Which of the following are true for electrolysis of aqueous solution…

The correct answer is Option (2) → A, C and D only

In the electrolysis of an aqueous solution, water competes with the ions of the salt at each electrode.

• A. Hydrogen is evolved at cathode. — Correct. Water (E⁰ = −0.83 V) is reduced in preference to \( \ce{Na^{+}} \) (−2.71 V), giving H₂.

• B. Oxygen is evolved at anode. — Incorrect. Because of the overpotential for oxygen, chlorine is evolved at the anode.

• C. The solution becomes alkaline after electrolysis. — Correct. \( \ce{OH^{-}} \) ions formed at the cathode stay with \( \ce{Na^{+}} \), so the solution becomes NaOH.

• D. The reaction with higher value of E° (reduction potential) is preferred at cathode. — Correct. A higher reduction potential means a greater tendency to be reduced, which is why \( \ce{H2O} \) is reduced rather than \( \ce{Na^{+}} \).

NCERT (2026-27 edition):
Products of Electrolysis – “… For example, if we use molten NaCl, the products of electrolysis are sodium metal and Cl₂ gas. … During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl₂ and H₂. In this case besides Na⁺ and Cl^– ions we also have H⁺ and OH^– ions along with the solvent molecules, H₂O.”

“At the cathode there is competition between the following reduction reactions:

    Na⁺(aq) + e^– → Na(s)    E^o = –2.71 V
    H⁺(aq) + e^– → ½H₂(g)    E^o = 0.00 V

The reaction with higher value of E^o is preferred …

    Net reaction: NaCl(aq) + H₂O(l) → Na⁺(aq) + OH^–(aq) + ½H₂(g) + ½Cl₂(g)”