The products of molten sodium chloride and aqueous solution of sodium chloride are not the same. In molten NaCl electrolysis, the products are only Na and Cl₂. In the electrolysis of aqueous NaCl, apart from Na⁺ and Cl^- ions, H⁺ and OH^- ions also exist because of presence of H₂O. At the cathode, there is competition between two reduction reactions –

(i) Na⁺(aq) + e^–→ Na(s) where standard electrode potential E_cell = –2.71 V

(ii) H⁺(aq) + e^– → 1/2 H₂(g) E_cell = 0.00 V

The reaction with higher potential value is preferred, hence the second equation takes place. But H⁺ is produced as a product of dissociation of H₂O, which is given by H₂O(l) → H⁺ + OH^-. The net reaction at the cathode is written as H₂O (l) + e^–→ 1/2H₂ (g) + OH^–.

At anode, two oxidation reactions are possible.

(i) Cl^– (aq) → 1/2 Cl₂ (g) + e^– E_cell = 1.36 V

(ii) 2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– E_cell = 1.23 V

Here, the issue of overpotential arises. Some electrochemical processes which are possible at an electrode, are so slow kinetically that at lower voltages these cannot take place and an extra potential (called overpotential) needs to be applied. The reaction at anode with lower value of E �_cell is preferred and accordingly, water should get oxidised in preference to Cl^–(aq). However, because of overpotential of oxygen, second reaction is preferred. Hence, the correct option is (ii).