While studying electrochemistry, I came across two key points that I'm unable to understand.

why does DC alone break down the electrolytic liquid

and b) Why doesn't AC do the same?

  • $\begingroup$ Could you explain what you mean by "break down the electrolytic liquid" $\endgroup$ Sep 29, 2022 at 4:18
  • $\begingroup$ My book says "it alters the composition". SO I'm going to assume it means ionisation $\endgroup$ Sep 29, 2022 at 4:32
  • 1
    $\begingroup$ The assumption that AC voltages do not lead to electrolytic effects is false, but the effects are non-linear. Interest in AC electrolysis is also not new, see e.g. "THE ALTERNATING CURRENT ELECTROLYSIS OF WATER", J. W. Shipley, Canadian Journal of Research, October 1929. If I remember correctly for small to intermediate voltages the theory predicts the formation of a charged double layer on the electrode materials which has to be dissolved first by the current, which takes time, hence the frequency dependence. $\endgroup$ Sep 29, 2022 at 4:33
  • $\begingroup$ I see, thanks ! $\endgroup$ Sep 29, 2022 at 4:39

1 Answer 1


I'm going to assume breakdown means the reassociation of the dissolved analyte in solution. Suppose the dissolved salt is potassium. The redox potential for K is:

K+ + e− ⇌ K(s) at -2.93 V

Any voltage less than -2.93 V will create solid potassium at the electrode and reduce the number of K ions near the surface. This creates a concentration gradient at the electrode surface. The concentration gradient promotes diffusion of K ions to the electrode surface for reduction to a solid. Since K is conductive, over time, all the K ions will be reduced leaving only water.

Conversely, any voltage greater than -2.93 V oxidizes any K on the electrode surface back into solution and changes the concentration gradient by increasing the ions at the electrode surface.

If an AC voltage is introduced where the voltage oscillates about the redox potential, the result is oxidation and reduction about the electrode and no net change in ion concentration at the electrode surface.

  • 1
    $\begingroup$ But this would only be valid if the reaction rates were identical (they are not because the current density changes) and if the reduce/oxidized compounds would not be removed from the back-reaction, which in case of gas release during water electrolysis is the case. Once molecular hydrogen forms, it will not participate in the reactions any longer, for instance. $\endgroup$ Sep 29, 2022 at 5:20
  • 1
    $\begingroup$ @FlatterMann You are correct. My example ignores hydrolysis or hydrogen overpotential. I should have chosen an analyte whose redox potential is greater than -1.29 V for my example. I should also have mentioned my example is only valid for reversible reactions whose reaction rate is faster than the applied frequency. Also my electrodes are chemically inert and only a source/sink of electrons. $\endgroup$ Sep 29, 2022 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.