Derive Nernst equation for the following galvanic cell.
`Ni_((S))|Ni_((aq))^(2+)||Ag_((aq))^(+)|Ag`

Explanation

* The cell reaction is as follows:
Anode: `Ni_((S))toNi_((aq))^(2+)+2e^(-)` . .. (Reduction)
Cathode: `2Ag_((aq))^(+)+2e^(-) to 2Ag_((S))` . . . (Oxidation)
Redox reaction: `Ni_((S))+2Ag_((aq))^(+) to Ni_((aq))^(2+)+2Ag_((S))`
If `E_(cell)`=Non standard electrode potential.
`E_(cell)^(Theta)`=Standard electrode potential.
n=Number of electrons taking part in the reaction at equilibrium.
* `E_(cell)=E_(cell)^(Theta)-(RT)/(2F)ln([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))`
`therefore E_(cell)=E_(cell)^(Theta)-(0.0591)/(2)"log"_(10)([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))`.