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(B) According to Faraday's law of electrolysis,the number of moles of metal deposited is given by $n = \frac{Q}{zF}$,where $Q$ is the charge,$z$ is th

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$1 : \frac{1}{2} : \frac{1}{3}$

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(B) According to Faraday's law of electrolysis,the number of moles of metal deposited is given by $n = \frac{Q}{zF}$,where $Q$ is the charge,$z$ is the valency (charge on the ion),and $F$ is Faraday's constant.Since the same quantity of electricity $(Q)$ is passed,the number of moles is inversely proportional to the valency $(z)$.For $Ag^{+}$ $(z=1)$,moles $= \frac{1}{1} = 1$.For $Cu^{2+}$ $(z=2)$,moles $= \frac{1}{2}$.For $Fe^{3+}$ $(z=3)$,moles $= \frac{1}{3}$.Therefore,the ratio of moles deposited is $1 : \frac{1}{2} : \frac{1}{3}$.

When the same quantity of electricity is passed through solutions of $Ag^{+}$,$Cu^{2+}$,and $Fe^{3+}$ ions,what is the ratio of the number of moles of each metal deposited?

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