I'm not sure if my understanding is correct, please correct me if there is something wrong.

Considering a 2 electrodes system with a constant voltage applied between both electrodes and with no oxydoreduction happening, only some initial electromigration of the ions in solution under the effect of the electric field. As long as the electrodes are not completely covered with ions with a certain thickness, the potential of the electrodes is not screened and ions in the bulk will continue to migrate to the electrodes? And when finally there are enough ions to screen them, all the remaining ions in the bulk are staying in the bulk?

And assuming everything above is correct, I am running an experiment where the ion concentration is really low and I suspect there is no ion in the bulk, is it experimentally provable somehow?

  • 2
    $\begingroup$ You do not consider the effect of the counter-potential gradient due the net charge offset within the electrolyte. Electromigration of ions would quickly slow down and finally stop, as the potential gradient would converge to zero. $\endgroup$
    – Poutnik
    Commented Apr 18 at 13:42

1 Answer 1


If we consider no Faradaic reactions happening on either electrode then all you will have is a capacitor.

Imagine you have some mobile ions in solution, then these ions will feel an electric field and start to migrate. There will be an opposite driving force of diffusion against the concentration gradient that has built up (on the negatively polarized electrode positive ions will accumulate and vice versa on the positively polarized electrode). This is what we call an electrical double layer, some charge on the electrode which is compensated by mobile ions in the electrolyte. At some point there will be a balance between the driving force of diffusion, the charge separation creating an electric field countering the electric field you applied, and the applied electric field leading to no net movement of the ions. The total electric field will not be constant between the electrodes as near the electrodes significant charge separation takes place which will strongly counteract the applied electric field. Making the electric field in between the electrodes outside the double-layer region near 0. The characteristic time to reach this situation is the same as for a normal capacitor.

One thing to consider is how strong all of these effects are, charge separation at a large distance generates an incredibly strong electric field: Poisson's equation: $\nabla\cdot \vec{E}=\frac{\rho}{\epsilon}$, with $\vec{E}$ the electric field, $\rho$ the charge density in C/m³ (to convert from moles to C use Faraday's constant 96485 C/mol), and $\epsilon$ the permittivity of the medium. Try plugging in µM concentration differences over cms and you will get very high voltages. So when you say the ion concentration in the bulk will be 0, it is very unlikely unless your concentration is very small and the distance between the electrodes is very short.

If you would like to know if ions are remaining in between the electrodes it depends on what accuracy you need. You can simply sample some liquid and measure the conductivity, but depending on your application it might not suffice.

Finally, if you have no ions at all then you will simply have a regular capacitor with a constant electric field between the electrodes.

I hope this answers your question.

  • 1
    $\begingroup$ "charge separation at a large distance generates an incredibly strong electric field". Isn't it the other way around? A short separation of charges at the electrodes suffices to counter the applied field? The closer the charges the harder to keep them apart (larger the attractive force). $\endgroup$
    – Buck Thorn
    Commented Apr 19 at 9:23
  • $\begingroup$ Yes I must've made a mistake whilst writing down my reasoning, thanks for letting me know! $\endgroup$
    – Noah
    Commented Apr 19 at 12:35

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