Question

During the electrolysis of brine, by using inert electrodes,

• A. O₂ liberates at anode
• B. H₂ liberates at anode
• C. Na deposits on cathode
• D. Cl₂ liberates at anode

Answer 1

The Correct Option is D

During the electrolysis of brine (aqueous NaCl solution) using inert electrodes, the ions present are:

• At Cathode (Reduction): Na⁺, H₂O molecules
• At Anode (Oxidation): Cl⁻, H₂O molecules

At Cathode (negative electrode):

Reduction potentials:

• Na⁺(aq) + e⁻ → Na(s) (Highly negative potential)
• 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) (Lower negative potential)

H₂ gas is liberated at cathode since reduction of water is more favorable than Na⁺ reduction.

At Anode (positive electrode):

Oxidation potentials:

• 2Cl⁻(aq) → Cl₂(g) + 2e⁻ (Favorable oxidation)
• 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ (Less favorable than chloride oxidation in concentrated solutions)

Cl₂ gas is liberated at anode, as oxidation of chloride ions is preferred due to the higher concentration and easier oxidation potential.

Overall reaction for brine electrolysis:

2NaCl(aq) + 2H₂O(l) → Cl₂(g) + H₂(g) + 2NaOH(aq)

Conclusion:

The correct option is: (D) Cl₂ liberates at anode

Correct Answer: Option (D)