electrochemistry - What are possible side reactions for the generation of hydrogen via electrolysis?

I recently completed making a simple water electrolysis setup to generate hydrogen which uses two graphite rods from a standard AA battery. The power source is a $24~\mathrm{V}$ $6~\mathrm{A}$ adapter. I am using salt water as electrolyte (The salt is iodized). Strangely, a minute or so after running, a black precipitate starts to form on the negative terminal.
Also, besides the hydrogen, another gas is generated which smells sour, not very sour, but somewhat like lemonade. I have no idea what is causing this.
I thoroughly cleaned the graphite rods before using them. Has anybody an idea about what reaction might be happening here?

Answer

First of all, check again your electrodes to see if this 'black precipitate' is formed at the negative electrode as you state; I predict it's actually happening at the positive electrode for the following reason:

Since you're using two graphite electrodes, it isn't a black precipitate that's forming, but is indeed a side reaction as you state. I suggest what you're actually making is oxygen gas at the positive terminal (as one would predict) which then reacts with the graphite electrode surface to generate $\ce{CO2}$. To verify this,take a look at your electrode—it should have started to break up, now having a non-uniform surface. Perhaps even a bit crumbly.

As LDC3 points out in his comment, you're forming $\ce{Cl2}$ as well as $\ce{O2}$ which is the characteristic smell you're noticing. To understand this we need to look at the standard electrode potential for each of these reactions. Forming $\ce{O2}$ from water at constant $\mathrm{pH}$ (compared to the standard hydrogen electrode) requires a potential difference of $-1.23~\mathrm{V}$ (this value is negative since this is the reverse of the tabulated version). Chlorine evolution on the other hand requires a p.d. of $-1.36~\mathrm{V}$. Remember that the larger (the less negative in this case) the reduction potential, the higher the tendency for reduction. So since $\ce{Cl-}$ reduction potential is more negative, it's less likely to undergo reduction and hence more likely to be oxidised (lose electrons) than water, which agrees with our chemical intuition.

So what we say here is that $\ce{Cl-}$ is preferentially oxidised over water, and as a consequence $\ce{Cl2}$ is formed preferentially over $\ce{O2}$. The reason why you have any $\ce{O2}$ forming at all is due to the electrolyte being the limiting reagent. Your supporting solvent (water) is in such abundance, essentially completely surrounding the electrode, that the system simply has to oxidise some water molecules some of the time.

Of course, experiment should rule the day, so revise your experiment to provide evidence for/refute my theoretical framework. Replace your positive terminal with a high purity metal electrode (copper should work and be readily available at high purity). As for the choice of electrolyte, test a couple of salts whose reduction potentials are less negative than water. Hint: perhaps look at sulphate salts and try and find a further reason why these are very useful in electrochemistry.