Electrolysis and actual electrode potential (using a supercapacitor)

Question

For an electrolytic cell, the voltage source Vs, does not contribute directly to the potential on the electrolytic electrodes. Vs determines the current and can add charges to the electrodes. This can be observed when charging supercapacitors and is discussed in these threads : https://www.physicsforums.com/threads/supercapacitor-charging-voltage.594372/ https://www.physicsforums.com/threads/ultracapacitor-charging-voltage.680719/

Question: What is the actual potential on the electrolytic electrodes for the electrolysis of water?

Example: An electrolytic cell has carbon electrodes (make them identical commercially available activated carbon supercapacitor electrodes, so a charge has to build up for a few seconds) and sulfuric acid, $H_2SO_4$, electrolyte.
Positive charge electrode reaction: $ 2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^- $ 1.23V
Negative charge electrode reaction: $ 2H^+_{(aq)}+2e^- \rightarrow H_{2(g)} $ 0.00V

1.23V is the minimum Vs should be for hydrogen and oxygen generation to start in this cell. There are many factors that increase this value, but use 1.23V for this example. The positive charge electrode is the electrolytic anode and the negative charged electrode is electrolytic cathode.

Charge is added to each electrode in equal amounts. The actual potential on each electrode for this cell is (1.23V/2)=0.615V. The positive charge electrode is +0.615V and the negative charge electrode is -0.615V. The electrodes are is series, so the potentials add to 1.23V. Breaking the charge down further: the positive charge electrode has (+0.3075 charge and -0.3075 ionic charge from $HSO_4^{-}$ (or $SO_4^{2-}$) and the negative charge electrode has (-0.3075 charge and +0.3075 ionic charge from $H^+$)

Is this correct?

Answer 1

$2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$.

You assign a value of $+1.23 \text{ V}$ to this half-reaction (the oxidation of water) but that is incorrect. $+1.23 \text{ V}$ is in fact the value for the reduction of water, not the oxidation of water. So for:

$2H_2O_{(l)} \rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$, the half-potential is actually $-1.23 \text{ V}$.

One can understand this also as follows: the cell potential is calculated as $E=E_{Ox}+E_{Red}$ with $E_{Ox}$ the half-potential of the oxidation reaction and $E_{Red}$ the half-potential of the reduction reaction $(2H^+_{(aq)}+2e^- \rightarrow H_{2(g)})$.

So in our case we have $E=-1.23 \text{ V} + 0.0 \text{ V}=-1.23 \text{ V}$. A negative value conventionally in electrochemistry means that these reactions will not spontaneously occur. Well, that's precisely what we observe: water does not auto-oxidise!

So now we know that in order to make these reactions proceed, we need to apply at least $+1.23 \text{ V}$ (disregarding any overpotentials that might arise) across the electrodes.

The second thing to understand is that the half-potentials of half-cells are always relative to the half-potential of:

$2H^+_{(aq)}+2e^- \rightarrow H_{2(g)}$

Purely conventionally, the half-potential of a platinum/hydrogen half-cell has been assigned the value of $0.00 \text{ V}$. The half-potential of some half-reduction, say:

$O + n e^- \rightarrow R^{n-}$.

... is then determined relative to the platinum/hydrogen half-cell.

This means that in reality we never know the 'real' value of a half-cell, we can only know the potential difference between the two electrodes (in a closed circuit).

Charge is added to each electrode in equal amounts. The actual potential on each electrode for this cell is (1.23V/2)=0.615V. The positive charge electrode is +0.615V and the negative charge electrode is -0.615V. The electrodes are is series, so the potentials add to 1.23V. Breaking the charge down further: the positive charge electrode has (+0.3075 charge and -0.3075 ionic charge from HSO−4 (or SO2−4) and the negative charge electrode has (-0.3075 charge and +0.3075 ionic charge from H+)

... is therefore a misinterpretation, albeit quite a smart one.