How exactly does the electrolysis of water work?

Question

When we pass electricity through water it breaks into hydrogen and oxygen gas.

$$\ce{2H2O -> 2H2 + O2}$$

What I couldn't understand is how the electrical energy is converted to breaking the bonds of water molecule. I've searched a lot but couldn't find any satisfactory answers on how the energy conversion happens.

Few thoughts :

1. The fast moving electrons from the battery hit the water molecules breaking them into hydrogen and oxygen ions.
2. I think above idea is incorrect as pure water is an insulator and blocks passage of electrons. Then it might have to do with the electric field that exists through out the water between the anode and cathode. Does this electric field somehow transfer energy to the water molecules and split them ?
3. Above idea makes sense to me as we can think of each $\ce{H2O}$ molecule as a dipole $\ce{2H+-O^2-}$, so it can interact with electric field.
4. I feel all above ideas are silly and there is some sophisticated way to reason this which I don't know yet.

Answer 1

Electrolysis of pure water is very difficult, but adding only a small amount of ions make the process easily achieved. In most places, there are enough minerals in the water that the ionic strength or conductivity of the water is great enough for electrolysis to effectively occur without needing to add additional ions to the water

Water dissociates into H+ ions and $\ce{OH–}$ ions; the $\ce{H+}$ ions are attracted to the negative electrode (the cathode) and are converted (reduced) to a hydrogen atom ($\ce{H}$) (i.e. $\ce{e- + H+ -> H}$). This is highly unstable and immediately reacts with another hydrogen atom to produce $\ce{H2}$, molecular hydrogen gas. At the other electrode (the anode), oxidation occurs. The $\ce{OH-}$ ions are attracted to the positive electrode where they are oxidized to form oxygen gas ($\ce{O2}$) and hydrogen ions ($\ce{H+}$). However, if chloride is present, it will oxidize (instead of the $\ce{OH–}$ ions) and form chlorine gas, which will then react with the water to form hypochlorous acid.

The cathode reaction is: $\ce{2H+ + 2e- (cathode) →H2 (g)}$
The anode reaction is: $\ce{2OH– → 4e-(anode) +O2 + 2H+}$
The overall reaction is: $\ce{2H2O → 2H2 (g) + O2 (g)}$

Answer 2

I think I can help with this question. Electrolysis is different than the incandenscence effect of light bulbs.

Water electrolysis requires an added chemical potential (a battery, say). This causes one electrode to be more prone to having excess electrons and behave as a cathode, reducing water to H2:
M + 2e- --> M(2-)
M(2-) + 2 H2O --> MH2 + 2 OH-
MH2 --> M + H2
The metal M acts only as a catalyst (could be Pt for instance). Note that M here is not a single atom, but a surface of many atoms, and MH2 represents a state in which there are two more-or-less covalent M-H bonds right beside each other. A quick electron bond rearrangement is all that is needed in the last step, e.g. imagine a box shape for HMMH and merely moving the bonds around: H-M M-H making M-M and H-H. No "hitting" or vibration/heat effect; just chemical potential at work.

The other electrode is, due to the battery, more prone to having electrons drawn away from it, behaving like an anode, oxidizing water to O2:
M --> M(4+) + 4e-
M(4+) + 2 H2O --> MO2 + 4 H+
MO2 --> M + O2
The quick electron bond rearrangement here can be imagined as O=M M=O making O=O and M=M (though O=O has a more complicated electronic description than a simple double bond).

--Allan in Regina