= Solution
The key is coupling <redox equilibrium> to <hydrolysis> and <precipitation (chemistry)>. 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 is
$$
4\mathrm{Ce^{3+}}+\mathrm{O_2}+6\mathrm{H_2O}
\rightleftharpoons4\mathrm{CeO_2(s)}+12\mathrm{H^+}.
$$
Removing $\mathrm{H^+}$ 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 $\mathrm{CeO_2+4H^++e^-\to Ce^{3+}+2H_2O}$ 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>. \b[<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.
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