[Ch1]

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, H₂SO₄, electrolyte.
Positive charge electrode reaction: 2H₂O(l) O₂(g) + 4H⁺(aq) + 4e^- 1.23V.
Negative charge electrode reaction: 2H⁺(aq) + 2e^- H₂(g) 0.00 V.
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 SO₄^-2) and the negative charge electrode has (-0.3075 charge and +0.3075 ionic charge from H⁺)

Is this correct?

[Enthalpy]

No reason to split the voltage in equal halfs 0.615V. The mere contact of the electrodes with the electrolyte can create a voltage between the electrolyte and the circuit.