# Intuition for relation between electrode potential and reaction quotient

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:

1. 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.

2. 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?

Let's consider an example of your galvanic cell : the Daniell cell, made of a zinc anode ($$\ce{Zn}$$) and a copper cathode ($$\ce{Cu}$$). When zinc $$\ce{Zn}$$ is in contact with water, it "prefers" being transformed into the cation $$\ce{Zn^{2+}}$$ in order to be dissolved in water. In your language, you say that they are more "comfortable" in water. Of course these new ions would repel one another in solution, and prevent new zinc atoms to produce the same operation. The reaction could not take place with more than $$1$$ or $$2$$ ions in solution. But it seems to me that you forget what happens in solution. These newly created $$\ce{Zn^{2+}}$$ ions are attracting negative ions from where they are in excess, for example in the cathode region, where they are repelled. Matter of fact, if the cathode zone contains a solution of $$\ce{CuSO4}$$ where $$\ce{Cu^2+}$$ ions are discharged by the arriving electrons, the remaining sulfate ions $$\ce{SO4^{2+}}$$ are repelled by these same arriving electrons, and are attracted by the anode zone. As a result, these "unwanted" $$\ce{SO4^{2+}}$$ cross the membrane between the two electrodes, so as to neutralize electrically the newly created $$\ce{Zn^{2+}}$$ ions in the anode zone. Is it what you expected to hear ?