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(B) According to Faraday's $1^{st}$ law of electrolysis,the mass deposited is proportional to the equivalent weight. For the same amount of electricit

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$x = 2y$

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(B) According to Faraday's $1^{st}$ law of electrolysis,the mass deposited is proportional to the equivalent weight. For the same amount of electricity $(Q)$,the number of moles of metal deposited is given by $n = Q / (n_f 	imes F)$,where $n_f$ is the valency factor. For $Ag^+$: $Ag^+ + e^- ightarrow Ag$,so $n_f = 1$. Thus,$x = Q / (1 	imes F) = Q / F$. For $Cu^{2+}$: $Cu^{2+} + 2e^- ightarrow Cu$,so $n_f = 2$. Thus,$y = Q / (2 	imes F)$. Comparing $x$ and $y$: $x / y = (Q / F) / (Q / 2F) = 2$. Therefore,$x = 2y$.

Same amount of electricity is passed through aqueous solutions of $AgNO_3$ and $CuSO_4$. The number of $Ag$ and $Cu$ atoms deposited are $x$ and $y$ respectively. The correct relationship between $x$ and $y$ is

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