Recently I was trying to solve a redox chemical equation:

$\ce{CuSCN + KIO3 + HCl <=> CuSO4 + KCl + HCN + ICI + H2O}$

The equation is in principle simple to solve; 1 element (I) is reduced and 2 elements (Cu, S) get oxidized. Number of exchanged electrons in both cases must be the same. However my problem is that I don't know how to assign the oxidation number to sulfur in a thiocyanate anion.

Later I solved the equation using matrix method and via obtained reaction coefficients ($4,7,14,4,7,4,7,5$) concluded that oxidation state of sulfur in thiocyanate is zero. Result is very surprising, because oxidation number zero in heteroatomic molecules is not common.

My questions are: how to assign the oxidation number in such cases? How is it possible that oxidation number of an atom bound to a more eletropositive atom is zero?

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    $\begingroup$ If you assume that carbon isn't reduced that you'll get such result, but it's wrong assumption. $\endgroup$ – Mithoron Mar 3 '16 at 20:04
  • $\begingroup$ Thanks for the idea! Only now oxidation states in relation to atom's valency make more sense. In $SCN^{-} $ I assigned oxidation states as S=-2, C=4, N=-3 and in $CN^{-}$ as C=2, N=-3 $\endgroup$ – Rok Narobe Mar 3 '16 at 20:39

In the end the oxidation state of sulfur does not really matter for balancing equations. Rather, iddntifying the real oxidizing and reducing agents is.

Clearly the iodine in the iodate ion is the oxidizing agent, dropping from +5 to +1. But copper, sulfur, and even carbon are all being reduced from the cuprous thiocyanate. We have multiple parts if the CuSCN participating in the electron exchange.

Why not then identify the whole CuSCN molecule as the reducing agent, thus covering all the reducing-agent atoms in their proper proportions? With this approach you see that the sum of oxidation states for this reductant goes from 0 to +7 (+2 for Cu in the copper sulfate, +6 for the sulfur and -1 for CN in the hydtogen cyanide). Then CuSCN gives 7 electrons, iodate takes 4, and thus you get 4 CuSCN + 7 KIO3 just like the matrix method.


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