Basicall...u must be familiar with the diagram right?
so, there is an electrolyte of NaOH, anode made up of a porous material impregnated with platinum catalyst, and the cathode is made up of a porous material with either cobalt oxide or platinum catalyst. There is an inlet on the anode side for hydrogen and on the cathode side for the oxygen.
Then the reactions are as follows:-
At anode: H2 + 2OH(-) ----> 2H2O + 2e(-)
At cathode: O2 + 4e(-) +2H2O ---->4OH(-)
because, loss and gain of electrons should be same,
therefore,
[ H2 + 2OH(-) ----> 2H2O + 2e(-) ]*2
Which gives us:
2H2 + 4OH(-) --->4H2O + 4e(-)
O2 + 4e(-) + 2H2O ---> 4 OH(-)
from the above two equations, we cancel out 4 e(-) from both side, 2H2O from the equations and 4OH(-) also from the equations
So, we are left with 2H2 + O2 ---->2H2O
the electrons flow to complete the circuit and in this way we also prove the water forming equation.
All understood?
?i hope that it helps......
