How would the reaction of a metal oxide with hydrogen gas occur?

Jan 3, 2017

This kind of reaction may be backwards (i.e. it may be naturally nonspontaneous). That would be the case if the metal (the pure metal) is more easily oxidized than hydrogen gas is.

However, if we supply heat and a steady, pressurized flow of ${\text{H}}_{2} \left(g\right)$, we can force this reaction to occur. even if that is the case. An example is:

$\textcolor{b l u e}{\text{MgO"(s) + "H"_2(g) stackrel(Delta, "Pressure"" ")(->) "Mg"(s) + "H"_2"O} \left(g\right)}$

We can see that ${\text{H}}_{2}$ was oxidized to water ($0 \to + 1$), and $\text{MgO}$ was reduced to $\text{Mg}$ ($+ 2 \to 0$).

Normally, ${\text{H}}_{2}$ is less reactive than $\text{Mg}$ (less easily oxidized), so that is why this reaction had to be catalyzed by heat and excess ${\text{H}}_{2} \left(g\right)$.

CHALLENGE: How would this reaction be written for sodium oxide instead? Assume high heat and pressure are still required. Would heat be required if it was silver oxide instead?

Normally, the reaction that is naturally spontaneous would be:

${\text{Mg"(s) + "H"_2"O"(g) -> "MgO"(s) + "H}}_{2} \left(g\right)$

which is why you should never put out magnesium fires with water! (Hydrogen gas is flammable.)