Two voltameters,one of copper and another of | 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 masses deposited are equal,we have $m_1 = m_2$,which implies $z_1 q_1 = z_2 q_2$.From this,the ratio of charges is $\frac{q_1}{q_2} = \frac{z_2}{z_1}$.We know that the total charge $q = q_1 + q_2$.We want to find the charge $q_2$ flowing through the silver voltameter.From the ratio,$q_1 = q_2 \frac{z_2}{z_1}$.Substituting this into the total charge equation: $q = q_2 \frac{z_2}{z_1} + q_2 = q_2 (1 + \frac{z_2}{z_1})$.Solving for $q_2$,we get $q_2 = \frac{q}{1 + \frac{z_2}{z_1}}$.

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