${\mathrm{I}}^{-}+{\mathrm{S}}_2{\mathrm{O}}_8^{-2}\stackrel{{\mathrm{Fe}}^{+3}}{\to }{\mathrm{I}}_2+{\mathrm{SO}}_4^{-2}$  Element undergoing oxidation is  ------ and element undergoing reduction is----                                

$\stackrel{-1}{{\mathrm{I}}^{-}}\to \stackrel{0}{{\mathrm{I}}_2} ; Iodine undergoes Oxidation$

Two of the peroxy oxygen atoms(-1 oxidation state) of ${\mathrm{S}}_2{\mathrm{O}}_8^{-2}$  reduce to oxygen atoms with ‘-2’ state in sulphate