Hexavalent
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Help with equilibria
In a hydrogen fuel cell at the anode, the following reaction takes place:
O2 + 4H+ + 4e- ↔ 2H2O
What drives the reaction forward, and minimises the reverse reaction?
"Success is going from failure to failure without loss of enthusiasm." Winston Churchill
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Oscilllator
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I don't think its an equilibrium reaction, I'm pretty sure the hydrogen just reacts with the oxygen.
If not, my guess is the removal of the water may drive the reaction to the right
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Vargouille
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A continuous supply of electrons from the conversion of hydrogen to protons. As for what causes the process initially, I would have to brush back up
on electrochemistry for that.
EDIT: As for what prevents the back-reaction, that's caused by the reduction potential of the half-reaction (Eo = 1.23 V). To get the water
to split into oxygen and protons, about 1.23 V would need to be pumped in (in an "ideal" case).
[Edited on 13-12-2012 by Vargouille]
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Hexavalent
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Thanks.
I also believe that the gaseous reactants are pressurised, meaning the equilibrium is further shifted to the right, aided by the cooling of the vessel
(as the reaction itself is exothermic).
P.S. Oscillator, the reaction would be non-reversible if the hydrogen gas itself was combusted in the oxygen, but I don't believe that holds in this
case.
"Success is going from failure to failure without loss of enthusiasm." Winston Churchill
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Nicodem
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It is chemical equilibria, hence the origin of the overvoltage at the equilibrium point. The reaction is driven in either direction by the addition or
removal of electrons to/from the electrode.
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scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
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