The following galvanic cell
`Zn|Zn(NO_3)_2(aq)||Cu(NO_3)_2(aq)|Cu`
`Anode (100mL,1M)(100mL,1M) cathode`
was operated as an electrolysis cell as Cu as the anode and Zn as the cathode. A current of 0.48 ampere was passed for 10 hours and then the cell was allowed to function to galvanic cell.What would be the emf of the cell at `25^@C` Assume that the only electrode reactions occurring were those involving
`Cu//Cu^(2+) and Zn//Zn^(2+)`
`(E_(Cu^(2+),Cu)^@=+0.34V, E_(Zn^(2+),Zn)^@=-0.76V)`

Explanation

When the cell acts as electrolysis cell, Cu being anode and Zn being cathode the concentration of `Cu^(2+)` will increase due to dissolution `(Cu to Cu^(2+)+2e)` and `Zn^(2+)` concentration will decrease due to deposition `(Zn^(2+)+2e to Zn)`
Eq. of Zn deposited= no. of faraday of electricity passed
(at cathode)
`(n o . of cou lombs)/96500`
`=(0.48 times (10 times 60 times 60))/96500`
`=0.18`
`therefore ` eq. of Cu dissolved =0.18
(at anode)
Thus, mole of `Zn^(2+)` removed from the cathodic compartment
`=0.18/2=0.09`
and mole of `Cu^(2+)` gone to anodic compartment=0.09
`therefore {(mol e of Zn^(2+) I nitially present =1 times 0.1=0.1), (and mol e of Cu^(2+) I nitially present =1 times 0.1=0.1):}` `therefore {(mol e of Zn^(2+) after el ectrolysis =0.1-0.09=0.01), (and mol e of Cu^(2+) present after el ectrolysis =0.1+0.09=0.19):}`
`[Zn^(2+)] =0.1M and [Cu^(2+)]=1.9M`
As the electrolysis cell is now allowed to act as a galvanic cell as represented below.
`Zn|Zn^(2+)(0.1M)||Cu^(2+) (1.9M)|Cu`,
`E_(cell)=E_(Cu^(2+),Cu)-E_(Zn^(2+)Zn)`
`={E_(Cu^(2+),Cu)^@+0.0591/2log[Cu^(2+)]}-{E_(Zn^(2+),Zn)^@+0.0591/2log[Zn^(2+)]}`
`={+0.34+0.0591/2log1.9}-{0.76+0.0591/2log0.1}`
`=1.37` volts