Why is the cathode reaction required in a fuel cell?
Why do we need two half reactions, one at the anode and another at the cathode? 


The whole thing eventually grinds to a halt without the other reaction. The electrons generated at the anode travel through the circuit to the cathode. What happens then? Something needs to happen to the electron. 

Additionally, what happens to that proton? As the concentration of protons increases, what will balance the charge that is produced in the cell, given that the electron is elsewhere? 

We can imagine that we can set up a cathode in which the reverse reaction of the anode occurs:


Cathode: $\ce{H2 -&gt; 2H+ + 2e-}$
Anode: $\ce{2H+ +2e- -&gt; H2}$


However, in this case, there is overall no net reaction, which means that $\Delta G =0$ and since $\Delta G = nFE$, then the cell potential $E=0$. We need a different reaction for the cathode to consume electrons.

For example:

$$\ce{O2 + 2H2O +4e- -&gt; 4OH-}$$

Since this reaction produces hydroxide anions, then we have something for the proton to do:

$$\ce{H+ + OH- -&gt; H2O}$$

Now we have the following combination of reactions:


Anode: $\ce{H2 -&gt; 2H+ + 2e-}$
Cathode: $\ce{O2 + 2H2O +4e- -&gt; 4OH-}$
Neutralization $\ce{H+ + OH- -&gt; H2O}$
Net: $\ce{2H2 + O2 -&gt; 2H2O}$


This net reaction is exergonic, and thus produces a potential.