Question

Electrolysis of an aqueous solution of $Na_{2}SO_{4}$ between Pt electrodes liberate a gas X at anode and gas Y at cathode. X and Y respectively are}

• A. $H_{2}, O_{2}$
• B. $O_{2}, H_{2}$
• C. $SO_{2}, H_{2}$
• D. $H_{2}, SO_{2}$

Hint:

P-A-O: Positive Anode Oxidation. In aqueous salt solutions, water often "wins" and gives $O_{2}$ at the anode.

Answer 1

The Correct Option is B

Step 1: Concept
In the electrolysis of aqueous $Na_{2}SO_{4}$, water is oxidized at the anode and reduced at the cathode because $Na^{+}$ and $SO_{4}^{2-}$ have higher discharge potentials than water.

Step 2: Meaning
Anode (Oxidation): $2H_{2}O \to O_{2} + 4H^{+} + 4e^{-}$. Cathode (Reduction): $2H_{2}O + 2e^{-} \to H_{2} + 2OH^{-}$.

Step 3: Analysis
Gas X (Anode) is Oxygen ($O_{2}$). Gas Y (Cathode) is Hydrogen ($H_{2}$).

Step 4: Conclusion
The gases are $O_{2}$ and $H_{2}$ respectively.
Final Answer: (B)