 Solution

The key is coupling redox equilibrium to hydrolysis and precipitation. Ce(IV) has a much greater tendency than Ce(III) to form a hydrated cerium dioxide solid. Raising pH therefore preferentially stabilizes the oxidized form. In the presence of dissolved oxygen a representative net reaction isRemoving ${\mathrm{H}}^{+}$ drives this reaction towards Ce(IV). Writing the solid as hydrated oxide changes its water content, not this redox argument.

The Nernst equation makes the pH effect explicit. At $25^{\circ }\mathrm{C}$, the reduction $\mathrm{Ce}{\mathrm{O}}_2+4{\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to \mathrm{C}{\mathrm{e}}^{3+}+2{\mathrm{H}}_2\mathrm{O}$ has a pH slope approximately $-0.236\mathrm{V}$ per pH unit at fixed Ce(III) activity. The oxygen/water reduction has slope about $-0.059\mathrm{V}$ per pH unit. Raising pH thus makes oxidation of Ce(III) by oxygen increasingly favourable within the relevant phase-stability region. Conversely strong acid favours soluble Ce(III) relative to the oxide and makes Ce(IV) a stronger oxidizing agent. pH changes the redox driving force; an electron acceptor or donor is still required, and slow oxygen evolution or oxidation kinetics can prevent immediate equilibration.