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

Given `I=1.5A,W=1.45g` of `Ag,t=?,E=108,n=1`
Using faraday's 1 st law of electrolysis W=Zit
or `W=(E)/(nF)It`
or `t=(1.45xx96500)/(1.5xx108)=863.73` seconds
Now for `Cu,W_(1)=1.45g` of `Ag,E_(1)=108,W_(2)=?`
From faraday's 2nd law of electrolysis `(W_(1))/(W_(2))=(E_(1))/(E_(2))`
`(1.45)/(W_(2))=(108)/(31.75) therefore W_(2)=(1.45xx31.75)/(108)`
`=0.426` g of Cu.
similarly for `Zn,W_(1)=1.45g` of Ag, `E_(1)=108`
`W_(2)=?,E_(2)=32.65`
Using formula `(W_(1))/(W_(2))=(E_(1))/(E_(2))`
`(1.45)/(W_(2))=(108)/(32.65)`
`thereforeW_(2)=(1.45xx32.65)/(108)=0.438` of Zn.