If in a voltaic cell 5 g of zinc is consumed | vedclass

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(C) According to Faraday's law of electrolysis,the mass $m$ consumed is given by $m = Z 	imes q$,where $Z$ is the $E.C.E.$ and $q$ is the total charg

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$4.1$

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(C) According to Faraday's law of electrolysis,the mass $m$ consumed is given by $m = Z 	imes q$,where $Z$ is the $E.C.E.$ and $q$ is the total charge in Coulombs.Given $m = 5 \ g = 5 	imes 10^{-3} \ kg$ and $Z = 3.387 	imes 10^{-7} \ kg/C$.$q = \frac{m}{Z} = \frac{5 	imes 10^{-3}}{3.387 	imes 10^{-7}} \ C$.To convert Coulombs to Ampere-hours $(Ah)$,we divide by $3600$ $(1 \ Ah = 3600 \ C)$.$q 	ext{ (in } Ah) = \frac{5 	imes 10^{-3}}{3.387 	imes 10^{-7} 	imes 3600} \approx 4.1 \ Ah$.

If in a voltaic cell $5 \ g$ of zinc is consumed,then how many ampere-hours are produced? (Given that the $E.C.E.$ of $Zn$ is $3.387 \times 10^{-7} \ kg/C$)

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