Question

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

Answer 1

$F_2$ can oxidize $Cl ^-$ to $Cl_2$, $Br ^-$ to $Br_2$, and $I ^-$ to $I_2$ as:

$F_2(aq)+2Cl(s)\rightarrow 2 F(aq)+Cl(g)$
$F_2(aq)+2Br^-(aq)\rightarrow2F^-(aq)+Br_2(l)$
$F_2(aq)+2I^-(aq)\rightarrow2F^-(aq)+I_2( s)$

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 $I_2 < Br_2 < Cl_2 < F_2$.
Hence, fluorine is the best oxidant among halogens.
$HI$ and $HBr$ can reduce $H_2SO_4$ to $SO_2$, but $HCl$ and $HF$ cannot.
Therefore, $HI$ and $HBr$ are stronger reductants than $HCl$ and $HF$

$2HI+H_2SO_4\rightarrow I_2+SO_2+2H_2O_2$
$2HBr+H_2SO_4\rightarrow Br_2+SO_2+2H_2O$
Again, $I ^-$ can reduce $Cu ^{2+}$ to $Cu ^+$ , but $Br ^-$ cannot.
$4I^-(aq)+2Cu^{2+}(aq)\rightarrow Cu_2I_2(s)+I_2(aq)$
Hence, hydroiodic acid is the best reductant among hydrohalic compounds.

Thus, the reducing power of hydrohalic acids increases in the order of $HF < HCl < HBr < HI$.