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I am reading the following paper in preperation for a research project on the theoretical energy density of batteries: Lithium - Air Battery: Promise and Challenges and on the second page they describe the following calculation:

The oxidation of 1 kg of lithium metal releases 11,680 Wh/kg

I was wondering specifically how this number was obtained. I think that this would be the right chemical equation to use for this type of approximation but I am not sure:

$$\ce{4Li (s) + O2 (g) -> 2Li2O}$$

I am wondering if anyone can explain to me how this number was produced.

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    $\begingroup$ Nice question! Have a look here and here, both of which indicates that the reaction in question is $2\ce{Li^{+} + 2e^{-} + O2 \leftrightharpoons Li2O2}$. $\endgroup$
    – Todd Minehardt
    Commented Sep 28, 2015 at 19:27
  • $\begingroup$ Ahhh thank you! That certainly makes a lot more sense. So using that I could say that 1 kg of $Li^{+}$ is 144.1 mol of $Li^{+} $ Which would produce 144.1 mol of electrons right? So I guess where I am confused is how to go from there to the Wh/kg which is a unit of energy. I guess you would have to know the voltage of the cell that the electrons would be crossing and use that potential difference to try to get to energy? $\endgroup$
    – sTr8_Struggin
    Commented Sep 28, 2015 at 19:38
  • $\begingroup$ Ok So I did a few calculations that get me close to the number in the paper and I am wondering if this is the right direction. 144.1 mol of electrons is $8.7 \times 10^{25}$ electrons. Assuming that the cell for this lithium ion battery is 3.2 volts then it would take 3.2eV to bring one electron through this potential. So 3.2 eV * $8.7 \times 10^{25}$ electrons = $2.8 \times 10^{26}$. Then converting eV to Wh I see that 1 eV = $4.45 \times 10^{-23}$. This is 12388.8 Wh/kg. Perhaps they assume that they are working with a lower voltage cell to get the lower number? Does this look right? $\endgroup$
    – sTr8_Struggin
    Commented Sep 28, 2015 at 19:51
  • $\begingroup$ Ok I think that this is about right. It at least kind of convinces me for the moment... I am coming from a physics background so this electrochemistry stuff has had my head spinning. $\endgroup$
    – sTr8_Struggin
    Commented Sep 28, 2015 at 19:59
  • $\begingroup$ I think you're on the right track, if only because you get an answer that isn't crazy and I think your process is sound. That said, I am pretty out of my depths with electrochemistry and, more specifically, batteries (google "specific energy" for more on what they're reporting, for instance). Hopefully some more knowledgeable folks will weigh in here. $\endgroup$
    – Todd Minehardt
    Commented Sep 28, 2015 at 20:31

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