The major product from my photocatalyst is methane. But the valence band edge of the photocatalyst is less positive than the oxidation potential of water. It means that the holes produced in the valence band of photocatalyst won't oxidize water to produce H+ which would then facilitate the CO2 reduction to CH4. Can anyone please help me how to justify this or if you can provide with some suitable literature addressing such case. I would be grateful!
James,
Thank you for the reply! Yes it is water, photocatalyst and CO2.
From my understanding, there still need H+ in the reaction of CO2 to CH4..Correct me if I'm wrong.
There is always H+ in the solution even if the water dose not split. the reason is simple, water gets ionized by itself, that's why we can measure the pH in neutral water, although the concentration is very low (pH = 7 which means [H+] = 10^-7 moles per liter or ). But when you dissolve CO2 in H2O the pH gets to even lower values (easily below 5, 400 ppm CO2 lower pH of natural water from ~8 to 5.6 [http://ion.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Carbonate/CO2.html]) by dissolution process which has nothing to do with electrochemistry. This means practically you have orders of magnitude more H+ available for CO2 reduction without the need for water to get splitted by electrical potential driving force. So basically if you detect low amounts of CH4 is pretty normal, although forming high share of your products. From thermodynamics you can calculate how much of CH4 is producable through this source of H+.
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