I have a redox reaction I found from a chemistry book from the 1970's. Two chemical species are oxidized, the Chromium and Iodine.

$\ce{CrI3 + KOH + Cl2 -> K2CrO4 + KIO4 +KCl + H2O}$

Hope someone can help on how to balance this.


1 Answer 1


Your best bet is to consider the entire reactant $\ce{CrI3}$ as the reducing agent.

The chromium (III) iodide starts out as that compound and gets oxidized to a combination of chromium (VI) (in the chromate) and iodine (VII) (in the periodate) with a ratio of chromium to iodine matching the original compound:

$\ce{CrI3 -> Cr^{VI} + 3 I^{VII}}$

Clearly the originally neutral reactant has been oxidized to atoms having a combined oxidation state of $+27$, so we balance this reaction with $27$ electrons:

$\ce{CrI3 -> Cr^{VI} + 3 I^{VII} + 27 e^-}$

Each chlorine molecule takes up two of the electrons so the redox ratio must be $\ce{2 CrI3 : 27 Cl2}$. Put those in and balance the remaining compounds as you normally would.

  • 1
    $\begingroup$ Hi Oscar, you are correct i believe. I have the answer for this and the lone Cl(2) has mole coefficient of 27. Thanks for these details. I am trying to do this via oxidation number method, so not doing the explicit electron counting. So you are assuming the chromium and Iodine are one compound and it is neutral, but it breaks apart into two things that are charged and are treated together in terms of adding their charges. $\endgroup$
    – Palu
    Jun 17, 2018 at 20:53
  • $\begingroup$ Yup, that's how it works. In effect the relevant "oxidation number" in tbe oxidation half reaction is the combination of chromium and iodine (which adds up to 0 before but +27 after the reaction), not either element alone. $\endgroup$ Jun 17, 2018 at 20:56
  • $\begingroup$ I have done disproportionation reactions, its kind of similar, but this one is making my head spin a bit. $\endgroup$
    – Palu
    Jun 17, 2018 at 20:59

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