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 1.50 amperes 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 ? [At. weights of Cu = 63.5, Zn = 65.3, Ag = 108]

Explanation

`Ag^(+) + e^(-) to Ag`
1 mole or 108 g of Ag are deposited by 96500 C of electricity
`therefore 1.45`g of Ag will be deposited by `(96500)/108 xx 1.45C = 1295.6 `C
Time of flow of current is obtained as under:
`Q = l xx t`
or `t = Q//l = 1295.6//1.50 = 863.7 s = 14` min 24 s
Masses of Cu and Zn deposited can be obtained as under:
`Cu^(2+) + 2e^(-) to Cu`
`2 xx 96500 C ` deposit = 63.5 g Cu
`therefore 1295.6` C will deposit `=63.5/(2 xx 96500) xx 1295.6 = 0.438 g`.