Standard Potential of the Oxygen–Water Couple

E∘(O2/H2O)=1.229 VE^{\circ}(\mathrm{O_2}/\mathrm{H_2O}) = 1.229\ \text{V}
Value1.229 V
StatusMeasured: ± 0.001 V (8.1 × 10⁻⁴ relative)
SourceUSGS PHREEQC 3 database minteq.v4.dat (MINTEQA2 version 4), the file the site's water pages use: O₂(g), O₂ + 4H⁺ + 4e⁻ = 2H₂O, log K 83.0894 at 25 °C, so E° = (2.302585 RT/4F) × 83.0894 = 1.229 V
CategoriesChemistry
Standard Potential of the Oxygen–Water Couple in every voltage unit
UnitValue
volt1.229 V
millivolt1,229 mV
kilovolt0.001229 kV
picovolt1,229,000,000,000 pV
nanovolt1,229,000,000 nV
microvolt1,229,000 μV
megavolt0.000001229 MV
gigavolt1.2290000e-09 GV

Learning zone

At +1.229 V, dissolved oxygen is a powerful enough oxidant to corrode almost every structural metal, and it is the cathodic reaction in nearly all aqueous corrosion: iron dissolves at the anode, oxygen is reduced at the cathode, and the loop closes through the water. That is why deaeration is the first line of defence in boiler feedwater treatment, and why oxygen scavengers are dosed to chase the last few parts per billion.

The same 1.229 V is the thermodynamic minimum for splitting water, which is where every green-hydrogen cost model starts. Real electrolysers need 1.8–2.0 V because the four-electron oxygen evolution reaction is kinetically sluggish and demands a substantial overpotential: the gap between 1.23 and reality is the efficiency loss the entire catalysis field is trying to close. Note that the potential is pH-dependent, falling by 59.16 mV per pH unit, so in neutral water it is nearer +0.815 V.

Revised 1 November 2026
Cite this page

formula.expert. (2026, November 1). Standard Potential of the Oxygen–Water Couple. https://formula.expert/constants/standard-potential-oxygen-water