Answer
$E^{o}_{cell}=2.89\,V$
The reaction is product-favored.
Work Step by Step
Anode: $Mg(s)\rightarrow Mg^{2+}(aq)+2e^{-}$
Cathode: $I_{2}(s)+2e^{-}\rightarrow 2I^{-}$
$E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$
$= (+0.535)-(-2.356)=2.89\,V$
As $E^{o}_{cell}$ is positive, the reaction is product-favored.
