Impure copper containing Fe,Au,and Ag as impu | vedclass

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(D) Total charge passed $Q = I 	imes t = 140 	imes 482.5 = 67550 \ C$. Number of Faradays $= \frac{67550}{96500} = 0.7 \ F$. At the cathode,only $Cu

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None of the above

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(D) Total charge passed $Q = I 	imes t = 140 	imes 482.5 = 67550 \ C$. Number of Faradays $= \frac{67550}{96500} = 0.7 \ F$. At the cathode,only $Cu^{2+}$ is reduced: $Cu^{2+} + 2e^- ightarrow Cu$. Equivalents of $Cu$ deposited $= 0.6928$. Mass of $Cu$ deposited $= 0.6928 	imes \frac{63.54}{2} = 22.011 \ g$. At the anode,$Cu ightarrow Cu^{2+} + 2e^-$ and $Fe ightarrow Fe^{2+} + 2e^-$. Total equivalents dissolved $= 0.7$. Let $x$ be equivalents of $Cu$ and $y$ be equivalents of $Fe$. $x + y = 0.7$. Mass of $Cu$ dissolved $+ 	ext{Mass of } Fe$ dissolved $= 22.26 \ g$. $x 	imes 31.77 + y 	imes 27.75 = 22.26$. Substituting $x = 0.7 - y$: $(0.7 - y) 	imes 31.77 + 27.75y = 22.26$. $22.239 - 31.77y + 27.75y = 22.26$. $-4.02y = 0.021 \implies y \approx 0.00522$. Mass of $Fe = 0.00522 	imes 27.75 = 0.1448 \ g$. Total mass of impure copper $= 22.26 \ g$. $\% 	ext{ of } Fe = \frac{0.1448}{22.26} 	imes 100 \approx 0.65 \%$. Since this value is not in the options,the correct answer is $D$.

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