Oxidization of Copper In Electrolysis
Sodium compounds:


$\ce{Na2C2}$ is sodium carbide. It reacts violently with water. $\ce{Na2CO3}$ is sodium carbonate. $\ce{NaO}$ is not a valid formula for any of the oxides of sodium: $\ce{Na2O},\ \ce{Na2O2},$ and $\ce{NaO2}$ are all real compounds, and they all react with water to produce $\ce{NaOH}$, which is probably what is formed.


  "Hydrocarbons"


If indeed you had $\ce{Na2C2}$, then you would form $\ce{H2C2}$ and $\ce{NaOH}$ when it was placed in water. However, you would not need electrolysis to do so. This is an acid-base reaction that is spontaneous under normal circumstances. $\ce{H2C2}$ cannot be the brown sludge on your cathode. It is a gas.

$$\ce{Na2C2 + 2H2O -&gt; 2NaOH + H2C2 ^}$$


  What actually happened?


In an electrolytic cell, the process that occurs (at least under standard conditions) is the one with the least negative standard potential $E^o_{cell}$.

The possible oxidations are either the oxidation of copper or the oxidation of water. Sodium carbonate is the supporting electrolyte.


$\ce{2Cu + 2OH- -&gt; Cu2O + H2O +2e-}\ \ \ E^o=+0.36 \ \text{V}$
$\ce{Cu -&gt; Cu^{2+} + 2e-} \ \ \ E^o=-0.34 \ \text{V}$
$\ce{4OH- -&gt;O2 + 2H2O +4e-}\ \ \ E^o=-0.40 \ \text{V}$
$\ce{2H2O -&gt; O2 + 4H+ +4e-} \ \ \ E^o=-1.23\ \text{V}$


Since the oxidation of copper to $\ce{Cu^{2+}}$ has the least negative $E^o$, it is the most likely oxidation. $\ce{Cu^{2+}}$ also explains the color, since aqueous solutions of $\ce{Cu^{2+}}$ are blue. The first equation is also likely, since it has a positive $E^o$, however it does not explain the color. 

The possible reductions are the reduction of water/hydroxide and the reduction of $\ce{Cu^{2+}}$, since we know that it is being formed.


$\ce{2H2O + 2e- -&gt; H2 + 2OH-} \ \ \ E^o=-0.83 \ \text{V}$
$\ce{Cu^{2+} + e- -&gt; Cu+} \ \ \ E^o=+0.16 \ \text{V}$


The reduction of $\ce{Cu^{2+}}$ to $\ce{Cu^{+}}$ is most likely, but not initially. Initially, the is no $\ce{Cu^{2+}}$, so the only reduction that can happen at the cathode is the first one, which produces hydrogen gas. The redox then might be: $$\ce{Cu + 2H2O -&gt; H2 + Cu^{2+} + 2OH-} \ \ \ E^o =-1.19 \ \text{V}$$

Once the $\ce{Cu^{2+}}$ diffuses over to the cathode, a different reduction starts to compete, with a different net redox reaction: $$\ce{2Cu + H2O -&gt; H2 + Cu2O} \ \ \ E^o=-0.47 \ \text{V}$$

This redox reaction still produces hydrogen, but it has a much less negative $E^o_{cell}$. The other peculiarity is that $\ce{Cu2O}$ is produced at the cathode after $\ce{Cu^{2+}}$ has diffused through the cell. $\ce{Cu2O}$ is your brown sludge, since like rust, it is both brown and completely insoluble in water.