Question

Three electrolytic cells $A,B,C$ containing solutions of $ZnS{O}_4,AgN{O}_3$ and $CuS{O}_3$, respectively are connected in series. A steady current of $1.5$ amperes was passed through them until $1.45g$ of silver deposited at the cathode of cell $B$. How long did the current flow? What mass of copper and zinc were deposited?

Answer 1

Cell $B$: $A{g}^{+}+e^{-}\rightleftharpoons Ag$ at cathode.

$1$ mole ($108$ g) of Ag is deposited by $96500$ $C$.

$1.45$ g of Ag will be deposited by $\frac{96500\times 1.45}{108}=1295.6C$.
Now,
$Q=It$
$1295.6=1.5\times t$
$t=864s$
Cell $A$: $Z{n}^{2+}+2e^{-}\to Zn$
$2$ moles of electrons ($2\times 96500$ C of current) produces $1$ mole ($63.5$ g) of zinc.
$1295.6$ $C$ of electricity will deposit $\frac{65.3}{2\times 96500}\times 1295.6=0.438$ g of zinc
Cell $C$: $C{u}^{2+}+2e^{-}\to Cu$
$2$ moles of electrons ($2\times 96500$ $C$) of current will produce $1$ mole ($63.5$ g) of $Cu$.
$1295.6$ $C$ of current will deposit $\frac{63.5\times 1295.6}{2\times 96500}=0.426g$ of copper