Three electrolytic cells A, B, C containing solutions of ZnSO_4, … - Vidyadip

Question

Three electrolytic cells A, B, C containing solutions of $ZnSO_4, AgNO_3$ and $CuSO_4$, 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?

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Answer

According to the reaction:
$\text{Ag}^{+}_\text{(aq)}+\text{e}^-\rightarrow\text{Ag}_\text{(s)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 108\text{g}$
i.e., 108 g of Ag is deposited by 96487 C
Therefore, 1.45 g of Ag is deposited by $=\frac{96487\times1045}{108}\text{C}$
= 1295.43 C
Given,
Current = 1.5 A
Therefore, Time $=\frac{1295.43}{105}\text{s}$
= 863.6 s
= 864 s
= 14.40 min
Again,
$\text{Cu}^{2+}_\text{(aq)}+2\text{e}^-\rightarrow\text{Cu}_\text{(s)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 63.5\text{g}$
i.e., 2 × 96487 C of charge deposit = 63.5g of Cu
Therefore, 1295.43 C of charge will deposit $=\frac{63.5\times1295.43}{2\times96487}\text{g}$
= 0.426 g of Cu
Therefore, 1295.43 C of charge will deposit $=\frac{65.4\times1295.43}{2\times96487}\text{g}$
= 0.439 g of Zn

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