= Solution
The discharge reaction is $2\mathrm{Li}+\mathrm I_2\to2\mathrm{LiI}$. It transfers two <electrons> and has $\Delta G^\circ=-539.4\,\mathrm{kJ\,mol^{-1}}$ per reaction as written. The reversible <open-circuit voltage> is
$$
\boxed{E^\circ=-\frac{\Delta G^\circ}{2F}=\frac{539400}{2(96500)}=2.795\,\mathrm V.}
$$
The supplied <free energy> is already at approximately the requested temperature, so an additional temperature factor is unnecessary. Actual activities or discharge polarization could alter the terminal <voltage>.
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