Answer
See the answer below.
Work Step by Step
As $E^{\circ}_{cell}=E^{\circ}_{cathode}- E^{\circ}_{anode}$, the value of $E^{\circ}_{cell}$ is largest when $E^{\circ}_{cathode}$ has the largest positive value and $E^{\circ}_{anode}$ has the largest negative value.
From table 17.1, we can write the half-reaction at the cathode (reduction), the reaction with large positive $E^{\circ}$ value as:
$F_{2}(g)+2e^{-}\rightarrow 2F^{-}(aq)$
The half reaction at the anode (oxidation), the reaction with large negative $E^{\circ}$ value is:
$2Li(s)\rightarrow 2Li^{+}(aq)+2e^{-}$
The overall reaction is:
$F_{2}(g)+2Li(s)\rightarrow 2F^{-}(aq)+2Li^{+}(aq)$
$E^{\circ}_{cell}= E^{\circ}_{cathode}-E^{\circ}_{anode}$
$=+2.87\,V-(-3.045\,V)=+5.92\,V$
