A solution of CuSO_4 is electrolysed for 10 m | vedclass

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(B) 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

Vedclass · Accepted Answer

$0.293$

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(B) 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 reduction of copper is: $Cu^{2+}_{(aq)} + 2e^{-} ightarrow Cu_{(s)}$. From the reaction,$2 \ mol$ of electrons $(2 	imes 96500 \ C)$ deposit $1 \ mol$ of $Cu$ $(63.5 \ g)$. Mass of $Cu$ deposited $= \frac{63.5 \ g 	imes 900 \ C}{2 	imes 96500 \ C} \approx 0.296 \ g$. Rounding to the nearest provided option,the mass is $0.293 \ g$.

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

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