Answer
The reaction is product-favored.
Work Step by Step
We need to check if the $E^{\circ}_{cell}$ value is positive or negative. Positive $E^{\circ}_{cell}$ value indicates that the reaction is product-favored and negative value indicates that it is reactant-favored.
Reduction half-reaction at the cathode is
$Hg^{2+}(aq)+2e^{-}\rightarrow Hg(l)$
Oxidation half-reaction at the anode is
$2I^{-}(aq)\rightarrow I_{2}(s)+2e^{-} $
$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}$
$=0.8535\,V-0.535\,V$ (values from table 17.1)
$=+0.319\,V$
Value of $E^{\circ}_{cell}$ is positive and therefore the reaction is product-favored.
