Topic 16 · page 19 of 19

The hydrogen-oxygen fuel cell

Hydrogen is oxidised at the negative electrode and oxygen is reduced at the positive electrode. The electrode equations differ in acidic and alkaline electrolytes, but the overall reaction and the emf are the same.

12 min read16.19

Read first: Fuel cells, Building redox equations

With an acidic electrolyte 16.19

In a hydrogen-oxygen fuel cell, hydrogen is supplied to one electrode and oxygen (or air) to the other. Two porous electrodes dip into an electrolyte, and an external circuit joins them.

A hydrogen-oxygen fuel cell with acidic electrolyte. Hydrogen enters at the negative electrode on the left. Electrons flow through the external load to the positive electrode on the right. H+ ions move through the electrolyte to the positive electrode, where water forms and leaves on the oxygen side.
Acidic electrolyte. H⁺ ions carry the charge to the positive electrode, and water forms on the oxygen side.
Electrode Half-equation Type
negative (hydrogen) H₂ → 2H⁺ + 2e⁻ oxidation
positive (oxygen) O₂ + 4H⁺ + 4e⁻ → 2H₂O reduction

Multiply the first by 2 so that the electrons match, then add:

2 HX2(g)+OX2(g)→2 HX2O(l) \ce{2H2(g) + O2(g) -> 2H2O(l)}

The cell voltage is Ecell⊖E^\ominus_\text{cell} = E⊖E^\ominus(O₂/H₂O) − E⊖E^\ominus(H⁺/H₂) = +1.23 − 0.00 = +1.23 V.

Hydrogen is oxidised at the electrode, releasing electrons into the external circuit. That is why the hydrogen electrode is the negative electrode.

With an alkaline electrolyte 16.19

The overall reaction is the same, but the ions present are different. There is almost no H⁺ in an alkaline solution, so the half-equations use OH⁻ and H₂O instead.

A hydrogen-oxygen fuel cell with alkaline electrolyte. Hydrogen enters at the negative electrode on the left. Electrons flow through the external load to the positive electrode on the right. OH minus ions move through the electrolyte to the negative electrode, where water forms and leaves on the hydrogen side.
Alkaline electrolyte. OH⁻ ions carry the charge to the negative electrode, and water forms on the hydrogen side.
Electrode Half-equation Type
negative (hydrogen) H₂ + 2OH⁻ → 2H₂O + 2e⁻ oxidation
positive (oxygen) O₂ + 2H₂O + 4e⁻ → 4OH⁻ reduction

The voltage is Ecell⊖E^\ominus_\text{cell} = +0.40 − (−0.83) = +1.23 V, using E⊖E^\ominus(O₂/OH⁻) = +0.40 V and E⊖E^\ominus(H₂O/H₂) = −0.83 V in alkali. It is the same as in acid because the overall reaction is the same.

MethodChecking an alkaline equation

  1. Double the hydrogen half-equation: 2H₂ + 4OH⁻ → 4H₂O + 4e⁻
  2. Add the oxygen half-equation: O₂ + 2H₂O + 4e⁻ → 4OH⁻
  3. Cancel the 4e⁻, the 4OH⁻ on each side and 2H₂O from the 4H₂O. You should get back to 2H₂ + O₂ → 2H₂O.

To convert an acid half-equation to alkali, add the same number of OH⁻ to both sides as there are H⁺, and combine H⁺ + OH⁻ into H₂O. For hydrogen: H₂ → 2H⁺ + 2e⁻ becomes H₂ + 2OH⁻ → 2H₂O + 2e⁻.

What changes and what does not 16.19

Acidic electrolyte Alkaline electrolyte
negative electrode H₂ → 2H⁺ + 2e⁻ H₂ + 2OH⁻ → 2H₂O + 2e⁻
positive electrode O₂ + 4H⁺ + 4e⁻ → 2H₂O O₂ + 2H₂O + 4e⁻ → 4OH⁻
overall 2H₂ + O₂ → 2H₂O 2H₂ + O₂ → 2H₂O
ion that carries the charge H⁺ moves to the positive electrode OH⁻ moves to the negative electrode
water forms at the positive electrode (oxygen side) at the negative electrode (hydrogen side)
Ecell⊖E^\ominus_\text{cell} +1.23 − 0.00 = +1.23 V +0.40 − (−0.83) = +1.23 V

Check yourself

  1. Write the electrode equations for a hydrogen-oxygen fuel cell with an acidic electrolyte, and state the type of reaction at each electrode.
  2. Write the electrode equations for the same cell with an alkaline electrolyte.
  3. Which electrode is the hydrogen supplied to, and why is it the negative electrode?
  4. State which ion moves through the electrolyte and in which direction, for each type of cell.
  5. Calculate Ecell⊖E^\ominus_\text{cell} for the alkaline cell, given +0.40 V for O₂/OH⁻ and −0.83 V for H₂O/H₂.
  6. Show that the alkaline half-equations combine to give 2H₂ + O₂ → 2H₂O.

Answers cover these until you have tried the questions

  1. Negative: H₂ → 2H⁺ + 2e⁻ (oxidation). Positive: O₂ + 4H⁺ + 4e⁻ → 2H₂O (reduction).
  2. Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻. Positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻.
  3. The negative electrode. Hydrogen is oxidised there, so it releases electrons into the external circuit.
  4. Acidic: H⁺ moves to the positive electrode. Alkaline: OH⁻ moves to the negative electrode.
  5. Ecell⊖E^\ominus_\text{cell} = +0.40 − (−0.83) = +1.23 V.
  6. 2 × (H₂ + 2OH⁻ → 2H₂O + 2e⁻) gives 2H₂ + 4OH⁻ → 4H₂O + 4e⁻. Add O₂ + 2H₂O + 4e⁻ → 4OH⁻: 2H₂ + 4OH⁻ + O₂ + 2H₂O + 4e⁻ → 4H₂O + 4e⁻ + 4OH⁻. Cancel 4e⁻, 4OH⁻ and 2H₂O: 2H₂ + O₂ → 2H₂O.