# Fuel cells

With galvanic cells, we assume that the redox reaction is kinetically inhibited so that equilibrium takes a long time to reach, so we can make a good measurement of the potential difference V. I thought fuel cells were the same originally, except that we recycle in reactants and products to make sure that there's enough of each to maintain a high potential V, so that we produce power (equal to IV where I is the current).

But now I came across the idea of using a catalyst with the fuel cell. That just doesn't make sense to me. The more the reaction occurs, the closer the system will get to equilibrium where V=0 and thus the power output is 0, so why would we want that? OK, so we are cycling in new reactants anyway, so the reaction will never be at equilibrium - but still, why would we want to catalyse it? What's the benefit in that, when the power output is based specifically on potential difference (which is a function of how much of the reaction is still left to go at any given moment in time)?

## Answers and Replies

Borek
Mentor
the power output is based specifically on potential difference

No.

Apparently you still don't understand how the cell works and (hint) why do we have to close the circuit.

No.

Apparently you still don't understand how the cell works and (hint) why do we have to close the circuit.

Ok, so is it because we have two conflicting interests - keeping a large potential difference V for which we need the reaction nowhere near to equilibrium, and getting the reaction to go reasonably fast so that we can get a decent current, I, out of it - and we maximize power (P=VI) by trying to make the reaction go fast (maximize I) and pumping in new reactants constantly (maximize V)? Or am I still not understanding?

The confusion was caused because my book said this was "just like the galvanic cell" except that reactants are pumped in continuously. I think this was slightly misleading because there is a crucial difference - in galvanic cells, we don't want current to flow (even though it is inevitable that a little does) because we want to measure V without the redox reaction getting any (significantly) closer to equilibrium.

Borek
Mentor
Ok, so is it because we have two conflicting interests - keeping a large potential difference V for which we need the reaction nowhere near to equilibrium, and getting the reaction to go reasonably fast so that we can get a decent current, I, out of it - and we maximize power (P=VI) by trying to make the reaction go fast (maximize I) and pumping in new reactants constantly (maximize V)?

That's it.