Arrange the following in increasing order of the mass of the metal deposited when 1 amp of current is passed through their electrolytic solutions for 1 minute.
(A). Al (atomic mass = \( 27\, g\, mol^{-1} \))
(B). Mg (atomic mass = \( 24\, g\, mol^{-1} \))
(C). Ag (atomic mass = \( 108\, g\, mol^{-1} \))
(D). Cu (atomic mass = \( 63.5\, g\, mol^{-1} \))
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B), (D), (C)
By Faraday’s laws, the mass of metal deposited depends on the charge passed and on the metal’s molar mass M and the charge n of its ion. Depositing one mole of \( \ce{M^{n+}} \) needs n mol of electrons:
\( \displaystyle \ce{M^{n+} + n e- -> M} \)
\( \displaystyle m = \frac{M}{n} \times \frac{Q}{F} \)
Here the same current flows for the same time through all four solutions, so the charge is the same for all:
\( \displaystyle Q = I \times t = 1\ \mathrm{A} \times 60\ \mathrm{s} = 60\ \mathrm{C} \)
So the mass deposited is proportional to M/n, the mass deposited per mole of electrons:
(A) Al: \( \ce{Al^3+ + 3e- -> Al} \): \( \frac{27}{3} = 9\ \mathrm{g} \) per mole of electrons.
(B) Mg: \( \ce{Mg^2+ + 2e- -> Mg} \): \( \frac{24}{2} = 12\ \mathrm{g} \)
(C) Ag: \( \ce{Ag+ + e- -> Ag} \): \( \frac{108}{1} = 108\ \mathrm{g} \)
(D) Cu: \( \ce{Cu^2+ + 2e- -> Cu} \): \( \frac{63.5}{2} = 31.75\ \mathrm{g} \)
Shortcut: Silver has the largest atomic mass and needs only one electron per atom, so it ends the order: Options (2) and (3). Al (9) is below Mg (12), so the order starts with (A), which leaves Option (2).