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In electrochemical impedance spectroscopy experiments (EIS) the diffusion of electroactive species at flat, macroscopic electrodes is commonly described by the Warburg element with the impedance expression

$$Z_W = \frac{σ}{\sqrt{w}} - i\frac{σ}{\sqrt{w}}$$

This means that the values have a phase angle of $-45^\circ$.

What is an intuitive way to understand the fact, that the Nernst-diffusion layer contributes both a real and imaginary part to the impedance of the system?

I find that it is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. This would explain the real part. I am unsure however, where the imaginary part, or energy-storing component, fits into this picture.

It is very hard to find any information in this regard, as practically all literature treats this just as a result of the mathematical derivation of the faradaic impedance, but does not go into any kind of intuitive explanation.

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  • $\begingroup$ Could it be the double layer capacitance that is responsible for the imaginary part? Impedance of capacitors decreases with frequency just as the 2nd term in the given equation $\endgroup$ – ChemEng Oct 16 '19 at 22:40
  • $\begingroup$ @ChemEng Thanks for the idea, but no, the double-layer is modeled separately. A simple model for an electrochemical reaction is the Randles circuit, which places the double-layer capacitance in parallel with the Faradaic impedance (series connection of charge transfer resistor and Warburg element). $\endgroup$ – Jens Oct 17 '19 at 7:12
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I am sorry, this is not an answer but a comment, too long to be edited in the comments section.

I don't agree with "straightforward" in the sentence : " It is relatively straightforward to understand that the more the diffusion layer grows, the shallower the concentration gradient gets and therefore that the current decreases. "

This would be an oversimplified view which implicitly considers that the phenomena involved are only on the energy dissipative kind (commonly modelized with resistors elements in the equivalent circuits). This cannot be so in case of Warburg impedance. Necessarily the phenomena involved includes electrical energy storage elements which are stocking energy during a part of the period and destocking during the other part of the period of the alternative current (commonly modelized with capacitor elements). One cannot forget the part of the alternative current due to the phenomena of energy storage kind. If we forget that, there is no phase shift and no complex impedance, thus no Warburg element.

In an intuitive way, but discutable because alternative models can be proposed, if we think that the more the diffusion layer grows, the shallower the concentration gradient gets also the shallower the electrical gradient of potential gets. So, the energy storage phenomena also decreases and the "out of phase" part of alternative current decreases as well. This is a loosely way for an intuitive approach.

The case of Warburg element (which is a particular case of CPE : http://www.consultrsr.net/resources/eis/cpe1.htm) can be related to dissipative energy phenomena and storage energy phenomena $\textbf{homogeneously}$ distributed. A possible equivalent circuit can be made of repeated identical pattern of resistor-capacitor. This is a particular case of network for example of the kind represented page 10, figure 7(a) in the paper : https://fr.scribd.com/doc/71923015/The-Phasance-Concept , only with the part noted $P_\varphi$, without $R_p$. Of course different interpretations and equivalent circuits can be proposed, depending on the kind of electrochemical system. This is briefly discussed page 26 in the above referenced paper.

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  • $\begingroup$ Thanks for your input JJacquelin. I understand that you need to have energy storing components to get the Warburg behavior and that there are correlations with e.g. Transmission Line Models (especially for bounded diffusion). I'll have to think about what you wrote in regards to the potential gradient and will also take a look at the paper. Thanks. $\endgroup$ – Jens Oct 5 '19 at 8:27
  • $\begingroup$ A small remark : The models cited are not Transmission Line Models which supposes four terminals, two for the input and two for the output. They have only two electrical terminals. This is a bipolar component in which the whole electric current is passed through it $\endgroup$ – JJacquelin Oct 5 '19 at 9:37

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