# Feasible Energy Density Pushes of Li-Metal vs. Li-Ion Cells
## 3 Excess Lithium and Energy Density  
As a first approach, we will compare a Li-metal battery (LMB) cell and a LIB cell only from the point of view of the active materials. For this purpose, we start with the reaction equations between the cathode NMC 811 and a graphite anode and Li-metal anode. Here we assume 0.5 Li extraction from the cathode (see introduction). As a result, we obtain the following discharge reactions for two different anode materials:  
\[\text{LiNi}_{0.8}\text{Co}_{0.1}\text{Mn}_{0.1}\text{O}_{2}\to 0.5\text{ Li}+\text{Li}_{0.5}\text{Ni}_{0.8}\text{Co}_{0.1}\text{Mn}_{0.1}\text{O}_{2}\text{ (Li-Metal)} \tag{4}\]  
\[\text{C}_{6}+2\text{LiNi}_{0.8}\text{Co}_{0.1}\text{Mn}_{0.1}\text{O}_{2} \to\text{Li}\text{C}_{6}+2\text{Li}_{0.5}\text{Ni}_{0.8}\text{Co}_{0.1}\text{ Mn}_{0.1}\text{O}_{2}\text{ (Graphite)} \tag{5}\]  
The amount of charge (Q) is obtained by multiplying the electrons transported in relation to the equations (\(n\)), by the Faraday constant (F = 96,485 C/mol or 26,801 \(\text{Ah}\)/mol). Specific capacity (\(\text{C}_{\text{sp}}\)) of the cell and the ED can be calculated by the following equations:  
\[\text{C}_{\text{sp}}\left(\text{Ah}/\text{g}\right)=\frac{n\times\text{F}}{ \text{Mw}} \tag{6}\]  
\[\text{ED }\left(\text{Wh}/\text{kg}\right)=\frac{\left(\text{Csp }\times\text{ Vcell}\right)}{1000} \tag{7}\]  
Figure 1: Illustration of graphite vs. Li-metal anode with combination of NMC 811 cathode. It should be noted that the molecular weight (Mw) values correspond to the Mw of the discharge products as a whole. Table 2 compares the zero-excess (anode-free) LMB and LIB according to the given discharge reactions 5 and 6. This theoretical capacity is difficult to obtain, because it is not feasible to recover the in-situ deposited Li-ions 100% from the anode due to formation of the SEI [32]. In order to overcome loss of Li, the excess amount of Li is introduced to the cell. Assuming that NMC 811 exhibits 2.2 mAh/cm\({}^{2}\) of areal capacity [33], and combining it with a zero-excess anode, namely only a copper current collector, this cathode would deposit dense Li-metal layer with a volumetric capacity of 2061 mAh/cm\({}^{3}\) (C\({}_{\rm Li}\)*density\({}_{\rm Li}\)). [PERSON] et al., reported that zero-excess is defined as the initial lithium thickness at the anode and corresponds to the areal capacity of the anode [32]. In this case, when the NMC 811 cathode is delithiated, the thickness of the Li-metal would be 2.2 mAh/cm\({}^{2}\): 2061 mAh/cm\({}^{3}\cong\) 10 \(\upmu\)m. Hence, if we initially use 10 \(\upmu\)m of Li-foil on the copper it corresponds to a one times excess of Li, 20 \(\upmu\)m of Li-foil with two times the excess Li, and so on. As the thickness of Li-metal increases, the volumetric ED decreases which is one of the main reasons to limit thickness of the Li-metal anode. However, the following question should be answered: How much excess provides benefits? Depending on the excess factor (X\({}_{\rm Li}\)) used to get a complete cell, one can calculate change in energy densities and bring these values into relation with LIBs in a very good approximation. The Figure 2 shows the maximum theoretical limit of the excess lithium in order not to obtain an ED lower than that of the cell with a graphite anode (423 Wh/kg). The calculations of the ED values for different excess amounts of lithium can be found in the Supplementary Materials. In theory, if the X\({}_{\rm Li}\) is less than 6, the ED of the cell will be already more than graphite-based LIBs. Every excess lithium factor after 6 will deliver less than 367 Wh/kg which \begin{table}
\begin{tabular}{c c c} \hline \hline
**System** & **LMB** & **LIB** \\ \hline Discharge products & LiNi\({}_{0.8}\)Co\({}_{0.1}\)Mn\({}_{0.1}\)O\({}_{2}\) & C\({}_{6}\) + 2 LiNi\({}_{0.8}\)Co\({}_{0.1}\)Mn\({}_{0.1}\)O\({}_{2}\) \\ Mw (g/mol) & 97.28 & 266.63 \\ \(n\) & 0.6 & 1 \\ Cell Potential (V) & 3.8 & 3.7 \\ C\({}_{\rm sp}\) (Ah/g) & 165.31 & 114.28 \\ ED (Wh/kg) & 628.15 & 422.84 \\ \hline \hline \end{tabular}
\end{table}