I am modelling lithium-sulfur batteries ($\ce{LiS}$ batteries) and a strange phenomenom happens during charge, named infinite charging, which I show below:

enter image description here

The factor $f_\mathrm{C}$ that appears in the description doesn't matter, and the compound is one of the electrolytes that are used in the $\ce{LiS}$ battery. As we can see, while the current $I$ goes down, it takes more time to recharge. If low enough, the battery doesn't charge at all, as the voltage enters a plateau and never gets out.

The reaction mechanism in $\ce{LiS}$ batteries is challenging, due to the many species during the electrochemical reactions. We also have disproportionations, precipitations, and dissolutions. However, from a modelling perspective, we can summarize it as follows during discharge.

  1. Positive electrode (cathode): solid sulfur $\ce{S8(s)}$ is dissolved into liquid $\ce{S8(l)}$. After that, a huge number of reductions take place to $\ce{S8^2-}$, $\ce{S7^2-}$, $\ce{S6^2-}$, up to $\ce{S^2-}$. This last one is insoluble in the electrolyte, so $\ce{Li2S(s)}$ appears at the end of the discharge. The charge transfer occurs at the carbon/electrolyte interface. We add carbon, or similar forms, because $\ce{S8(s)}$ is a horrible conductor and charge transfer will not occur over a sulfur/electrolyte interface.
  2. Negative electrode (anode): the lithium is just oxidized $$ \ce{Li(s)} \rightarrow \ce{Li^+(l)} + \ce{e^-}$$ and the charge transfer occurs at the lithium/electrolyte interface. The symbol $\ce{(l)}$ here denotes always the electrolyte in this post, for all species.

During charge, there is a consensus in the community that the reaction mechanism changes. However, we may simplify matters and just say that the reactions are inverted. In addition, we are sure that the shuttle effect takes place, where the infinite charging is one of its consequences. This shuttle effect I also summarize below:

  1. $\ce{Li2S(s)}$ starts dissolving into $\ce{S^2-}$ and $\ce{Li+}$. After that, the oxidations start to take place at the positive electrode, so we start getting $\ce{S_2^2-}$, $\ce{S_3^2-}$, $\ce{S_4^2-}$, $\ce{S_5^2-}$, etc.
  2. Many of these guys, called the polysulfides, suffer electromigration to the negative electrode.
  3. These tranported guys suffer reduction. For example, $\ce{S_5^2-}$ reduces to $\ce{S_4^2-}$, or $\ce{S_7^2-}$ to $\ce{S_6^2-}$, etc.
  4. These reduced polysulfides suffer electromigration to the positive electrode and are re-oxidized.

Funnily, the same electrochemical reactions take place in both electrodes, but in the opposite direction. If we combine these points, it is easy to understand why we obtain the infinite charging in the upper image. The compounds that are being oxidized at the positive electrode, are then reduced in the negative electrode. The net effect is that the charge is just circulating back and forth, and we cannot go back to the starting place. Again, there isn't a consensus about the exact behavior of the whole shuttle effect in the community, but this will suffice.

From a modelling perspective, however, a simple approach may function to capture the infinite charging: once charging initiates, we just add another reaction in the negative electrode, that mimics its counterpart in the positive electrode. For example, while the following happens in the positive electrode (oxidation) $$ \ce{\frac{1}{2} S8^2-(l) \rightarrow \frac{1}{2} S8(l) + e^- (carbon/elyte interface)} $$ we just include the opposite in the negative electrode (reduction) $$ \ce{\frac{1}{2} S8(l) + e^-(lithium/elyte interface) \rightarrow \frac{1}{2} S8^2-(l)} $$

Question It is not important the details of how to mathematically model this. Enough is to say that, in order to implement it, we need to know the standard reduction potential $E^\circ$ of these reactions in order to use Nernst's equation. Since they are the same reaction, we may assume that $E^\circ$ is the same. However, both happen at different interfaces, since one is carbon and the other is lithium. Is there any evidence that a same reaction happening at a different interface may have a different standard reduction potential, or different $E^\circ$? Is it a conceptual mistake to assume that they are different?

If the $E^\circ$ was different in both reactions, that would allow to implement this computationally. If they are equal, it can be shown that many problems happen. I am looking forward to either experimental or theoretical proofs, if possible.

References Image from:

  • Mikhaylik, Yuriy & Akridge, James. (2004). Polysulfide Shuttle Study in the Li/S Battery System. Journal of The Electrochemical Society. 151. A1969-A1976. 10.1149/1.1806394.
  • $\begingroup$ I was curious when you will pull up this topic. :-) Interesting stuff. $\endgroup$
    – Poutnik
    Jul 1, 2023 at 5:06
  • $\begingroup$ You were curious? Oh thanks, Poutnik. Unfortunately, it is heavy stuff for me. Do you think the post is clear? $\endgroup$ Jul 1, 2023 at 12:03


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.