Q.
In the disproportionation reaction
$ 3HCl{{O}_{3}}\xrightarrow{{}}HCl{{O}_{4}}+C{{l}_{2}}+2{{O}_{2}}+{{H}_{2}}O, $
the equivalent mass of the oxidizing agent is (molar mass of $ HCl{{O}_{3}}=84.45 $ )
KEAMKEAM 2011
Solution:
This is a disproportionation reaction since here the oxidation state of chlorine decreases from +5 to 0 in $ C{{l}_{2}} $ as well as increases from +5 to +7 in $ HCl{{O}_{4}} $ . Thus, $ HCl{{O}_{3}} $ acts as oxidizing as well as reducing agent. Equivalent mass of oxidizing agent (i.e., $ HCl{{O}_{3}} $ ) $ =\frac{molar\,mass}{Change\,in\,oxidation\,number} $
$ =\frac{84.45}{(5-0)}=16.89 $
