Three electrolytic celss A,B,C containing solutions of `ZnSO_(4),AgNO_(3)` and `CuSO_(4)`, respectively are connected in series. A steady current of `1.5` amperes was passes 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`
`because` 108 g of Ag is deposited by `1F=96500C`
`therefore` 1.45g of Ag is deposited by `=(96500)/(108)xx1.45=1295.6C`
`thereforet=(Q)/(I)=(1295)/(1.5)=863.75sec`
Now, `Cu^(2+)+2e^(-)toCu`
`therefore` Copper (Cu) deposited `=(63.5)/(2xx96500)xx1295.6=0.426g`
`therefore` Zinc (Zn) deposited `=(65.4)/(2xx96500)xx1295.6=0.439g`