Methanol fuel cell anode - why not oxygen instead of water?

The methanol fuel cell:


\begin{align}
\text{Overall}&:& \ce{CH3OH + 3/2 O2 &-> 2H2O + CO2}\\
\text{Cathode}&:& \ce{O2 + 4H^+ + 4e^- &-> 2H2O}\\
\text{Anode} &:& \ce{CH3OH + H2O &-> CO2 + 6H^+ + 6e^-}
\end{align}


All the equations balance and still work in theory if the methanol reacts with oxygen at the anode instead of water:


$$\text{Anode}: \ce{CH3OH + 1/2 O2 -> CO2 + 4H^+ + 4e^-}$$


So why is the first anode equation (with water) the correct one? How would a student know to use water at the anode and not oxygen?




The answer is rather simple. A student is not supposed to know beforehand. The anode and cathode reaction will depend *on what is being fed to the anode and cathode compartments* in a fuel cell. In the anode section, the only components which are injected there are methanol + water along. Oxidation has to occur at the anode-*by definition*. If water and methanol are present, under those circumtances, methanol gets oxidized to carbon dioxide.


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The student is supposed to use the chemical principles they've learned and come up with reasonable answers. A [redox, really any] reaction separated into its oxidation and and reduction moieties can possibly be used in a cell or fuel cell. The question was to separate the oxidation and reduction half reactions.
Overall: CH3OH + 3/2O2 = CO2 + 2H2O nicely balanced!
Reduction is easy: 3/2O2 + 6e- = 3O= This is the cathode and a source of H+ is needed to neutralize the oxide ion the logical source is H2O altho I don't know if a weak acid could be used.
The cathode reaction: 3/2O2 + 3H2O + 6e- = 6OH-
The oxidation reaction is a bit trickier; I will break it into simple steps!
CH3OH = C+4 + 4H+ + O= +6e-; what a mess! = CO+2 + 4H+ + 6e- The problem is now evident! There are just not enough O= ions to go around. A source of O= ions is needed. H2O is a source of oxide ions; O2 is NOT a source of oxide ions. The anode reaction is: CH3OH + H2O = CO2 + 6H+ + 6e-. [The next project is to get the H+ and OH- ions together and possibly capture that energy.]
Please don't get all literal and object that O= and C+4 ions don't exist. The trick is to make the reactions as electrical as possible and keep the oxidation and reduction separate, then try to work out the chemistry; thinking about it outside the box might even elicit some fruitful ideas.


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