Question

Write the cathodic and anodic cell reactions of a Hydrogen – Oxygen fuel cell.

Answer 1

A fuel cell can be understood as a device that can produce electrical energy using some non – conventional sources as fuel. A fuel cell is used to convert chemical potential energy into electrical energy. The most important part of a fuel cell is the PEM or the Proton exchange membrane. You can visualise a proton exchange membrane as a sieve which only let’s sub – atomic particles to pass through. This fuel cell uses hydrogen and oxygen gas as fuel to generate electricity. Also, platinum is used as a catalyst in this fuel cell. This catalyst is coated on one side of the PEM.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
Like any electrolytic cell, the fuel cell too has an anode and a cathode. An anode is basically the negative electrode of the cell where the conduction of electrons takes place. On the other hand, the cathode is the positive electrode of the fuel cell. In a fuel cell, the cathode conducts the electrons to the catalyst via an external circuit.
Now, the two fuel gases are used on either electrode of the cell. Hydrogen gas is pressurized and is made to enter the cell via the anode. Hydrogen gas is forced through the catalyst by pressure. This causes the hydrogen atom to split into protons and electrons and it goes directly to the cathode side via an external circuit. While moving through the external circuit, these electrons can be used for useful work like powering a machine. After this, the electrons move towards the cathode.
On the other hand, oxygen gas is pressurized and forced through the catalyst from the cathode. This causes the diatomic oxygen molecule to divide into two atoms. Both these atoms are negatively charged and attract the positive hydrogen ions formed at the anode. These ions now combine to form water molecules.
The electrode reactions for this process can be given as:
Anode: $2{H_2} \to 4{H^ + } + 4{e^ - }$
Cathode: ${O_2} + 4{H^ + } + 4{e^ - } \to 2{H_2}O$
Overall reaction: $2{H_2} + {O_2} \to 2{H_2}O$

Note: By converting chemical potential energy directly into electrical energy, fuel cells avoid the “thermal bottleneck” (a consequence of the ${2^{nd}}$ law of thermodynamics) and are thus inherently more efficient than combustion engines, which must first convert chemical potential energy into heat, and then mechanical work.