An example of reaction:

$$\ce{2Al + 2KOH + 6H2O->2 K[Al(OH)4] + 3H2 ^}$$

Aluminium is not ionic, how then does it attract the $\ce{OH-}$ groups to bond with them into the complex ion $\ce{[Al(OH)4]-}$?

I started re-reading non-organic chemistry at a slightly deeper level recently after going through the basics of organic chemistry. This reaction stumped me. I've re-read some stuff about aluminium and complex ions, but can't seem to understand it yet.

There's no positive charge on $\ce{Al}$, yet it reacts with hydroxide groups floating around. Where will the electron from a non-ionic aluminium atom go?

I cannot 'imagine' the reaction, visualize its mechanism, and this makes it hard to remember it.

P.S. Aluminium hydroxide seems to be mysterious in its structure. I quote Chemguide:

The chemistry textbooks that I have to hand aren't too clear about the structure of aluminium hydroxide as far as the degree of covalent character is concerened, and a web search (until I got totally bored with it!) didn't throw up any reliable chemistry sites which discussed it. Several geology sites describe the mineral gibbsite (a naturally occurring form of aluminium hydroxide) in terms of ions, but whether it actually contains ions or whether this is just a simplification as a convenient way of talking about and drawing a complicated structure, I don't know.

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    $\begingroup$ P.S. I try to edit the text using the $$ enclosed notation, but my old PC literally goes into heart attack mode, to the extent that I have to type in a word and then wait a second before it appears on the screen. $\endgroup$ – CowperKettle May 1 '16 at 5:49
  • $\begingroup$ P.P.S. I've just found one possible explanation (in Russian), but it's not completely clear either. Stashing it here to read it later. $\endgroup$ – CowperKettle May 1 '16 at 5:58
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    $\begingroup$ To put things into perspective: you might have heard that many metals react with acids. (In most inorganic curriculums, this is a pretty heavily used example, and even the defining feature of acids, at least until you get to Brønsted–Lowry). All your questions are applicable to that reaction as well. Indeed, it is just the neutral metal with protons floating around, and all of a sudden they just up and react. $\endgroup$ – Ivan Neretin May 1 '16 at 6:47
  • $\begingroup$ It trumped you?? What card did it play? Or has the presidential race given us a new definition for that word... $\endgroup$ – foobarbecue May 2 '16 at 1:14
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    $\begingroup$ The rest of your post was so fluent that it did not even occur to me English might not be your first language. I'm guessing you meant to say "The reaction stumped me" rather than "The reaction trumped me." The world "trump" comes from card games and also happens to be the name of a particular presidential candidate... $\endgroup$ – foobarbecue May 2 '16 at 13:00

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.

  • $\begingroup$ I guess, the part of the question for this answer would be: why "Apparently, that changes in the course of the reaction".. $\endgroup$ – YakovL May 1 '16 at 8:43

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).

  • $\begingroup$ I wonder if the reaction $\ce{Al2O3 + Ca(OH)2}$ is possible, along the same lines as with $\ce{KOH}$. $\endgroup$ – CowperKettle May 12 '16 at 7:37

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