I've seen papers use oxidation and dissolution of pyrite. Are these two processes the same? I was always under the impression that pyrite does not dissolve easily. What is the difference between oxidation and dissolution of pyrite?
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$\begingroup$ this is a google question, google "oxidization" and google "dissolution" they are very different and distinct basic chemistry concepts. $\endgroup$– JohnJan 27, 2018 at 14:10
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Oxidation and dissolution are not the same processes, but they are related when discussing the breakdown of pyrite in near surface environments.
Dissolution is a process(*1) in which the chemical components of a solid become part of a liquid, in the liquid state. For example, take fresh water and salt. Now dissolve the salt in the fresh water. Instead of having NaCl and H2O, you now have H2O with Na+ and Cl- ions floating in it, not solid anymore. This is not something happens easily with pyrite. You can't just dump FeS2 in water and expect to have Fe2+ and S(2)- in the water. This is because the sulfur is in the reduced state, as sulfide. The more soluble form of sulfur is sulfate, the oxidised form.
Oxidation is a process in which a compound or element bonds with (more) oxygen(*2). In order to dissolve pyrite, you have to convert the insoluble sulfide to soluble sulfate. In your previous question you noted the oxidation reaction for pyrite:
$\ce{FeS2(s) + 15/4 O2(aq) + 7/2 H2O → Fe(OH)3(s) + 4H+(aq) + 2SO4^2-(aq)}$
You start with pyrite in the solid form, and you end up with sulfate (sulfur bonded to four oxygens). The leftover iron is deposited as insoluble iron oxides and hydroxides (basically - rust), and you're also generating a bunch of protons that make everything acidic (i.e. acid mine drainage).
*1 - strictly speaking, dissolution doesn't have to be a solid in a liquid. It can be anything in anything.
*2 - actually oxidation is an electron transfer reaction and can occur with no oxygen involved at all.
