= Solution
During discharge, the graphite-based electrode is the \b[<anode>], where lithium is deintercalated and electrons are released; the cobalt-oxide electrode is the \b[<cathode>], where lithium and electrons are inserted. Adding the stated half-reactions gives
$$
\mathrm{LiC_6+CoO_2\longrightarrow C_6+LiCoO_2}.
$$
The supplied potentials are not, strictly, the potentials of those two insertion reactions. Approximating the negative-electrode potential by $\mathrm{Li^+/Li}$ and using the proton-coupled cobalt reduction as a proxy for the positive electrode gives the intended rough estimate
$$
\boxed{E_{\rm cell}\simeq(+1.1)-(-3.10)=4.2\,\mathrm V.}
$$
\b[This 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 $\mathrm{CoO_2}$ 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.
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