Why can't Al3+ be reduced to solid, pure aluminum?

There are efforts underway to develop ionic liquids for the purpose of reducing bauxite using exotic intermediates such as aluminum fluoride. However, the aqueous $\ce{Al^3+}$ ion is tantalizingly close to pure aluminum. What prevents it from being reduced?

• Nothing prevents it from being reduced other than the conditions of the aqueous solution (electric potential, pH, and so on): Are you asking about the more specific case of why aluminum metal is not produced this way commercially and is instead produced (usually) by the Hall-Heroult process? – Todd Minehardt Nov 4 '15 at 23:17
• @ToddMinehardt Yes, I am interested in why aluminum is not produced this way. – Dale Nov 4 '15 at 23:24

It is aluminium’s standard electrode potential that prevents it from easily being reduced. For the half-cell $\ce{Al^3+ + 3e- <=> Al(s)}$, the standard potential is $E^0 = -1.66~\mathrm{V}$. For comparison a few other values:

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

1. 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.

2. 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.

3. 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:

$$\ce{Fe2O3 + 2Al -> 2Fe + Al2O3}$$

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

• I thought that maybe the Al+3 ion would be not bonded to oxygen. Is Al+3 surrounded by water ligands or is it an oxide? MaxW's answer suggests that reducing aluminum may not be the problem, but preventing the reduced aluminum from being oxidized. This makes me wonder what the ionic liquid activity coefficients for the nernst equation must be for aluminum reduction. – Dale Nov 5 '15 at 23:18
• @Dale For the standard potential, I believe we are dealing with $\ce{[Al(H2O)6]^3+}$. But it doesn’t really matter regardless, because the point is water (hydrogen) will be instantly reduced by aluminium metal. – Jan Nov 6 '15 at 15:28
• This answer misses the fact that there is an en.wikipedia.org/wiki/Overpotential for the electrochemical evolution of a gas at a metal surface. I've not found a number for H2 at Al, though. The theoretically necessary pH level would however be quite a bit lower. Still impossible, of course. – Karl Jan 5 '17 at 3:10
• Works for sodium, for example: commons.wikimedia.org/wiki/… , due to amalgamation AND overpotential. – Karl Jan 5 '17 at 3:16
• @Karl Amalgamation introduces an entire additional layer of complexity to the question. – Jan Jan 5 '17 at 9:23

According to Wikipedia "elemental aluminium cannot be produced by the electrolysis of an aqueous aluminium salt because hydronium ions readily oxidize elemental aluminium." So the aluminium oxide is reduced via the Hall–Héroult process.

The alumina itself is produced from mined bauxite via the Bayer process.

So the gist is that industry spends many many millions of dollars converting bauxite too aluminium metal. No doubt they do it the cheapest way possible.

• So it is not the Al+3 that can't be reduced, it is the solid aluminum oxide coating that forms AFTER the Al+3 has been reduced (assuming non-aluminum electrodes).? – Dale Nov 5 '15 at 23:14
• In general it wouldn't matter what metal the "initial" electrode was made from. Once the electrode gets coated with Al metal then it would form an oxide coating which would "ruin" the electrochemical reaction. The only exception would be a Hg electrode. The Al would dissolve in the Hg rather than forming a coating. But you couldn't get a lot of Al in Hg, and cycling the Hg would be a problem. So that isn't practical. – MaxW Nov 6 '15 at 0:21
• Mercury does some damage to aluminum. youtube.com/watch?v=Z7Ilxsu-JlY Also, see: en.wikipedia.org/wiki/Aluminium_amalgam – Dale Nov 7 '15 at 1:24