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.
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Sign up to join this communityI 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.
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.