Aqueous CuSO_4 solution was electrolysed by p | vedclass

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(B) Given: Current $(I) = 2 \ A$,Time $(t) = 10 \ min = 600 \ s$,Molar mass of $Cu = 63 \ g \ mol^{-1}$,Faraday constant $(F) = 96500 \ C \ mol^{-1}$.

Vedclass · Accepted Answer

$0.39$

Vedclass · Answer

(B) Given: Current $(I) = 2 \ A$,Time $(t) = 10 \ min = 600 \ s$,Molar mass of $Cu = 63 \ g \ mol^{-1}$,Faraday constant $(F) = 96500 \ C \ mol^{-1}$.The reaction at the cathode is: $Cu^{2+} + 2e^{-} ightarrow Cu(s)$.Number of electrons transferred $(n) = 2$.Total charge $(Q) = I 	imes t = 2 \ A 	imes 600 \ s = 1200 \ C$.Mass of copper deposited $(m) = \frac{M 	imes Q}{n 	imes F} = \frac{63 	imes 1200}{2 	imes 96500}$.$m = \frac{75600}{193000} \approx 0.3917 \ g$.Rounding to two decimal places,the mass is $0.39 \ g$.

Aqueous $CuSO_4$ solution was electrolysed by passing $2 \ A$ of current for $10 \ min$. What is the weight (in $g$) of copper deposited at the cathode? (Given: $Cu = 63 \ u$; $F = 96500 \ C \ mol^{-1}$)

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