During electrolysis of aqueous sodium chloride
A. Sodium metal is obtained at anode
B. Hydrogen is evolved at cathode
C. Oxygen is evolved at anode
D. Oxygen is evolved at cathode
E. Chlorine is evolved at anode
Choose the correct answer from the option given below.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → B, E only
In the electrolysis of an aqueous solution, water molecules compete with the ions of the salt at both electrodes. The species that is reduced (at the cathode) or oxidised (at the anode) most easily gives the product.
At the cathode (reduction), \( \ce{Na^{+}} \) and \( \ce{H2O} \) compete:
\( \displaystyle \ce{Na+(aq) + e- -> Na(s)} \quad E^\circ = -2.71\ \mathrm{V} \)
\( \displaystyle \ce{H2O(l) + e- -> 1/2H2(g) + OH-(aq)} \quad E^\circ = -0.83\ \mathrm{V} \)
Water has the higher (less negative) reduction potential, so hydrogen is evolved and sodium is not deposited.
At the anode (oxidation), \( \ce{Cl^-} \) and \( \ce{H2O} \) compete.
Oxidation of water looks easier on E° alone, but it needs a large overpotential, so in practice chlorine is evolved.
A. Sodium metal is obtained at anode — Incorrect. Sodium ions are reduced only in molten NaCl, and even then at the cathode, not the anode.
B. Hydrogen is evolved at cathode — Correct. Water is reduced in preference to \( \ce{Na^{+}} \).
C. Oxygen is evolved at anode — Incorrect. Because of the overpotential of oxygen, \( \ce{Cl^{-}} \) is oxidised instead.
D. Oxygen is evolved at cathode — Incorrect. The cathode is where reduction occurs; oxygen is a product of oxidation.
E. Chlorine is evolved at anode — Correct. Because of the overpotential of oxygen, \( \ce{Cl^-} \) is oxidised: \( \ce{Cl^- -> 1/2Cl2 + e^-} \). The NaOH formed stays in the solution.
NCERT (2026-27 edition):
Products of Electrolysis – …
