Match List-I (element deposited from its ions: Cu2+, Ag+, Zn2+, Al3+)…

The correct answer is Option (2) → A-r, B-p, C-s, D-q

To deposit one mole of a metal from its ion Mⁿ⁺, n moles of electrons (n F) are needed. So the mass deposited by 1 C is the molar mass divided by nF:

    \( \displaystyle m=\frac{M}{n\times96500}\,\mathrm{g} \)

(A) Copper — This matches (r) 0.000329. \( \ce{Cu^2+ + 2e^- -> Cu} \), so n = 2:

    \( \displaystyle m=\frac{63.5}{2\times96500}=3.29\times10^{-4}\,\mathrm{g} \)

(B) Silver — This matches (p) 0.00112. \( \ce{Ag^+ + e^- -> Ag} \), so n = 1:

    \( \displaystyle m=\frac{108}{1\times96500}=1.12\times10^{-3}\,\mathrm{g} \)

(C) Zinc — This matches (s) 0.000339. \( \ce{Zn^2+ + 2e^- -> Zn} \), so n = 2:

    \( \displaystyle m=\frac{65.4}{2\times96500}=3.39\times10^{-4}\,\mathrm{g} \)

(D) Aluminium — This matches (q) 0.000093. \( \ce{Al^3+ + 3e^- -> Al} \), so n = 3:

    \( \displaystyle m=\frac{27}{3\times96500}=9.3\times10^{-5}\,\mathrm{g} \)

NCERT (2026-27 edition):
Faraday’s Laws of Electrolysis – “(ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).”

“… It is obvious that one mole of Mg²⁺ and Al³⁺ require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively.”