A cell, `Ag|Ag^(o+)||Cu^(2+)|Cu` , initially contains `1 M Ag^(o+)` and `1M Cu^(2+)` ions. Calculate the change in the cell the potential after the passage of `9.65A` of current for `1h.`

Explanation

The given cell will not work as electrochemical cell since `E_(Ag^(+)//Ag)^(0) gt E_(Cu^(2+)//Cu)^(0)`
The equation for electrochemical cell will be
`Cu_((s)) to Cu^(2+) +2e`
`2Ag^(+) +2e to Ag_((s))`
Emf of cell `Cu|Cu^(2+)|Ag^(+)|Ag` will be
`E_("cell")=E_("cell")^(0)-(0.059)/(2) log ([Cu^(2+)])/([Ag^(+)]^(2))`
`[Ag^(+)]=1M, [Cu^(2+)]=1M`
`E_("cell")=E_("cell")^(0)-(0.059)/(2)log 1/1`
`E_("cell")=E_("cell")^(0)`
After passage of 9.65 ampere for 1. hour `(9.65 xx 60 xx 60" coloumb charge")` during which the cell reaction is reversed thus `Cu^(2+)` are discharged from solution and Ag metal passes to ionic state. The reaction during passage of current are
`Cu^(2+) +2e to Cu`
`2Ag to 2Ag^(+)+2e`
`Ag^(+)` ions formed `=(9.65 xx 60 xx 60)/(96500)=0.36=0.36" mole"`
`Cu^(2+)" discharged"=(9.65 xx 60 xx 60)/(96500)=0.36 eq=0.18" mole"`
Finally `[Ag^(+)]=1+0.36=1.36M`
`[Cu^(2+)]=1-0.8=0.82M`
New cell: `Cu|Cu_((0.82M))^(2+) ||Ag_((1.36M))^(+)|Ag`
`E_("cell")=E_("cell")^(0)-(0.059)/(2) log((0.82))/((1.36)^(2))`
`=E_("cell")^(0)+0.010 volt`
Thus `E_("cell")` increased by 0.010 V.