For a metal like $\ce{Mg}$ to corrode into $\ce{MgO}$, the double bond in $\ce{O2}$ must break. Since the dissociation energy of $\ce{O2}$ is 500 kJ/mol, using estimates from the Boltzmann distribution, it looks like there would not be any $\ce{O2}$ molecules in the entire atmosphere with enough energy to break this bond at room temperature: $$\text{Fraction of $\ce{O2}$ with at least $\pu{500 kJ}$ energy} \approx \mathrm e^\left(\dfrac{-500\,000\ \mathrm{J/mol}}{298\ \mathrm K(8.3145\ \mathrm{J/(mol\ K)})}\right) \lt 10^{-261}.$$

But my understanding is that $\ce{MgO}$ does form at room temperature. In fact, it seems that even stable $\ce{Pt}$ can break the $\ce{O2}$ bond at room temperature (for example in the context of catalyzing the ignition of hydrogen).

It seems that something like this must be happening: an $\ce{O2}$ molecule collides with $\ce{Mg}$ metal and gets stuck onto the surface, temporarily forming $\ce{MgO2}$. And once this bond is formed, the remaining bond between the two oxygen atoms suddenly becomes easy to break, and one of the oxygen atoms falls off and bonds with a different $\ce{Mg}$ atom. Is this the case?

I'd like to understand this mechanism better. How weak does the $\ce{O2}$ bond become after latching onto magnesium, and why does it become weaker? Why are metals able to split the $\ce{O2}$ bond this way, compared to other materials like hydrocarbons?

  • $\begingroup$ Pt is not breaking the oxygen bonds. Pt only interacts by adsorbing hydrogen. I am sure the mechanism has been very very well studies. Use Google Scholar this time. Pt is just helping by lowering the kinetic barrier of the reaction. $\endgroup$
    – AChem
    Jun 9, 2019 at 19:11
  • 1
    $\begingroup$ Humidity is the catalyst. You need water, then it happens electrochemically, see DOI: 10.1039/QR9672100029 'The Mechanism of Rusting'. Or more specifically: doi.org/10.1002/… 'Corrosion Mechanisms of Magnesium Alloys' $\endgroup$ Jun 18, 2019 at 15:38

2 Answers 2


The corrosion of metals like iron is essentially an electrochemical process.

In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of $\ce{Fe}$ (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and we can write the reaction.

\begin{align} \tag{Anode} \ce{2Fe (s) &-> 2 Fe^{2+} + 4 e-} & E^\circ_{\ce{Fe^{2+}/Fe}} &= \pu{– 0.44 V} \end{align}

Electrons released at the anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $\ce{H+}$ (which is believed to be available from $\ce{H2CO3}$ formed due to dissolution of carbon dioxide from air into water. Hydrogen ions in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction

\begin{align} \tag{Cathode} \ce{O2(g) + 4 H+(aq) + 4 e- &-> 2 H2O (l)} & E^\circ_{\ce{H^+/O2/H2O}} &= \pu{1.23 V} \end{align}

The overall reaction being:

\begin{align} \ce{2Fe(s) + O2(g) + 4H+(aq) &-> 2Fe^{+2}(aq) + 2 H2O (l)} & E^\circ_{\text{cell}} &= \pu{1.67 V} \end{align}

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide $\ce{(Fe2O3. x H2O)}$ and with further production of hydrogen ions.

  • $\begingroup$ I think more important is the reaction $\ce{O2 + 2 H2O + 4 e- -> 4 OH-}$ and consequently the formation of iron hydroxides, which may age to iron oxides. $\endgroup$ Jun 19, 2019 at 13:31

MgO2 is easy to visualize, but is probably not an intermediate in the oxidation of magnesium. In fact, magnesium is not well described for this situation as Mg, but rather as (Mg)n, because we are looking at bulk material, with different surfaces and facets and irregularities on the surface.

If the corrosion continues in the presence of water, then an electrochemical explanation is appropriate.

But magnesium does not continue to corrode in dry O2. A surface oxide forms which is quite inert; further oxidation is stifled. Aluminum is similar, but continues to oxidize slower and slower, building up a thick oxide coat. Stainless steel oxidizes to a very passive state, with a very thin - very thin! - oxide layer. Zinc is similar to aluminum.

Atoms which are adsorbed on the surface of metals may be bonded securely, as in the case of stainless steel, or less so, and in the case of metals which continue to oxidize, the adsorbed atoms are evidently moving around.

You could make the adsorbed atoms move around faster by heating the system. Magnesium will ignite in hot O2. Many metals will. If you get a magnesium fire going, it will even burn in nitrogen (https://video.search.yahoo.com/yhs/search?fr=yhs-Lkry-SF01&hsimp=yhs-SF01&hspart=Lkry&p=magnesium+fire+nitrogen#id=2&vid=eaf040c65f5c489889407fb36f263ece&action=click).

Combustion is a free radical process; perhaps oxidation of metals could be viewed in the same light. Things that burn develop into pieces of a larger molecule, but with a free electron. Oxygen is electronegative; it likes to grab electrons, even if only one at a time. Then the oxygen becomes part of a chain --O-O-, and the O2 bond is weakened. On a magnesium surface, this could lead to a splitting of the O2 bond and formation of MgO, except that this "MgO" molecule is still part of the bulk metal, until you scrape it off.

So corrosion of metals in dry air is not electrochemical, but could be some type of chain reaction which has a fairly high, but not super-high activation energy. Older cars had lots of iron in the engine compartment which corroded by the high temperatures there; parts of grills used for cooking may similarly corrode at high temperatures.


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