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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2023/paper-1/section-b/11/vi/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 1 / Section B / 11 / vi / Solution by

Codex  0 2026-10-20

Each formula unit of

CF_{0.8}

accepts

0.8

electrons. Using the supplied Faraday constant and the periodic-table masses,

M = 12.011 + 0.8 (18.998) = 27.2094 g mo l^{- 1},

(1)

q_{active} = \frac{0.8 F}{3.6 M} ≃ 788.5 mAh g^{- 1} .

(2)

The factor

3.6

converts coulombs to mAh. This value is per gram of the active fluorinated carbon, not per gram of the entire electrode.

The composite value is underdetermined: the mass of active CF₀.₈ is not supplied. If that mass is

m

grams, and the carbon black and binder are inactive in the stated conversion reaction, then

q_{composite} = 788.5 \frac{m}{m + 2.0 + 0.2} = 788.5 \frac{m}{m + 2.2} mAh g^{- 1} .

(3)

This uses cathode composite mass and excludes the lithium counterelectrode and hardware. No unique numerical composite capacity follows from the two additive masses alone; assuming an unprovided active mass would change the requested quantity.

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