Electrolysis: Water to Hydrogen gas? I recently conducted an experiment where I electrolyzed a solution of sodium chloride. I was using a copper coin as the anode, and I observed a pale blue precipitate in the cup when I was done with the experiment. I then deduced the following:

The chloride ions get oxidized at the anode: 

$$\ce{2Cl- -&gt; Cl2 + 2e-}$$

The chlorine gas then reacts with the copper anode to form copper(II) chloride which goes into solution as it is completely soluble.

At the cathode, the hydrogen in the water gets reduced to form hydrogen gas: 
$$\ce{2H2O + 2e- -&gt; 2H2 + 2OH-}$$

I know the above half reactions exist from the list of standard reduction potentials.

Next, the copper(II) ions react with the hydroxide ions being formed to produce copper(II) hydroxide, which is both insoluble and pale blue.

However, this source mentions that the hydronium ions in the water get reduced at the cathode: 

$$\ce{2H+ + 2e- -&gt; H2}$$

Is this correct? In my understanding, the concentration of hydronium ions ($\pu{1e-7 M}$) would be much too low, and where would the hydroxide ions come from to form copper(II) hydroxide?
In answer to your last statement about the formation of Hydrogen. The actual amount of hydronium ions doesn't matter because it is at equilibrium ($\ce{H2O &lt;=&gt; OH- + H+}$) and so once you use an $\ce{H+}$ you'll simply have more $\ce{H2O}$ dissociate into $\ce{OH-}$ and $\ce{H+}$ (Le Chatlier states) so if you have battery current running through the system you should experience a healthy, visible stream of hydrogen bubbles. I built a similar thing myself back in high school. 

The hydroxide ions will come from the water also. As mentioned above, the $\ce{H2O}$ will dissociate into $\ce{H+}$ and $\ce{OH-}$ and you are using both. The $\ce{H2O}$ will keep dissociating and try to keep the $\ce{H+}$ and $\ce{OH-}$ concentrations at $10^{-7} \, \small\text{M}$  as you keep using them. 

Feel free to question my knowledge.