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_2gt Br_2gt I_2` , Therefore , `F_2` can oxidise `Cl^(-),Br^(-) and I^(-)` to respective `Cl_2 , Br_2, I_2` .
`F_2(g) +2Cl^(-)(aq) rarr 2F^(-) (aq) +Cl_2(g)`
`F_2(g) +2Br^(-) (aq) rarr 2F^(-) (aq) +Br_2(l)`
`F_2(g) +2I^(-)(aq) rarr 2F^(-)(aq) +I_2(s)`
`Cl_2` can oxidise `Br^(-)` to `Br_2` and `I^(-)` to `I_2`
`Cl_2(g) +2I^(-)(aq) rarr 2Cl^(-) (aq) +I_2(s)`
However , `Cl_2` cannot oxidise `F^(-)` to `F_2` .
`Br_2` can oxidise only `I^(-)(aq)` to `I_2` and not `F^(-) " or " Cl^(-)` to `F_2 " or " Cl_2` .
`Br_2(l)+2I^(-)(aq)rarr 2Br^(-) (aq) +I_2(s)`
Thus, `F_2` is the best oxidant
Among hydrophilic acids , the reducing power decreases as : `HI gt HBr gt HCl gt HF` . For example , HI and HBr reduce `H_2SO_4 and " to " SO_2` while HCl and HF. For example , HI and HBr reduce `H_2SO_4" to " SO_2` while HCl and HF donot.
`2HBr+H_2SO_4 rarr Br_2+SO_2+2H_2O`
`2HI +H_2SO_4 rarr I_2+SO_2+2H_2O`
Similarly , `I^-` reduces `Cu^(2+) " to "Cu^(2+)` but `Br^(-)` does not.
`2Cu^(2+)(aq) + 4I^(-) (aq) rarr Cu_2I_2(s) + I_2(aq)`
Further among HCl and HCl is a stronger reducing agnet than HF because HCl can reduce `MnO_2 " to " Mn^(2+)` .
`MnO_2(s) +4HCl (aq) rarr MnCl_2(aq) +Cl_2(g) +2H_2O`
But HF does not reduce `MnO_2` . Therefore , among hydrohalic acid HI is the strongest reducing agent.