 Solution

During discharge, the graphite-based electrode is the anode, where lithium is deintercalated and electrons are released; the cobalt-oxide electrode is the cathode, where lithium and electrons are inserted. Adding the stated half-reactions givesThe supplied potentials are not, strictly, the potentials of those two insertion reactions. Approximating the negative-electrode potential by $\mathrm{L}{\mathrm{i}}^{+}/Li$ and using the proton-coupled cobalt reduction as a proxy for the positive electrode gives the intended rough estimateThis is a proxy estimate, not an exact thermodynamic derivation for the specified lithium-ion cell. The graphite insertion chemical potential differs from lithium metal, and lithium insertion into $\mathrm{Co}{\mathrm{O}}_2$ is not the supplied proton-insertion reaction. Actual voltage depends on lithium composition, activities and the nonaqueous electrolyte; the source's aqueous notation does not make a practical graphite–cobalt-oxide cell an aqueous battery. Exact insertion potentials or free energies would be needed for a more rigorous calculation.