Chemistry - How does aluminium react with bases to form aluminates?

Solution 1:

There's no positive charge on Al, yet it reacts with hydroxide groups floating around.

Your observation is correct. In the beginning, aluminium in its elemental state has the oxidation state 0. Apparently, that changes in the course of the reaction: $$\ce{Al <=> Al^3+ + 3e-}$$

Where will the electron from a non-ionic aluminium atom go?

Again, your observation is correct. If one species is oxidized, another one must get reduced. In this case, protons from the water are reduced to form hydrogen gas:

$$\ce{ 2H2O + 2e- <=> 2OH- + H2 ^}$$

If you combine both half reactions and balance the stoichiometry, you end up with the initial equation for the whole redox reaction.

Solution 2:

There are multiple layers here. Let’s initially substitute aluminium for something as reactive as lithium. Throw lithium into water and it will react forming hydrogen according to the following equation:

$$\ce{2Li + 2 H2O -> 2 LiOH + H2 ^}$$

Technically, this is nothing else than the typical metal dissolution in acidic media, which typically follows the equation:

$$\ce{2 M + 2$n$ H+ -> 2 M^{$n$+} + $n$ H2 ^}$$

Albeit here we have a less acidic media, so instead of adding $\ce{H+}$ to the reaction, we add water and remove hydroxide:

$$\ce{2 M + 2$n$ H2O -> 2 M^{$n$+} + 2$n$ OH- + $n$ H2 ^}$$

Basically, this is equivalent to taking water as a Brønsted acid.

Metals are divided into noble and non-noble metals depending on whether they should react with $1~\mathrm{M}\,\ce{HCl}$ solution, i.e. their standard potential. (Aluminium would love to react with acids because it is a very non-noble metal.) But that is relatively irrelevant, we’re in water (or base) so the standard potential shifts and not everybody knows (or cares) what it shifs to. Rather, it is important to note that aluminium is doing nothing else than what I wrote up there for a general metal, namely dissolving in water with water being the acidic species:

$$\ce{2Al + 6 H2O -> 2 Al^3+ + 6 OH- + 3 H2 ^}\\\ce{2Al^3+ + 6 OH- -> 2 Al(OH)3 v }$$

Or at least that is what should happen. Taking your random tin foil, the aluminium has created a substantial layer of aluminium oxide and hydroxyide on the top which is difficult to penetrate for water and creates a passivation. So any type of aluminium you put into a jar of water should really be considered aluminium coated with aluminium oxide and put into water.

This is where the hydroxide of $\ce{KOH}$ comes into play. It is able to coordinate the aluminium cations in the oxide/hydroxide, turning them into aluminates $\ce{[Al(OH)4]-}$ which can dissolve in water. This destroys the passivation layer, allows fresh aluminium to be reached, which can again react with the remaning water to form more aluminium oxide/hydroxide which is attacked … etc.

This reaction happens because aluminium is so non-noble that the oxidation to $\ce{Al^3+}$ is favourable even under basic conditions (whereas some other non-noble metals are stable under basic conditions because there is not enough potential to oxidise them).