Two electrolytic cells one containing $A l C l_{3}$ solution and the other containing $Z n S O_{4}$ solution are connected in series. The same quantity of electricity is passed between the cells. Calculate the amount $Z n$ deposited in $Z n S O_{4}$ cell if 1.2g of $A l$ are deposited in $A l C l_{3}$ cell. The molar masses of $A l$ and $Z n$ are 27g $m o l^{- 1}$ and 65.4g $m o l^{- 1}$ respectively.

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Solution

i) According to Faraday's second law,
$\frac{W e i g h t o f A l}{W e i g h t o f Z n} = \frac{E q u i v a l e n t w e i g h t o f A l}{E q u i v a l e n t w e i g h t o f Z n}$
ii) Equivalent weight of $A l = \frac{A t o m i c m a s s o f A l}{V a l e n c y} = \frac{27}{3} = 9$
Equivalent weight of $Z n = \frac{A t o m i c m a s s o f Z n}{V a l e n c y} = \frac{65.4}{2} = 32.7$
iii) $∴$ $\frac{1.2 g}{x} = \frac{9}{32.7}$
$∴$ $x = \frac{1.2 \times 32.7}{9} = 4.36 g$

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i) According to Faraday's second law, WeightofAlWeightofZn=EquivalentweightofAlEquivalentweightofZnii) Equivalent weight of Al=AtomicmassofAlValency=273=9 Equivalent weight of Zn=AtomicmassofZnValency=65.42=32.7iii) ∴ 1.2gx=932.7 ∴ x=1.2×32.79=4.36g