Answer
$E^{o}_{cell}=0.65\,V$
The reaction is product-favored.
Work Step by Step
Anode: $Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^{-}$
Cathode: $2Ag^{+}(aq)+2e^{-}\rightarrow 2Ag(s)$
$E^{o}_{cell}=E^{o}_{cathode}-E^{o}_{anode}$
$= (+0.7991)-(+0.15)=0.65\,V$
As $E^{o}_{cell}$ is positive, the reaction is product-favored.
