Answer
$E^{o}_{cell}=1.164\,V$
The reaction is product-favored.
Work Step by Step
Anode: $2Zn(s)\rightarrow 2Zn^{2+}(aq)+4e^{-}$
Cathode: $O_{2}(g)+2H_{2}O(l)+4e^{-}\rightarrow 4OH^{-}$
$E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$
$= (+0.401)-(-0.763)=1.164\,V$
As $E^{o}_{cell}$ is positive, the reaction is product-favored.
