Answer
$E^{o}_{cell}=-0.028\,V$
This reaction is not product-favored.
Work Step by Step
Anode: $Ag(s)\rightarrow Ag^{+}(aq)+e^{-}$
Cathode: $Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)$
$E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$
$= (+0.771)-(+0.7991)=-0.028\,V$
As $E^{o}_{cell}$ is negative, the reaction is not product-favored.
