Electrolysis of water and actual electrode potential.

In summary, the voltage source Vs in an electrolytic cell does not directly contribute to the potential on the electrolytic electrodes. It determines the current and can add charges to the electrodes, as seen in the charging of supercapacitors. The potential on each electrode in this cell is 0.615V, with the positive charge electrode at +0.615V and the negative charge electrode at -0.615V. However, the charges on the electrodes do not correspond to the ions in the solution and instead form a double layer. This can be compared to charging a regular capacitor, where the potential difference is the sum of the potentials on each plate.
  • #1
Phy1
3
0
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: 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 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.)

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?
 
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  • #2
Phy1 said:
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.
 
  • #3
Borek said:
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.
Picture1.GIF


Borek said:
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.
 
  • #4
Phy1 said:
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.
 

1. What is electrolysis and how does it work?

Electrolysis is a chemical process that uses an electric current to break down a compound into its individual elements. In the case of water, electrolysis involves passing an electric current through the water to separate it into its component gases, hydrogen and oxygen.

2. What is the purpose of performing electrolysis on water?

The purpose of performing electrolysis on water is to produce hydrogen and oxygen gas, which can then be used as a source of fuel or in other chemical reactions.

3. What are the two electrodes used in electrolysis of water?

The two electrodes used in electrolysis of water are the anode and the cathode. The anode is the positively charged electrode and the cathode is the negatively charged electrode.

4. How does the electrode potential affect the rate of electrolysis in water?

The electrode potential, also known as the redox potential, is a measure of the tendency of an electrode to gain or lose electrons. In electrolysis of water, the higher the electrode potential, the faster the rate of electrolysis will be. This is because a higher potential difference between the electrodes leads to a stronger electric current.

5. What is the actual electrode potential and how is it determined?

The actual electrode potential is the measured potential difference between an electrode and a reference electrode. It is determined by measuring the voltage between the two electrodes in an electrochemical cell. This value can be used to calculate the standard electrode potential, which is the potential difference under standard conditions.

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