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I am confused as to the meaning of fully oxidized, in the context of environmental chemistry (what is water soluble). While I believed that CO$_2$ was fully oxidized after reading http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/redoxea.html, I also read that SO$_2$ is a reducing agent. Additionally, looking at the electronegativities of carbon versus sulfur, they are the same. Thus, what types of chemicals would fully oxidized refer to?

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  • $\begingroup$ Highest oxidation state $\endgroup$
    – Mithoron
    Commented Mar 14, 2017 at 19:19
  • $\begingroup$ A couple notes: carbon and sulfur do have similar electronegativities, but that is not relevant to their possible oxidation states. Also, $\ce{SO2}$ is amphoteric in that it can act as an oxidant or reductant. The S in $\ce{SO2}$ is in the +4 oxidation state, but S can exist in oxidation states from +6 to -2. So, if $\ce{SO2}$ becomes oxidized to $\ce{SO3}$ for example (S goes from +4 ox state to +6 ox state), then $\ce{SO2}$ indeed would have acted as the reducing agent for that reaction (S was oxidized, another reactant was reduced). $\endgroup$
    – airhuff
    Commented Mar 14, 2017 at 20:35

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It comes down to the oxidation number of the element in the compound. From the link you provided, $\ce{CH4}$ has carbon in its most reduced form, with an oxidation number of $-4$. Carbon can't become any more reduced because a $-5$ state would imply that carbon had $9e^-$ in its valence shell, which would be highly unfavorable. Similarly, $\ce{CO2}$ has carbon in its most oxidized form, with an oxidation number of $+4$. Carbon can't become any more oxidized because a $+5$ state would imply that carbon lost an electron from its inner shell, which would be highly unfavorable.

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  • $\begingroup$ My only minor nitpick with this answer is that there is no tie in to the OP's (rather confusing) request for an answer in the context of environmental chemistry and water soluble compounds (which methane is obviously not). Still, I thought this was overall a straight-forward, solid +1 answer to the title question ;) $\endgroup$
    – airhuff
    Commented Mar 15, 2017 at 0:36

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