The aqueous solution of \( ZnSO_4, AgNO_3, FeCl_3 \) and \( WCl_6 \) are electrolyzed for 2 hours using a current of 5 amperes. The decreasing order of mass deposited of the following metals at cathode will be.........
(A) Zn
(B) Ag
(C) Fe
(D) W
(atomic mass: \( Fe =56\, g\, mol^{-1}, Zn =65\, g\, mol^{-1}, Ag = 108\, g\, mol^{-1}, W = 184\, g\, mol^{-1} \))
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (A), (D), (C)
By Faraday’s laws, the mass of a metal deposited by a charge Q is:
\( \displaystyle m = \frac{M \times Q}{n \times F} \)
where M is the molar mass and n the number of electrons needed to deposit one atom (the charge on the metal ion). The same current flows for the same time through all four solutions, so Q is the same and the mass is proportional to M/n.
\( \displaystyle Q = I \times t = 5\ \mathrm{A} \times (2 \times 3600\ \mathrm{s}) = 36000\ \mathrm{C} \)
\( \displaystyle \text{moles of electrons} = \frac{36000}{96500} = 0.373\ \mathrm{mol} \)
The charge on each ion follows from its salt (Cl and \( \ce{NO3} \) are −1, \( \ce{SO4} \) is −2):
(A) \( \ce{ZnSO4} \): \( \ce{Zn^2+ + 2e^- -> Zn} \), n = 2. \( M/n = 65/2 = 32.5 \); mass = 0.373 × 32.5 = 12.1 g.
(B) \( \ce{AgNO3} \): \( \ce{Ag^+ + e^- -> Ag} \), n = 1. \( M/n = 108/1 = 108 \); mass = 0.373 × 108 = 40.3 g.
(C) \( \ce{FeCl3} \): \( \ce{Fe^3+ + 3e^- -> Fe} \), n = 3. \( M/n = 56/3 = 18.7 \); mass = 0.373 × 18.7 = 7.0 g.
(D) \( \ce{WCl6} \): \( \ce{W^6+ + 6e^- -> W} \), n = 6. \( M/n = 184/6 = 30.7 \); mass = 0.373 × 30.7 = 11.4 g.
In decreasing order of mass deposited:
\( \displaystyle \text{Ag}\ (40.3\ \mathrm{g}) \gt \text{Zn}\ (12.1\ \mathrm{g}) \gt \text{W}\ (11.4\ \mathrm{g}) \gt \text{Fe}\ (7.0\ \mathrm{g}) \)
Shortcut: Only the ratio M/n matters: 108, 32.5, 30.7, 18.7. Silver has by far the largest value, so only Option (3), which starts with (B), can be right.