# Thinking of reducing aluminum hydroxide with magnesium

Is it possible to reduce aluminum hydroxide to aluminum metal by using a sacrificial magnesium electrode in a single displacement reaction?

I checked my revision of redox charts and it looks like it should work.

For more detail, consider: $$\ce{Al}$$ is very fond of bonding with oxygen. It's such a strong reaction that the leading industrial process to refine it involves adding $$\ce{Na3AlF6}$$ (cryolite) to lower its melting point, and using sacrificial carbon electrodes, generating large amounts of carbon dioxide.

I was curious as to whether anyone has tried using magnesium metal instead, since $$\ce{Mg}$$ is easier to obtain electrochemically, Together, they form a eutectic that melts over 200 °C lower than either metal alone, and $$\ce{Mg}$$ seems to support reacting more favorably to oxygen and hydroxide in comparison to $$\ce{Al}.$$

So, what am I missing? Does it work?

The reaction $$\ce{Al2O3 + 3 Mg -> 3 MgO + 2 Al}$$ is slightly exothermic, with a $$\Delta H = \pu{-130 kJ mol^-1}.$$ It is also exoergic at all reasonable temperatures, as $$\Delta S$$ is very small $$(\pu{-1.5 J mol^-1}).$$ So, the reaction is in principle feasible. The trouble is that $$\ce{Al2O3}$$ and $$\ce{Mg}$$ are solid at room temperature. And the reactions between solids are never efficient, if no gas is produced. They only happen at the points of contact of the solids. So, the yield is poor.
The only way to improve the contacts between reagents is to use liquid reagents. Here the first substance to melt when heated is magnesium metal, which melts at $$\pu{654 °C}.$$ So, the reaction should be pretty efficient above $$\pu{654 °C}.$$ But magnesium takes fire in air much before $$\pu{600 °C},$$ and it reacts violently with $$\ce{O2}$$ and with $$\ce{N2}.$$ So, this reaction should be carried out in argon gas or in a vacuum. This prevents this process to become commercial.