In an electrolysis experiment,a current i pas | vedclass

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(C) According to Faraday's second law of electrolysis,when the same current passes through cells in series,the mass $m$ deposited is proportional to t

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In the ratio of half the atomic weight of $Cu$ to the atomic weight of $Ag$

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(C) According to Faraday's second law of electrolysis,when the same current passes through cells in series,the mass $m$ deposited is proportional to the equivalent weight $E$ of the substance.Thus,$\frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}}$.The equivalent weight $E$ is defined as $\frac{	ext{Atomic weight}}{	ext{Valency}}$.For $Cu$ in $CuSO_4$,the valency is $2$,so $E_{Cu} = \frac{	ext{Atomic weight of } Cu}{2}$.For $Ag$ in $AgNO_3$,the valency is $1$,so $E_{Ag} = \frac{	ext{Atomic weight of } Ag}{1}$.Therefore,the ratio of the rates of increase of weight is $\frac{m_{Cu}}{m_{Ag}} = \frac{\frac{1}{2} 	imes 	ext{Atomic weight of } Cu}{	ext{Atomic weight of } Ag}$.

In an electrolysis experiment,a current $i$ passes through two different cells in series,one containing a solution of $CuSO_4$ and the other a solution of $AgNO_3$. The rate of increase of the weight of the cathodes in the two cells will be

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