What happens to the hydroxide anion during the electroreduction of water? The reduction of water at a cathode is represented by the following equation:

Cathode (reduction): $\ce{2 H2O_{(l)} + 2e^{−} → H2_{(g)} + 2 OH^{-}_{(aq)}}$

If this is done in a divided cell, what would happen to the $\ce{OH-}$ anions? The hydrogen gas would accumulate and bubble would form. Would the $\ce{OH-}$ anions just build up in the solution? Do they interact with each other naturally forming $\ce{H2}$ and $\ce{O2}$?
You only show one a half-reaction, reduction at the cathode.  Write the other half-reaction, oxidation at the anode to give oxygen.  A real world water electrolysis cell requires a salt electrolyte for conductivity.  $\ce{NaCl}$ could be electrolyzed to $\ce{Cl2}$ and $\ce{Na}$, the latter reacting with water to give $\ce{H2}$ and $\ce{NaOH}$.

http://www.ineris.fr/ippc/sites/default/interactive/brefca/image16.gif

If you want $\ce{H2}$ and $\ce{O2}$, make the electrolyte $\ce{NaOH}$.  $\ce{Na2CO3}$ can be interesting.