= Solution
A lithium–oxygen system can have \b[much greater theoretical active-material <specific energy>] because it avoids the heavy transition-metal host and carbon host of the preceding insertion cell. Oxygen can be taken from the surroundings, but a mass comparison must say whether that oxygen is counted.
The peroxide reaction transfers two electrons per $\mathrm{Li_2O_2}$ unit. If oxygen is counted, the reactant mass per two electrons is
$$
M=2(6.94)+32.00=45.88\,\mathrm{g\,mol^{-1}},
$$
compared with $2(169.93)=339.86\,\mathrm g$ for two electrons in the idealized lithium-ion reaction. At the same average voltage, the active-material specific energy would therefore be larger by $339.86/45.88\simeq7.4$. If only stored lithium is counted, the ideal mass advantage is larger still, but this excludes the oxygen and is not a like-for-like closed-cell mass basis.
The source supplies no lithium-peroxide formation free energy or oxygen-electrode potential, so it does \b[not] determine an absolute lithium–oxygen voltage or specific energy. Real devices require an oxygen electrode, gas-management and electrolyte; insulating discharge products, parasitic reactions, charging overpotential and mass transport reduce the practical advantage. The question's reaction is chemical oxygen reduction, not a biological energy pathway.
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