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Q. An electric charge of $5$ Faradays is passed through three electrolytes $AgNO_{3},CuSO_{4}$ and $FeCl_{3}$ solution. The grams of each metal liberated at cathode will be
[Atomic weight; $Fe - 56 \,g/mol, Cu - 63.5\, g/mol, Ag - 108 \,g/mol]$

NTA AbhyasNTA Abhyas 2022

Solution:

$Ag^{+}+\underset{1 \text{ mol } \equiv \text{ 1F}}{e^{-}} \rightarrow Ag$
$Cu^{2 +}+\underset{2 \text{ mol } \equiv \text{ 2F}}{2 e^{-}} \rightarrow Cu$
$Fe^{3 +}+\underset{3 \text{ mol } \equiv \text{ 3F}}{3 e^{-}} \rightarrow Fe$
Grams of Ag liberated $=\frac{108}{1}\times 5=540\text{ g}$
Grams of Cu liberated $=\frac{63.5}{2}\times 5=158.75\text{ g}$
Grams of Fe liberated $=\frac{56}{3}\times 5=93.3\text{ g}$