If a current of 5 A is passed through a zinc | vedclass

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(B) According to Faraday's first law of electrolysis,the mass $w$ deposited is given by $w = \frac{Z 	imes I 	imes t}{96500}$,where $Z = \frac{	ext

Vedclass · Accepted Answer

$4.065$

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(B) According to Faraday's first law of electrolysis,the mass $w$ deposited is given by $w = \frac{Z 	imes I 	imes t}{96500}$,where $Z = \frac{	ext{Molar mass}}{	ext{Valency factor}}$.For zinc $(Zn^{2+} + 2e^- ightarrow Zn)$,the molar mass is $65.38 \ g/mol$ and the valency factor $n = 2$.So,$Z = \frac{65.38}{2 	imes 96500} \ g/C$.Given $I = 5 \ A$ and $t = 40 \ 	ext{minutes} = 40 	imes 60 = 2400 \ s$.Substituting the values: $w = \frac{65.38 	imes 5 	imes 2400}{2 	imes 96500} \approx 4.065 \ g$.

If a current of $5 \ A$ is passed through a zinc sulfate solution for $40 \ \text{minutes}$,what is the mass of zinc deposited at the cathode (in $g$)?

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