A copper voltameter and a silver voltameter a | vedclass

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(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the

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$\frac{q}{1 + \frac{z_2}{z_1}}$

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(D) According to Faraday's law of electrolysis,the mass $m$ deposited is given by $m = zq$,where $z$ is the electrochemical equivalent and $q$ is the charge.Since the mass deposited in both voltameters is equal,we have $m_1 = m_2$.Therefore,$z_1 q_1 = z_2 q_2$,where $q_1$ and $q_2$ are the charges flowing through the copper and silver voltameters respectively.This implies $\frac{q_1}{q_2} = \frac{z_2}{z_1}$.The total charge $q$ is the sum of charges flowing through the parallel branches: $q = q_1 + q_2$.We want to find $q_2$. From $q = q_1 + q_2$,we have $q_1 = q - q_2$.Substituting this into the ratio: $\frac{q - q_2}{q_2} = \frac{z_2}{z_1}$.$\frac{q}{q_2} - 1 = \frac{z_2}{z_1} \Rightarrow \frac{q}{q_2} = 1 + \frac{z_2}{z_1}$.Thus,$q_2 = \frac{q}{1 + \frac{z_2}{z_1}}$.

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