A current of 0.4 A is passed through excess | vedclass

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

Vedclass · Accepted Answer

$4.8$

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(C) The reaction at the cathode is: $Mg^{2+} + 2e^- ightarrow Mg(s)$.According to Faraday's law,the mass deposited $w = \frac{I 	imes t 	imes M}{n 	imes F}$.Given: $I = 0.4 \ A$,$t = 9.65 	imes 10^4 \ sec$,$M(Mg) = 24 \ g/mol$,$n = 2$,and $F = 96500 \ C/mol$.Substituting the values: $w = \frac{0.4 	imes 9.65 	imes 10^4 	imes 24}{2 	imes 96500}$.$w = \frac{0.4 	imes 96500 	imes 24}{2 	imes 96500} = \frac{0.4 	imes 24}{2} = 0.2 	imes 24 = 4.8 \ g$.

$A$ current of $0.4 \ A$ is passed through excess of molten $MgCl_2$ for $9.65 \times 10^4 \ sec$. The mass of $Mg$ deposited at the cathode is .............. $gm$

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