My understanding is that electrolytes conduct electricity by:

1) movement of ions, the negative ions migrate to the positive electrode and vice versa. This is however not enough, if these ions just remain ions, the migration and separation of charges will create a potential difference in the solution counteracting the applied potential and at that moment migration of ions will stop. This means no more current will flow.

So therefore:

2) For current to keep flowing the ions must undergo electrolysis at the electrodes to form their neutral forms. In this way their is no build-up of counterpotential, ions keep flowing and the electrons exchanged in the electrolysis can flow from one electrode to the other. (This argument is often given as the reason electrolytes are conductive when I was doing an internet search, also on this site)

Now if we have say a NaCl solution, this would mean that for electrolysis Na+ needs to be converted in to Na, and Cl- into Cl2-gas. The potential needed for the electrolysis of NaCl as calculated from my tables is more than 4V.

So does this mean that if the potential difference is less than 4 V a salt solution does not conduct electricity, ie has very low or zero conductance? If not, how exactly is the current created?

I've read that conductivity is measured in a way that electrolysis is avoided (alternating voltage, wikipedia), so I guess then only the migration of ions is accounted for, but the very definition of conductivity is the ability of a substance to conduct electricity. According to my understanding to maintain a current this requires electrolysis.

  • 1
    $\begingroup$ Na+ doesn't get reduced in solution. $\endgroup$
    – Mithoron
    Commented Nov 12, 2018 at 18:00
  • $\begingroup$ Ok, so say another reduction takes place, and say H2 is formed, the argument remains the same, you'd need a certain voltage to maintain the current, wouldn't you? $\endgroup$
    – Stikke
    Commented Nov 12, 2018 at 19:36
  • $\begingroup$ Lower potential makes reaction's K lower then 1 - still non-zero. $\endgroup$
    – Mithoron
    Commented Nov 12, 2018 at 19:39
  • $\begingroup$ Yes, in order to have a continuous current flow there must be both an oxidation and a reduction reaction. $\endgroup$
    – MaxW
    Commented Nov 12, 2018 at 19:55
  • 1
    $\begingroup$ Ok, I should have said: Yes, in order to have a continuous direct current flow there must be both an oxidation and a reduction reaction. // A conductivity meter uses a low voltage and AC current so that it is measuring the "capacitance" of a solution. If a low DC voltage was used the electrodes would charge up and the current flow would stop, just like charging a capacitor. $\endgroup$
    – MaxW
    Commented Nov 12, 2018 at 21:07

1 Answer 1


Under low potential differences, electrolytes conducts electricity in a similar way that they does at high potentials differences. I think, after reading the question and the comments, that the key point to understand this is: electric potential difference between which spatial points?

The conductivity of a solution is measure of its ability to conduct electricity when the solution is under a difference of electric potential. That is: the solution must be under the influence of an electric field (remember that $\boldsymbol E = - \nabla \boldsymbol \phi$) for having a non zero flux of charges.

You have to look the potential differences inside the solution and not at the electrodes.

You are right in that in order to maintain for long time a continuous current electrolysis is required (at least for simple experimental devices). If there is not electrolysis you won't see a current. However, this is not because the solution (lets say the bulk, that is where the solution as such is (homogeneous)) does not have the ability of conduct electricity, but because there would not be a net external electric field at all in that region.

In short, the conductivity of the solution is not dependent of your ability for creating an external electric field on it.


But then again, what if the potential difference passes a treshold and at a certain point allows electrolysis, wouldn't the conductivity increase very steeply at that point?

If the potential difference between electrodes is low enough to ignore electrolysis, then there will be a charge accumulation in their surroundings that generates a contrary effect in the potential of the solution "far" from electrodes. So, the mean electric field in the bulk will be null.

According to Ohm's law:

$$ \boldsymbol J = \sigma \, \boldsymbol E$$

For the reasons stated above: $\boldsymbol E = \boldsymbol 0$, and so $\boldsymbol J = \boldsymbol 0$. What is the value of $\sigma$ for the solution? As any value satisfies the equation: We can say nothing from Ohm's law ! (technically, it can be computed even if no electric field is present, but it is not important here).

If the potential allows electrolysis, then a current appears so in principle you can measure the conductivity of the solution using Ohm's law. Notice that it is not an step function the current vs. potential. Due to this, there is not a truly step, but it seems because the growing is almost exponential, so we do not notice the increment till we notice a big increment. You can check out the Butler-Volmer equation for more details.

Once the current due to migration (allowed by electrolysis) is measurable, you could measure the solution conductivity. BUT it is a very bad idea, as normally the solutes concentrations are very small and the process of electrolysis change them! (and of course, many times (but not always) we want to know the conductivity to estimate concentrations)

  • $\begingroup$ But then again, what if the potential difference passes a treshold and at a certain point allows electrolysis, wouldn't the conductivity increase very steeply at that point? $\endgroup$
    – Stikke
    Commented Nov 15, 2018 at 13:52
  • $\begingroup$ No, I am going to edit the answer to clarify that when I finish something that I have to resolve right now. $\endgroup$ Commented Nov 15, 2018 at 13:58
  • $\begingroup$ @Stikke I added more details, I hope them help $\endgroup$ Commented Nov 15, 2018 at 22:58

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