Question

One of the fuel cells uses the reaction of hydrogen and oxygen to form water. What is the cell reaction taking place in the anode and cathode of that fuel cell?

Answer 1

We know that at anode, oxidation occurs and at cathode, reduction occurs. Oxidation process is basically addition of oxygen atoms and reduction process if addition of hydrogen. So, find out which molecules need to be oxidized and which molecules need to be reduced in order to produce water.

Complete answer:
In the given fuel cell, we have hydrogen molecules and oxygen molecules and these are being converted to water.
For hydrogen molecules, in order to convert these to water we need to add oxygen atoms. This process is done by oxidation of hydrogen molecules. We can write this oxidation reaction as follows
$\rm{{H_2}(g) + 2O{H^ - }(aq) \to 2{H_2}O(l) + 2{e^ - }}$
Since this reaction is an oxidation reaction, it occurs at the anode part of the cell.
For oxygen molecules, In order to convert them to water we need to add hydrogen atoms. This process can be done by reducing the molecules using water and electrons. We can write the equation for this reaction as follows
$\rm{{O_2}(g) + 2{H_2}O(l) + 4{e^ - } \to 4O{H^ - }(aq)}$
Since this reaction is a reduction reaction, this occurs at the cathode part of the cell.
By balancing and adding these two equations, we get the net equation of the redox reaction as
$\rm{2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l)}$

Note:
In order to find the equations for the reactions occurring at the cathode and the anode, we find out the equations for the oxidation and reduction reactions and their occurrence at each part of the cells. This way, we get the equations for individual reactions as well as the net reaction.