A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively

Options

A. `H_2,O_2`
B. `O_2, H_2`
C. `O_2, Na`
D. `SO_2`

Explanation

To solve the problem of determining the products at the cathode and anode during the electrolysis of a sodium sulfate solution using inert electrodes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electrolyte**: The solution is sodium sulfate (Na2SO4) dissolved in water. This dissociates into sodium ions (Na⁺) and sulfate ions (SO₄²⁻) in solution.

2. **Understand the Electrolysis Process**: Electrolysis involves the movement of ions towards the electrodes. The cathode is negatively charged, attracting cations (positive ions), while the anode is positively charged, attracting anions (negative ions).

3. **Identify Ions Present**: In addition to Na⁺ and SO₄²⁻, water (H₂O) also dissociates into H⁺ (hydrogen ions) and OH⁻ (hydroxide ions). Therefore, the ions present in the solution are Na⁺, SO₄²⁻, H⁺, and OH⁻.

4. **Determine the Reaction at the Anode**: At the anode, negatively charged ions will migrate. The possible anions are OH⁻ and SO₄²⁻. Since OH⁻ is less reactive than SO₄²⁻, OH⁻ will be oxidized at the anode to produce oxygen gas (O₂) and water (H₂O):
\[ 4OH⁻ \rightarrow 2H₂O + O₂ + 4e⁻ \]

5. **Determine the Reaction at the Cathode**: At the cathode, positively charged ions will migrate. The possible cations are Na⁺ and H⁺. Since H⁺ is less reactive than Na⁺, H⁺ will be reduced at the cathode to produce hydrogen gas (H₂):
\[ 2H⁺ + 2e⁻ \rightarrow H₂ \]

6. **Summarize the Products**: The products formed during the electrolysis are:
- At the anode: Oxygen gas (O₂)
- At the cathode: Hydrogen gas (H₂)

7. **Final Answer**: Therefore, the products at the cathode and anode are respectively:
- **Anode**: O₂ (Oxygen)
- **Cathode**: H₂ (Hydrogen)