1. Electrolysis
Electrolysis is the term used to describe any chemical reaction that is powered by an electric current. The reaction you are seeing is the breaking up of water into its two component elements: hydrogen and oxygen.
The reaction can be written this way: H2O(l)  H (g) + O (g)
Since water molecules contain twice as many H’s as O’s, we need to put a 2 in front of the H in the reaction above. This is called a coefficient.
Both hydrogen and oxygen are “diatomic.” That means that whenever they occur as pure elements they are most stable in pairs – two atoms bonded together:
So that means there need to be little 2’s after the H and O.
These are called “subscripts.”
Finally, to balance the whole equation, we need another “2” coefficient in front of the H2O.
That ensures that there are the same number of H’s and O’s on both sides.
But how is this actually occurring in the Hoffman apparatus you are looking at?
It turns out that it is a little more complicated, and that the reaction actually occurs as two separate half reactions: one at each of the two electrodes. 2 2 2 2 O H O H

2. First let’s understand what these electrodes are all about
First let’s understand what these electrodes are all about. They are simply metal strips through which electricity (a flow of electrons) is being pumped.
The electrons are being pumped toward one electrode. That gives it a surplus of electrons, so it is negatively charged. (Recall that electrons are negatively charged).
This negatively charged electrode is often referred to as the “cathode.”
The other electrode is having electrons pumped away so it has a deficit of electrons and this gives it a positive charge.
[It’s hard to draw a deficit of electrons, so we’ll just label it “+”] - + - - - - - - - - - -

3. - H H O O - - H H - - - - - H - O O H - - H H - - -
In the water molecule, the H’s all have a sort of 1+ charge and the O’s all have a sort of 2- charge.
So let’s take a look at what happens at the cathode:
Since the cathode is neg. charged, it will attract the H ends of the water molecules.
Once there, the H’s will break off of their molecules, grab one electron each, and then bond together into an H2 molecule.
Thus, we form two H2 molecules and we also form four negatively charged OH- ions in solution: 4 H2O(l)  2 H2(g) + 4 OH-(aq) - + H + H O 2- O - - H + 2- H + - - - - + - H + - O O 2- H 2- - - + H H + - - -

4. + H O H - - - - H O H Now let’s look at what happens at the anode.
Since the anode is pos. charged, it will attract the O ends of the water molecules.
Once there, the O’s will break off of their molecules, give up two electrons to the “electron hungry” anode, and then bond together into an O2 molecule.
Thus, we form one O2 molecule and we also form four positively charged H+ ions in solution: 2 H2O(l)  O2(g) + 4 H+(aq) + + H O 2- H + - - - - O 2- + H H +

5. - + H H O H O - - - H - - - H - H O H O - - - H
Although you do not see it, there is another reaction that takes place in the cross-bar between the two electrodes.
The four OH- ions from the cathode reaction, come together with the four H+ ions from the anode reaction…
And form 4 H2O molecules! OH-(aq) H+(aq)  4 H2O(l)
If we add in the other two reactions: H2O(l)  2 H2(g) + 4 OH-(aq)
2 H2O(l)  O2(g) + 4 H+(aq)
And cancel out the repeats on each side:
It gives us the overall reaction we already stated: 2 H2O(l)  2 H2(g) + O2(g)
TA DA! - + H + H O 2- + H O + 2- - - - H + - - - H + - H O 2- + H O + 2- - - - H +