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

Note that given cell will not work as electrochemical cell since `E_(OP_(Cu))^(@)gtE_(OP_(Ag))^(@)`. The equation electrochemical cell will be:
`CurarrCu^(2+)+2e`
`2Ag^(+)+2erarr2Ag`
Thus, e.m.f of cell `Cu|Cu^(2+)"||"Ag^(+)|Ag` will be
`E_(cell)=E_(OP_(Cu))^(@)+E_(RP_(Ag))^(@)+(0.059)/(2)log_(10).([Ag^(+)]^(2))/([Cu^(2+)])`
`because[Ag^(+)]="1M and"[Cu^(2+)]=1M`
`thereforeE_(cell)=E_(cell)^(@)+(0.059)/(2)log_(10).(1)/(1)`
`E_(cell)=E_(cell)^(@)("where E"_(cell)^(@)=E_(OP_(Cu))^(@)+E_(RP_(Ag))^(@))`
After the passage of `9.65` ampere for 1 hr, i.e., `9.65xx60xx60` coulomb charge, during which the cell reaction it reversed thus, `Cu^(2+)` are discharged from solution and AG metal passes to ionic state.
The reaction during passage of current are:
`Cu^(2+)+2erarrCu`
`2Agrarr2Ag^(+)+2e`
`Ag^(+)` ions formed
`=(9.65xx60xx60)/(96500)eq.=0.36eq. =0.18"mole"`
Thus, `[Ag^(+)]_("left")=1+0.36=1.36M`
`[Cu^(2+)]_("left")=1-0.18=0.82M`
Thus, new cell is `Cu|{:(Cu^(2+)),(0.82M):}||{:(Ag^(+)),(1.36M):}|Ag`
Thus, `E_(cell)=E_(cell)^(@)+(0.059)/(2)log_(10).((1.36)^(2))/((0.82))`
`=E_(cell)^(@)+0.010"volt"`
Thus, `E_(cell)` increases by `0.010V`