Which species is oxidized in the reaction below?

#I^(-)(aq) + ClO^(-) (aq) -> IO^(-) (aq) + Cl^(-) (aq)#

1 Answer
anor277
Oct 31, 2016

Iodine is oxidized to the conjugate base of hypiodous acid, #HOI#.

Explanation:

The iodide ion, #I^-# clearly has a formal oxidation number of #-I#. On the other hand, iodine in hypoiodous acid, clearly has a formal oxidation number of #+I#. If this is not clear to you ask for clarification; iodine is certainly LESS electronegative than oxygen, and oxygen is presumed to get the electron when we distribute the bonding electron.

So #"oxidation:"#

#I^(-) + H_2O rarr IO^(-) +2H^(+) +2e^-#

And #"reduction:"#

#ClO^(-)+ 2H^(+)+ 2e^(-) rarr Cl^(-) +H_2O#

And #"overall:"#

#ClO^(-)+ I^(-) rarr Cl^(-) +IO^(-) #

Which species is reduced in the above reaction?

Related questions
See all questions in Oxidation and Reduction Reactions
Impact of this question
9461 views around the world
You can reuse this answer
Creative Commons License