The aqueous solutions of the following salts were electrolyzed for 30 min with a current of 20 amp. Arrange the following in the increasing order of the amount of metal deposited at the cathode.
\( \text{(A) }\mathrm{WCl}_6 \)
\( \text{(B) }\mathrm{ZnSO}_4 \)
\( \text{(C) }\mathrm{HfCl}_4 \)
\( \text{(D) }\mathrm{AgNO}_3 \)
(Atomic mass of: W =184 u, Zn = 65 u, Hf = 178 u and Ag = 108u)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B), (C), (D)
All four solutions receive the same current for the same time, so the same charge passes through each. By Faraday’s laws, the same charge deposits masses in the ratio of the equivalent masses, molar mass divided by the number of electrons needed per metal ion:
\( \displaystyle m = \frac{M}{n} \times \frac{Q}{F} \)
The charge of the metal ion in each salt gives n:
(A) \( \ce{WCl6} \): \( \ce{W^6+} \), n = 6; \( M/n = 184/6 = 30.7 \)
(B) \( \ce{ZnSO4} \): \( \ce{Zn^2+} \), n = 2; \( M/n = 65/2 = 32.5 \)
(C) \( \ce{HfCl4} \): \( \ce{Hf^4+} \), n = 4; \( M/n = 178/4 = 44.5 \)
(D) \( \ce{AgNO3} \): \( \ce{Ag+} \), n = 1; \( M/n = 108/1 = 108 \)
With \( Q = 20\ \mathrm{A} \times 1800\ \mathrm{s} = 36000\ \mathrm{C} \), \( Q/F = 36000/96500 = 0.373 \) mol of electrons, so the masses deposited are about 11.4 g W, 12.1 g Zn, 16.6 g Hf and 40.3 g Ag. In increasing order:
\( \displaystyle \ce{W}\,(30.7) \lt \ce{Zn}\,(32.5) \lt \ce{Hf}\,(44.5) \lt \ce{Ag}\,(108) \)
Shortcut: The charge is the same for all four, so only M/n needs comparing; the current and time do not have to be used at all.