During electrolysis of an aqeous solution of sodium sulphate if 2.4 Lof oxygen at STP was liberated at anode. The volume of hydrogen at STP, liberated at cathode would be:

Options

A. 1.2L
B. 2.4L
C. 2.6L
D. 4.8L

Answer

To solve the problem, we need to analyze the electrolysis of an aqueous solution of sodium sulfate (Na2SO4) and determine the volume of hydrogen gas liberated at the cathode when 2.4 L of oxygen gas is liberated at the anode.

### Step-by-Step Solution:

1. **Identify the Electrolysis Reaction**:
- During electrolysis of Na2SO4, the solution dissociates into sodium ions (Na⁺) and sulfate ions (SO₄²⁻). Water (H₂O) also dissociates into hydrogen ions (H⁺) and hydroxide ions (OH⁻).
- The half-reactions at the electrodes are:
- **Anode (oxidation)**: 4OH⁻ → O₂ + 2H₂O + 4e⁻
- **Cathode (reduction)**: 2H⁺ + 2e⁻ → H₂

2. **Determine the Molar Ratio of Gases**:
- From the half-reactions, we can see that for every 1 mole of O₂ produced at the anode, 2 moles of H₂ are produced at the cathode. This gives us a molar ratio of:
- H₂ : O₂ = 2 : 1

3. **Calculate the Volume of Hydrogen**:
- We are given that 2.4 L of O₂ is liberated at the anode. Using the molar ratio, we can find the volume of H₂ produced at the cathode.
- Since the volume ratio is the same as the mole ratio, we can set up the following relationship:
- Volume of H₂ = 2 × Volume of O₂
- Volume of H₂ = 2 × 2.4 L = 4.8 L

4. **Conclusion**:
- Therefore, the volume of hydrogen gas liberated at the cathode is **4.8 L**.

### Final Answer:
The volume of hydrogen gas liberated at the cathode is **4.8 L**.

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