On electrolysis of a solution of dilute `H_(2)SO_(4)` between platinum electrodes , the gas evolved at the anoe is

Options

A. `SO_(2)`
B. `SO_(3)`
C. `O_(2)`
D. `H_(2)`

Explanation

To determine the gas evolved at the anode during the electrolysis of a dilute solution of sulfuric acid (H₂SO₄) between platinum electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolysis Process**:
- Electrolysis involves the breakdown of a compound or solution into its components using electrical energy. In this case, we are dealing with a dilute solution of H₂SO₄.

2. **Understand the Electrolytic Cell**:
- In an electrolytic cell, there are two electrodes: the anode (positive electrode) and the cathode (negative electrode). Oxidation occurs at the anode, and reduction occurs at the cathode.

3. **Determine the Reactions at the Electrodes**:
- At the anode, oxidation reactions occur. In dilute H₂SO₄, water (H₂O) is present, and we need to consider the oxidation of water versus the oxidation of sulfate ions (SO₄²⁻) or hydrogen ions (H⁺).

4. **Consult the Electrochemical Series**:
- The oxidation potential of water is higher than that of sulfate ions and hydrogen ions. This implies that water will be oxidized at the anode rather than the sulfate ions or hydrogen ions.

5. **Write the Half-Reaction for Water Oxidation**:
- The oxidation of water can be represented by the half-reaction:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
- This reaction shows that oxygen gas (O₂) is produced at the anode.

6. **Conclusion**:
- Therefore, during the electrolysis of dilute H₂SO₄, the gas evolved at the anode is oxygen (O₂).

### Final Answer:
The gas evolved at the anode during the electrolysis of a dilute solution of H₂SO₄ is **Oxygen (O₂)**.

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