Question

Three electrolytic cells $A,B$ and $C$ containing electrolytes $Zn{SO}_4$, $Ag{NO}_3$ and $Cu{SO}_4$ respectively were connected in series. A steady current of 1.50 ampere was passed through them until 1.45 of $Ag$ were deposited at the cathode of cell $B$. How long did the current flow? What mass copper and zinc were deposited? (Atomic weight of $Cu=63.5,Zn=65.3,Ag=108$)

Answer 1

${Ag}^{+}\left(aq\right)+e^{(-)}⟶Ag\left(s\right)\left(108g\right)$

i.e, $108g$ of $Ag$ is deposited by $96487C$

Therefore, $1.45g$ of $Ag$ is deposited by $=\frac{96487\times 1.45}{108}C$

$=1295.43C$

given,

current $=1.5A=\frac{1295.43}{1.5}S$

$\therefore$ Time $=863.65=14.40$ min.

Again, ${Cu}^{2+}\left(aq\right)+2e^{-}⟶Cu\left(s\right)\left(63.5g\right)$

i.e, $2\times 96487C$ of charge deposit $=63.5g$ of $Cu$

Therefore, $1295.43C$ of charge will deposit $=\frac{63.5\times 1295.43}{2\times 96487}g$

$=0.426g$ of $Cu$.

${Zn}^{2+}\left(aq\right)+{2e}^{-}⟶Zn\left(s\right)$

i.e, $2\times 96487C$ of charge deposit $=65.4g$ of $Zn$

Therefore, $1295.43C$ of charge will deposit $=\frac{65.4\times 1295.43}{2\times 96487}g$

$=0.439g$ of $Zn$