Fuel cells

  • Thread starter Big-Daddy
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  • #1
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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)?
 

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  • #2
Borek
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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.
 
  • #3
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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.
 
  • #4
Borek
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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.
 

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