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

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$KI$

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(A) According to Faraday's first 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 one mole of metal ion).For the same amount of charge $(Q)$,the number of moles deposited is inversely proportional to the valence factor $(n_f)$: $n \propto \frac{1}{n_f}$.Let us determine $n_f$ for each metal ion:$A$: $K^+$ $(n_f = 1)$$B$: $Zn^{2+}$ $(n_f = 2)$$C$: $Au^{3+}$ $(n_f = 3)$$D$: $Ca^{2+}$ $(n_f = 2)$Since $K^+$ has the lowest valence factor $(n_f = 1)$,it will result in the maximum number of moles of metal deposited.

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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