Question

Three faradays of electricity are passed through molten $A{l}_2{O}_3$, aqueous solution of $CuS{O}_4$ and molten $NaCl$ taken in three different electrolytic cells. The amount of $Al,Cu$ and $Na$ deposited at the cathodes will be in the ratio of

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

Answer 1

The correct option is B $1$ mole : $1.5$ mole : $3$ mole

$A{l}^{3+}+3e^{-}\to Al$

$\frac{3}{2}C{u}^{2+}+3e^{-}\to \frac{3}{2}Cu$

$3N{a}^{+}+3e^{-}\to 3Na$

$\therefore MolesofAl:Cu:Na=1:1.5:3$

Option B is correct.