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

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., $2 \times 96487\ C$ 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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