Match List-I with List-II.List-IList-IIProcessMain products of electr…

The correct answer is Option (1) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)

The products of electrolysis depend on the state of the electrolyte and its concentration. When several species can react at an electrode, the one that needs the least energy reacts: the species with the higher reduction potential is reduced at the cathode, and the one most easily oxidised reacts at the anode (allowing for overpotential).

• (A) Electrolysis of molten NaCl — This matches (II) Sodium metal and Cl₂ gas. Only \( \ce{Na^{+}} \) and \( \ce{Cl^{-}} \) ions are present, so \( \ce{Na^{+}} \) is reduced and \( \ce{Cl^{-}} \) is oxidised.

      \( \displaystyle \text{Cathode: }\ce{Na^+ + e^- -> Na}\quad\text{Anode: }\ce{2Cl^- -> Cl2 + 2e^-} \)

• (B) Electrolysis of NaCl (aq) — This matches (III) Sodium hydroxide, Cl₂ and H₂ gas. At the cathode water is reduced instead of \( \ce{Na^{+}} \), because E°(\( \ce{H2O} \)/\( \ce{H2} \)) = −0.83 V is much higher than E°(\( \ce{Na^{+}} \)/Na) = −2.71 V; \( \ce{Cl^{-}} \) is oxidised at the anode (overpotential of oxygen), and \( \ce{Na^{+}} \) and \( \ce{OH^{-}} \) stay in solution as NaOH.

      \( \displaystyle \text{Cathode: }\ce{2H2O + 2e^- -> H2 + 2OH^-}\quad\text{Anode: }\ce{2Cl^- -> Cl2 + 2e^-} \)

      \( \displaystyle \ce{2NaCl + 2H2O -> 2NaOH + Cl2 + H2} \)

• (C) Electrolysis of dilute \( \ce{H2SO4} \) — This matches (IV) O₂ gas. In dilute solution water is oxidised at the anode (\( \ce{H2} \) forms at the cathode).

      \( \displaystyle \text{Cathode: }\ce{4H^+ + 4e^- -> 2H2}\quad\text{Anode: }\ce{2H2O -> O2 + 4H^+ + 4e^-} \)

      \( \displaystyle \text{Cell reaction: }\ce{2H2O -> 2H2 + O2} \)

• (D) Electrolysis of concentrated \( \ce{H2SO4} \) — This matches (I) S₂O₈^2- (aq). At high concentration the sulphate ions are oxidised to peroxodisulphate.

      \( \displaystyle \text{Cathode: }\ce{2H^+ + 2e^- -> H2}\quad\text{Anode: }\ce{2SO4^2- -> S2O8^2- + 2e^-} \)

Shortcut: The pair (C)–(IV) appears in Options (1) and (3), and (A)–(II) then leaves only Option (1).

NCERT (2026-27 edition):
Products of Electrolysis – “… For example, if we use molten NaCl, the products of electrolysis are sodium metal and Cl₂ gas. … During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl₂ and H₂. … During the electrolysis of sulphuric acid, the following processes are possible at the anode:

    2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^–    E^o = +1.23 V    (2.38)

    2SO₄^2–(aq) → S₂O₈^2–(aq) + 2e^–    E^o = 1.96 V    (2.39)

For dilute sulphuric acid, reaction (2.38) is preferred but at higher concentrations of H₂SO₄, reaction (2.39) is preferred.”