Justify,by giving reactions,that among halogens,fluorine is the best oxidant and among hydrohalic compounds,hydroiodic acid is the best reductant.

(N/A) $F_{2}$ can oxidize $Cl^{-}$ to $Cl_{2}$,$Br^{-}$ to $Br_{2}$,and $I^{-}$ to $I_{2}$ as:
$F_{2(aq)} + 2Cl^{-}_{(aq)} \longrightarrow 2F^{-}_{(aq)} + Cl_{2(g)}$
$F_{2(aq)} + 2Br^{-}_{(aq)} \longrightarrow 2F^{-}_{(aq)} + Br_{2(l)}$
$F_{2(aq)} + 2I^{-}_{(aq)} \longrightarrow 2F^{-}_{(aq)} + I_{2(s)}$
On the other hand,$Cl_{2}$,$Br_{2}$,and $I_{2}$ cannot oxidize $F^{-}$ to $F_{2}$. The oxidizing power of halogens increases in the order: $I_{2} < Br_{2} < Cl_{2} < F_{2}$. Hence,fluorine is the best oxidant among halogens.
$HI$ and $HBr$ can reduce $H_{2}SO_{4}$ to $SO_{2}$,but $HCl$ and $HF$ cannot. Therefore,$HI$ and $HBr$ are stronger reductants than $HCl$ and $HF$.
$2HI + H_{2}SO_{4} \longrightarrow I_{2} + SO_{2} + 2H_{2}O$
$2HBr + H_{2}SO_{4} \longrightarrow Br_{2} + SO_{2} + 2H_{2}O$
Again,$I^{-}$ can reduce $Cu^{2+}$ to $Cu^{+}$,but $Br^{-}$ cannot.
$4I^{-}_{(aq)} + 2Cu^{2+}_{(aq)} \longrightarrow Cu_{2}I_{2(s)} + I_{2(aq)}$
Hence,hydroiodic acid $(HI)$ is the best reductant among hydrohalic compounds. Thus,the reducing power of hydrohalic acids increases in the order: $HF < HCl < HBr < HI$.