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

Solution:

F_{2} can oxidize C l^{-}to C l_{2}, B r^{-}to B r_{2}, and I^{-}to I_{2} as:

F_{2(a q)}+2 C l_{(s)}^{-} \rightarrow 2 F_{(a q)}^{-}+C l_{2(g)}

F_{2}(a q)+2 B r_{(a q)}^{-} \rightarrow 2 F_{(a q)}^{-}+B r_{2(l)}

F_{2(a q)}+2 I_{(a q)}^{-} \rightarrow 2 F_{(a q)}^{-}+I_{2(s)}

Be that as it may, C l_{2}, B r_{2}, and I_{2} can’t oxidize F^{-}to F_{2}. The oxidizing force of incandescent light expansions in the request as given underneath:

I_{2}<B r_{2}<C l_{2}<F_{2}

Accordingly, fluorine is the best oxidant among incandescent light.

H I and H B r can decrease H_{2}, S O_{4} to S O_{2}, yet H C l and H F can’t. Thus, H I and H B r are more grounded reductants contrasted with H C l and H F.

2 \mathrm{HI}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{I}_{2}+\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}

2 \mathrm{HBr}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Br}_{2}+\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}

I^{-}can diminish C u^{2+} to C u^{+}, yet B r^{-}cannot.

4 I_{(a q)}^{-}+2 C u_{(a q)}^{2+} \rightarrow C u_{2} I_{2}(s)+I_{2(a q)}

Along these lines, hydrochloric corrosive is the best reductant among hydrohalic compounds.

Thus, the decreasing force of hydrohalic acids increments as given underneath:

H F<H C l<H B r<H I