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

Explanation

Halogens have a strong tendency to accept electrons therefore they are strong oxidising agents their relative oxidising power is however measured in terms of their electrode potential since the eletrode potential of halogens decrease in the order `F_(2)(+2.87 v) CI_(2)(+1.36 v) gt Br_(2)(+1.09 v)gtI_(2)(+0.54v)` therefore their oxidising power decrease in the same order
This is evident from the observation that `F_(2)` oxidises `CI^(-)` to `CI_(2) Br^(-)` to `Br_(2) I^(-)` to `I_(2),CI_(2)` oxidises `Br^(-)` to `Br_(2)` and `I^(-)` to `I_(2)` but not `F^(-)` to `F_(2). Br_(2)` howver oxidises `I^(-)` to `I_(2)` but not `F^(-)` to `F_(2)` and `CI^(-)` to `CI_(2)`
`F_(2)(g)+2CI^(-)(aq)rarr2F^(-)(aq)+CI_(2)(g),F_(2)(g)+2 Br^(-)(aq)rarr2F^(-)(aq)+Br_(2)(l)`
`F_(2)(g)+2CI^(-)(aq)rarr2F^(-)(aq)+I_(2)(a),CI_(2)(g)+2Br^(-)(aq)rarr2CI^(-)(aq)+Br_(2)(l)`
`CI_(2)(g)+2I^(-)(aq)rarr2cI^(-)(aq)+I_(2)(s)` and `Br_(2)(l)+2I^(-)rarr 2Br^(-)(aq)+I_(2)(s)`
Thus `F_(2)` is the best oxidant
conversely halide ions have a tendency to lose electrons and hence can act as reducing agent since the electrode potentials of hlide ions decrease in the order therefore the reducing power of the halide ions or their corresponding hydrohalic acids decrease in the same order `HIgt HBrgt HCIgtHF` thus hydroidic acid is the best reductant this is supported by the following reaction for example HI and HBr reduce `H_(2)SO_(4)` to `SO_(2)` while HCI and HF do no
Thus HI is a stronger reductant than HBr
further among HCI and HF ,HCI is a stronger reducing agent than HF because HCI reduces `MnO_(2)` to `Mn^(2+)` but HF does not
`MltnO_(2)(s)+4 HCI(aq)rarrMnCI(2)(aq)+CI_(2)(g)+2H_(2)O`
`MnO_(2)(s)+4Hf (l)rarrF_(2)` is not formed
Thus the reducing character of hydrohalic acids decrease in the order `HI gt HBr HCI gt HF`