Question

Two different electrolytic cells filled with molten $Cu{\left(N{O}_3\right)}_2$ and molten $Al{\left(N{O}_3\right)}_3$ respectively are connected in series. When electricity is passed $2.7g$ $Al$ is deposited on electrode. Calculate the weight of $Cu$ deposited on cathode.
$[Cu=63.5;Al=27.0g{mol}^{-1}]$

• A. $190.5g$
• B. $9.525g$
• C. $63.5g$
• D. $31.75g$

Answer 1

The correct option is B $9.525g$

$C{u}^{+2}+2e^{-}\to Cu\left(s\right)$
$A{l}^{+3}+3e^{-}\to Al\left(s\right)$

When same current is passed through two electrolytic solution containing copper nitrate $\left[Cu{\left(N{O}_3\right)}_2\right]$ and aluminium nitrate $\left[{Al\left(N{O}_3\right)}_3\right]$, connected in series, the weight of copper and silver deposited are

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

Now, equivalent weight of metal $=\frac{\text{Molecular mass}}{\text{Charge on its ions}}$

Put the given values in equation (i)
$\frac{\text{Weight of Cu deposited}}{\text{2.7g Al deposited}}=\frac{63.5/2}{27/3}=\frac{31.75}{9}$

$\therefore$ Weight of $Cu$ deposited $=\frac{31.75\times 2.7}{9}$
$=\frac{85.725}{9}$

Weight of $Cu$ deposited $=9.525g$.