I have a pure hunk of magnesium metal submerged in water. After several hours/days, the metal becomes coated with something.

Question 1: What is the coating?

I've been told it's magnesium oxide. This would make sense to me if it was exposed to air, but it's submerged in water. Is it still MgO? I'll assume there is enough free oxygen in the water to react slowly.

If I add an acid to the mix, the coating gradually disappears. I use some malic acid crystals. Where does the MgO go?

Question 2: What is in the water when the coating disappears?

As the coating disappears to expose the metal, the magnesium reacts more and more vigorously offgassing hydrogen, which I understand would leave magnesium malate in the water. But what about the MgO? Does the acid react with MgO? And either way, where does it go? Is the MgO just dissolved invisibly in the water or is it reacting to form something else?

  • 2
    $\begingroup$ Consider that the coating is magnesium hydroxide, $\ce{Mg(OH)2}$. Does the whole picture make more sense now? $\endgroup$ Commented Jan 3, 2017 at 9:00

2 Answers 2

  1. What is the coating?

As pointed by Klaus Warzecha, there is no magnesium oxide. It is magnesium hydroxide, $\ce{Mg(OH)2}$ and that's the coating.

$$\ce{Mg + H2O -> Mg(OH)2 + H2}$$

  1. What is in the water when the coating disappears?

The magnesium hydroxide reacts with malic acid to form magnesium malate which will be precipitated at the bottom.

$$\ce{Mg(OH)2 + malic~acid -> Mg-malate + H2O}$$

This is a classic acid-base reaction.

Thus, the coating gets dissolved, the magnesium is now exposed to water, it reacts vigorously with it offgasing more and more hydrogen.

  • $\begingroup$ There's some problems with the first equation. $\endgroup$
    – DHMO
    Commented Jan 3, 2017 at 23:58
  • $\begingroup$ @DHMO Yes, i did not wrote the hydrogen gas. Editing the answer. $\endgroup$ Commented Jan 4, 2017 at 13:05
  • $\begingroup$ You don't even need an "acid" to set off the reaction. Some transition metal salts, like copper(II) salts, hydrolyze to make enough acid to get things going. $\endgroup$ Commented Jan 11 at 15:26

In water Mg oxidizes and the half-reactions are: $$\ce{Mg -> Mg^{2+} + 2e-}$$ $$\ce{2H2O + 2e- → H2 + 2OH-}$$

As the E($\ce{Mg}$/$\ce{Mg^{2+}}$)>E($\ce{H2O}$/$\ce{H2}$), Mg oxidizes and water reduces.

Metal ions can change their speciation depending on the pH. In acidic conditions $\ce{Mg^{2+}}$ dissolves and associates with malate (which in solution is deprotonated) and end up precipitating because the malate is in high concentration and the solution is saturated. And, in neutral o basic conditions they precipitate as metal hydroxides which are quite insoluble.

What you have to take into account is that metal ions can precipitate in basic conditions and dissolve in acidic ones. Also, water has a reduction potential and has to be taken into account in rusting processes with metals. You can look for more information with Pourbaix diagrams where you can see the changes in speciation of metals in front of pH and the reduction potential of several metals.

Here's the Pourbaix diagram of Mg: enter image description here

I hope this helps!


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