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Related (very similar, but here I want a mechanism) https://physics.stackexchange.com/q/21827/7433

By the Kohlrausch law, all ions contribute to the conductivity of an electrolyte.

Now, as I understand it, the mechanism of conduction in an electrolyte is thus:

  • Ions migrate in solution
  • These ions get reduced or oxidized at the electrodes and converted to electrons
  • These electrons continue down the wire, leading to an increased/maintained conductivity/current

But, this mechanism doesn't work for ions which do not get redoxed--movement of ions in the solution cannot be translated to movement of electrons in the wire and thus it seems (to me) that conductivity should not increase.

But ions like $\ce{NO3-}$ have a comparable $\lambda$ (ionic molar conductivity)--so they experimentally do increase conductivity.

What is the mechanism for conduction via these ions?

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  • $\begingroup$ Thermodynamically nitrogen dioxide is more accessible from nitrate than water is from dioxygen. I think many conductivity meters apply a potential that would make this chemistry possible. This is a separate issue than the interdependence of ionic drifts. If you would like, I can talk about the latter. $\endgroup$
    – Chris
    May 7, 2012 at 2:08
  • $\begingroup$ @Chris you could, but that would make $\lambda_{\ce{NO2}}$ dependant on the potential, wouldn't it? $\endgroup$
    – ManishEarth
    May 7, 2012 at 2:34
  • $\begingroup$ Experimentally, how do you determine the the conductivity? What physically happens during this measurement? $\endgroup$
    – Chris
    May 7, 2012 at 5:23
  • $\begingroup$ @Chris: Run current, measure voltage/current, use $V=IR$, use the dimensions of cell to get conductivity from conductance $\endgroup$
    – ManishEarth
    May 7, 2012 at 5:29
  • $\begingroup$ Right. A solution of 0.5 $\mu$ moles of acetic acid will require ~4 V potential drop across the electrodes to give 20 $\mu$A. An even smaller concentration will result in an even smaller current at that potential, requiring even greater sensitivity/more expensive hardware. $\endgroup$
    – Chris
    May 7, 2012 at 6:48

1 Answer 1

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This can arise due to ion-ion interactions brought on by Coulombic fields.

A proper explanation is rooted in transport phenomena. Unfortunately chemists are largely not taught this, but chemical engineers do get the opportunity.

Under standard conditions, ions in a solution will take a random walk during diffusion. This results in no net movement of our ions over some span of time. Such a system perturbed by an applied electric field will cause ions to preferentially move in one direction, we say the ions will drift. We say they have a drift velocity. These velocities are affected by a number of parameters, as there is still continual collision, but we can say that the mobilities will be unequal for ions of unequal sizes.

Different mobilities mean concentration gradients form and charge separation too. Typically we use an electroneutrality field as a reasonable approximation to allow analytic solutions for various equations. This field depends on all the ionic fluxes present and as such will affect the total measurable current.

The result is simple: the introduction of any ion will cause a perturbation in the field that all the ions encounter in solution.

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  • $\begingroup$ But how does this translate to electrons in the wire? Or does the ion "pull and hold" the electron in place (basically the Coulomb force bring some extra electrons to the electrode as a once-per-ion effect?) Transfer phenomena sound interesting, thanks for the rest! $\endgroup$
    – ManishEarth
    May 7, 2012 at 17:33
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    $\begingroup$ Oh I see.. You're saying that the nonredox ions affect the field and thus improve mobilities-->conductance. $\endgroup$
    – ManishEarth
    May 7, 2012 at 18:32
  • $\begingroup$ This is a good answer (upvoted!) but is very "atom-focused" instead of "field-focused". Maybe it's a chemist vs. physicist thing or something. Anyway, ions diffusing while under the influence of electric fields are governed by the Nernst-Planck equation. This is true of any ion, whether or not they participate in chemical reactions at electrode surfaces. Electric fields that emanate from the electrodes of an electrochemical cell affect the behavior of every charged particle (i.e. ion) that's nearby. $\endgroup$
    – Curt F.
    Apr 9, 2015 at 3:24

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