The electrochemical equivalents of Cu and Ag | vedclass

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(C) According to Faraday's laws of electrolysis,the mass $m$ of a substance deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalen

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$2.4$

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(C) According to Faraday's laws of electrolysis,the mass $m$ of a substance deposited is given by $m = ZIt$,where $Z$ is the electrochemical equivalent,$I$ is the current,and $t$ is the time.For two different substances deposited by the same current for the same time,the ratio of their masses is equal to the ratio of their electrochemical equivalents:$\frac{m_{Cu}}{m_{Ag}} = \frac{Z_{Cu}}{Z_{Ag}}$Given: $m_{Cu} = 14 \ g$,$Z_{Cu} = 7 	imes 10^{-6}$,$Z_{Ag} = 1.2 	imes 10^{-6}$.Substituting the values:$\frac{14}{m_{Ag}} = \frac{7 	imes 10^{-6}}{1.2 	imes 10^{-6}}$$m_{Ag} = \frac{14 	imes 1.2}{7} = 2 	imes 1.2 = 2.4 \ g$.

The electrochemical equivalents of $Cu$ and $Ag$ are $7 \times 10^{-6}$ and $1.2 \times 10^{-6}$ respectively. If a certain current deposits $14 \ g$ of $Cu$,then the amount of $Ag$ deposited is .......... $g$.

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