Is the reversible voltage the same as a working cell’s voltage?
The ideal H₂/O₂ reversible voltage with liquid water product is approximately 1.229 V at standard 25°C conditions. It is an equilibrium reference. A working-cell comparison needs measured voltage and current, along with temperature, gas conditions and the cell definition.
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EXPLORE THE MODEL
Fuel-cell reaction and flow account
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What this representation means
H₂/O₂ PEM reaction; 100% Faradaic accounting. 1.229 V is the standard reversible liquid-water value at 25 °C. Assumed voltage loss is not a measured polarization curve. Consumed flow differs from supplied flow.
- Remaining reactants
- 2 H₂ + 1 O₂
- Water formed
- 2 H₂O
- Conserved inventory
- 8 hydrogen atoms + 4 oxygen atoms
- Paths
- Protons through membrane; electrons through external circuit
2H₂ + O₂ → 2H₂O; ṅ(H₂) = I/(2F); P = IV
The pictured inventory and the flow calculation answer different questions. Protons cross the membrane, electrons travel through the external circuit, and oxygen is reduced at the cathode. The assumed loss changes power; it does not change the stoichiometric reactant flow at fixed current. At zero loss the power is a reversible bound, not attainable finite-rate operation.
What this model can establish
H₂/O₂ PEM reaction; 100% Faradaic accounting. 1.229 V is the standard reversible liquid-water value at 25 °C. Assumed voltage loss is not a measured polarization curve. Consumed flow differs from supplied flow.
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What does selecting a 0.30 V loss establish?
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HISTORY THROUGH EVIDENCE
A gas battery in the archive
Historical apparatus linked a gas-consuming reaction to a sustained current.
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