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

Explanation

`F_(2)` can oxidize `Cl^(-)` to `Cl_(2) , Br^(-)` to `Br_(2)` and `l^(-)` to `I_(2)` as :
`F_(2 (aq)) + 2 Cl_((s))^(-) to 2 F_((aq)) + Cl_(g)`
`F_(2 (ag)) + 2Br_(aq)^(-) to 2 F_(aq)^(-) + Br_(2 (l))`
`F_(2 (aq)) + 2I_(aq)^(-) to 2 F_(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 of `l_(2) lt Br_(2) lt Cl_(2) lt F_(2)` . Hence , fluorine is the best oxidant among halogens .
Hl and HBr can reduce `H_(2)SO_(4)` to `SO_(2)` , but HCl and HF cannot . Therefore , Hl and HBr are stronger than HCl and HF.
`2HI + H_(2)SO_(4) to I_(2) + SO_(2) + 2H_(2)O`
`2HBr + H_(2)SO_(4) to Br_(2) + SO_(2) + 2H_(2)O`
Again `l^(-)` can reduce `Cu^(2+)` to `Cu^(+)` , but `Br^(-)` cannot .
`4I_(aq)^(-) + 2Cu_(aq)^(2+) to Cu_(2)I_(2 (s))_(2 (aq))`
Hence , hydroiodic acid is the best reductant among hydrohalic compounds .
Thus , the reducing power of hydrohalic acids increases in the order of `HF lt HCl lt HBr lt Hl`.