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The actual value for open circuit voltage of an operating electrolyzer will lie between the 1.23 V and 1.48 V depending on how the cell/stack design utilizes the thermal energy inputs. This is however quite difficult to determine or measure because an operating electrolyzer also experiences other voltage losses from internal electrical resistances, proton conductivity, mass transport through the cell and catalyst utilization to name a few.

Cathode reaction

The half reaction taking place on the cathode side of a PEM electrolyzer is commonly referred to as the Hydrogen Evolution Reaction (HER). Here the supplied electrons and the protons that have conducted through the membrane are combined to create gaseous hydrogen.

${\displaystyle {\ce {4H+ (aq) + 4 e^- -> 2H2 (g)}}}$

The illustration below depicts a simplification of how PEM electrolysis works, showing the individual half-reactions together along with the complete reaction of a PEM electrolyzer. In this case the electrolyzer is coupled with a solar panel for the production of hydrogen, however the solar panel could be replaced with any source of electricity.

Diagram of PEM electrolyzer cell and the basic principles of operation.

Second law of thermodynamics

As per the second law of thermodynamics the enthalpy of the reaction is:

${\displaystyle \Delta H=\underbrace {\Delta G} _{\textrm {elec.}}+\underbrace {T\Delta S} _{\textrm {heat}}}$

Where ${\displaystyle \Delta G}$ is the Gibbs free energy of the reaction, ${\displaystyle T}$ is the temperature of the reaction and ${\displaystyle \Delta S}$ is the change in entropy of the system.

${\displaystyle {\ce {H2O (l) + \Delta H -> H2 + 1/2 O2}}}$

The overall cell reaction with thermodynamic energy inputs then becomes:

${\displaystyle {\ce {H2O(l)->[+\overbrace {237.2\ {\ce {kJ/mol}}} ^{\ce {electricity}}][+\underbrace {48.6\ {\ce {kJ/mol}}} _{\ce {heat}}]{H2}+1/2O2}}}$

The thermal and electrical inputs shown above represent the minimum amount of energy that can be supplied by electricity in order to obtain an electrolysis reaction. Assuming that the maximum amount of heat energy (48.6 kJ/mol) is supplied to the reaction, the reversible cell voltage ${\displaystyle V_{\textrm {rev}}^{0}}$ can be calculated.

Open circuit voltage (OCV)

${\displaystyle V_{\textrm {rev}}^{0}={\frac {\Delta G^{0}}{n\cdot F}}={\frac {237\ {\textrm {kJ/mol}}}{2\times 96,485\ {\textrm {C/mol}}}}=1.23{\textrm {V}}}$

where ${\displaystyle n}$ is the number of electrons and ${\displaystyle F}$ is Faraday's constant. The calculation of cell voltage assuming no irreversibilities exist and all of the thermal energy is utilized by the reaction is referred to as the lower heating value (LHV). The alternative formulation, using the higher heating value (HHV) is calculated assuming that all of the energy to drive the electrolysis reaction is supplied by the electrical component of the required energy which results in a higher reversible cell voltage. When using the HHV the voltage calculation is referred to as the thermoneutral voltage.

${\displaystyle V_{\textrm {th}}^{0}={\frac {\Delta H^{0}}{n\cdot F}}={\frac {285.9\ {\textrm {kJ/mol}}}{2\times 96,485\ {\textrm {C/mol}}}}=1.48{\textrm {V}}}$

Voltage losses

The performance of electrolysis cells, like fuel cells, is typically compared through polarization curves, which are obtained by plotting cell voltages against current densities. The primary sources of increased voltage in a PEM electrolyzer (the same also applies for PEM fuel cells) can be categorized into three main areas, Ohmic losses, activation losses and mass transport losses. Due to the reversal of operation between a PEM fuel cell and a PEM electrolyzer, the degree of impact for these various losses is different between the two processes.^[1]

${\displaystyle V_{\textrm {cell}}=E+V_{\textrm {act}}+V_{\textrm {trans}}+V_{\textrm {ohm}}}$

A PEM electrolysis system's performance can be compared by plotting overpotential versus cell current density. This essentially results in a curve that represents the power per square centimeter of cell area required to produce hydrogen and oxygen. Conversely to the PEM fuel cell, the better the PEM electrolyzer the lower the cell voltage at a given current density. The figure below is the result of a simulation from the Forschungszentrum Jülich of a 25 cm² single cell PEM electrolyzer under thermoneutral operation depicting the primary sources of voltage loss and their contributions for a range of current densities.

Polarization curve depicting the various losses attributed to PEM electrolysis cell operation.

Faradaic losses and crossover

Faradaic losses describe the efficiency losses that are correlated to the current, that is supplied without leading to hydrogen at the cathodic gas outlet. The produced hydrogen and oxygen can permeate across the membrane, referred to as crossover.^[13] Mixtures of both gases at the electrodes result. At the cathode, oxygen can be catalytically reacted with hydrogen on the platinum surface of the cathodic catalyst. At the anode, hydrogen and oxygen do not react at the iridium oxide catalyst.^[13] Thus, safety hazards due to explosive anodic mixtures hydrogen in oxygen can result. The supplied energy for the hydrogen production is lost, when hydrogen is lost due to the reaction with oxygen at the cathode and permeation from the cathode across the membrane to the anode corresponds. Hence, the ratio of the amount of lost and produced hydrogen determines the faradaic losses. At pressurized operation of the electrolyzer, the crossover and the correlated faradaic efficiency losses increase.^[13]

Hydrogen compression during water electrolysis

Hydrogen evolution due to pressurized electrolysis is comparable to an isothermal compression process, which is in terms of efficiency preferable compared to mechanical isotropic compression.^[13] However, the contributions of the aforementioned faradaic losses increase with operating pressures. Thus, in order to produce compressed hydrogen, the in-situ compression during electrolysis and subsequent compression of the gas have to be pondered under efficiency considerations.

PEM efficiency

When determining the electrical efficiency of PEM electrolysis, the HHV can be used.^[15] This is because the catalyst layer interacts with water as steam. As the process operates at 80 °C for PEM electrolysers the waste heat can be redirected through the system to create the steam, resulting in a higher overall electrical efficiency. The LHV must be used for alkaline electrolysers as the process within these electrolysers requires water in liquid form and uses alkalinity to facilitate the breaking of the bond holding the hydrogen and oxygen atoms together. The lower heat value must also be used for fuel cells, as steam is the output rather than input.

PEM electrolysis has an electrical efficiency of about 80% in working application, in terms of hydrogen produced per unit of electricity used to drive the reaction.^[16][17] The efficiency of PEM electrolysis is expected to reach 82-86%^[18] before 2030, while also maintaining durability as progress in this area continues at a pace.^[19]