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 for electrochemical cell will be:
`Cu rarr Cu^(2+) + 2e`
`2Ag^(+) + 2e rarr 2Ag`
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`
`:. E_(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., `9.65 xx 60 xx 60` coulomb 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 rarr Cu`
`2Ag rarr 2Ag^(+) + 2e`
`Ag^(+)` ions formed `= (9.65 xx 60 xx 60)/(96500)eq`.
`= 0.36 eq. = 0.36` mole
`Cu^(2+)` ions discharged `= (9.65 xx 60 xx 60)/(96500)eq. = 0.36 eq`.
`= 0.18` mole
Thus, `[Ag^(+)]_("left") = 1 + 0.36 = 1.36 M`
`[Cu^(2+)]_("left") = 1 - 0.18 = 0.82 M`
Thus, new cell is `Cu|(Cu^(2+)),(0.82 M)||(Ag^(+)),(1.36 M)|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 `0.010 V`