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))` 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 change as : `E^(@)=-(DeltaG^(@))/(nF)` . At equilibrium, the Nernst equation is given as :`E^(@) = -(0.059)/(n) log K`
At equilibrium , when K = 1 the correct relation is

Options

A. `DeltaG ne 0 , DeltaG^(@) ne 0`
B. `DeltaG = 0 , DeltaG^(@) ne 0`
C. `DeltaG = 0 , DeltaG^(@) = 0`
D. `DeltaG ne 0 , DeltaG^@ = 0`

Answer

C

Explanation

`DeltaG = 0 - DeltaG^(@) = 2.303 RT log K_(e) implies DeltaG^(@)=0`