Question

Impure copper containing $Fe,Au,Ag$ as impurities is electrolytically refined. A current of $140A$ for $482.5s$ decreased the mass of the anode by $22.26g$ and increased the mass of cathode by $22.011g$. Percentage of iron in impure copper is
(Given molar mass $Fe=55.5g$ $mo{l}^{-1}$, molar mass $Cu=63.54gmo{l}^{-1}$)

• A. $0.85$
• B. $0.80$
• C. $0.95$
• D. $0.90$

Answer 1

Answer: (D)

Amount of impurity

=decreased mass of anode

- increased mass of cathode

$=22.26-22.011$

$=0.249g$

Amount of pure Cu deposited,

$W=Zit$

$=\frac{E}{96500}\times it$

$=\frac{63.54}{2\times 96500}\times 140\times 482.5$

$=22.239g$

But increased mass of cathode $=22.011g$

$\therefore$ Amount of impurity (Fe)

$=22.239-22.011=0.228g$

Now, from Faraday's second law of electrolysis

$\frac{\text{ Wt. of Fe deposited }}{\text{ Wt. of Cu deposited }}=\frac{\text{ Eq. wt. of }Fe}{\text{ Eq. wt. of }Cu}$

(\approx amount of impurity)

$\frac{\text{ Wt. of Fe deposited }}{0.228}=\frac{27.55}{31.77}$

$\therefore$ Wt. of $Fe=\frac{27.75\times 0.228}{31.77}=0.199$

$\therefore \%$ of iron in impure copper

$=\frac{0.199}{22.26}\times 100$

$=0.89\approx 0.90$