Ni_((S)) |Ni_((aq))^(2+)||Ag_((aq))^(+)|Ag_((S)) is a non-standard cell, in which concentration of ion is less than 1 M, then determine correct formula to calculate potential of non-standard cell.

Ni_((S)) |Ni_((aq))^(2+)||Ag_((aq))^(+)|Ag_((S)) is a non-standard cel

1.	Ni_((S)) \|Ni_((aq))^(2+)\|\|Ag_((aq))^(+)\|Ag_((S)) is a non-standard cell, in which concentration of ion is less than 1 M, then determine correct formula to calculate potential of non-standard cell.
Answer» `E_(cell)=E_(cell)^(THETA)-(0.059)/(2)"log"([A_(G)^(+)])/([Ni^(2+)])` `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)])/([AG^(+)])` `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)]^(2))/([Ag^(+)]^(2))` `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)])/([Ag^(+)]^(2))` Solution :The following cell reaction is occurred `Ni_((S)) + 2Ag_((aq))^(+) to Ni_((aq))^(2+)+2Ag_((S))"Where, "n=2` `K=(["Product"])/(["Reactant"])=([Ni_((aq))^(2+)][Ag_((S))]^(2))/([Ni_((S))][Ag_((aq))^(+)]^(2))=([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))` So, `E_(cell)=E_(cell)^(theta)-(0.059)/(2)"log"([Ni_((aq))^(2+)])/([Ag_((aq))^(+)]^(2))`.