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: $\ce{H2 -> 2H+ + 2e-}$
• Anode: $\ce{2H+ +2e- -> H2}$

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:

$$\ce{O2 + 2H2O +4e- -> 4OH-}$$

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

$$\ce{H+ + OH- -> H2O}$$

Now we have the following combination of reactions:

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

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

• Wow thanks, this answer takes me to the next level. However I still have that dangling drawback: what makes the proton to actually travel through the membrane? The fact that on the cathode anions are produced? On the other hand, shouldn't the anode be positively charged for the circuit to work? – Leonid Usov Aug 9 '15 at 21:19
• @LeonidUsov electrons always flow into the cell at the cathode and out at the anode. So when a cell is being charged the + plate is the anode & the - plate is the cathode, but when a cell is being discharged (as here) the + plate is the cathode & the - plate is the anode. So it's the cathode that's positively charged: electrons flow out of the anode, round the circuit back to the cathode . For current to flow it's essential for a charged species (in this case H+) to pass through the membrane, but really it's the sum of all reaction steps that provide the energy that drives the current. – Level River St Aug 9 '15 at 22:30
• @steveverrill I just noticed that I made a mistake in my comment above saying that anode should be positively charged, but I meant cathode of course. The paradox that I was trying to describe is how a positively charged cathode attracts ions+ through the membrane. There should be some non-electrostatic force that makes them come. Apparently this force is somehow a result of the cathode reaction... – Leonid Usov Aug 10 '15 at 8:14
• @LeonidUsov at the membrane itself, there is no paradox: electrostatic forces pull the H+ ions through, where they combine with the OH- ions. (This is relevant to the need to have two half reactions. If there were only one half reaction, static electricity would build up very quickly and block the current.) At the surface of the electrodes, what we see is a jump in electrical potential, due to the energy released by the half reactions. This is analogous to the jump in electrical potential at the junction of two dissimilar metals in a thermocouple. Does that seem paradoxical too? – Level River St Aug 10 '15 at 8:34