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. `10^(-37)`
B. `10^(37)`
C. `10^(-39)`
D. `10^(39)`

Answer

B

Explanation

`E^(@) = (0.0591)/(n) log K_(c) , log K_(c) = (2 xx 1.1)/(0.0591) =37.225 , K_(c) = (10)^(37.225)`