Question

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

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

Answer 1

Answer: (D)

$0.90$

Answer 2

Amount of impurity

=decreased mass of anode

- increased mass of cathode

$=22.26-22.011$
$=0.249\, g$

Amount of pure Cu deposited,

$W =Z i t$
$=\frac{E}{96500} \times i t$
$=\frac{63.54}{2 \times 96500} \times 140 \times 482.5$
$=22.239\, g$

But increased mass of cathode $=22.011\, g$
$\therefore$ Amount of impurity (Fe)

$=22.239-22.011=0.228\, g$

Now, from Faraday's second law of electrolysis

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

(\approx amount of impurity)

$\frac{\text { Wt. of Fe deposited }}{0.228}=\frac{27.55}{31.77}$
$\therefore$ Wt. of $F e=\frac{27.75 \times 0.228}{31.77}=0.199$
$\therefore \%$ of iron in impure copper

$=\frac{0.199}{22.26} \times 100$
$=0.89 \approx 0.90$