Match List I with List II:
| List I | List II |
| A. aq. \(NaCl\) solution with \(Pt\) electrodes | I. Oxygen gas evolved at anode |
| B. Dilute solution of \(H_2SO_4\) with \(Pt\) electrodes | II. Metal loss at anode |
| C. aq. \(CuCl_2\) with \(Pt\) electrode | III. Hydrogen gas evolved at cathode |
| D. aq. \(CuCl_2\) with \(Cu\) electrodes | IV. Metal deposited at cathode |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-I, C-IV, D-II
(A) aq. NaCl solution with Pt electrodes — This matches (III) Hydrogen gas evolved at cathode. Water is reduced in preference to \( \ce{Na+} \) (chlorine is given off at the anode):
\( \ce{H2O(l) + e- -> 1/2H2(g) + OH^-(aq)} \)
(B) Dilute solution of H2SO4 with Pt electrodes — This matches (I) Oxygen gas evolved at anode. Water is oxidised in preference to the sulphate ion:
\( \ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-} \)
(C) aq. CuCl2 with Pt electrode — This matches (IV) Metal deposited at cathode. \( \ce{Cu^2+} \) (E° = +0.34 V) is reduced more easily than water; chlorine is given off at the inert anode:
Cathode: \( \ce{Cu^2+(aq) + 2e- -> Cu(s)} \) Anode: \( \ce{2Cl^-(aq) -> Cl2(g) + 2e-} \)
(D) aq. CuCl2 with Cu electrodes — This matches (II) Metal loss at anode. The copper anode itself is oxidised, and copper deposits on the cathode:
Anode: \( \ce{Cu(s) -> Cu^2+(aq) + 2e-} \) Cathode: \( \ce{Cu^2+(aq) + 2e- -> Cu(s)} \)