The reduction potentials of

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

Hence,

Cu|Cu²⁺||Ag⁺|Ag

Cu + 2Ag⁺......>Cu²⁺ + 2Ag

 

here, n=2

 

and, E_o=.80-.34=.46V

 

and by using the formula:

 

E _cell = E_o - (.0592/2) log₁₀ (Cu²⁺/(Ag+)² )

 

0=.46V-.0592/2 log(.1/(Ag+)²)

 

Ag⁺=5*10^-9 M