Statement I: During the electrolysis of aqueous sodium chloride solution, chlorine (and not oxygen) is liberated at the anode.
Statement II: The liberation of oxygen at the anode in the electrolysis of aqueous sodium chloride needs an overpotential.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
In aqueous NaCl two oxidations compete at the anode: oxidation of \( \ce{Cl^-} \) (E° = 1.36 V) and oxidation of water to \( \ce{O2} \) (E° = 1.23 V). On E° values alone, water should be oxidised first.
Statement I: During the electrolysis of aqueous sodium chloride solution, chlorine (and not oxygen) is liberated at the anode. — True. The products are NaOH, \( \ce{H2} \) at the cathode and \( \ce{Cl2} \) at the anode:
\( \displaystyle \ce{Cl^-(aq) -> 1/2Cl2(g) + e^-} \)
Statement II: The liberation of oxygen at the anode in the electrolysis of aqueous sodium chloride needs an overpotential. — True. The oxidation of water to \( \ce{O2} \) is feasible but kinetically slow, so an extra potential, the overpotential, must be applied before it takes place.
Why Statement II explains Statement I?
Because of the overpotential of oxygen, the oxidation of water needs more than its E° suggests, and the oxidation of \( \ce{Cl^-} \) takes place first.
That is why chlorine, and not oxygen, is liberated at the anode.
NCERT (2026-27 edition):
2.5.1 Products of Electrolysis – “… During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2.”
“… Moreover, some of the electrochemical processes although feasible, are so slow kinetically that at lower voltages these do not seem to take place and extra potential (called overpotential) has to be applied, which makes such process more difficult to occur.”
“… At the anode the following oxidation reactions are possible:
\( \displaystyle \ce{Cl^-(aq) -> 1/2Cl2(g) + e^-}\qquad E^{\circ}_{(cell)}=1.36\,\mathrm{V} \qquad (2.36) \)
\( \displaystyle \ce{2H2O(l) -> O2(g) + 4H^+(aq) + 4e^-}\qquad E^{\circ}_{(cell)}=1.23\,\mathrm{V} \qquad (2.37) \)
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.”