# PEM electrolyzer - How does platinum reduce activation energy?

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:

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!

Platinum does not reduce activation energy it is only used for electrodes because it resists oxidation better then other materials.

Borek
Mentor
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.

Astronuc