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

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Solution

$F_{2}$ can oxidize $C l^{-} t o C l_{2}, B r^{-} t o B r_{2}, a n d I^{-} t o I_{2}$ as:
$F_{2 (a q)} + 2 C l_{(s)}^{-} ⟶ 2 F_{(a q)}^{-} + C l_{(g)} F_{2 (a q)} + 2 B r_{(a q)}^{-} ⟶ 2 F_{(a q)}^{-} + B r_{2 (I)} F_{2 (a q)} + 2 I_{(a q)}^{-} ⟶ 2 F_{(a q)}^{-} + I_{2 (s)}$
On the other hand, $C l_{2}, B r_{2},$ and $I_{2}$ cannot oxidize $F^{-} t o F_{2}$ . The oxidizing power of halogens increases in the order of $I_{2} < B r_{2} < C l_{2} < F_{2}$ . Hence, fluorine is the best oxidant among halogens.
HI and HBr can reduce $H_{2} S O_{4} t o S O_{2}$ , but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.
$2 H l + H_{2} S O_{4} ⟶ I_{2} + S O_{2} + 2 H_{2} O 2 H B r + H_{2} S O_{4} ⟶ B r_{2} + S O_{2} + 2 H_{2} O$
Again, $I^{-}$ can reduce $C u^{2 +}$ to $C u^{+}$ , but $B r^{-}$ cannot.
$4 I_{(a q)}^{-} + 2 C u_{(a q)}^{2 +} ⟶ C u_{2} I_{2 (s)} + I_{2 (a q)}$
Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of HF < HCl < HBr < HI.

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