In electrolysis of dilute `H_(2)SO_(4)` what is liberated at anode?

Options

A. `H_2`
B. `SO_4^(2-)`
C. `SO_2`
D. `O_2`

Explanation

To determine what is liberated at the anode during the electrolysis of dilute sulfuric acid (H₂SO₄), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Electrolysis Process**:
- Electrolysis involves passing an electric current through a solution, leading to chemical reactions at the electrodes (anode and cathode).
- In the case of dilute H₂SO₄, the solution contains H⁺ ions (from the acid) and OH⁻ ions (from the water).

2. **Identify the Reactions at the Electrodes**:
- At the **anode** (positive electrode), oxidation occurs. This means that electrons are lost.
- At the **cathode** (negative electrode), reduction occurs, meaning that electrons are gained.

3. **Determine the Products at the Anode**:
- The possible reactions at the anode can involve the oxidation of hydroxide ions (OH⁻) or water (H₂O).
- The reaction can be represented as:
\[ 4 OH^- \rightarrow 2 H_2O + O_2 + 4 e^- \]
- This indicates that oxygen gas (O₂) is produced at the anode.

4. **Determine the Products at the Cathode**:
- At the cathode, the reaction involves the reduction of hydrogen ions (H⁺):
\[ 2 H^+ + 2 e^- \rightarrow H_2 \]
- This indicates that hydrogen gas (H₂) is produced at the cathode.

5. **Conclusion**:
- Since the question specifically asks for what is liberated at the anode, we conclude that **oxygen (O₂)** is the gas that is produced at the anode during the electrolysis of dilute H₂SO₄.

### Final Answer:
**Oxygen (O₂) is liberated at the anode during the electrolysis of dilute H₂SO₄.**