I'm in the process of tutoring a close friend of mine in basic redox chemistry, and this problem shows a little bit of my chemistry atrophy (It's been two years since I've been active in chemistry work).

The problem stated is the decomposition of potassium chlorate into potassium chloride and oxygen gas, so

$$\ce{2KClO3 -> 2KCl + 3O2 ^}$$

Obviously, the half-reactions of this process can be written as the following:

$$\ce{2Cl^5+ + 12e^- -> 2Cl^-}$$ $$\ce{6O^2- -> 3O2 + 12e^-}$$

The issue I'm encountering is in successfully demonstrating the identity of the reducing agent and oxidizing agent. It's clear from the half-reactions which entity is oxidized versus reduced (and the fact that $\ce{K^+}$ is a spectator ion), but my friend has prior worked examples that appear to show that the identities of the reducing/oxidizing agent are that of the reactants. If so, can $\ce{KClO3}$ be both the reducing and oxidizing agent? If not, would this information be stated as the individual ionic entities listed above?

  • 1
    $\begingroup$ Yes, it is both the reducing and oxidising agent. $\endgroup$ Nov 23, 2015 at 9:53
  • $\begingroup$ Indeed, there is nothing exceptional in having both the reductant and oxidant within one compound. $\endgroup$ Nov 23, 2015 at 9:54

2 Answers 2


If I remember my introductory lab course correctly, there are two reactions going on, one after the other. First, $\ce{ClO3-}$ disproportions according to the following equation:

$$\ce{4 ClO3- -> Cl- + 3 ClO4-}$$

Which has these two half-equations:

$$\ce{ClO3- + 6 e- + 6 H+ -> Cl- + 3 H2O}\\ 3(\ce{ClO3- + H2O -> ClO4- + 2 e- + 2 H+})$$

Where the formally required water is merely spectator.

Only upon further heating does the perchlorate anion decompose according to:

$$\ce{ClO4- -> Cl- + 2 O2}$$

Which again can be broken down to:

$$\ce{ClO4- + 8 e- + 8 H+ -> Cl- + 4 H2O}\\ 2(\ce{2H2O -> O2 + 4 e- + 4 H+})$$

And again formally requires water as a spectator.

Note that it is not at all unusual for one entity to be both; the simplest example is probably the disproportion of hydrogen peroxide according to:

$$\ce{2 H2O2 -> O2 + 2 H2O}$$

Where one molecule is reduced by another of the same type.

  • $\begingroup$ Great answer - was not aware of the perchlorate steps involved. So in the circumstance that one entity is both reductant and oxidant; it is that of cooperation - that is, one serves temporarily as the oxidant and vice-versa? $\endgroup$
    – ecfedele
    Nov 23, 2015 at 20:49
  • $\begingroup$ Sorry, not sure what you mean there @ecfedele =C $\endgroup$
    – Jan
    Nov 24, 2015 at 12:38
  • $\begingroup$ I'm trying to grasp how one molecule can reduce/oxidize itself in a reaction such as this. To give an example, above you give the decomposition of hydrogen peroxide - what makes one of those hydrogen peroxides become a reducing agent toward itself? $\endgroup$
    – ecfedele
    Nov 24, 2015 at 23:56
  • $\begingroup$ That’s an interesting question that I can’t answer out of the blue. From a thermodynamic, macroscopic point of view it is certainly that the combined stabilities of $\ce{O2}$ and $\ce{H2O}$ are better than that of 2 $\ce{H2O2}$. So maybe it’s just random deviation; by chance one oxygen is almost losing its electron to give to another or something. $\endgroup$
    – Jan
    Nov 25, 2015 at 9:48

If you consider the oxidation states of chlorine and oxygen, oxygen is REDUCING the chlorine as it goes from +5 to -1 (oxygen in turn is OXIDISED from -2 to 0). This makes no sense as O is more electronegative than Cl and I can only assume that since the reaction requires heat, this energy is driving a non-spontaneous reaction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.