I'm trying to understand how/if the Warburg impedance, $Z_w$, of an electrode-electrolyte interface dominates the impedance at low frequencies. I cannot understand how this model can predict a DC current flow.

A commonly used electrical model of a Faradic electrode is from e.g. 1: Interface with charge transfer resistance R_ct in series with Warburg impedance Z_w

where $R_{ct}$ is the charge transfer resistance, $C_d$ is double layer capacitance and $R_b$ is the bulk resistance.

I found an approximation of $Z_w$ here.

$Z_w = \sigma/\omega^{1/2} - j\sigma/\omega^{1/2} \\ |Z_w| = \sqrt{2}\sigma/\omega^{1/2}$

According to this, the Warburg impedance approaches infinity as frequency approaches zero. This means that in the model in ref 1, there would be no DC current flow. Yet in ref 1 they say "At low frequencies, the effect of Warburg impedance can be neglected and the element in series with the diffusion impedance is merely $R_{ct}$..."

This is where I get confused. I would be grateful for any help/guidance on this.

1 Comprehensive study of noise processes in electrode electrolyte interfaces Arjang Hassibi,a) Reza Navid, Robert W. Dutton, and Thomas H. Lee, Journal of Applied Physics 96, 1074–1082 (2004)

  • 2
    $\begingroup$ On the face of it, there is the contradiction you have described: the DC limit of the Warburg impedance would be infinite and there would be no DC current flow. Presumably, the approximation of the Warburg impedance is simply not applicable in the neighborhood of zero Hz. You might be better off asking this at the electrical engineering stack exchange and providing more information about the circuit model(s) used for Warburg impedances. The EE folks are great at doing circuit modeling and can make short work of this if you meet them half way. Good luck! $\endgroup$
    – Ed V
    May 13, 2023 at 11:32
  • $\begingroup$ I removed the link you provided to the equations as it currently leads to a strange page (surfing site?). The equations are available at the Wikipedia. $\endgroup$
    – Buck Thorn
    Jan 20 at 8:01

1 Answer 1


As Ed V mentioned, the Warburg approximation may be not applicable. This can be partially correct.

At least, it is worth to know that even for the Warburg element you should specify its type. You can find that there are 3 types of Warburg element:

  1. Warburg 0.5-CPE: semi-infinite diffusion element; (I like to consider it as the constant phase element with an exponent n)
  2. Warburg Short: finite-length diffusion element with transmissive boundary;
  3. Warburg Open: finite-length diffusion element with reflective boundary.


For the 2nd type (finite-length diffusion element with transmissive boundary), the impedance value is described by the equation $$ {\displaystyle Z_{W_{\mathrm {S} }}={\frac {A_{\mathrm {W} }}{\sqrt {j\omega }}}\tanh \left(B{\sqrt {j\omega }}\right)} $$ Specifically this type does have a typical 45° slope at high frequencies in Nyquist plot, but it gradually transforms into an arc. You can find an image describing it in the article Electrochemical Impedance Spectroscopy─A Tutorial, Alexandros Ch. Lazanas and Mamas I. Prodromidis (open access). https://pubs.acs.org/doi/10.1021/acsmeasuresciau.2c00070

So, coming back to the equation, you can calculate that at low frequencies you get a real number: $$ {\displaystyle \lim_{\omega \to 0} Z_{W_{\mathrm {S} }}=\lim_{\omega \to 0} {\frac {A_{\mathrm {W} }}{\sqrt {j\omega }}}\tanh \left(B{\sqrt {j\omega }}\right) = A_{\mathrm {W} }B } $$ Therefore, in the scape of your concern, I prefer to consider that the authors assumed that $A_{\mathrm {W} }B \ll R_{ct}$, i. e., the effect of the Warburg impedance (Warburg Short element) is neglected at low frequencies.

  • $\begingroup$ (+1) Informative contribution! Welcome to the Chemistry Stack Exchange! $\endgroup$
    – Ed V
    Jan 13 at 0:18
  • $\begingroup$ This answers my question. Thanks for the reference, I will study it. $\endgroup$
    – bes
    Jan 20 at 7:26

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