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(B) The reaction at the cathode is: $Zn^{2+} + 2e^- ightarrow Zn(s)$.According to Faraday's law,the mass deposited $w = \frac{M 	imes I 	imes t}{n

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$4.065$

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(B) The reaction at the cathode is: $Zn^{2+} + 2e^- ightarrow Zn(s)$.According to Faraday's law,the mass deposited $w = \frac{M 	imes I 	imes t}{n 	imes F}$.Here,$M$ (molar mass of $Zn$) $= 65.38 \ g/mol$,$I = 4 \ A$,$t = 50 \ min = 50 	imes 60 \ s = 3000 \ s$,$n = 2$,and $F = 96500 \ C/mol$.Substituting the values: $w = \frac{65.38 	imes 4 	imes 3000}{2 	imes 96500}$.$w = \frac{784560}{193000} \approx 4.065 \ g$.

$A$ $4 \ A$ current is passed through a solution of zinc sulphate for $50 \ min$. Find the amount of zinc deposited at the cathode in $g$.

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