Aqueous \(NaCl\) solution is electrolyzed. The product formed at the anode is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(Cl_2\) gas
In aqueous NaCl, besides \( \ce{Na^{+}} \) and \( \ce{Cl^{-}} \), water can also be reduced or oxidised. The product at each electrode is the one formed most easily.
At the cathode, \( \ce{H2O} \) is reduced rather than \( \ce{Na^{+}} \) (E° of \( \ce{Na^{+}} \)/Na is −2.71 V):
\( \ce{H2O(l) + e^- -> 1/2H2(g) + OH^-(aq)} \)
At the anode two oxidations compete:
\( \ce{Cl^-(aq) -> 1/2Cl2(g) + e^-} \quad E^\circ_{(\text{cell})} = 1.36\,\mathrm{V} \)
\( \ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-} \quad E^\circ_{(\text{cell})} = 1.23\,\mathrm{V} \)
From E° values alone water should be oxidised, but the evolution of oxygen needs a large overpotential. So in practice chloride is oxidised and chlorine gas is liberated at the anode. The net reaction is:
\( \ce{2NaCl(aq) + 2H2O(l) -> 2NaOH(aq) + H2(g) + Cl2(g)} \)
Why the other options are incorrect
Option (1) \(H_2\) gas — Incorrect. Hydrogen is formed at the cathode, by reduction of water.
Option (2) \(O_2\) gas — Incorrect. Oxygen would be expected from the E° values, but its overpotential makes \( \ce{Cl^{-}} \) oxidise first.
Option (3) \(Na\) metal — Incorrect. Sodium is deposited only in the electrolysis of molten NaCl, and at the cathode, not the anode.
NCERT (2026-27 edition):
Products of Electrolysis – “… During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2.”
Products of Electrolysis – “… At the anode the following oxidation reactions are possible:
Cl– (aq) → ½ Cl2 (g) + e– E°(cell) = 1.36 V (2.36)
2H2O (l) → O2 (g) + 4H+(aq) + 4e– E°(cell) = 1.23 V (2.37)
The reaction at anode with lower value of E° is preferred and therefore, water should get oxidised in preference to Cl– (aq). However, on account of overpotential of oxygen, reaction (2.36) is preferred.”