Timeline for Electrolysis and actual electrode potential (using a supercapacitor)

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Oct 10, 2015 at 21:24	comment	added	Gert		@user76330: You are right that it’s not $\Delta V$ that causes the chemical reactions to take place: it’s the flow of electrons, i.e. current. But no current w/o $\Delta V$, though.
Oct 10, 2015 at 21:24	comment	added	Gert		@user76330: Unfortunately it isn’t. Firstly you confound voltage (the difference in electric potential energy, in Volt) and charge (in Coulomb). Secondly you still assume potential is distributed between the electrodes in a certain way. In reality we cannot know this and do not need to either. We only need to (and can) know the needed potential difference $\Delta V$.
Oct 10, 2015 at 20:18	comment	added	MarkJanus1		Maybe this is better notation: $ 2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^- $ -1.23V and $ O_{2(g)}+4H^+_{(aq)}+4e^- \rightarrow 2H_2O_{(l)}$ +1.23V. Actual is not a good word. Potentials are relative values. The surprising thing to me was that Vs does not contribute to the chemical reaction. Vs adds charge equally to each electrode and their relative potentials add in series. Logically, I think, their relative potentials break up as described.
Oct 10, 2015 at 20:00	comment	added	MarkJanus1		Perhaps this is better:
Oct 9, 2015 at 20:12	history	edited	Gert	CC BY-SA 3.0	added 4 characters in body
Oct 9, 2015 at 20:06	history	answered	Gert	CC BY-SA 3.0