Relationship between K and the Nernst Equation

[705961146]

Hi,
I just wanted some clarification on the relationship between K and the Nernst Equation. In discussion section, we did practice problems about this connection. I know that the Nernst equation E = (-RT)/(nF) ln(K) depends on the term ln(K), so does this mean that if E is negative, K is greater than 1 and if E is positive, K is less than 1? Does this always hold true?
Thanks!

[Kyle_Phong_2J]

Hi,

I believe it is the other way around where if E cell is negative, K is less than 1 and if E cell is positive, K is greater than 1.

[405993361]

For a spontaneous reaction, Q would be less than K because products are favored over reactants. Therefore, ln(K) would be negative, and K would be greater than 1. If you have a positive value of E, then the reaction is not spontaneous. In a nonspontaneous reaction, Q would be greater than K because reactants are favored over products. So if E is positive, ln(K) would be positive, and K would be less than 1. I am pretty sure that this would hold true for any reaction at standard conditions and since we are looking at K instead of Q, it would be at standard conditions.