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(A) According to Faraday's law of electrolysis,the number of moles of metal deposited $(n)$ is given by the formula: $n = \frac{Q}{n_f 	imes F}$,wher

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$KI$

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(A) According to Faraday's law of electrolysis,the number of moles of metal deposited $(n)$ is given by the formula: $n = \frac{Q}{n_f 	imes F}$,where $Q$ is the charge passed,$F$ is Faraday's constant,and $n_f$ is the valence factor (number of electrons involved in the reduction of the metal ion).For a constant charge $Q$,the number of moles $n$ is inversely proportional to the valence factor $n_f$ $(n \propto \frac{1}{n_f})$.Let us determine the $n_f$ for the metal ions in each case:$A$: $K^+$ $(n_f = 1)$$B$: $Zn^{2+}$ $(n_f = 2)$$C$: $Au^{3+}$ $(n_f = 3)$$D$: $Ca^{2+}$ $(n_f = 2)$Since $n_f$ is minimum for $K^+$ $(n_f = 1)$,the number of moles deposited will be maximum for $KI$.

Which of the following aqueous solutions will result in the maximum number of moles of metal deposited on passing the same amount of charge?

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