Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, $1.45 \mathrm{~g}$ of $\mathrm{Ag}$ is deposited by $=\frac{96487 \times 1.45}{108} \mathrm{C}$
= 1295.43 C
Given,
Current = 1.5 A
$\therefore$ Time $=\frac{1295.43}{1.5} \mathrm{~s}$
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, $1295.43 \mathrm{C}$ of charge will deposit $=\frac{63.5 \times 1295.43}{2 \times 96487} \mathrm{~g}$
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, $1295.43 \mathrm{C}$ of charge will deposit $=\frac{65.4 \times 1295.43}{2 \times 96487} \mathrm{~g}$
= 0.439 g of Zn
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