= Solution
Write both <electrode> reactions first as <reductions>:
$$
\mathrm{CH_3COO^-+5H^++4e^-\longrightarrow C_2H_5OH+H_2O},
$$
$$
\mathrm{CeO_2+4H^++e^-\longrightarrow Ce^{3+}+2H_2O}.
$$
The \b[conventional cell reaction] runs with <oxidation> on the left and <reduction> on the right. Reverse the first equation and multiply the second by four:
$$
\boxed{\mathrm{C_2H_5OH+4CeO_2+11H^+
\longrightarrow CH_3COO^-+4Ce^{3+}+7H_2O}}.
$$
Four <electrons> pass through the external circuit per reaction as written. This direction consumes the oxide; deposition requires the reverse <electrolysis> direction.
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