Why does NaCl act as a catalyst for the reaction of CuSO4 + H2O + Al?

I am analyzing a reaction between water sulfate copper and aluminum. A chunk of $\ce{Al}$, e.g., aluminum foil. I realize that a layer of $\ce{Al2O3}$ forms on the surface of the $\ce{Al}$, passivizing it, and I realize that the $\ce{NaCl}$ (or rather just the $\ce{Cl^-}$) cleans the aluminum oxide off the surface of the chunk of aluminum, but I don't understand why. I'd appreciate greatly if somebody could offer me an explanation.

Furthermore, (I didn't exactly include this in the topic of my question,) why is it that the reaction looks like this:

$$\ce{6H2O + 3CuSO4 + 5Al \rightarrow 3H2 + Al2(SO4)3 + 3Al(OH)2 + 3Cu}$$

(or something along the lines of that) rather than:

$$\ce{3CuSO4 + 2Al -> Al2(SO4)3 + 3Cu}$$ ?

Something along these lines has to be the case, because the reaction yields gas..

• You should split your first equation into two, one where Al reduces water, one where Al reduces Cu. Both reactions happen more or less independently, there is no absolute stochiometric relation between them. – Karl Nov 15 '16 at 21:44
• Would that imply that clean Al would react with H2O to produce H2 gas equally as vigorously as when it's reacting with CuSO4 + H2O solution and forming H2 gas, if at the same temperature? Also, thanks for the suggested edit. – Jake Waitze Nov 15 '16 at 21:49
• No, that's what I meant by "more or less". I would not remove the first equation, just add a third that describes the evolution of H2 from water + Al. There is surely some interesting interaction between the two parts. Btw. Al-sulfate is soluble in water. – Karl Nov 15 '16 at 22:00