Consider the reaction, ${\mathrm{S}}_2{\mathrm{O}}_8^{-2}+{\mathrm{I}}^{-}\stackrel{{\mathrm{Fe}}^{+3}}{\to }{\mathrm{SO}}_4^{-2}+{\mathrm{I}}_2 ;$The element undergoing oxidation is ------- and the element undergoing reduction is------

Iodide oxidizes to iodine;

In persulphate ion there are six oxygen atoms with -2 state and 2 oxygen atoms with -1 state. In sulphate ion all oxygen atoms are in -2 state;

Thus persulphate reduces to sulphate where oxygen atom undergoes reduction;