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 '14 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. $\endgroup$ – Jason Patterson Oct 7 '14 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 '14 at 20:19
  • $\begingroup$ @tschoppi: Only the product ions are absent. So, in the Nernst eq. their activity would be zero. $\endgroup$ – Davide La Vardera Oct 7 '14 at 20:26
  • $\begingroup$ @DavideLaVardera Yes, everything starts to be clear now. $\endgroup$ – tschoppi Oct 7 '14 at 20:34

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.


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