Derive Nernst equation for the following cell:
`NiNi^(2+)(0*2M)||Cu^(2+)(0*2M)|Cu`.

Explanation

Cell representation is as follows: `Ni|Ni^(2+)(0.2M)||Cu^(2+)(0.2M)|Cu`
For this cell, the electrode potential for cathode can be written as :
`E_(Cu^(2+)//Cu)=E_(Cu^(2+)//Cu)^(@)-(RT)/(2F)l n(1)/([Cu^(2+)])("for n=2")`
Similarly for anode `E_(Ni^(2+)//Ni)=E_(Ni^(2+)//Ni)^(@)-(RT)/(2F)l n(1)/([Ni^(2+)])`
The overall cell potential is given by
`E_("cell")=E_("right")-E_("left")=E_(Cu^(2+)//Cu)-E_(Ni^(2+)//Ni)`
`E_("cell")=E_(Cu^(2+)//Cu)^(@)-(RT)/(2F)l n(1)/([Cu^(2+)])-E_(Ni^(2+)//Ni)^(@)+(RT)/(2F)l n(1)/([Ni^(2+)])`
`=(E_(Cu^(2+)//Cu)^(@)-E_(Ni^(2+)//Ni)^(@))-(RT)/(2F)[l n(1)/([Cu^(2+)])-l n(1)/([Ni^(2+)])]=E_("cell")^(@)-(RT)/(2F)l n([Ni^(2+)])/([Cu^(2+)])`
`E_("cell")=E_("cell")^(@)-(0.059)/(2)log""([Ni^(2+)])/([Cu^(2+)])`