How do you account for the strong reducing power of lithium in aqueous solution?

(N/A) The reducing power of an element is determined by the standard electrode potential $(E^{\circ})$,which depends on three factors: sublimation enthalpy,ionization enthalpy,and hydration enthalpy.
$Li(s) \rightarrow Li(g)$ (Sublimation enthalpy)
$Li(g) \rightarrow Li^{+}(g) + e^{-}$ (Ionization enthalpy)
$Li^{+}(g) + H_2O \rightarrow Li^{+}(aq)$ (Hydration enthalpy)
Although $Li$ has the highest ionization enthalpy among alkali metals,it has the smallest size,which results in the highest hydration enthalpy. This large negative hydration enthalpy compensates for the high ionization enthalpy,making the overall process highly favorable. Consequently,$Li$ has the most negative standard electrode potential $(E^{\circ} = -3.04 \ V)$,making it the strongest reducing agent in aqueous solution.