 Solution

ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1/section-b/11/ii/b/solution

One mole of the overall discharge reaction transfers one mole of electrons. Count both active electrode reactants, including the graphite host:

M_{active} = M_{Li C_{6}} + M_{Co O_{2}} = 6.94 + 6 (12.01) + 58.93 + 2 (16.00) = 169.93 g mo l^{- 1} .

Using the preceding constant-voltage approximation,

W = FE ≃ (96500) (4.2) = 405300 J mo l^{- 1} = 112.58 Wh mo l^{- 1},

so the ideal active-material specific energy is

ε_{sp} ≃ \frac{112.58}{0.16993} = 663 Wh k g^{- 1} .

This assumes complete one-electron utilization and the proxy

4.2 V

throughout discharge. It is not a packaged-cell value or a cathode-only value. A varying equilibrium voltage requires integrating

\int V d Q

; using a maximum voltage as a constant average overestimates the delivered energy.

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