Question

Three faradays of electricity are passed through molten $Al_{2}O_{3}$, aqueous solution of $CuSO_{4}$ and molten $NaCl$taken in different electrolytic cells. The amount of Al Cu and Na deposited at the cathodes will be in the ratio of

• A. $1mole:2mole:3mole$
• B. $3mole:2mole:1mole$
• C. $1mole:1.5mole:3mole$
• D. $1.5mole:2mole:3mole$

Answer 1

Answer: (C)

$1mole:1.5mole:3mole$

Answer 2

Eq. of Al = Eq. of Cu = Eq. of Na, or $\frac{1}{3}$ mole Al$= \frac{1}{2}$mole Cu =1 mole Na

or $2:3:6$ or $1:1.5:3$mole ratio.