Electrolysis of water and actual electrode potential.

[Phy1]

For an electrolytic cell, the voltage source V_s, does not contribute directly to the potential on the electrolytic electrodes. V_s 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: 2 H2O(l) → O2(g) + 4 H+(aq) + 4e− 1.23 V
Negative charge electrode reaction: 2 H+(aq) + 2e− → H2(g) 0.00 V
1.23V is the minimum V_s 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.)

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?

 

[Borek]

Is this correct?

No.

You are more or less right about 1.23 V being the potential difference between the electrodes when the gas starts to evolve. After that you are mixing several different concepts in a way that I fail to understand (but the mix is definitely wrong).

Just because the total voltage is 1.23 V doesn't mean absolute voltages are ±0.0615 V.

Charges on electrodes are not these of the ions in the solution. Ions are attracted to the charged electrodes, that's one of the reasons why they migrate in the solution, but their charge is not part of the charge of the electrodes. If anything they neutralize the electrode charge becoming part of a double layer on the surface.

 

[Phy1]

You are more or less right about 1.23 V being the potential difference between the electrodes when the gas starts to evolve. After that you are mixing several different concepts in a way that I fail to understand (but the mix is definitely wrong).

Hi Borek,
If you like, the terms can be changed from positive charge electrode to electrolytic anode and negative charged electrode to electrolytic cathode.

Charges on electrodes are not these of the ions in the solution. Ions are attracted to the charged electrodes, that's one of the reasons why they migrate in the solution, but their charge is not part of the charge of the electrodes. If anything they neutralize the electrode charge becoming part of a double layer on the surface.

Think of charging a regular capacitor. If you charge a capacitor to 2V then the positive plate gets +1V and the negative plate gets -1V. When the capacitor discharges, the potential difference is 2V. The electric double layer forms a capacitor and the ions form half of the charge.

 

[Borek]

If you charge a capacitor to 2V then the positive plate gets +1V and the negative plate gets -1V.

Depends on what you assume to be the reference point. What if one plate is at 60 V (measured against the ground) and the other at 62 V (again, measured against the ground)?

There are two double layers, one on each electrode, so if anything, the solution acts as a two capacitors in the row.