Match List-I with List-II.
| List-I (Substance electrolysed) | List-II (Product at the cathode; product at the anode) |
|---|---|
| (A) Molten sodium chloride | (I) H2; O2 |
| (B) Aqueous sodium chloride, with platinum electrodes | (II) Na; Cl2 |
| (C) Dilute sulphuric acid, with platinum electrodes | (III) H2; S2O82− ion |
| (D) Sulphuric acid at a high concentration, with platinum electrodes | (IV) H2; Cl2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-IV, C-I, D-III
The products of electrolysis depend on the species present and their electrode potentials. At the cathode the reduction with the higher E° is preferred; at the anode the oxidation with the lower E° is preferred, unless an overpotential makes it slower.
(A) Molten sodium chloride — This matches (II) Na; Cl2. The melt has only one cation and one anion, so Na+ is reduced at the cathode and Cl− is oxidised at the anode:
\( \ce{Na+ + e^- -> Na} \)
\( \ce{Cl^- -> 1/2Cl2 + e^-} \)
(B) Aqueous sodium chloride — This matches (IV) H2; Cl2. At the anode water should be oxidised first, but because of the overpotential of oxygen, Cl2 is formed. At the cathode water is reduced in preference to Na+ (E° 0.00 V against −2.71 V), giving H2, and NaOH stays in the solution:
\( \ce{H2O(l) + e^- -> 1/2H2(g) + OH^-(aq)} \)
(C) Dilute sulphuric acid — This matches (I) H2; O2. H+ ions give H2 at the cathode, and in dilute acid water is oxidised at the anode:
\( \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e^-} \) (E° = +1.23 V)
(D) Sulphuric acid at a high concentration — This matches (III) H2; S2O82− ion. At higher concentrations of the acid the sulphate ion is oxidised at the anode instead:
\( \ce{2SO4^2-(aq) -> S2O8^2-(aq) + 2e^-} \) (E° = 1.96 V)
NCERT (2026-27 edition):
2.5.1 Products of Electrolysis – “For example, if we use molten NaCl, the products of electrolysis are sodium metal and Cl2 gas. Here we have only one cation (Na+) which is reduced at the cathode (Na+ + e– → Na) and one anion (Cl–) which is oxidised at the anode (Cl– → ½Cl2 + e–). During the electrolysis of aqueous sodium chloride solution, the products are NaOH, Cl2 and H2.”
“…At the anode the following oxidation reactions are possible:
Cl–(aq) → ½Cl2(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.”
“…During the electrolysis of sulphuric acid, the following processes are possible at the anode:
2H2O(l) → O2(g) + 4H+(aq) + 4e– E°(cell) = +1.23 V (2.38)
2SO42–(aq) → S2O82–(aq) + 2e– E°(cell) = 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.”