Two electrolytic cells, one containing AlCl _3 solution and the o… - Vidyadip

Question

Two electrolytic cells, one containing $AlCl _3$ solution and the other containing $ZnSO _4$ solution are connected in series. The same quantity of electricity is passed through the cells. Calculate the amount of $Zn$ deposited in $ZnSO _4$ cell if $1.2 g$ of Al are deposited in $AlCl _3$ cell. The molar masses of Al and $Zn$ are $27 g mol ^{-1}$ and $65.4 g mol ^{-1}$ respectively.

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Answer

Given: Mass of Al deposited $=1.2 g$
Molar mass of $Al =27 g mol ^{-1}$
Molar mass of $Zn =65.4 g mol ^{-1}$
Mass of zinc deposited $=\omega_{ Zn }=$ ?
Reduction reactions in electrolysis :

Number of moles of Al deposited $=\frac{1.2}{27}$
$=0.04444 \text { mol. }$
From reaction (i),
$\because 1$ mol Al requires 3 mol electrons
$\therefore 0.04444 mol Al$ requires $3 \times 0.04444$
$=0.1333$ mol electrons
Hence 0.1333 moles of electrons are passed through both the cells in the series.
From reaction (ii),
$\because 2$ moles of electrons deposit $1 mol Zn$
$\therefore 0.1333$ moles of electrons will deposit, $\frac{0.1333}{2}=0.06665 mol Zn$
Mass of $Zn$ deposited $=0.06665 \times 65.4=4.36 g$
Mass of $Zn$ deposited $=4.36 g$

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