Three electrolytic cells A, B, C containing `ZnSO_(4), AgNO_(3)` and `CuSO_(4)` solution 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 How long did the current flow ? What mass of copper and zinc were deposited ?

Explanation

Reaction for deposition of silver : `underset("1 mol (1F) 1 mol(108g)")(AG^(+)+e rarr Ag)`
1 F(or 96500 C) of electricity deposited 108 g silver
`therefore` 1.45 g Ag is deposited by `= (96500)/(108)xx1.45=1295.6C`
Since, `Q =It, t=(Q)/(l)=(1295.6)/(1.5)=863.7s = 14.39` min
`therefore` The current will flow for 863.7 s = 14.39 min
Reaction for deposition of zinc : `Zn^(2+)+underset(2xx96500 C)underset("2 mol")(2e)rarr underset(65.3 g)underset("1 mol")(Zn)`
`therefore 2xx96500 C` electricity deposited 65.3 g Zn
`therefore 1295.6 C` electricity deposit `=(65.3xx1295.6)/(2xx96500)=0.438 g Zn`
Reaction for deposition of copper : `Cu^(2+)+underset(2xx96500 C)underset("2 mol")(2e)rarr underset(63.5 g)underset("1 mol")(Cu)`
`therefore 2xx96500 C` electricity deposited 63.5 g Cu
`therefore 1295.6 C` electricity deposit `=(63.5xx1295.6)/(2xx96500)=(0.426 g Cu)`.