Answer
$Mg(s) -> Mg^{2+}(aq) + 2e^{-}$
This reaction at the anode would produce a battery with the highest voltage.
Work Step by Step
At cathode, $Ni^{2+}$ reduces to Ni. So, we need metal that will be oxidized i.e. more active than Ni in the metal activity series i.e. above Ni in the metal activity series for a spontaneous reaction. Also, to produce the highest voltage in the battery we need metal that is most active (farthest above Ni in the metal activity series). Here, Mg is more active than the other three metals if we look at the metal activity series.
So, $Mg(s) -> Mg^{2+}(aq) + 2e^{-}$ This reaction at the anode would produce a battery with the highest voltage.
