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As I understand it, a displacement reaction is where a more reactive element kicks out a less reactive element from a compound.

For example, chlorine is more reactive than iodine, so chlorine will displace iodine from potassium iodide, yielding potassium chloride and iodine.

If magnesium reacts with water you get magnesium oxide and hydrogen. But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards? The above paragraph predicts that hydrogen would displace the magnesium from magnesium oxide.

So why does it actually happen the other way round?

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    $\begingroup$ ‘But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards?’ — no. $\endgroup$ – Jan Jan 6 '16 at 19:18
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    $\begingroup$ Magnesium is much more reactive than hydrogen. $\endgroup$ – orthocresol Jan 6 '16 at 19:47
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    $\begingroup$ Geez, couldn't we simply edit out all mentions about "displacement" ? You're asking about redox reactions. $\endgroup$ – Mithoron Jan 6 '16 at 20:45
  • $\begingroup$ "Reactive" is a rather unfortunate choice of word in the given situation. You are asking about electropositivity, not reactivity. $\endgroup$ – Greg Jan 7 '16 at 1:18
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If magnesium reacts with water you get magnesium oxide and hydrogen.

True. So you can't put out a magnesium fire with water. It just burns hotter.

$$ \ce{Mg(s) + H2O ->T[aqueous] MgO(s) + H2(g) ^}$$

But hydrogen is generally a lot more reactive than magnesium, so isn't this backwards?

No, in the above reaction you're reacting water with magnesium. The hydrogen is given off as a gas, though a tiny amount will stay dissolved in the water.

So why does it actually happen the other way round?

It will under suitable conditions. So with heat and under a pressurized flow of hydrogen gas the reverse reaction will occur.

$$ \ce{ MgO(s) + H2(g) ->T[heat, pressure] Mg(s) + H2O(g) ^} $$

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Hydrogen definitely will reduce metal oxides in the right environment. Metal oxides tend to be very stable compounds, and hydrogen, though it is flammable, is not terribly reactive stuff either at room temperature. Simply combining the two substances is unlikely to yield any real results. Most metal/metal oxide combinations won't react vigorously at room temperature either.

However, at higher temperatures, when a reasonable fraction of the collisions between reactants involved will overcome the activation energy of the reaction, hydrogen readily reduces some metal oxides. A fairly common lab involves the reduction of copper(II) oxide by hydrogen. When my students do this, they first use zinc to reduce the hydrogen in sulfuric acid to hydrogen gas, then that is piped to a tube containing copper(II) oxide powder that is being strongly heated. The reaction is strongly exothermic; it can be thought of as a thermite reaction, so this isn't particularly surprising. Water and metallic copper are produced.

$\ce{H2 + CuO -> H2O + Cu}$

The activity/reactivity series of metals can be used to predict which metals will be reduced by hydrogen.

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  • $\begingroup$ Thanks, I'm still a little confused -- magnesium burns if you light it with a blue flame, and hydrogen explodes at an orange flame. Care to comment on where my understanding of reactivity is going wrong? $\endgroup$ – spraff Jan 8 '16 at 17:19
  • $\begingroup$ @spraff Are you confused about why magnesium is emitting higher energy visible photons than hydrogen when it burns? It's difficult to make generalizations about reactivity, but we can say that magnesium donates electrons much more readily than hydrogen does. Magnesium and hydrogen will even react with one another to form magnesium hydride, in which hydrogen has a -1 oxidation state rather than its usual +1. In this type of reaction that is what matters, which species wants to donate electrons more. $\endgroup$ – Jason Patterson Jan 9 '16 at 1:57

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