Solution
= Solution
$$
E^\circ=E^\circ_{\rm right}-E^\circ_{\rm left}
=1.660-(-0.285)=\boxed{1.945\,\mathrm V}.
$$
With $n=4$,
$$
\Delta_rG^\circ=-nFE^\circ
=-4(96485)(1.945)
=\boxed{-7.51\times10^5\,\mathrm{J\,mol^{-1}}}.
$$
Thus the \b[conventional forward reaction is spontaneous under standard conditions], dissolving CeO$_2$ while oxidizing <ethanol>. Oxide deposition is its reverse and needs external electrical work; its intention does not change the sign convention.