The cell reaction occurring at anode in the electrolysis of aqueous \(NaCl\) solution is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(Cl^-(aq) \longrightarrow \frac{1}{2}Cl_2(g) + e^-,\ E^\circ_{cell} = 1.36\ V\)
At the anode oxidation takes place. In aqueous NaCl both \( \ce{Cl^-} \) and water can be oxidised. By E° values alone water would be oxidised more easily, but oxygen has a large overpotential, so in practice chlorine is liberated at the anode.
Why the other options are incorrect
Option (1) \(H^+(aq) + e^- \longrightarrow \frac{1}{2}H_2(g),\ E^\circ_{cell} = 0.00\ V\) — Incorrect. This is a reduction; it takes place at the cathode, not at the anode.
Option (3) \(Na^+(aq) + e^- \longrightarrow Na(s),\ E^\circ_{cell} = -2.71\ V\) — Incorrect. A reduction, and in water it does not happen at all, because \( \ce{H2O} \) is reduced far more easily than \( \ce{Na^{+}} \) (Na is obtained only from molten NaCl).
Option (4) \(2H_2O(l) \longrightarrow O_2(g) + 4H^+(aq) + 4e^-,\ E^\circ_{cell} = -1.23\ V\) — Incorrect. This oxidation is possible on paper, but the overpotential of oxygen makes \( \ce{Cl^{-}} \) oxidation the reaction that actually occurs.
NCERT (2026-27 edition):
Products of Electrolysis – …
