- #1
Ronin2004
- 18
- 0
I am currently looking into photocatalytic materials. Now I might be over simplifying this process but it my observation that I am just looking at electrolysis but instead of electricity provided by a battery I am using electron hole pairs to generate a potential to do my splitting. My question is about the cathode side. If I am splitting distilled water what happens at the cathode to the OH radicals if they have no impurities in the water do combine back with the hydrogen and form water or do they just stay in solution turning it more basic. It just would be easier and cheaper to be able to determine the amount of O2 and H2 produced from being able to determine the amount of OH in solution
Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)
Cathode (reduction): 2 H2O(l) + 2e− → H2(g) + 2 OH-(aq)