So if we have a galvanic cell obviously our Ecell is positive meaning its a spontaneous reaction thus deltaG would be negative.
But I am getting hung on the concept surrounding Equilibrium constant and why it has to be greater than 1. Yes I know you can plug in a negative deltaG into the equation: DeltaG = -RT ln Keq and obtain a K greater than 1
but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1.
Lets back up a couple steps because this can get tricky:
The determine delta G for any rxn that isn't at equalibrium use the equation: delG = del Gstand + RT lnQ.
So above we have del G or rxn. del G standard (which comes from a book, never changes and is delG when standard conditions are present ie 1atm 298K and 1M concentration of all products and reactants), R is const, T is temp, and Q is rxn quotient (concentration of products and reactants at a givin instant.)
Now you can see why this equation is useful. Because for a given rxn, the stand del G is fixed! If you add RT ln Q to that (rem Q is intantaneous rxn concentrations) you will get the del G under those condidtions.
Lets take note of a few things here bf continuing.
( At
standard conditions all concentrations of products and reacants = 1M)
Consider rxn A-----> B. If at this moment you have way more reactants than product, the result will be a Q that is very small (smaller than K more than likely). So intuitively you can see that with lots of reactant, the rxn will be proceeding toward the right (LeChatlie's principle). If you were to plug in the numbers for Q ( a very small number seeing as how there are many more reactants than products), you would indeed get a negative numerical value for del G indicating a spontaneous process for A---->B occuring in the direction written.
But your question is about K. Well if we are discussing K we must be talking about equilibrium right? So here are a few things you NEED to know about Equilibrium, del G, and Ecell.
@Equilibrium:
1)Q=K
2)del G = 0
3)Ecell = 0
You have to know that @ equilibrium, ALL of these statements are true. Furthermore if you know any of these statements are true, then they ALL are true and you are @ equilbrium. It also important to note that the del G and Ecell = 0 does NOT refer to standard values; remember those come from a book and do not change. So knowing this lets take a look at the original equation and manipulate it to derive an equation that represents equilibrium condititions.
We started with delG = delGstand + RT ln Q. At equilibrium, del G = 0
and Q=K. Sub in del G = 0 and Q=K and you have new equation that relates del G standard to equilibrium constant and is as follows:
delG stand = -RTlnK.
Here is where I think you may be a little mixed up:
"
but concept wise wouldn't the Equilibrium constant be less than one because its a spontaneous reaction and we would want it to go forward so the reactants concentration would be higher than the products concentration making the K less than 1 instead of greater than 1"
Remember that we are at EQUILIBRIUM. That means there is no more going forward and backwards to approach equilibrium since we are already there. Think about that for a second of two. That's why del G and Ecell = 0 at this point; bc there is no more "free energy" or "electromotive force" left to move rxn.
So if we are at equilibrium (where conc of products and reactants is not changing any more) and delGstand is < 0; indicating that the reaction progressed in the forward direction forming lots of product. And when you have lots of product at equilibrium, K>>1.
I think that's your hangup. You are thinking as if the rxn is still moving but it isn't. Consider rxn A--->B at
equilibrium. Lets look at some possible values for K keeping in mind K = {Prod}^x/(React)^y.
K>>1 means lots of product at equilib
K<<1 means lots of reactant at equilib
K = 1 means prod and reactant concentrations are equal at equilib.