Question

One Faraday of electricity is passed through molten $Al_2O_3$ aqueous solution of $CuSO_4$ and molten $NaCl$ taken in three different electrolytic cells connected in series. The mole ratio of $Al,Cu$ and $Na$ deposited at the respective cathode is

• A. $2 : 3 : 6$
• B. $6 : 2 : 3$
• C. $6 : 3 : 2$
• D. $1 : 2 : 3$

Answer 1

Answer: (A)

$2 : 3 : 6$

Answer 2

$Al^{3+} + 3e^- {->} Al$ $Cu^{2+} + 2e^- {->} Cu$ $Na^+ + e^- {->} Na$ Hence, $1 \,F$ will deposit $1/3$ mole of $Al$, $1/2$ mol of $Cu$ and $1$ mole of $Na$ $\therefore$ molar ratio $= \frac{1}{3} : \frac{1}{2} : 1$ $= 2 : 3 : 6$