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A fixed quantity of current for a fixed time is passed through two solutions of ${AgNO}_{3}$ and ${CuSO}_{4}$ in two different cells connected in series. If 0.20 g of Cu is deposited, the mass of silver deposited will be (Given: $M (Cu) = 63.5 g {mol}^{- 1}$ and $M (Ag) = 108 g {mol}^{- 1})$

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0.68 g

0.42 g

0.86 g

1.20 g

answer is B.

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Detailed Solution

Since $m = (I t / F) (M / v_{e})$ , we have

$\frac{m_{Ag}}{m_{Cu}} = (\frac{M_{Ag}}{v_{Ag}}) (\frac{v_{Cu}}{M_{Cu}}); m_{Ag} = (\frac{108 g {mol}^{- 1}}{1}) (\frac{2}{63.5 g {mol}^{- 1}}) (0.2 g) = 0.68 g$

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