Yes, I have what it takes, I realize that tonight, and I pass you the video
Look what I just found on another forum:
I just post the questions below
If you had really electrolysed water, you would have observed a gas release at both electrodes (H2 on one side, O2 on the other).
The problem is not simple
From the moment you want to use an aqueous solution to conduct the current, you will always have problems, unless you use a potential difference between the electrodes which is weak, less than 1,23V (electrochemical window of the water, without consider kinetic effects at the electrodes).
Then pure water is not conductive (or very little) must use electrolytes that conduct the current effectively. There, no choice, the two best are H3O + and OH- in the water. Prefer an acid solution, more conductive.
Your solution must be electrically neutral, an anion must be put with H3O + and have a potential for oxidation-reduction greater than 1,23V, Cl- for example.
Finally, last point, it is imperative to use an AC voltage, to avoid the polarization of the electrodes (accumulation of charged species on the electrodes and therefore decrease in conductivity).
So, a solution of HCl 0,1mol.l-1 (pH slightly higher than 1), lower AC voltage to 1,23V if you use fast electrodes
Hello sburmate
The potential for oxidation reduction of water is 0.41V or 1.23V?
Reaction at the anode (Oxidation): 2H²O (liquid) -> O² (gas) + 4H + (aqueous) + 4e- + 0,82V
Cathode reaction (Reduction): 4H²O (liquid) + 4e- -> 2H² (gas) + 4OH- (aqueous) -0,41V
Can we really perform an electrolysis at a voltage lower than 1.23V?
What is this accumulation of species around the electrodes if we do not use alternating current?
What is the ideal frequency for electrolysis of water?
Thank you in advance for your answers
Pascal