When dilute `H_(2)SO_(4)` is electrolyzed between `Pt` electrodes, the gas liberated at the anode will be`………………………….`

Explanation

To determine the gas liberated at the anode during the electrolysis of dilute \( H_2SO_4 \) between platinum electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolyte**:
- The electrolyte in this case is dilute sulfuric acid (\( H_2SO_4 \)). This solution contains both sulfuric acid and water.

2. **Understand the Electrolysis Process**:
- During electrolysis, oxidation occurs at the anode and reduction occurs at the cathode. We need to focus on the oxidation reaction at the anode.

3. **Identify Possible Oxidation Reactions**:
- In dilute \( H_2SO_4 \), the two main species that can undergo oxidation are water (\( H_2O \)) and sulfate ions (\( SO_4^{2-} \)).

4. **Determine Oxidation Potentials**:
- The standard oxidation potential for water is approximately \( +0.83 \, V \).
- The standard oxidation potential for sulfate ions is approximately \( -0.20 \, V \).

5. **Compare Oxidation Potentials**:
- Since the oxidation potential of water (\( +0.83 \, V \)) is greater than that of sulfate ions (\( -0.20 \, V \)), water is more likely to be oxidized.

6. **Write the Oxidation Reaction**:
- The oxidation of water can be represented by the following half-reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
- Here, water loses electrons to produce oxygen gas (\( O_2 \)) and hydrogen ions (\( H^+ \)).

7. **Conclusion**:
- Therefore, the gas liberated at the anode during the electrolysis of dilute \( H_2SO_4 \) is oxygen (\( O_2 \)).

### Final Answer:
The gas liberated at the anode will be **oxygen (\( O_2 \))**.

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