Calculate the emf of the cell in which the following reaction takes place

Please try to type the question  in right way and the equation should be like this:

Ni+2Ag+-------------> Ni²⁺ +2Ag

Oxidation is taking place at Ni electrode and reduction is at Ag.

Hence,

Ni|Ni²⁺||Ag⁺ | Ag

Ni + 2Ag⁺......>Ni²⁺ + 2Ag 

here, n=2

 and, E_o=1.05 V

and by using the formula:

_E = E_o - (.0592/2) log₁₀ ((Ni²⁺)/(Ag⁺)² )

 

E=1.05-.0592/2 log(.160/(.002)²)

 

Solve for ' E'