The electrode reactions in an alkaline hydrogen–oxygen fuel cell.

The idea

In the alkaline cell the electrolyte supplies hydroxide ions, so both half-equations must contain OH⁻. At the negative electrode hydrogen is oxidised: H₂ + 2OH⁻ → 2H₂O + 2e⁻, for which E⦵ = −0.83 V. At the positive electrode oxygen is reduced: ½O₂ + H₂O + 2e⁻ → 2OH⁻, E⦵ = +0.40 V. Adding them cancels the electrons, one water and the hydroxide to give the overall reaction H₂ + ½O₂ → H₂O, and the EMF is +0.40 − (−0.83) = +1.23 V. Electrons leave the negative electrode and travel round the external circuit; the hydroxide ions travel the other way through the electrolyte, from the oxygen electrode to the hydrogen electrode, so charge does not build up at either end.

What earns the marks

• H₂ + 2OH⁻ → 2H₂O + 2e⁻ as the oxidation at the negative electrode.
• ½O₂ + H₂O + 2e⁻ → 2OH⁻ as the reduction at the positive electrode.
• The overall equation H₂ + ½O₂ → H₂O with the electrons cancelling.
• An EMF of +1.23 V from the E⦵ values −0.83 V and +0.40 V.

Common mistakes and how to put them right

• Using the acidic half-equations H₂ → 2H⁺ + 2e⁻ and O₂ + 4H⁺ + 4e⁻ → 2H₂O when the electrolyte is stated to be alkaline.

• Leaving OH⁻ out of the half-equations, so the charges do not balance.

• Writing hydrogen as reduced at the negative electrode instead of oxidised.

• Adding the two E⦵ values to give −0.43 V instead of subtracting to give +1.23 V.

Exam tips

• Read the electrolyte before writing anything; alkaline and acidic fuel cells have completely different half-equations.
• Check each half-equation for charge as well as atoms; in the alkaline cell it is the OH⁻ that makes the charges work.