Question

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

Answer 1

Fluorine oxidizes chloride ion to chlorine, bromide ion to bromine and iodide ion to iodine respectively.
$F_2+2C{l}^{-}\to 2F^{-}+C{l}_2$
$F_2+2B{r}^{-}\to 2F^{-}+B{r}_2$
$F_2+2I^{-}\to 2F^{-}+I_2$.
Chlorine oxidizes bromide ion to bromine and iodide ion to iodine.
$C{l}_2+B{r}^{-}\to 2C{l}^{-}+B{r}_2$
$C{l}_2+I^{-}\to 2C{l}^{-}+I_2$
Bromine oxidizes iodide ion to iodine.
$B{r}_2+I^{-}\to 2B{r}^{-}+I_2$
But bromine and chlorine cannot oxidize fluoride to fluorine. Hence, fluorine is the best oxidizing agent amongst the halogens. The decreasing order of the oxidizing power of halogens is $F_2>C{l}_2>B{r}_2>I_2$.
HI and HBr can reduce sulphuric acid to sulphur dioxide but $HCl$ and $HF$ cannot. Thus, $HI$ and $HBr$ are stronger reducing agents than $HCl$ and $HF$.
$2HI+H_{2}S{O}_4\to I_2+S{O}_2+2H_{2}O$
$2HBr+H_{2}S{O}_4\to B{r}_2+S{O}_2+2H_{2}O$
Iodide ion can reduce $Cu\left(II\right)$ to Cu(I) but bromide cannot.
$4I^{-}+2C{u}^{2+}\to C{u}_2{I}_2+I_2$
Hence, among the hydrohalic compounds, hydroiodic acid is the best reductant. The reducing power of hydrohalic acids is $HF<HCl<HBr<HI$.