Recently I was trying to balance a redox chemical equation:

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

The equation is in principle simple to solve; one element ($\ce{I}$) is reduced and two elements ($\ce{Cu}$, $\ce{S}$) are oxidized. The 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 the matrix method and via obtained reaction coefficients ($4$, $7$, $14$, $4$, $7$, $4$, $7$, $5$) and concluded that the oxidation state of sulfur in thiocyanate is zero. This result is very surprising, because the 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?
  • 1
    $\begingroup$ If you assume that carbon isn't reduced that you'll get such result, but it's wrong assumption. $\endgroup$
    – Mithoron
    Commented Mar 3, 2016 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
    Commented Mar 3, 2016 at 20:39

1 Answer 1


In the end the oxidation state of sulfur does not really matter for balancing equations. Rather, identifying 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 oxidized or reduced from the cuprous thiocyanate. We have multiple parts of the $\ce{CuSCN}$ participating in the electron exchange.

Why not then identify the whole $\ce{CuSCN}$ 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 $\ce{Cu}$ in the copper sulfate, $+6$ for the sulfur in the same compound, and $-1$ for $\ce{CN}$ in the hydrogen cyanide). Then $\ce{CuSCN}$ gives $7$ electrons, iodate takes $4$ electrons, and thus you get $\ce{4 CuSCN + 7 KIO3}$, just like with the matrix method.

  • $\begingroup$ Could you add a bit on the oxidation states of S and N in thiocyanate, anyway? I came here looking for that. $\endgroup$
    – harry
    Commented May 15, 2021 at 13:06

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