Why can't Al3+ be reduced to solid, pure aluminum?
It is aluminium’s standard electrode potential that prevents it from easily being reduced. For the half-cell $\ce{Al^3+ + 3e- &lt;=&gt; Al(s)}$, the standard potential is $E^0 = -1.66~\mathrm{V}$. For comparison a few other values:


$\ce{Fe2O3 + 3 H2O + 2e- &lt;=&gt; 2 Fe(OH)2(s) + 2 OH-}$ and $\ce{Fe(OH)2 + 2 e- &lt;=&gt; Fe(s) + 2 OH-}$ have potentials of $E^0 = -0.86~\mathrm{V}$ and $-0,89~\mathrm{V}$, respectively;
$\ce{Zn^2+ + 2 e- &lt;=&gt; Zn(s)}$ has a potential of $E^0 = -0.76~\mathrm{V}$;
$\ce{Cr^3+ + 3 e- &lt;=&gt; Cr(s)}$ has a potential of $E^0 = -0.74~\mathrm{V}$;
$\ce{Co^2+ + 2 e- &lt;=&gt; Co(s)}$ has a potential of $E^0 = -0.28~\mathrm{V}$;
$\ce{Cu^2+ + 2 e- &lt;=&gt; Cu(s)}$ has a potential of $E^0 = +0.34~\mathrm{V}$.


(All values taken from Wikipedia)

The higher the standard potentials are, the easier the ion can be reduced. Hydrogen, of course, has a potential of $E^0 = 0~\mathrm{V}$ at pH 0 by definition. So in acidic media, hydrogen is reduced first. So let’s lower the pH to impede the formation of hydrogen. According to the Nernst equation:

$$E = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\mathrm{Ox}}{a_\mathrm{red}}$$

Where $z$ is the number of electrons transferred in the half-cell, and $a$ is the activity of the reduced/oxidised species. $a_\mathrm{red}$ is $1$ by definition and $z = 2$ for the hydrogen half-cell. So we need to solve the following simplified equation to determine the acid concentration that will allow reduction of aluminium:

$$-1.66~\mathrm{V} = 0~\mathrm{V} + \frac{0.059~\mathrm{V}}{2} \lg \frac{[\ce{H+}]}{1}\\
\frac{0.059}{2} \mathrm{pH} = 1.66 \\
0.059 \times \mathrm{pH} = 3.32 \\
\mathrm{pH} = 3.32 / 0.059 = 56$$

Whoops, so we would need a pH of $56$ to allow the reduction of aluminium in the presence of water. This calculation was highly theoretical, because:


A pH of 56 is impossible in anything that once could have been water. Wikipedia gives the $\mathrm{p}K_\mathrm{b}(\ce{O^2-}) \approx -38$; because $\mathrm{p}K_\mathrm{a}(\ce{HA}) + \mathrm{p}K_\mathrm{b}(\ce{A-}) = 14$ in water that must mean that $\mathrm{p}K_\mathrm{a}(\ce{OH-}) \approx 52$ — thus we would be dealing with an ‘oxide solution’ rather than water.
Although it is nice that both equations (pH and Nernst) use acitivities, we cannot really say anything about the activities at this large a concentration range.
The redox system of both species would be totally different in these environments.


Note that you can reduce most air-stable metals in thermite reactions from their oxides thereby creating aluminium oxide. An example reaction would be:

$$\ce{Fe2O3 + 2Al -&gt; 2Fe + Al2O3}$$

That alone should show the unwillingness of aluminium to be reduced in aquaeous solutions; think about the heat that the thermite reactions generate.