Why is the cathode reaction required in a fuel cell? So, at the anode of a fuel cell hydrogen is split into a positively charged proton and an electron. Then, there is the membrane which can let only protons through to the cathode, whereas electrons need to follow external circuit (effectively powering it).

Question: why do we need any other reaction on the cathode? Isn't the fact that positively charged ions travel through the membrane to the cathode enough to create the EMF?

Trying to solve it I assume that the reaction is there to actually apply some kind of force to move the ions though the membrane to the cathode, otherwise they'd just stay and reunite with electrons on the anode. What is that force and how does it work?
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.