, it is indeed a disproportionation reaction. Therefore, **Statement 1 is correct**.

### Step 6: Analyzing Statement 2
Statement 2 claims that H₂O₂ can act as both an oxidizing agent and a reducing agent.

- As an oxidizing agent, H₂O₂ can oxidize other substances while itself being reduced (as seen in the formation of H₂O).
- As a reducing agent, H₂O₂ can reduce other substances while itself being oxidized (as seen in the formation of O₂).

### Step 7: Conclusion on Statement 2
Since H₂O₂ can act as both an oxidizing agent and a reducing agent, **Statement 2 is also correct**.

### Final Conclusion
Both statements are true:
- **Statement 1**: H₂O₂ to H₂O + (1/2)O₂ is an example of a disproportionation reaction.
- **Statement 2**: H₂O₂ can act as an oxidizing as well as a reducing agent.