![Picture from [1]](https://s3-us-west-2.amazonaws.com/secure.notion-static.com/508f05e4-efa5-4054-80c3-920b0504cb56/Untitled.png)
Part of Electrode lithiation stages.
When looking at the open-circuit potential graph like the one below, it is very easy to fall into the trap of thinking that the plateau voltage levels (the rightmost plateau is at about 0.09V vs. Li/Li+ and the second plateau is at about 0.13V) correspond to the chemical potentials of Lithium intercalation at stage I and stage II, respectively: $E_i = \mu_i/e$, where $e$ is an elementary charge (of a Li+ ion), $\mu_i$ is the chemical potential of intercalation at a certain stage, and $E_i$ is a plateau voltage. But this is wrong!
Plateau labels "Stage-I", "Stage-II", "Stage-III" on the picture below also nudge towards this confusion:
Picture from [1]
Moving an electrode particle from full stage II to full stage I does not amount to just some intercalating Lithium atoms while expending $\mu_1$ energy on each, where $\mu_1$ is the chemical potential of intercalation at stage I. Analytically, it's rather like removing all ions at stage II from the particle, then pushing them back again into the particle at stage I, and only after that adding the remaining Lithium ions at stage I.
If $N$ is the total number of lattice slots for Lithium in a graphite particle to 100% stoichiometry, then the total energy expenditure for moving a particle from 50% to 100% stoichiometry is approximately $-0.5N\mu_2 + 0.5N\mu_1 + 0.5N\mu_1 = N\mu_1 - 0.5N\mu_2$. Since the total number of Lithium atoms intercalated during this process is $0.5N$, then the voltage of the rightmost plateau is approximately $E_1 = (N\mu_1 - 0.5N\mu_2)/0.5N/e = (2\mu_1 - \mu_2)/e$. Similar reasoning provides that the second from the right plateau's voltage should be approximately $E_2 = (3\mu_2 - 2\mu_3)/e$, and the third plateau's voltage is $E_3 = (4\mu_3 - 3\mu_2)/e$. Compare these results with the incorrect formula given at the beginning of this page.
Yet another confusing factor is that at certain stoichiometry (lithiation) ranges, the surface of electrode particles is at a certain stage, for example, stage I ($\mathrm{LiC_{6}}$) on the picture below. Vacancies appear in the stage structure only occasionally. These vacancies (i. e., quasiparticles) move outside-in the electrode during lithium de-intercalation and inside-out the electrode during lithium intercalation. Within the stoichiometry range corresponding to some voltage plateau, the change in the electrode stoichiometry only affects the position of the stage boundary within the particle, but not the lithiation stage at the surface of the particle.
The chemical energy expended to intercalate Lithium atoms on the surface is $\mu_1$ on average, but the plateau's voltage also "accounts for" the phase transition happening within the particle.
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These phase transitions are quite similar to the classical phase transitions such as ice melting when the temperature of the liquid and ice mixture stays at the melting point as long as there is some ice left to melt (assuming a perfectly stirred mixture).
Note that we don't need to consider Lithium atom over-concentration responsible for the Cell diffusion voltage component of the cell voltage because we are analysing equilibrium (half-)cell open-circuit voltage functions here.
The width of plateaus decreases roughly logarithmically towards lower stoichiometry because a transition of all graphite between $\mathrm{LiC_{18}}$ and $\mathrm{LiC_{12}}$ stages, for instance, takes 3 times fewer Lithium ions than a transition between $\mathrm{LiC_{12}}$ and $\mathrm{LiC_{6}}$ stages. This is also one of the reasons why Electrode voltage curves are steeper when the electrode has little lithium. However, the $\mathrm{LiC_{6}}$ stage is not three times wider than $\mathrm{LiC_{12}}$ but less than that because what is actually happening is more complicated than what is described above (see [1] for details).
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