Question

Two electrolytic cells are connected in series containing $CuSO_4$ solution and molten $AlCl_3$. If in electrolysis $0.4$ moles of $'Cu'$ are deposited on cathode of first cell. The number of moles of $'Al'$ deposited on cathode of the second cell is

• A. $0.6$ moles
• B. $0.27$ moles
• C. $0.18$ moles
• D. $0.4$ moles

Answer 1

Answer: (B)

Key Idea First find weight of copper deposited, by using the formula number of moles $=\frac{\text { weight }}{\text { molecular weight }}$ and then calculate weight of $Al$ deposited and number of moles of $Al$ by using second law of Faraday's.

Given, Number of moles of $Cu$ deposited $=0.4$ moles

According to Faraday's second law,

$\frac{\text {weight of Cu deposited}}{\text {weight of Al deposited}}=\frac{\text {Eq wt. of Cu}}{\text { Eq wt. of Al }}$...(i)

$\because$ No. of moles $=\frac{\text {weight}}{\text {molecular weight}}$

$\therefore $ Weight of $C u=0.4 \times 63.5$

Now, from Eq. (i),

$=\frac{0.4 \times 63.5}{\text {weight of Al deposited}}=\frac{\frac{63.5}{2}}{\frac{27}{3}}$

$\therefore $ Weight of $Al$ deposited

$=\frac{0.4 \times 63.5 \times 9}{31.75}=7.2 g$

Now, number of moles of $Al$ deposited

$=\frac{7.2}{27}=0.27$ moles.