Consider the following reaction in a galvanic cell: $$\ce{X(s) + Y+(aq) -> Y(s) + X+(aq)}$$
According to the Nernst equation, $$E =E_\circ - \dfrac{RT}{F}\ln\left(\dfrac{\ce{[X+]}} {\ce{[Y+] }} \right)$$
Thus, the lower the concentration of the product ion, the higher the electrode potential.
I am trying to figure out why this is true.
According to me, there are two factors at play:
Some metals are more "comfortable" being cations than others are, due to solvation effects, low effective nuclear charge leading to high electropositivity, and so on. The higher the "comfort difference" between the metals in the cell, the higher the electrode potential.
As electrons leave the metals that are comfortable being cations, they leave mutually repelling positively charged particles in their wake that increase the energy of the system. This is undesirable for stability, so the higher the product cations' concentration, the lower the electrode potential.
I believe taking the fraction of the concentrations of product and reactant ions rather than their absolute figures nicely captures this tradeoff. This is because even though increasing the product concentration will increase repulsion between cations, a proportionate increase in the reactant concentration will increase the "discomfort" mentioned in point 1. Electrons will again flow to the discomfited cations at the same rate.
Is this model correct? Is it complete?