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?


1 Answer 1


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 ?

  • $\begingroup$ Ah, then the reasoning in my post is incorrect. If there is no "cost" for electrons to migrate from anode to cathode (in the form of increased repulsion between zinc cations) why does reaction quotient matter while calculating electrode potential? I know that the straightforward answer is that it is because the Nernst equation is just the Gibbs-Helmhotz in disguise, but I want to get some intuition. $\endgroup$ Jul 17, 2021 at 9:14
  • $\begingroup$ @Ray Bradbury. Why do you want more than Nernst equation ? $\endgroup$
    – Maurice
    Jul 17, 2021 at 13:58
  • $\begingroup$ I figured it out myself, and it was elementary. Obviously electrons will go where there are more positively charged particles (cations), and obviously they will move faster if there's a great difference between cation concentrations. If there are much more cations in the anode than in the cathode, the electrons won't ever want to leave. But thank you for your answer, you pointed out a pretty significant error in my thinking. $\endgroup$ Jul 17, 2021 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.