I am studying electrochemical water splitting and I have a huge doubt. The standard redox potential for the total reaction is 1.23 V, but this is in standard conditions, where the hydrogen and oxygen partial pessures are 1 bar. Now, if I stick two metallic electrodes in an open pool of water and apply a voltage between them to make water splitting, isn't the redox potential required (even not considering the overpotentials that have to be added to make the kinetics faster and compensate resistive losses) going to be different because of the contribution of the partial pressures of the gases that are not the standard ones, but the ones in the real atmosphere? Or am I getting something wrong? Do they compensate this somehow in electrolyzers? But then why do many experimental groups studying materials for water splitting simply put the electrodes in an open becher? I made a rough calculation using Dalton's principle for their partial pressures and the contribution to the Nernst equation due to the partial pressures of these gases doesn't look negligible compared to the standard redox potential, as the actual partial pressure of hydrogen in the atmosphere is a lot smaller than 1 bar. Thanks in advance for the answer
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$\begingroup$ Ongoing electrolysis very quickly establishes the dissolved gas activities corresponding to the external pressure that is very close to the standard one. So the major deviation from standard potentials are overpotentials (especially oxygen one) and nonstandard ion activities. // Commercial electrolyzers control rather the optimal current, and adjust the electrolytic voltage accordingly, ignoring assumed theoretical values. $\endgroup$– PoutnikCommented Feb 25 at 13:31
1 Answer
In a real electrolysis setup, the electrode lives in its own cloud of bubbles filled with the corresponding gas.
The gas in the bubbles has almost atmospheric pressure because of hydrostatic reasons and is either pure hydrogen or pure oxygen.
One may assume the bubbles being in near-equilibrium with the dissolved gas in the water around the electrode, but pretty much not with the atmosphere. Bubbles are one-way transport.
If you try to really run the electrolysis as a reversible process...
Well, you can abuse the same equation and calculate how much hydrogen partial pressure we could have at 0V between the electrodes and 20kPa oxygen pressure. Because at 0V one does not really need electrodes, this is the lowest possible hydrogen we could have in the atmosphere (we have somewhat more because we do have other sources of hydrogen around)
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$\begingroup$ Thanks! What you said makes a lot of sense in my head. So, can I imagine that in the first instants of the reaction the potential required would be different because I still have no hydrogen and oxygen bubbles, then I change it gradually until the partial pressures are in hydrostatic equilibrium with the external atmosphere? Or did I get it wrong? $\endgroup$ Commented Feb 24 at 20:40
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$\begingroup$ The second part of my answer implies that in the first instants the potential required will be virtually zero. It will go up as both hydrogen and oxygen build up around the electrodes and will stabilize as the concentration of both gases rises to the point of forming bubbles and growing these bubbles against the pressure of the surrounding liquid. $\endgroup$– fraxinusCommented Feb 24 at 20:49