Oxidation of aluminum and manufacturing of electrolytic capacitors
You are performing an electrolysis.
This means that on the (+) electrode de reaction is Al $\rightarrow$ Al$^{3+}$ and on the (-) electrode the reverse occurs.


Distilled water is required to avoid the presence of chloride ions which will form chlorine at the (+) electrode and attack your aluminum foil.
Sodium bicarbonate is used to raise the pH so that the formed aluminum ions precipitate as the oxide instead of dissolving as Al$^{3+}$ ions.
The oxidation is Al $\rightarrow$ Al$^{3+}$. Then, the basic pH due to the dissolution of bicarbonate: 

NaHCO$_3$ $\rightarrow$ Na$^+$ + HCO$_3^-$

HCO$_3^-$ + H$_2$O $\rightarrow$ OH$^-$ + H2CO$_3$

the aluminium oxide precipitates:

2 Al$^{3+}$ + 6 OH$^-$ $\rightarrow$ Al$_2$O$_3$ + 3 H$_2$O


The electric current forces the oxidation reaction to continue. 
In air a thin layer of oxide is formed and then the rest of the metal is not in contact with the air anymore. 
This is called passivation.

Since the aluminium ions are precipitating as oxide, in the (-) electrode the reduction is of sodium ions:

$\hspace{1cm}$Na$^{+}$ $\rightarrow$ Na

which reacts with water to form hydrogen and OH$^-$ so that the cycle is complete:

$\hspace{1cm}$2 Na + 2 H$_2$O $\rightarrow$ 2 Na$^{+}$ + 2 OH$^-$ + H$_2$