Voltage of a Battery

A battery is thermodynamically similar to a fuel cell with closed chambers. In a fuel cell, the chambers contain water and oxygen/hydrogen gases, whereas in a Li‑ion battery these gases are replaced with lithium. (Lithium does not easily exist as a gas at ambient temperature and pressure; instead it is stored as a solid or by intercalation into host compounds such as CoO2\text{CoO}_2, FePO4\text{FePO}_4, or graphite.)

In the fuel cell the mobile ion is O2\text{O}^{2-} (which, along with an electron and O2\text{O}_2, forms H2O\text{H}_2\text{O}). In a Li‑ion battery the mobile ion is Li+\text{Li}^+, which reacts with an electron and a metal oxide (e.g. CoO2\text{CoO}_2) to form LiCoO2\text{LiCoO}_2. This battery chemistry is now standard for portable electronics.

Schematic of a Li-ion battery under discharge.

Figure 4.3.1:Schematic of a Li-ion battery under discharge.

Figure 4.3.1 schematically shows a Li‑ion battery with different host materials on the two sides (implying different chemical potentials for Li\text{Li}). Let’s calculate the battery voltage using the same approach as in Module 2. In equilibrium, the two half‑reactions (as well as the mobile Li+\text{Li}^+ ion) must be balanced:

LiN++eN    Li\text{Li}^+_{\text{N}} + e^-_{\text{N}} \; \rightleftharpoons \; \text{Li}
LiCoO2CoO2  LiP++eP\text{LiCoO}_2 - \text{CoO}_2 \rightleftharpoons \; \text{Li}^+_{\text{P}} + e^-_{\text{P}}
LiN+    LiP+\text{Li}^+_{\text{N}} \; \rightleftharpoons \; \text{Li}^+_{\text{P}}

Since electrons cannot equilibrate between the two electrodes without shorting the battery, the net reaction is

(eNeP)    Li(LiCoO2CoO2)\bigl( e^-_{\text{N}} - e^-_{\text{P}} \bigr) \; \longleftrightarrow \; \text{Li} - \left( \text{LiCoO}_2 - \text{CoO}_2 \right)

Applying the equilibrium condition (similar to Equation 2.5.2) gives:

μeNμeP=(μLiμLiCoO2CoO2)\mu_{e^-_{\text{N}}} - \mu_{e^-_{\text{P}}} = \Bigl( \mu_{\text{Li}} - \mu_{\text{LiCoO}_2-\text{CoO}_2} \Bigr)

Here, μLi\mu_{\text{Li}} is the chemical potential of Li\text{Li} in metallic form and μLiCoO2CoO2\mu_{\text{LiCoO}_2-\text{CoO}_2} is the chemical potential of Li\text{Li} when inserted in the host. For simplicity, we denote the Li\text{Li} in LiCoO2\text{LiCoO}_2 as “Li” and write μ“Li”=μLiCoO2CoO2\mu_{\text{“Li”}} = \mu_{\text{LiCoO}_2-\text{CoO}_2}.

Recall that the battery voltage is given by the difference in chemical potential between the negative and positive electrodes:

V=1F(μeNμeP)=1F(μLiμ“Li”)V = \frac{1}{\mathscr{F}} \Bigl( \mu_{e^-_{\text{N}}} - \mu_{e^-_{\text{P}}} \Bigr) = \frac{1}{\mathscr{F}} \Bigl( \mu_{\text{Li}} - \mu_{\text{“Li”}} \Bigr)

A positive voltage implies that the chemical potential of Li\text{Li} is higher in the negative electrode (metallic Li\text{Li}) than in the positive electrode (where Li\text{Li} reacts with CoO2\text{CoO}_2 to form LiCoO2\text{LiCoO}_2). To discharge a battery, Li\text{Li} moves from a region of higher potential (negative electrode) to one of lower potential (positive electrode), producing electrical work.

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