First, we discuss the chemical potential of “Li”, or Li ions in the positive electrode. The “reaction” in Li-ion batteries consists of inserting Li atoms into a host lattice, in this case CoO2. Therefore we want to investigate the Gibbs free energy per lattice site (specific Gibbs free energy, Gˉ). This will depend on the composition X of Li in LiXCoO2, where X is the fraction of lithium sites that are occupied. The X is related, but not exactly equal, to the state of charge of the battery. For positive electrodes, an X of 1 usually indicates a fully discharged battery. That being said, many battery electrodes cannot be cycled stably from X=0 to X=1. Thus in practice, a state of X=0.5 may indicate that a battery is fully charged.
At constant temperature and pressure the chemical potential is defined as
μ=(∂N∂G)T,P where N is the number of Li atoms. Writing the composition as X=N/J, with J the number of “lattice sites” within a CoO2 host lattice, we have
μ=∂N∂G×(1/J1/J)=∂X∂Gˉ Then, in integral form,
Gˉ=μX(4.9) For a completely filled site (X=1) the specific Gibbs free energy per filled lattice site should equal the chemical potential of a purely lithiated (LCO) lattice, μLCO0, while for an empty site (X=0) it equals the chemical potential of the empty host, μCO0. A simple weighted-average model gives
Gˉ“Li”=μLCO0X+μCO0(1−X)(4.10) or, rearranged,
Gˉ“Li”=(μLCO0−μCO0)X+μCO0(4.11) However, because G=H−TS and there is a configurational entropy associated with mixing Li in the host lattice, the dependence on X is modified. In fact, one writes
Gˉ“Li”=Gˉunmixed+ΔGˉmix(4.12) with
ΔGˉmix=−TΔSˉmix(4.13) For a lattice gas the entropy of mixing is given by
ΔSˉmix=−R{XlnX+(1−X)ln(1−X)}(4.19) Thus, the full expression for the specific Gibbs free energy in the positive electrode becomes
Gˉ“Li”=(μLCO0−μCO0)X+μCO0+RT{XlnX+(1−X)ln(1−X)}(4.20) Graphically (see Figure 4.8), the instantaneous chemical potential of Li in the positive electrode is the derivative of Gˉ“Li” with respect to X, yielding
μ“Li”=μ“Li”0+RTln1−XX(4.22) where
μ“Li”0=μLCO0−μCO0