The driving force `DeltaG` diminishes to zero on the way to equilibrium, just as in any other spontaneous process. Both `DeltaG` and the corresponding cell potential `(E=-(DeltaG)/(nF))` are zero when the redox reaction comes to equilibrium. The Nernst equation for the redox process of the cell may be given as :
`E=E^(@)-0.059/n log Q`
The key to the relationship is the standard cell potential `E^(@)`, derived from the standard free energy changes as :
`E^(@)=-(DeltaG^(@))/(nF)`
At equilibrium, the Nernst equation is given as :
`E^(@)=0.059/n log K`
The equilibrium constant `K_(c)` for the reaction :
`Cu(s)+2Ag^(+) (aq.)+2Ag(s)" "(E_(cell)^(@)=0.46 V)` will be :

Options

A. antilog 15.6
B. antilog 2.5
C. antilog 1.5
D. antilog 12.2

Answer

A