Which of the following are the products of electrolysis of aqueous so…

The correct answer is Option (3) → Cathode - \(H_2\) gas, anode - \(Cl_2\) gas

In the electrolysis of an aqueous solution, water can also be reduced or oxidised, so each electrode takes the reaction that is easier. At the cathode, \( \ce{H+} \) (E° = 0.00 V) is reduced in preference to \( \ce{Na+} \) (E° = −2.71 V); the \( \ce{H+} \) comes from water, so the net cathode reaction is:

    \( \displaystyle \text{Cathode: }\ce{H2O(l) + e^- -> 1/2H2(g) + OH^-(aq)} \)

At the anode, oxidation of water (E° = 1.23 V) should be preferred over \( \ce{Cl-} \) (1.36 V), but because of the overpotential of oxygen, chloride is oxidised:

    \( \displaystyle \text{Anode: }\ce{Cl^-(aq) -> 1/2Cl2(g) + e^-} \)

Adding the two (for two electrons) gives the cell reaction; NaOH remains in solution:

    \( \displaystyle \ce{2NaCl(aq) + 2H2O(l)} \xrightarrow{\text{electrolysis}} \ce{2NaOH(aq) + H2(g) + Cl2(g)} \)

So hydrogen is liberated at the cathode and chlorine at the anode.

Why the other options are incorrect

• Option (1) Cathode-Sodium metal, anode - \(Cl_2\) gas — Incorrect. Sodium is obtained only from molten NaCl; in water \( \ce{H^{+}} \) is reduced in preference to \( \ce{Na^{+}} \).

• Option (2) Cathode-Sodium metal, anode - \(H_2\) gas — Incorrect. \( \ce{H2} \) is a reduction product and cannot form at the anode, and no sodium is deposited.

• Option (4) Cathode - \(Cl_2\) gas, anode - Sodium metal — Incorrect. The electrodes are reversed; \( \ce{Cl^{-}} \) is oxidised at the anode, and reduction takes place at the cathode.

NCERT (2026-27 edition):
Products of Electrolysis – “… 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_cell = – 2.71 V; H⁺ (aq) + e^– → ½ H₂ (g) E^o_cell = 0.00 V. The reaction with higher value of E^o is preferred and therefore, the reaction at the cathode during electrolysis is: H⁺ (aq) + e^– → ½ H₂ (g)”