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Question

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


Solution

Fluorine oxidizes chloride ion to chlorine, bromide ion to bromine and iodide ion to iodine respectively.
$$\displaystyle F_2+2Cl^- \rightarrow 2F^- + Cl_2 $$
$$\displaystyle F_2+2Br^- \rightarrow 2F^- + Br_2  $$
$$\displaystyle  F_2+2I^- \rightarrow 2F^- + I_2$$.
Chlorine oxidizes bromide ion to bromine and iodide ion to iodine.
$$\displaystyle Cl_2 + Br^- \rightarrow 2Cl^- + Br_2 $$
$$\displaystyle Cl_2 + I^- \rightarrow 2Cl^- + I_2 $$
Bromine oxidizes iodide ion to iodine.
$$\displaystyle Br_2 + I^- \rightarrow 2Br^- + I_2 $$
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  $$\displaystyle F_2 > Cl_2 >
Br_2 > I_2 $$. 
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 $$HF$$.
$$\displaystyle 2HI+ H_2SO_4 \rightarrow I_2 +SO_2  + 2 H_2O $$
$$\displaystyle 2HBr+ H_2SO_4 \rightarrow Br_2 +SO_2  + 2 H_2O $$
Iodide ion can reduce $$Cu(II)$$ to Cu(I) but bromide cannot.
$$\displaystyle 4I^- + 2Cu^{2+} \rightarrow Cu_2I_2 + I_2 $$
Hence, among the hydrohalic compounds, hydroiodic acid is the best reductant. The reducing power of hydrohalic acids is $$\displaystyle HF < HCl <HBr < HI $$.

Chemistry
NCERT
Standard XI

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