Which species is oxidized in the reaction below?

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

1 Answer
Oct 31, 2016

Answer:

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?