Electrolysis of Water
Nobody has given an answer yet, so I'll give it a crack.
In electrolysis, as you probably know, there are two electrodes – the cathode (negative) and the anode (positive). We can deal with what happens at each separately, then tie them together.
At the cathode:
$$\ce{2H2O + 2e- -&gt; H2 + 2OH-}$$
You're interested in the mechanism of these reactions. We know that water exists in solution in an equilibrium:
$$\ce{H2O + H2O &lt;=&gt; H3O+ + OH-}$$
At the cathode, electrons are supplied to the $\ce{H3O+}$ ions, reducing them:
$$\ce{H3O+ + e- -&gt; H2O + H* }$$
With many reactions happening at once, the $\ce{H*}$ radicals react together, forming $\ce{H2}$. By removing the $\ce{H3O+}$ ions, it shifts the equilibrium of water to the right, to oppose the change – removing $\ce{H3O+}$ ions and increasing the concentration of $\ce{H2O}$:
$$\ce{H2O + H2O &lt;=&gt;&gt; H3O+ + OH-}$$
Because the concentration of $\ce{OH-}$ ions doesn't change, the local solution around the cathode becomes basic – this can be done in a school lab. Add indicator. Electrolyse. Observe the color change near the cathode.
At the anode:
$$\ce{2H2O -&gt; O2 + 4H+ + 4e-}$$
Here I suspect the mechanism is a bit similar:
$$\ce{H2O + H2O &lt;=&gt; H3O+ + OH-}$$
We take the $\ce{OH-}$ and oxidise it:
$$\ce{OH- -&gt; OH* + e-}$$
This results in a radical – the hydroyxl radical – being formed. The steps from here, to get to water being formed are probably quite complicated as a mechanism.