# Voltage potential in PEM electrolyzers

• HelloCthulhu
In summary, electrolysis using PEM fuel cells can be used to produce hydrogen. The cell voltage is 1.48 volts at standard conditions and the efficiency ranges from 57 to 69 percent.
HelloCthulhu
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:
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.

## 1. What is the purpose of voltage potential in PEM electrolyzers?

The voltage potential is necessary for the electrolysis process to occur in PEM electrolyzers. It provides the necessary energy for the splitting of water molecules into hydrogen and oxygen gas.

## 2. How does the voltage potential affect the efficiency of PEM electrolyzers?

The voltage potential plays a crucial role in the efficiency of PEM electrolyzers. Higher voltage potentials result in a faster rate of electrolysis, leading to higher efficiency. However, excessively high voltage potentials can also lead to energy loss and decreased efficiency.

## 3. What factors can impact the voltage potential in PEM electrolyzers?

The voltage potential in PEM electrolyzers can be affected by several factors, including the type and quality of the membrane, the temperature of the electrolyte, the concentration of the electrolyte, and the design of the electrodes.

## 4. How do you measure the voltage potential in PEM electrolyzers?

The voltage potential in PEM electrolyzers can be measured using a voltmeter or multimeter. These tools can be connected to the anode and cathode of the electrolyzer to measure the voltage difference between them.

## 5. What is the ideal voltage potential for PEM electrolyzers?

The ideal voltage potential for PEM electrolyzers depends on various factors and may vary for different applications. Generally, a voltage potential of 1.23 volts is considered ideal for the electrolysis of water, as this is the minimum energy required to split water molecules into hydrogen and oxygen gas.

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