Justify by giving reactions that among halogens,$F_{2}$ is the best oxidant and among hydrohalic acids,$HI$ is the best reductant.

(A) $1$. Oxidizing power of halogens: $F_{2}$ is the strongest oxidizing agent because it has the highest standard reduction potential. It can oxidize other halide ions $(Cl^{-}, Br^{-}, I^{-})$ to their respective halogens.
$F_{2(aq)} + 2Cl^{-}_{(aq)} \rightarrow 2F^{-}_{(aq)} + Cl_{2(g)}$
$F_{2(aq)} + 2Br^{-}_{(aq)} \rightarrow 2F^{-}_{(aq)} + Br_{2(l)}$
$F_{2(aq)} + 2I^{-}_{(aq)} \rightarrow 2F^{-}_{(aq)} + I_{2(s)}$
Since $Cl_{2}, Br_{2},$ and $I_{2}$ cannot oxidize $F^{-}$ to $F_{2}$,the order of oxidizing strength is $I_{2} < Br_{2} < Cl_{2} < F_{2}$.
$2$. Reducing power of hydrohalic acids: The reducing power increases as the bond dissociation enthalpy decreases down the group $(HF < HCl < HBr < HI)$. $HI$ is the strongest reducing agent.
$HI$ can reduce concentrated $H_{2}SO_{4}$ to $SO_{2}$:
$2HI + H_{2}SO_{4} \rightarrow I_{2} + SO_{2} + 2H_{2}O$
$HI$ can also reduce $Cu^{2+}$ to $Cu^{+}$:
$4I^{-} + 2Cu^{2+} \rightarrow Cu_{2}I_{2} + I_{2}$