Gibbs Free Energy and the Chemistry of Green Hydrogen
Every chemistry student learns ΔG = ΔH − TΔS as a spontaneity test: negative ΔG means the reaction can happen on its own, positive means it cannot. What is rarely shown is that the same equation, applied to one reaction — splitting water — produces a hard number that the entire green hydrogen industry is built around: 1.23 volts. This article derives that number from scratch, then explains honestly why no real electrolyser runs at it.
The two equations you need
ΔG = −nFE ⟹ E = −ΔG / (nF)
The first is the definition of Gibbs free energy change. The second links it to electrochemistry: any reaction driven or delivered by moving electrons has a voltage associated with its free energy change.
| Term | Meaning | Unit |
|---|---|---|
| ΔG | Gibbs free energy change — the maximum useful (non-expansion) work | J/mol or kJ/mol |
| ΔH | Enthalpy change — total heat content difference | kJ/mol |
| T | Absolute temperature | K |
| ΔS | Entropy change of the system | J/(mol·K) |
| n | Moles of electrons transferred per mole of reaction | — |
| F | Faraday constant = 96485 C/mol | C/mol |
| E | Cell potential (reversible) | V |
Worked example — the free energy of splitting water
The reaction is H₂O(l) → H₂(g) + ½O₂(g). Standard data at 298.15 K: ΔH° = +285.8 kJ/mol, and the standard molar entropies are H₂O(l) = 69.9, H₂(g) = 130.7 and O₂(g) = 205.2 J/(mol·K).
Step 1 — entropy change.
ΔS° = S(H₂) + ½S(O₂) − S(H₂O)
ΔS° = 130.7 + (0.5 × 205.2) − 69.9
ΔS° = 130.7 + 102.6 − 69.9 = 163.4 J/(mol·K)
Step 2 — the TΔS term. Convert to kJ before subtracting:
TΔS° = 298.15 × 0.1634 = 48.72 kJ/mol
Step 3 — Gibbs free energy.
ΔG° = 285.8 − 48.72 = +237.1 kJ/mol
Step 4 — the voltage. Each water molecule needs 2 electrons, so n = 2:
E = 237100 ÷ (2 × 96485) = 237100 ÷ 192970 = 1.229 V ≈ 1.23 V
ΔG° is positive, so water does not split by itself — which is exactly why oceans exist. To force it, you must supply at least 237.1 kJ of work per mole, and delivered electrically that is 1.23 V across the cell.
Where this is actually used
Green hydrogen means hydrogen made by electrolysing water using renewable electricity, rather than by steam reforming natural gas. The whole economics of that industry is a comparison between the theoretical energy requirement above and the electricity actually consumed. Three voltages matter, and they come straight from the thermodynamics:
| Voltage | Value | What it means |
|---|---|---|
| Reversible voltage, ΔG/nF | 1.23 V | Absolute minimum; the cell would absorb heat from its surroundings to make up the difference |
| Thermoneutral voltage, ΔH/nF | 1.48 V | 285800 ÷ 192970; at this voltage the cell neither heats nor cools itself |
| Typical operating voltage | roughly 1.7–2.0 V | What industrial alkaline and PEM cells actually run at, to get useful current |
The gap between 1.23 V and the operating voltage is overpotential. It has three main sources: the activation overpotential of the oxygen-evolution reaction, which is slow because it must rearrange four electrons and form an O=O bond; the activation overpotential of hydrogen evolution, which is much smaller; and ohmic losses through the electrolyte, membrane and hardware. This is why catalyst research in electrolysis is overwhelmingly aimed at the oxygen electrode rather than the hydrogen one — the thermodynamic requirement is fixed, so the only room to improve is in the overpotential.
Efficiency figures quoted for electrolysers are usually one of these ratios. A cell at 1.8 V has a voltage efficiency of 1.23 ÷ 1.8 = 68% measured against ΔG, or 1.48 ÷ 1.8 = 82% measured against ΔH. Both numbers describe the same cell. Whenever you read an efficiency claim, the first question is which denominator was used.
The honest limitation: ΔG sets the floor, never the rate
This is the single most important thing to carry away. Gibbs free energy is a state function. It depends only on the initial and final states, so it tells you nothing whatever about the path between them — no activation energy, no mechanism, no speed. A reaction with a hugely negative ΔG can sit unchanged for centuries; diamond converting to graphite is the classic example.
So thermodynamics can tell an engineer that 1.23 V is impossible to beat, and that any claim of hydrogen produced below that voltage is wrong. It cannot tell them how much current a given catalyst will pass, how fast the cell will degrade, or whether the process is economic. That is kinetics and materials science, and it needs the Arrhenius equation, Tafel analysis and real measurement — not this formula.
Mistakes to avoid
- Unit mismatch in ΔH − TΔS. ΔH is in kJ/mol, ΔS in J/(mol·K). Using 163.4 instead of 0.1634 gives ΔG = −48,437 kJ/mol — a nonsense answer that reverses the conclusion. Always convert first.
- Wrong n. For H₂O → H₂ + ½O₂ the electron count is 2, not 4. It is 4 if you write the balanced equation as 2H₂O → 2H₂ + O₂, and then ΔG is also doubled to 474.2 kJ — the voltage stays 1.23 V. Voltage is intensive; free energy is not.
- Liquid versus gaseous water. Starting from steam, ΔG° is 228.6 kJ/mol, giving 228600 ÷ 192970 = 1.18 V. High-temperature steam electrolysis exploits exactly this. Quoting 1.23 V for a steam cell is wrong.
- Assuming standard conditions apply. The 1.23 V figure is for 25 °C, 1 bar and unit activities. Real cells run hot and pressurised, so the true reversible voltage must be corrected with the Nernst equation before it is compared with anything.
- Reading ΔG < 0 as "will happen". It means "is allowed to happen". Those are different statements, and the difference is the entire field of catalysis.
Run the numbers yourself. Enter ΔH, T and ΔS and the Gibbs Free Energy calculator returns ΔG with the unit conversion handled for you, so the kJ/J trap above cannot bite.
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