0
$\begingroup$

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.

$\endgroup$

closed as off-topic by Mithoron, aventurin, Jon Custer, Tyberius, pentavalentcarbon Jun 18 '18 at 15:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

3
$\begingroup$

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.

$\endgroup$
  • 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 '18 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. $\endgroup$ – Oscar Lanzi Jun 17 '18 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 '18 at 20:59

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