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

Use a lithium-metal negative electrode and a carbon-monofluoride composite positive electrode, separated by a porous, electrically insulating battery separator wetted by a lithium-salt electrolyte in a suitable nonaqueous solvent. Conducting carbon and a binder provide electronic contact through the fluorinated-carbon electrode to its electric current collector; the lithium electrode has its own electric current collector. A sealed stacked or wound assembly supplies useful area and contains the electrolyte.

During discharge lithium is the anode and the carbon-monofluoride composite is the cathode. lithium ions move through the electrolyte while electrons flow through the external load. The battery separator prevents electronic shorting while permitting ionic transport. An aqueous electrolyte is unsuitable because metallic lithium reacts with water. The requested very long operating time requires enough active material and low self-discharge, not merely a high nominal voltage.

For the specified $x=1$, the discharge half-reactions and their sum areEach mole of CF formula units consumes one mole of lithium and transfers one mole of electrons. For general $x$, the corresponding formula-unit reaction would be ${\mathrm{CF}}_x+x\mathrm{Li}⟶\mathrm{C}+x\mathrm{LiF}$.

Using $365$ days per year, the required time is $2\times 365\times 24=17520\mathrm{h}$. Hencelithium transfers one electron per atom, soThis is the stoichiometric minimum for one cell, or one cell-equivalent parallel assembly, with complete utilization and no self-discharge. Practical loading must be larger. If a pack uses several cells in series, each series cell must carry the full delivered charge and the total lithium mass increases accordingly; no required pack voltage or series count is supplied.

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.

For the $x=1$ reaction there is one mole of transferred electrons per mole of reaction. Treating the supplied fresh-cell open-circuit voltage as the standard equilibrium cell potential givesA fresh-cell voltage alone would not normally establish standard conditions or equilibrium. The result above is the formal standard-free-energy estimate under the interpretation required to use the given datum; a metastable open-circuit voltage should not be mistaken for an independently measured standard thermodynamic potential.

This is a primary cell and should not be treated as rechargeable. Discharge converts the fluorinated-carbon electrode into carbon and lithium fluoride. It does not simply insert lithium into a host that can be reversibly deintercalated. Reversing electric current does not cleanly reconstruct the original distributed C–F bonding and electrode microstructure.

Insulating LiF, loss of the original reaction interfaces and unwanted electrolyte reactions obstruct reversibility. Replating lithium can also produce an irregular metallic deposit and internal shorting. A thermodynamically possible formal reverse reaction is therefore not evidence of a practical rechargeable cell.

If the voltage were held at $4.57\mathrm{V}$ for the entire ideal discharge, thenand the electrical power would beThese are constant-voltage estimates, not a determination of the actual energy from the fresh-cell open-circuit voltage. Under load the voltage depends on state of discharge, electric polarization and internal resistance. The physical result is $E=I{\int }_0^{t_f}{V}_{\mathrm{load}}(t)dt$, and $P(t)=I{V}_{\mathrm{load}}(t)$. The supplied single open-circuit value does not specify that voltage curve; practical delivered energy will generally be lower than the simple estimate.

Increase the electrochemically accessible area and reduce transport resistance rather than merely forcing more electric current through the same thick electrode. A thinner, porous composite distributed over a larger area, a conductive-carbon network, good electrolyte wetting and short ionic paths reduce local current density and electric polarization. Adequate battery separator conductivity and low-resistance collectors also help. Parallel cells can distribute total electric current while retaining the total active-material inventory.

The objective is to preserve usable capacity at a higher rate by avoiding premature voltage cutoff and inaccessible active material. It does not create more charge: a fixed inventory with $175.2\mathrm{Ah}$ delivers a greater electric current for a proportionally shorter time. Maintaining the original two-year duration at greater electric current requires proportionally more capacity.