# When water undergoes electrolysis where in the beaker does the water molecule break apart?

When I was thinking about it I remembered reading two things, one was that certain gases "form" on certain electrodes. The other was something like, such and such ion "travels" towards such and such electrode. Does the molecule break up on the electrodes? Or perhaps it breaks up at a random spot between?

• The water molecule gets split up into a cation and an anion in the solution, the ions then rush to their respective electrodes and finally gain or lose electrons to form the gases you're talking about. – Berry Holmes May 3 '17 at 20:15
• I think you do not know much about electrolysis. – Pritt says Reinstate Monica May 4 '17 at 2:38

The water molecules form long networks of hydrogens bonds; let's imagine one such H-bonded chain that stretches from the anode to the cathode. Reduction will occur at the cathode, so the nearest water molecule (on the end of our chain) will be deprotonated and that proton will be reduced to $H\cdot$ radical, which can combine with a nearby $H\cdot$ to form the reduction product, molecular hydrogen.
Now, the water molecule that was deprotonated can re-protonate itself with the hydrogen of its neighbor (the second $H_2O$ in our hypothetical chain), likely the very hydrogen that they were sharing as an H-bond. This leaves the second water molecule deprotonated, so it fully claims the proton the third molecule was offering in their H-bond.
At the anode, hydroxide will be oxidized to $O_2$ through a few steps that aren't necessarily relevant to your question. The big idea is that the hydroxide will be created from a water molecule at the anode end of the chain of H-bonded molecules, and it will probably happen nearly simultaneously as a water molecule at the cathode end is deprotonated to form the proton reactant. Most probably both ends of our H-bonded chain of water molecules will react at more or less the same time.