The Frost diagram for sulfur shows the relative stability (in terms of cell potentials) of the possible aqueous oxidation states. What is the reason for the instability of higher oxidation states in the acidic regime and their stability in the basic regime ? Something I have always noticed but never been able to apply chemical intuition to.

Frost diagram

  • 1
    $\begingroup$ Negative charges aren't happy in acidic media. $\endgroup$
    – Dissenter
    Mar 24, 2015 at 16:46
  • $\begingroup$ Yes, but the Frost diagram shows us the difference between the S oxidation states; degree of protonation doesn't affect that. $\endgroup$
    – J. LS
    Mar 24, 2015 at 17:14

1 Answer 1


Write out the half-reactions and you will see that oxidation of S almost always creates lots of protons. So you can think of Le Chatelier's principle: if there are already lots of protons around, it will be harder to create more of them. Here's the example for complete oxidation of sulfide to sulfate at pH ~7:

$$\ce{HS- + 4 H2O -> 8e- + 9H+ + SO4^{-2} }$$

Conversely, under alkaline conditions, the equilibrium favors the products even more, because of the "removal" of the protons by the alkaline conditions.


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