Which one of the following statements is correct for electrolysis of brine solution?

Solution

Chemistry ma’am · 3 steps

In brine, water is reduced at the cathode, so you get H₂ gas and OH⁻ ions there.

1. At the cathode. Both Na⁺ and water compete for reduction. Na⁺ has a very negative reduction potential (-2.71 V), so water is reduced instead, giving H₂ gas and leaving OH⁻ ions in solution.

   2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)

2. At the anode. Water should be easier to oxidise on paper, but oxygen evolution has a large overpotential. So in concentrated brine, Cl⁻ is oxidised: 2Cl⁻ → Cl₂(g) + 2e⁻.
3. So. H₂ and OH⁻ form at the cathode and Cl₂ forms at the anode. The OH⁻ joins with Na⁺ to give NaOH, so option (d) is correct.

The idea to keep. In electrolysis of brine, the cathode always gives H₂ and OH⁻ (water is reduced, not Na⁺), and the anode gives Cl₂ because of the overpotential of oxygen.

Answer from the NTA final answer key. Solution by PrepWiser.