What is the equation for the reduction of water?
I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!

Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann).
The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, it is an electrolysis of water experiment. What will see at the side connected to the negative terminal of the battery that it produces a gas, which is twice the volume as compared to the side connected to the positive terminal of the battery.
This gas which is very light, burns with a blue flame and it forms water again- so this must be hydrogen. Secondly, another interesting effect is that electrode region which is producing hydrogen becomes highly alkaline. If you put phenolphthalein indicator, the negative terminal becomes pink indicating an excess of OH(-) ions.



So three things can convince us with the following:
$\ce{2H2O + 2e- -&gt; H2 + 2OH-}$
a) The electron rich electrode is the cathode in electrolysis cell. Water can gain electrons here.
b) The cathode becomes highly alkaline during electrolysis
c) The cathode produces hydrogen gas, which is twice in volume as compared to oxygen.
d) You can even calculate the number of electrons used to produce a given volume of hydrogen from the measured current and the time.