Voltage potential in PEM electrolyzers

  • Thread starter Thread starter HelloCthulhu
  • Start date Start date
  • Tags Tags
    Potential Voltage
Click For Summary

Discussion Overview

The discussion revolves around the application of PEM (Polymer Electrolyte Membrane) technology in electrolysis, particularly focusing on the voltage potential, efficiency, and the underlying electrochemical processes involved. Participants explore concepts related to electronegativity, half-cell potential, and the relationship between voltage and the separation of hydrogen and oxygen during electrolysis.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes the historical use of PEM fuel cells for electrolysis and expresses confusion regarding the relationship between voltage, electronegativity, and efficiency.
  • Another participant clarifies the minimum voltage required for water electrolysis under standard conditions, stating it is approximately 1.23 V, and mentions that 1.48 V is necessary without heat withdrawal.
  • A participant questions how the voltage on the anode relates to the electronegativity on the cathode and seeks to understand the forces involved in the separation of hydrogen protons from oxygen.
  • One participant expresses confusion about the relationship between the dissociation energy required to split water and the electronic potential between electrodes, indicating a need for further research into electrode kinetics.
  • Another participant explains the electrochemical process of water decomposition, emphasizing the role of electron transfer and the migration of ions under an applied electric field in the context of PEM cells.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the electrochemical principles involved in PEM electrolysis. There is no consensus on the specific relationship between voltage, electronegativity, and efficiency, indicating ongoing exploration and debate.

Contextual Notes

Participants reference various voltage values and conditions for electrolysis, but there are unresolved questions regarding the implications of these values on efficiency and the mechanisms of ion migration and electron transfer.

HelloCthulhu
Messages
150
Reaction score
3
I've been studying PEM fuel cells for a while and wondered if they could be used for electrolysis. I was surprised at first to find out the technique had been used for decades, but after watching this animation the dynamics make much more sense:

https://en.wikipedia.org/wiki/Polym...lectrolysis#/media/File:PEM_Elektrolyse_5.gif

However, my understanding of electronegativity, voltage, and efficiency is still lacking for both fuel cells and electrolyzers. Wiki ranges the cell voltage for an electrolyzer at 1.75-2.20 V and the range of efficiency at 57-69%. I know that splitting a mole of liquid water to produce a mole of hydrogen at 25°C requires 285.8 kJ of input energy for a battery—237.2 kJ as electricity and 48.6 kJ as heat. And that cell voltage is 1.48V at standard atm and temperature:

E_o=\frac{(∆_f H)}{zF}=\frac{285,840 J/mol}{2*96,485 C/mol}=1.48Volts/cell

The question I have is regarding half-cell potential and electronegativity. In terms of voltage, why is the hydrogen attracted to the cathode side of the cell? How does this affect the efficiency of the cell? I'm trying to use a galvanic cell as reference, but, since hydrogen isn't losing it's proton in solution the same way copper and zinc do in the example, I'm still finding myself confused.

https://en.wikipedia.org/wiki/Galvanic_cell
 
Last edited:
Chemistry news on Phys.org
To clarify:

The minimum necessary voltage to start water electrolysis under standard conditions is ≈ 1.23 V.
The voltage necessary to start water electrolysis without withdrawing heat from the surroundings under standard conditions is ≈ 1.48 V.
 
Thanks for your input! But I'm still confused about how voltage on the anode side is related to the increased electronegativity on the cathode side of the cell. Any thoughts on what forces the hydrogen protons to separate from oxygen in terms of voltage? It might be something simple, but I'm just not seeing the connection yet.
 
I think the relationship that confuses me is between the dissociation energy necessary to split 1 mole of water and the electronic potential between the electrodes. Hopefully researching electrode kinetics will put me on the right track!
 
The decomposition of a water molecule H2O – to use the simplest picture – means to transfer two electrons from the oxygen ion O-- back to two hydrogen ions H+. To accomplish this you need a certain amount of energy input.

Neglecting every detail, equilibrium thermodynamics says: Such ions are always present, whether in liquid water or at metal surfaces in contact with water. The respective concentrations depend on the given conditions.

You can now use “electrolysis cells” to separate these ions under the influence of an applied electric field (in case of PEM cells you force H+ ions which exist at the three phase boundary “anode metal/water (liquid or gaseous)/PEM membrane” to migrate through the membrane to the cathode region, O-- ions "remain" at the three phase boundary in the anode region).

At the end, if the applied voltage is high enough, you can even “extract” the electrons from the oxygen ions at the anode region and “transfer” them to the hydrogen ions which arrive at the cathode. By this, you establish a continuous decomposition of water molecules.

Have a look at http://www1.lsbu.ac.uk/water/electrolysis.html for details with respect to classical water electrolysis.
 

Similar threads

Chemistry Amount of water electrolyzed during PEM electrolysis
  • · Replies 26 ·
Replies
26
Views
5K
How to Calculate the Theoretical Voltage of a PEM Fuel Cell?
  • · Replies 3 ·
Replies
3
Views
8K