[Generic Username]

Given the following reaction, I'm struggling to see how it's possible.

ClO^-(aq) + 3I^-(aq) + 2H⁺  Cl^-(aq) + I₃^-(aq) + H₂O(l)

(mod edit to put text into message)

[sjb]

Consider https://en.wikipedia.org/w/index.php?title=Triiodide&oldid=704463427 or similar. The oxidation state of -¹/₃ for the iodine atoms in your product is an average, just as e.g. in propane the oxidation state of carbon is probably formally +⁸/₃.

Oxidation numbers are useful in some instances (like here) but less so in others.

[Generic Username]

Consider https://en.wikipedia.org/w/index.php?title=Triiodide&oldid=704463427 or similar. The oxidation state of -¹/₃ for the iodine atoms in your product is an average, just as e.g. in propane the oxidation state of carbon is probably formally +⁸/₃.

Oxidation numbers are useful in some instances (like here) but less so in others.

Hi sjb,
Then if I understand you correctly, the oxidation value of the reactant I is -3/3 which during the reaction means: (-3/3) - (-1/3) = -2/3 or an oxidation value of 2/3. The book I'm following just throws new stuff out there without any explanation, making it hard to get your head around new ideas. Thank you.

[AWK]

2/3 is a number of electrons moved for 1 iodine atom. but I₃^- you have 3 atoms then number of moved electrons in reaction is 3·(2/3)=2.

Mean oxidation number for I in I₃^- is 1/3

[Generic Username]

2/3 is a number of electrons moved for 1 iodine atom. but I₃^- you have 3 atoms then number of moved electrons in reaction is 3·(2/3)=2.

Mean oxidation number for I in I₃^- is 1/3

I see. Hence them multiplying by 3 later on in the working in order to match the result for Cl, which is 2. Thank you.