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Thank you very much!

This is a very good question. First, I will say that I think most probably every single electrode, even those with what appear to have relatively uniform pore structures, are in reality more like what you described. That is, probably every porous electrode has a distribution of pore sizes and structures even if the average size is pretty close to being the same. So no matter what, everything is likely as you describe it, just to be clear. Following my above hypothesis, I would propose that the use of a single Warburg short is still probably reasonable in most cases because it is capturing some kind of average or aggregate finite-length diffusional behavior, and I would also propose that even in the case you wanted to construct a circuit with multiple Warburg short elements to capture the variety of pore sizes present (and, thus, the variety of diffusional behaviors going on), this would not be easy at all for a couple reasons. Firstly, the software probably cannot easily differentiate between these phenomena as you are envisioning as the conscious, intelligent researcher. The software fitting algorithm is not "intelligent" to the effect that it cannot know what should be applied to what. You would likely have to manually try to adjust parameters, which would make it very fraught. Additionally, it would be computationally very complex and might bog down the software itself. And secondly, even trying to apply multiple Warburg short elements implies your EIS data is sensitive and accurate enough that it is capturing the behavior from individual pore sizes and structures. To be honest, I highly doubt whether these kinds of processes would all be accurately captured in a single EIS test such that they could be analyzed via circuit fitting. Apologies for the lengthy reply, but I really do think this is a good question! Finally, the last thing I will say on this matter is that in the case you described, it may be more applicable to apply a Transmission Line model instead of a Warburg. The Transmission Line model is designed to capture a distribution of charge effective throughout a porous electrode microstructure. This is not exactly the same perhaps as a distribution in the microstructure and pore size itself, but the infinite sum structure of the Transmission Line element might be more ideal to capture this kind of heterogeneity in your porous electrode. If you want to have some more live discussion about porous electrodes and these kinds of circuit fitting models, I am happy to do so on one of our Livestreams we do every Friday at 1pm EST. Join me (Neil) then if you like and I can talk more about this.

​@@Pineresearch Thank you, I really appreciate the effort you put into your answer! This is way more detailed than I would have ever imagined.

@@lucaschuster2824 My pleasure. Thanks for commenting on our video, and feel free to keep in touch with us here on KZfaq, we enjoy engaging with the community on electrochemistry discussions!

This is a good question. First, I will just tell you I have added this question to be answered during our next livestream, which as of this comment will be Friday, April 19 2024 at 1pm EST. I encourage you to join or watch the replay (it will be episode #49 of Ask Us Anything About Electrochemistry). As far as a quick answer, it honestly depends. For a multi-layered electrode, theoretically every single interface (substrate/film, film 1/film 2, film 2/electrolyte, etc) can result in a unique impedance response. But this is not a guarantee. And as it relates to Warburg diffusional impedance, it is just tricky because you may not know what is diffusing, how far, and into which layers. For instance, there is likely diffusion in the electrolyte, but is there also diffusion through the outermost layer? The second layer too? Every layer? And you may only be able to resolve each of these phenomena if they happen to occur at different timescales, which is also not a guarantee and if this is the case, you might get sort of an aggregate diffusional measurement that cannot be easily resolved into the component parts specific to each layer in your electrode structure.

@@Pineresearch I will definitely catch the livestream. I would like to expand on my query (just to clarify it better with regard to the circuit). If, in a substrate/film1/film2/electrolyte multilayer coating scenario, the EIS circuit is given as R1+Q1/R2+Q2/R3, would the Warburg element be included so as to reframe the circuit as R1+Q1/(R2+W1)+Q2/R3 or as R1+Q1/(R2+W1)+Q2/(R3+W2)? Thank you for explaining the possible multi-diffusion happening across the layers, never thought about that. I will definitely catch the stream and am eager to learn more. Thank you again.

@@devxpentatom7932 Thanks for the follow up. I understand your question. In short, I would not anticipate adding a separate Warburg embedded within BOTH Randles interfaces to be accurate (or even likely to be physically occurring). Additionally, I think you're missing a Randles element representative of the last film/electrolyte interface. I will try to illustrate this tomorrow during the livestream for you.

Most probably, there is no easy way to definitively confirm whether the line would continue at 45° (infinite Warburg), curl down after the initial 45° (finite Warburg), or make an actual semicircle (Randles element). You would need to continue the EIS experiment to lower frequencies (and hope diffusion/drift doesn't cause instability to make the measurements invalid). I wouldn't say there's a "danger" in assuming an infinite Warburg, that might be too strong. Some of your assessment should be based on knowledge of the physical setup. If you have a small electrode with plenty of electrolyte, and not a rotating electrode or a microelectrode, it might be safe enough to assume the diffusion layer thickness is large in comparison and an infinite Warburg is valid.

I'm glad that you are enjoying the videos. That section of the Warburg coefficient has to do with the contribution to the Warburg impedance from both diffusion of C^*_A and C^*_B. However, when doing EIS you are probably applying a sinewave around a potential where A + e --> B, which means the current response (and hence the impedance response) is due to only the contribution of C^*_A and no contribution of C^*_B. Which is why we can eliminate the end term in the Warburg coefficient. Does that make sense?