Question

Two electrolytic cells are connected in series containing $CuS{O}_4$ solution and molten $AlC{l}_3$. If in electrolysis $0.4$ moles of ${}^{\prime }C{u}^{\prime }$ are deposited on cathode of first cell. The number of moles of ${}^{\prime }A{l}^{\prime }$ 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 $Cu=0.4\times 63.5$

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

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

Now, number of moles of $Al$ deposited

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