MnO4−+8H++5e−→Mn2++4H2O
EMn2+MnO4−=Eo−50.059log[MnO4−][H+]8[Mn2+]
=1.51−50.059log[10−3]81
(Assuming [MnO4−]=[Mn2+]=1M )
=1.51−50.059×24=1.51−0.28
=1.23V
E(red)(Mn2+MnO4−)o=1.23V>EredBr−Br2o>EredI−I2o
i.e. it will oxidise Br− and I−, only.