How does increasing the dissolved oxygen partial pressure, pO2, at fixed pH affect the electrode potential of the O2/H2O couple?

• A. Less positive
• B. More positive
• C. Unchanged
• D. Not defined

Answer: (B)

The test is about how the electrode potential responds to changes in the activities of species in a redox couple, which is governed by the Nernst equation. For the oxygen reduction reaction O2 + 4 H+ + 4 e- → 2 H2O, the potential depends on the partial pressure of O2 and the activity of H+. Water is a pure liquid, so its activity is essentially 1, and at fixed pH the proton activity is a constant. The Nernst form simplifies to E ≈ E° + (RT/4F) ln(pO2) plus a constant offset from [H+]. Because pO2 appears inside a logarithm, increasing pO2 makes the potential more positive. At 25°C, this relation is E ≈ E°' + 0.0591/4 × log10(pO2) ≈ E°' + 0.0148 × log10(pO2), with the constant term from pH fixed.

So increasing the dissolved oxygen partial pressure raises the O2/H2O electrode potential, making it more positive.