# Electrode polarity in fuel cells

I am studying fuel cells right now and I'm confused about the polarity of the two electrodes.

In the diagram below, the anode is shown as negative. However, why do $\ce{H+}$ ions move away from the anode, when they should be attracted to it because of its negative charge?

On the other hand, if I swapped the polarity around, movement of electrons from the positive to the negative terminal wouldn't work out.

Please explain which polarity is correct, and why.

Please explain which polarity is correct, and why.

In a fuel cell, the anode does sit at a more negative potential than the cathode. This polarization is opposite that of an electrolytic cell, where the anode is positive relative to the cathode.

The chemical composition of the fuel $(\ce{H2})$ and oxidant $(\ce{O2})$ feeds at the anode and cathode, respectively, cause this. Thermodynamically, in this system the product of the fuel cell reaction, $\ce{H2O}$, is much more stable than the reactants. Thus, the system biases itself to move in favor of the forward reaction:

$$\ce{2H2 + O2 <=> 2H2O}$$

As the $\ce{H2}$ is oxidized, the electrons from this half-reaction are passed into the anode:

$$\ce{2H2 -> 4H+ + 4e-}$$

These electrons have to move through the circuit in order to be able to supply the electrons needed for the oxygen reduction reaction:

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

Here's the key: In order for the electrons to move through the circuit in the correct direction, the potential has to be more-positive at the cathode: the negatively charged electrons are attracted toward the positively charged cathode, along the conductive pathway of the circuit.

However, why do $\ce{H+}$ ions move away from the anode, when they should by attracting because of negative charges?

For the most part, the circuit is going to be composed of (semi)conducting materials. Conductive materials in a circuit cannot sustain a very substantial charge polarization at all--as soon as they become very slightly polarized, they allow current to pass that equalizes the charge imbalance. (This is why they are used to carry current!)

As a result, the actual charging of the anode is negligible, even though it develops a finite electrical potential. Thus, the electrostatic attraction of the $\ce{H+}$ ions to the anode can in general be ignored, as it is orders of magnitude too small to keep the protons from diffusing away, toward the cathode where they are being consumed to produce water.