Let's say we have a stable solution of $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$ and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$ , with their respective counter-ions.

The associated redox reactions are:



where $\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}<\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$ , since no reaction between $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$ and $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$ is occurring.

The questions are:

  1. Provided that neither $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$ nor $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$ are present in the solution, how can we determine, form the Nernst equation, the $\mathcal{E}$ needed to start the electrolysis process $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$?
  2. In other words, how can the Nernst equation deal with activities that are equal to zero?

Thanks in advance!

  • 3
    $\begingroup$ Interesting question. At first glance this seems to involve taking the natural logarithm of inf or 0, since the Nernst equation involves a ln[red]/[ox] term. $\endgroup$
    – t.c
    Oct 7, 2014 at 16:07
  • $\begingroup$ I don't know that the Nernst equation can handle this kind of starting condition. That said, I would try to approach it from the point of view of Gibb's free energy, with $\Delta G = -nFE$, if possible. $\endgroup$ Oct 7, 2014 at 18:02
  • $\begingroup$ I don't get it completely... As far as I understand, you want to "do electrolysis" on a solution that does not contain the ions you need? $\endgroup$
    – tschoppi
    Oct 7, 2014 at 20:19
  • $\begingroup$ @tschoppi: Only the product ions are absent. So, in the Nernst eq. their activity would be zero. $\endgroup$ Oct 7, 2014 at 20:26
  • $\begingroup$ @DavideLaVardera Yes, everything starts to be clear now. $\endgroup$
    – tschoppi
    Oct 7, 2014 at 20:34

2 Answers 2


The Nernst equation only technically applies when the system is in electrochemical equilibrium—when there's no net current flow, but there's actually an exchange current. A system in which the reaction can only happen in one direction because there are no products wouldn't apply. Source: Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.

What I'm not sure about is the validity of ∆G values calculated from $∆G=∆G°+RT\ln Q$ at the limits where the system is all products or all reactants. ±∞ don't seem to be meaningful values, but the system should quickly move away from these boundary values, even if only by a small amount. These chemical thermodynamics methods are well-behaved when near equilibrium, but well away from equilibrium, they can produce invalid results.


$$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}+\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$$ If you don't have $\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}$ or $\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}$ you must be having atleast $\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}$ or $\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}$, the reactants to produce the products when there is no product the case is: $$\mathcal{E}=\mathcal{E}^\circ-\frac{\mathcal{R}T}{\mathcal{\nu F}}\ln{\frac{[\mathrm{A}_{\left(\mathrm{aq}\right)}^{ +}][\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}]}{[\mathrm{A}_{\left(\mathrm{aq}\right)}^{ 2+}][\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}]}}$$ when the $\ln$ term is zero, the $\mathcal{E}\to\infty$ implying an instantaneous reaction which requires no external force, also implies by the Le'Chatelirer's principe but note that when even a slightest amount of product is created the tendency to procees forward decreases until it reaches the equilibrium where the reaction attains the equilibrium.


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.