In this page we describe, briefly, the theoretical functioning of the “Hydrogen Fuel Cell” we use in our systems. An attentive reader may notice that the electrochemical reactions we describe in this page are the “inverse” of the reactions we describe at the page Electrolysis of the Water, indeed we can see these 2 reactions, together, form the “Hydrogen – Water” Energy Cycle.
Description of the Process
To simplify the concept, we can say that the core of the Fuel Cell is formed by “2 Electrodes”, separated by an “Electrolyte Membrane”.
The “Anode Electrode” contains a catalyst substance, that is able to break the Hydrogen (H2), that is forced to get in contact with this Electrode, producing positive ions (H+) and negative electrons (e–).
The “Electrolyte Membrane” is permeable only to the positive ions (H+), which therefore can cross it, arriving directly to the “Cathode Electrode”, but is not permeable to the electrons (e–), which so, have to “run” in the external circuit generating the electrical current we use for our applications.
The “Cathode Electrode”, that is forced to get in contact with the oxygen, and that receives the positive ions (H+) and the negative electrons (e–), allows these to form water, rebalancing the total reaction.
Description of the Electrochemical “Redox Reaction”
A “Redox Reaction” (reduction–oxidation reaction), to simplify the concept, can be divided in the 2 half reactions: “Reduction Reaction” and “Oxidation Reaction”.
Oxidation Reaction
At the “Anode Electrode” an oxidation reaction occurs, where 2 molecules of Hydrogen (H2) split in:
- 4 Ions of Hydrogen (H+).
- 4 Free electrons (e–).
Oxidation at Anode: 2 * H2 => 4 * H+ + 4e–
Reduction Reaction
At the “Cathode Electrode” a reduction reaction occurs, where an Oxygen molecule (O2) reacting with 4 Ions of Hydrogen (H+) and 4 Free electrons (e–) goes to form 2 molecules of Water (H2O).
Reduction at cathode: O2 + 4 * H+ + 4e– => 2 * H2O
Redox Reaction
So the whole process can be represented by the following expressions:
Oxidation at Anode: 2 * H2 => 4 * H+ + 4e–
Reduction at cathode: O2 + 4 * H+ + 4e– => 2 * H2O
that combined can give the final expression:
Overall reaction: 2 * H2 + O2 => 2 * h2O
