Question

Assume that impure copper contains iron, gold and silver as impurities. After passing a current of $140$ ampere for $482.5$ second, the mass of anode decreased by $22.260g$ and the cathode increased in mass by $22.011g$. Estimate the percentage of iron and copper originally present.

Answer 1

The increase at the cathode is due to copper only. Hence, there is $22.011g$ of copper and rest impurities of iron, gold and silver.
Mass of impurities $=(22.26-0-22.011)=0.249g$
At anode, only copper and iron are oxidised; the gold and silver collect below anode in the form of anodic mud.
$\underset{\left(Copperandiron\right)}{M}\to M^{2+}+2e^{-}$
No. of moles of metal oxidised $=\frac{140\times 482.5}{2\times 96500}=0.35$
No. of moles of copper $=\frac{22.011}{63.5}=0.3466$
No. of moles of iron $=(0.35-0.3466)=0.0034$.