`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.

Options

A. `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([A_(g)^(+)])/([Ni^(2+)])`
B. `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)])/([Ag^(+)])`
C. `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)]^(2))/([Ag^(+)]^(2))`
D. `E_(cell)=E_(cell)^(Theta)-(0.059)/(2)"log"([Ni^(2+)])/([Ag^(+)]^(2))`

Answer

D

Explanation

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))`.