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Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of \( 140 \mathrm{~A} \)
for \( 482.5 \mathrm{~s} \) decreased the mass of the anode by \( 22.26 \mathrm{~g} \) and increased the mass of cathode by
\( 22.011 \mathrm{~g} \). Percentage of iron in impure copper is
(Given molar mass \( \mathrm{Fe}=55.5 \mathrm{~g} \mathrm{~mol}^{-1} \), molar mass \( \mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1} \) )
Options:
for \( 482.5 \mathrm{~s} \) decreased the mass of the anode by \( 22.26 \mathrm{~g} \) and increased the mass of cathode by
\( 22.011 \mathrm{~g} \). Percentage of iron in impure copper is
(Given molar mass \( \mathrm{Fe}=55.5 \mathrm{~g} \mathrm{~mol}^{-1} \), molar mass \( \mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1} \) )
Solution:
1804 Upvotes
Verified Answer
The correct answer is:
None of the above
No. of gram equivalents of copper deposited $=\frac{22.011}{31.77}$
$=0.6298$
No. of gram equivalents from the current $(Q)$
$=I \times t=140 \times 482.5=67550 C$
No. of gram equivalents $=\frac{67550}{96500}=0.7$
Since only Cu and Fe are dissolved from the anode no. of gram equivalents of Fe
$=0.7-0.6928$
$=0.0072$
Therefore, mass of Fe $=0.0072 \times 27.75$
$=0.1998 g$
$\%$ of Fe $=\frac{\text { Mass of Fe }}{\text { Mass of impurities }} \times 100$
$=\frac{0.1998}{22.26}=0.89$ or $0.89 \%$
$=0.6298$
No. of gram equivalents from the current $(Q)$
$=I \times t=140 \times 482.5=67550 C$
No. of gram equivalents $=\frac{67550}{96500}=0.7$
Since only Cu and Fe are dissolved from the anode no. of gram equivalents of Fe
$=0.7-0.6928$
$=0.0072$
Therefore, mass of Fe $=0.0072 \times 27.75$
$=0.1998 g$
$\%$ of Fe $=\frac{\text { Mass of Fe }}{\text { Mass of impurities }} \times 100$
$=\frac{0.1998}{22.26}=0.89$ or $0.89 \%$
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