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

Explanation

The oxidising power of halogens decreases in the order: `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`. This is evident from he observation that `F_(2)` oxidises `Cl^(Θ)` to `Cl_(2), Br^(Θ)` to `Br_(2)` and `I^(Θ)` to `I_(2), Cl_(2)` oxidises `Br^(Θ)` to `Br_(2)` and `I^(Θ)` to `I_(2)` but not `F^(Θ)` to `F_(2)`. `Br_(2)`, however, oxidises `I^(Θ)` to `I_(2)` but not `F^(Θ)` to `F_(2), Cl^(Θ)` to `Cl_(2)`.
`F_(2)(g)+2Cl^(Θ)(aq)rarr 2F^(Θ)(aq)+Cl_(2)(g)`
`F_(2)(g)+2Br^(Θ)(aq)rarr 2F^(Θ)(aq)+Br_(2)(l)`
`F_(2)(g)+2I^(Θ)(aq)rarr 2F^(Θ)(aq)+I_(2)(s)`
`Cl_(2)(g)+2Br^(Θ)(aq)rarr 2Cl^(Θ)(aq)+Br_(2)(l)`
`Cl_(2)(g)+2I^(Θ)(aq)rarr 2Cl^(Θ)(aq)+I_(2)(s)` and
`Br_(2)(l)+2I^(Θ)rarr 2Br^(Θ)(aq)+I_(2)(s)`
Thus, `F_(2)` is the best oxidant
Among hydrohalic acids, the reducing power decreases in the order: `HI gt HBr gt HI gt HF`. Thus, `HI` and `HBr` reduce `H_(2)SO_(4)` to `SO_(2)` while `HCl` and `HF` do not.
`2HBr+H_(2)SO_(4)rarr Br_(2)+SO_(2)+2H_(2)O`
`2HI+H_(2)SO_(4)rarr I_(2)+SO_(2)+2H_(2)O`
Firther, `I^(Θ)` reduces `Cu^(2+)` to `Cu^(+)` but `Br^(Θ)` does not.
`2Cu^(2+)(aq)+4I^(Θ)(aq) rarr Cu_(2)I_(2)(s) + I_(2)(aq)`
`Cu^(2+)(aq) +2Br^(Θ) rarr "No reaction"`.
Furher, among `HCl` and `HF, HCl` is a stronger reducing agent than `HF` because `HCl` reduces `MnO_(2)` to `Mn^(2+)` but `HF` does not.
`MnO_(2)(s) + 4HCl(aq) rarr MnCl_(2)(aq0 + Cl_(2)(g) + 2H_(2)O`

Thus, the reducing character of hydrohalic acids decreases in the order : `HI gt HBr gt HCl gt HF`.