# Nernst equation and electrolysis

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:

$\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{A}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{A}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$

$\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}+\mathrm{e}^{-}\rightleftharpoons\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}\qquad\mathcal{E}_{\mathrm{C}_{\left(\mathrm{aq}\right)}^{2+}/\mathrm{C}_{\left(\mathrm{aq}\right)}^{+}}^{\circ}$

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?

• 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. – t.c Oct 7 '14 at 16:07
• 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. – Jason Patterson Oct 7 '14 at 18:02
• 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? – tschoppi Oct 7 '14 at 20:19
• @tschoppi: Only the product ions are absent. So, in the Nernst eq. their activity would be zero. – Davide La Vardera Oct 7 '14 at 20:26
• @DavideLaVardera Yes, everything starts to be clear now. – tschoppi Oct 7 '14 at 20:34

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.