# How to reduce Al2O3 in a H2 atmosphere on industrial scale?

As known, reducing $\ce{Al2O3}$, which is the main process of the aluminium production, isn't really cheap, mainly because its energy need. Its high energy need is coming from the high binding energy between $\ce{Al}$ and $\ce{O}$.

What differs from the common $\ce{Al}$ production: the whole process has to run in $\ce{H2}$ atmosphere, where the cost of the $H_2$ gas is zero.

Thus, we need a net result of $\ce{Al2O3 + 3 H2 \rightarrow 2 Al + 3 H2O}$.

Does it make the process easier? How could it work?

Remark: actually, the goal would be to extract the $\ce{O2}$ and to produce water from it, while we have a quasi-infinite $\ce{H2}$ source. The $\ce{Al}$ production here isn't important.

• The question came into my mind while thinking on terraforming. – peterh Jan 10 '17 at 13:33
• Also if this was viable, it would be probably used. – Mithoron Jan 10 '17 at 15:14
• No, in the real life the goal of the reduction is to produce Al. Here to goal would be to produce water. Furthermore, in the normal circumstances, there isn't a quasi-infinite H2 source, it had to be made from water decomposition, which is also costly. I think you can have right if the H2 production is much cheaper as the Al production (I don't have here pretty data about this). – peterh Jan 10 '17 at 15:17
• There's even policy of not using mathjax in titles, but because of searching issues. Some elements are produced via H2 reduction, but it would be hard to turn the equilibrium of this reaction backwards IMO. – Mithoron Jan 10 '17 at 15:24