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 $1.5 \ A$ 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?

(N/A) According to the reaction:
$Ag_{(aq)}^{+} + e^{-} \to Ag_{(s)}$
$108 \ g$ of $Ag$ is deposited by $96487 \ C$ of charge.
Therefore,$1.45 \ g$ of $Ag$ is deposited by charge $Q = \frac{96487 \times 1.45}{108} \ C = 1295.43 \ C$.
Given current $I = 1.5 \ A$,time $t = \frac{Q}{I} = \frac{1295.43}{1.5} \ s = 863.6 \ s \approx 14.40 \ min$.
For copper:
$Cu_{(aq)}^{2+} + 2e^{-} \to Cu_{(s)}$
$2 \times 96487 \ C$ of charge deposits $63.5 \ g$ of $Cu$.
So,$1295.43 \ C$ of charge deposits $\frac{63.5 \times 1295.43}{2 \times 96487} \ g = 0.426 \ g$ of $Cu$.
For zinc:
$Zn_{(aq)}^{2+} + 2e^{-} \to Zn_{(s)}$
$2 \times 96487 \ C$ of charge deposits $65.4 \ g$ of $Zn$.
So,$1295.43 \ C$ of charge deposits $\frac{65.4 \times 1295.43}{2 \times 96487} \ g = 0.439 \ g$ of $Zn$.