Q. In the reaction, $HASO_2 + Sn^{2+} \to As + An^{4+} + H_2O$ Oxidising agent is:
Delhi UMET/DPMTDelhi UMET/DPMT 2003
Solution:
Oxidising agent itself undergoes reduction during a redox reaction.
Oxidation state of $A$s in $ HAs{{O}_{2}} $ is $ =x $
$ \therefore $ $ 1+x+(2\times -2)=0 $
or $ x=3 $ Oxidation state of $ As=0 $
Oxidation state of $Sn$ in $ S{{n}^{2+}}=+2 $
Oxidation state of $Sn$ in $ S{{n}^{4+}}=+4 $
$ \therefore $ Increase in oxidation number is oxidation.
$ \therefore $ $ S{{n}^{2+}} $ is oxidised to $ S{{n}^{4+}} $
during reaction and it is reducing agent.
$ \because $ Decrease in oxidation number is reduction.
$ \therefore $ $ HAs{{O}_{2}} $ is reduced to $As$ and it is oxidising agent.
