PEM electrolyzer - How does platinum reduce activation energy?

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SUMMARY

The discussion centers on the role of platinum in PEM electrolyzers, specifically how it influences activation energy during the dissociation of water (H2O). While some participants argue that platinum does not reduce activation energy, others clarify that platinum black possesses catalytic properties that lower activation energy, facilitating more efficient electrolysis. The operating specifications of the Horizon Mini PEM electrolyzer include an input voltage of 1.8V to 3V and a hydrogen production rate of 7ml per minute at 1A. Understanding these dynamics is crucial for optimizing the performance of PEM electrolyzers.

PREREQUISITES
  • Understanding of PEM electrolyzer operation and specifications
  • Familiarity with the Arrhenius equation and its application in chemical kinetics
  • Knowledge of catalytic properties of materials, specifically platinum black
  • Basic principles of electrolysis and gas production rates
NEXT STEPS
  • Research the Arrhenius equation and its application in electrochemical reactions
  • Explore the catalytic properties of platinum black in electrolysis
  • Investigate the effects of overpotential on electrolysis efficiency
  • Learn about alternative electrode materials and their performance compared to platinum
USEFUL FOR

This discussion is beneficial for electrochemists, materials scientists, and engineers involved in the design and optimization of PEM electrolyzers and those interested in the catalytic properties of electrode materials.

HelloCthulhu
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I've been researching pem electrolyzers, but still don't understand how to mathematically express how platinum reduces the activation energy necessary to dissociate H2O. I've seen the Arrhenius equation solved before, but didn't understand how to get the values for it. Here are the operating specs for the electrolyzer:

https://www.fuelcellstore.com/manuals/horizon-mini-pem-electrolyzer-instructions-fcsu-010.pdf

- Input Voltage: 1.8V ~ 3V (D. C.)

- Input Current: 0.7A

- Hydrogen production rate: 7ml per minute at 1A

- Oxygen production rate: 3.5ml per minute at 1A

In a previous post, I tried to figure out what the volumetric flowrate would be. Not sure if the mass is necessary to find the activation energy, but I'll include it anyway:

https://www.physicsforums.com/threads/amount-of-water-electrolyzed-during-pem-electrolysis.992823/

2H2O —> 2H2 + O2

3.5 ml O2/min = 0.00015625 M O2(STP)

0.0003125 H2O(l) →0.0003125 H2(g)/1A*min + 0.00015625O2(g)/1A*min
0.0003125M H2O *18 g = 0.005625 g/1A*min H2O

Any help is greatly appreciated!
 
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Platinum does not reduce activation energy it is only used for electrodes because it resists oxidation better then other materials.
 
Miloje said:
Platinum does not reduce activation energy it is only used for electrodes because it resists oxidation better then other materials.

It is not that simple - platinum black has some catalytic properties which make it perfect for electrode material, as when used most processes work with lower overpotential. And - as it is a case with catalysis - it works by lowering activation energy.
 
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