Chemical Potential of "Li" (in Positive Electrode)

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\text{CoO}_2. Therefore we want to investigate the Gibbs free energy per lattice site (specific Gibbs free energy, Gˉ\bar{G}). This will depend on the composition XX of Li in LiXCoO2\text{Li}_X\text{CoO}_2, where XX is the fraction of lithium sites that are occupied. The XX is related, but not exactly equal, to the state of charge of the battery. For positive electrodes, an XX of 1 usually indicates a fully discharged battery. That being said, many battery electrodes cannot be cycled stably from X=0X=0 to X=1X=1. Thus in practice, a state of X=0.5X=0.5 may indicate that a battery is fully charged.

At constant temperature and pressure the chemical potential is defined as

μ=(GN)T,P\mu = \left( \frac{\partial G}{\partial N} \right)_{T, P}

where NN is the number of Li atoms. Writing the composition as X=N/JX = N/J, with JJ the number of “lattice sites” within a CoO2\text{CoO}_2 host lattice, we have

μ=GN×(1/J1/J)=GˉX\mu = \frac{\partial G}{\partial N}\times \left(\frac{1/J}{1/J}\right) = \frac{\partial \bar{G}}{\partial X}

Then, in integral form,

Gˉ=μX(4.9)\bar{G} = \mu X \qquad (4.9)

For a completely filled site (X=1X=1) the specific Gibbs free energy per filled lattice site should equal the chemical potential of a purely lithiated (LCO) lattice, μLCO0\mu^0_{\text{LCO}}, while for an empty site (X=0X=0) it equals the chemical potential of the empty host, μCO0\mu^0_{\text{CO}}. A simple weighted-average model gives

Gˉ“Li”=μLCO0X+μCO0(1X)(4.10)\bar{G}_{\text{“Li”}} = \mu^0_{\text{LCO}}\,X + \mu^0_{\text{CO}}\,(1-X) \qquad (4.10)

or, rearranged,

Gˉ“Li”=(μLCO0μCO0)X+μCO0(4.11)\bar{G}_{\text{“Li”}} = (\mu^0_{\text{LCO}} - \mu^0_{\text{CO}})X + \mu^0_{\text{CO}} \qquad (4.11)

However, because G=HTSG = H - TS and there is a configurational entropy associated with mixing Li in the host lattice, the dependence on XX is modified. In fact, one writes

Gˉ“Li”=Gˉunmixed+ΔGˉmix(4.12)\bar{G}_{\text{“Li”}} = \bar{G}_{\text{unmixed}} + \Delta \bar{G}_{\text{mix}} \qquad (4.12)

with

ΔGˉmix=TΔSˉmix(4.13)\Delta \bar{G}_{\text{mix}} = -T \Delta \bar{S}_{\text{mix}} \qquad (4.13)

For a lattice gas the entropy of mixing is given by

ΔSˉmix=R{XlnX+(1X)ln(1X)}(4.19)\Delta \bar{S}_{\text{mix}} = -R \{ X \ln X + (1-X) \ln (1-X) \} \qquad (4.19)

Thus, the full expression for the specific Gibbs free energy in the positive electrode becomes

Gˉ“Li”=(μLCO0μCO0)X+μCO0+RT{XlnX+(1X)ln(1X)}(4.20)\bar{G}_{\text{“Li”}} = (\mu^0_{\text{LCO}} - \mu^0_{\text{CO}})X + \mu^0_{\text{CO}} + RT \{ X \ln X + (1-X) \ln (1-X) \} \qquad (4.20)

Graphically (see Figure 4.8), the instantaneous chemical potential of Li in the positive electrode is the derivative of Gˉ“Li”\bar{G}_{\text{“Li”}} with respect to XX, yielding

μ“Li”=μ“Li”0+RTlnX1X(4.22)\mu_{\text{“Li”}} = \mu^0_{\text{“Li”}} + RT \ln \frac{X}{1-X} \qquad (4.22)

where

μ“Li”0=μLCO0μCO0\mu^0_{\text{“Li”}} = \mu^0_{\text{LCO}} - \mu^0_{\text{CO}}
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