Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2021 / Paper 1 / 11 / ii / b / Solution

The reaction consumes one mole of CF for each mole of electrons. Using $M_{\mathrm{CF}}=12.01+19.00=31.01\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$,This is active CF alone for the same single-cell-equivalent assumption, not the full positive-electrode or pack mass. Conductive additive, binder, electrolyte, battery separator, collectors and casing add mass, and incomplete utilization increases the required CF.