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Why is the oxidation of certain metals such as iron favorable from a thermodynamic standpoint? What is it about these metal oxides that makes them inherently more thermodynamically stable than their starting, elemental forms? The only reason I can imagine is that iron, being a d-block element, has a low electronegativity and cannot stabilize valence electron density as well as oxygen, the second most electronegative element on the table.

Also, is oxidation always thermodynamically favorable?

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The oxidation of gold is not favorable. For many other metals the metallic bonding is weaker than the bonding with oxygen.

Let's look at the steps in oxidation:

Breaking metallic bonds (endothermic).

Ionizing the metal atoms (endothermic).

Electron affinity of oxygen (exo for the first, endo for the second).

Lattice energy of the oxide (exothermic).

The lattice energy is so high that the process is favorable.

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When you think about it, metals tend to want to lose electrons to finish with full outer shells. When they bond with oxygen, they do lose those electrons, ending in a more stable electron configuration. These ions don't go off and violently react with other things because they're ionically attracted to the oxygens they've just given their electrons to. So all their electrons are happily occupied, meaning they have no need to react and share them off. Also, because of oxygen's high electronegativity, it is unlikely to give those electrons up to another element, causing a reaction.

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