Three electrolytic cell `A,B`, and `C` contaning solutions of `ZnSO_(4), AgNO_(3)`, and `CuSO_(4)`, respectively, are connected in series. A steady current of `1.5A` 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 ?

Explanation

`Ag^(+)+e^(-)toAg,` i.e., 108 g of Ag are deposited by 1F=`96500C`
`therefore1.45g` of Ag will be deposited by `(96500)/(108)xx1.45C=1295.6C`
`Q=Ixxt` or `t=Q//I=1295.6//1.50=863.7s=14`min, 24sec.
`Cu^(2+)+2e^(-)toCu`
i.e., `2xx96500C` deposit Cu=63.5g `therefore1295.6C` will deposit `Cu=(63.5)/(2xx96500)xx1295.6=0.426g`
`Zn^(2+)+2e^(-)toZn. therefore` Zn deposited`=(65.3)/(2xx96500)xx1295.6=0.438g`,