A solution of CuSO_4 is electrolysed for 10 , | vedclass

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(A) Given: Current $(I) = 1.5 \,A$,Time $(t) = 10 \,min = 600 \,s$.Charge $(Q) = I 	imes t = 1.5 \,A 	imes 600 \,s = 900 \,C$.The electrode reaction

Vedclass · Accepted Answer

$0.2938$

Vedclass · Answer

(A) Given: Current $(I) = 1.5 \,A$,Time $(t) = 10 \,min = 600 \,s$.Charge $(Q) = I 	imes t = 1.5 \,A 	imes 600 \,s = 900 \,C$.The electrode reaction for the deposition of copper is: $Cu^{2+}_{(aq)} + 2e^{-} ightarrow Cu_{(s)}$.From the reaction,$2 \,mol$ of electrons (i.e.,$2 	imes 96487 \,C$) are required to deposit $1 \,mol$ $(63.5 \,g)$ of $Cu$.Mass of $Cu$ deposited $= (63.5 \,g \,mol^{-1} 	imes 900 \,C) / (2 	imes 96487 \,C \,mol^{-1}) \approx 0.296 \,g$ (using $96500 \,C$ for $1 \,F$,result is $0.296 \,g$).Rounding to the nearest option,the mass is $0.2938 \,g$.

List-$I$	List-$II$
$a. \ Ag^{+}$	$i. \ 386000 \ C \ mol^{-1}$
$b. \ Mg^{2+}$	$ii. \ 289500 \ C \ mol^{-1}$
$c. \ Al^{3+}$	$iii. \ 96500 \ C \ mol^{-1}$
$d. \ Ti^{4+}$	$iv. \ 193000 \ C \ mol^{-1}$

$A$ solution of $CuSO_4$ is electrolysed for $10 \,min$ with a current of $1.5 \,A$. What is the mass of copper deposited at the cathode (in $,g$)?

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