Water Electrolysis and Hydrogen Production

Introduction

Electrolysis runs the overall hydrogen–oxygen reaction in the energy-storing direction: 2H₂O → 2H₂ + O₂ . Electricity supplies most or part of the required free energy, depending on temperature and heat input. Alkaline, PEM and solid-oxide electrolysers carry different ions through their separators and impose different materials demands. A fuel cell can convert stored hydrogen back to electricity, but the two steps each have losses, so the combination is not perfectly reversible in practice.

Core explanation

For acidic PEM electrolysis, anode water oxidation is 2H₂O → O₂ + 4H⁺ + 4e⁻ . Protons cross the membrane; at the cathode 4H⁺ + 4e⁻ → 2H₂ . For an alkaline cell, cathode hydrogen production can be written 4H₂O + 4e⁻ → 2H₂ + 4OH⁻ , and anode oxygen production is 4OH⁻ → O₂ + 2H₂O + 4e⁻ . OH⁻ travels toward the anode. In a common oxide-ion solid-oxide electrolyser, steam at the fuel electrode accepts electrons to form H₂ and O²⁻; O²⁻ moves through the ceramic and releases electrons while forming O₂ at the oxygen electrode. These architectures share an overall reaction but not an ion-transport pathway.

At 25 °C with liquid-water standard states, ideal reversible voltage is about 1.23 V from ΔG/(2F) per H₂. A larger thermoneutral voltage of about 1.48 V corresponds to reaction enthalpy on an HHV liquid-water basis. Operating voltage above reversible voltage supplies additional energy to overcome activation, ohmic and transport losses; it may also generate heat. High-temperature electrolysis can use supplied thermal energy to reduce the electrical fraction needed, but total heat plus electricity must still be counted. An electrical-only efficiency can therefore be misleading without heat accounting.

Alkaline electrolysers with liquid alkaline electrolyte have a long operating history and can use different catalyst materials than acidic PEM devices, but caustic electrolyte management and gas separation matter. PEM electrolysers use a proton-conducting membrane and can operate dynamically, but acidic oxygen evolution often needs durable precious-metal catalysts and suitable components. Solid-oxide electrolysers exploit high-temperature kinetics and steam heat, yet demand durable ceramics, seals and interfaces during thermal cycling. Water quality, gas purity, pressure and balance-of-plant loads affect all of them.

Faraday's law ties ideal hydrogen amount to charge: n(H₂)=Q/(2F) if every two electrons produce one H₂. At Faradaic efficiency φ F, multiply by φ F. This counts product molecules, not electrical energy efficiency. Energy efficiency additionally depends on cell voltage and auxiliary power. If produced H₂ later enters a fuel cell, round-trip electricity efficiency is approximately electrolyser electrical efficiency times fuel-cell electrical efficiency under consistent heating-value and boundary conventions, with storage and compression losses added.

Step-by-step reasoning

Choose the electrolyte and identify which ion crosses the separator. Balance each half-reaction and confirm the common overall water-splitting equation. Use Q=It and two electrons per H₂ to calculate theoretical yield; apply Faradaic efficiency separately. For energy, specify reversible, thermoneutral and operating voltage plus heat and auxiliary inputs. When evaluating hydrogen's climate value, include the source of electricity and the system boundary.

Visual explanation

Draw three electrolysers side by side. Mark H⁺ crossing PEM, OH⁻ crossing an alkaline separator and O²⁻ crossing a hot ceramic. Show hydrogen at each cathode/fuel electrode and oxygen at the opposite side. Underneath draw 1.23 V reversible, approximately 1.48 V thermoneutral and a higher example operating voltage, with arrows for kinetic and resistance losses. Connect stored H₂ by an arrow to a fuel cell with another voltage loss.

Real-world analogy

Pumping water uphill stores potential energy; releasing it through a turbine returns some electricity but friction and pumps prevent a perfect round trip. Electrolysis and fuel-cell reconversion similarly store and release chemical free energy with losses. The analogy is limited because electrolyser heat input and chemical products have value beyond a simple gravitational store.

Real-world example

A facility pairs variable renewable electricity with a PEM electrolyser, hydrogen storage and a fuel cell for later generation. Operators record kWh per kg H₂, compressor energy and fuel-cell net electrical output. Quoting only stack voltage would omit pumps, drying and storage losses. If the hydrogen is used as a chemical feedstock instead of reconverted to electricity, round-trip electrical efficiency is not the relevant performance metric.

Why?

Why does electrolysis need more than the reversible voltage at useful current? Electrode reactions and ion transport require driving overpotential. Why are hydrogen and oxygen separated? Mixing products wastes recoverable chemical energy and creates a safety hazard. Why is a high-temperature process potentially attractive? Some required energy can be supplied as heat while reaction kinetics may improve, although materials become harder to maintain.

Common misconception

The 1.23 V reversible number is not the electricity consumed per mole in a practical electrolyser at finite current. A second misconception is that hydrogen is automatically a low-emission fuel regardless of the electricity used to make it. A third is to multiply efficiency percentages that use inconsistent HHV/LHV or stack/system boundaries.

Worked example

Question: An electrolyser passes 19,297 C. At 90% Faradaic efficiency, how many moles of H₂ are produced? Use F=96,485 C mol⁻¹.

Reasoning: Ideal amount is Q/(2F)=19,297/192,970=0.100 mol H₂. Apply 0.90 product selectivity to obtain 0.0900 mol. This does not determine kWh consumed; the operating voltage and time-dependent current would be needed.

Answer: 0.0900 mol H₂.

Quick check

1. Which ion crosses the separator from cathode toward anode in a conventional alkaline electrolyser? Answer: Hydroxide ions, OH⁻, move toward the oxygen-producing anode.

Exam focus

Balance half-reactions and label the transported ion. Distinguish Faradaic yield, reversible voltage, thermoneutral voltage and system energy efficiency. Include heat input in high-temperature comparisons and use one consistent HHV or LHV basis throughout a round-trip calculation.

Advanced insight

An electrolyser can operate below the thermoneutral voltage while absorbing heat from its surroundings, or above it while rejecting heat, under suitable thermodynamic conventions. This does not violate energy conservation: electrical and thermal inputs together supply the enthalpy of water splitting. Dynamic operation can also change gas crossover, catalyst state and membrane hydration, so efficiency measured at one steady current is not the full story for variable-power systems.

Summary

Water electrolysis produces H₂ and O₂ using electrical driving force. Alkaline, PEM and solid-oxide designs differ in transported ion, temperature and materials. Faraday's law fixes ideal hydrogen yield per charge, while voltage, heat and auxiliary equipment set energy use. Fuel-cell reconversion closes a chemical storage loop with additional losses.

Practice questions

1. How many electrons are needed per H₂ molecule in water electrolysis? Answer: Two electrons per H₂ molecule.

2. Write the PEM electrolyser anode reaction. Answer: 2H₂O → O₂ + 4H⁺ + 4e⁻.

3. If ideal H₂ production is 1.0 mol and Faradaic efficiency is 80%, what is actual H₂ amount? Answer: 0.80 mol, assuming the efficiency applies over the whole charge interval.

4. Why can a solid-oxide electrolyser's electrical-only efficiency overstate its full energy performance? Answer: It may use substantial external steam or process heat, which must be included in total energy input.

Sources: US DOE, hydrogen production by electrolysis; US DOE NETL Fuel Cell Handbook; NREL, electrolyser technology comparison.