$\begin{aligned}
\because \mathrm{E}{\text {cell }}^{\mathrm{o}} & =\mathrm{E}{\text {cathode }}^{\mathrm{o}}-\mathrm{E}_{\text {Anode }}^{\mathrm{o}} \
1.21 & =1.229-\mathrm{E}_{\text {Anode }}^{\mathrm{o}}
\end{aligned}$
∵ Fuel cell involves oxidation of methanol which will occur at anode and reduction of O2 will occur at cathode.