A silver and zinc voltameter are connected in series and a current $i$ is passed through them for a time $t$ liberating $W\, gm$ of zinc. The weight of silver deposited is nearly ..............$W$
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(d) $\frac{{{m_{Zn}}}}{{{m_{Ag}}}} = \frac{{{E_{Zn}}}}{{{E_{Ag}}}}$ $ \Rightarrow $ ${m_{Ag}} = W\,\left( {\frac{{{E_{Ag}}}}{{{E_{Zn}}}}} \right)$= $3.3\, W$ = $3.5\, W$
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