[Big-Daddy]

If we put two electrodes in solution and each establish their own half-cell equilibrium, what does it mean to say that the reaction is or is not spontaneous? I know all the basics about what spontaneity entails in terms of ΔG° and Q/K, basically if K>1 than the reaction moves forward spontaneously (for the cell) but if K<1 then it does not. Then we accordingly can say that any cell set-up with positive E°_cell will have its reaction spontaneously move forward as K>1.

But for half-cells, if K<1, shouldn't this mean that the reaction moves backwards, i.e. in the opposite direction? The opposite reaction will be happening, but there will still be an ongoing reaction and exchange of electrons. So do we need to connect up the electrodes to the correct terminals or something, or just put them in the solution, and they will naturally form their own terminals, so that the electrode with more positive E° is on the right of the cell notation and reduction occurs there, whereas oxidation occurs at the other.

So is this right? E_cell and E°_cell change signs if you switch the positions/terminals of the electrodes, but the reaction continues to occur in the same direction (i.e. the electrode with more positive E° will by definition be undergoing more reduction, since the E° value is more positive if the K is higher for that half-cell equilibrium, whereas the electrode with more negative E° will be undergoing oxidation). Both directions head towards equilibrium for the individual E° values (which is the same as equilibrium for the cell, E°_cell).

A reaction is spontaneous and occurs whenever you put the electrodes in solution, but you will never get the opposite direction of reaction (more positive electrode's equilibrium will always have a high K and thus reactants will be pushed towards products, whereas more negative electrode's equilibrium will always have low K and thus products will be pushed towards reactants).

Sorry for the long post, I just want to be sure on this topic. My clarity developed as I went along.