• # question_answer 25) Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant?

$E_{{{F}_{2}}/{{F}^{-}}}^{{}^\circ }=++2.87\,\text{V};\,\,\,E_{C{{l}_{2}}/C{{l}^{-}}}^{{}^\circ }=(+1.36V);$ $E_{B{{r}_{2}}/Br}^{{}^\circ }=++1.09\,\text{V;}\,\,\,\,\,\,\text{E}_{{{I}_{2}}/{{I}^{-}}}^{{}^\circ }=0.54V$ We know that, more is the positive value of standard reduction potential greater the oxidising behaviour. Thus, oxidising power of halogens will lie in following sequence. ${{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}$ ${{F}_{2}}$is best oxidant as it displaces other halides from their salts e.g., $2NaCl+{{F}_{2}}\to 2NaF+C{{l}_{2}}$ $2KBr+{{F}_{2}}\to 2KF+B{{r}_{2}}$ $2KI+{{F}_{2}}\to 2KF+{{I}_{2}}$ Halide ions have tendency to loose electrons, thus they act as reducing agent. Reducing character of hydrohalic acids lies in following sequences. $HI>HBr>HCl>HF$ HI and $HBr$both reduce sulphuric acid to $S{{O}_{2}}$ but$~HF$ and $HCl$ are unable to do so. $2HI+{{H}_{2}}S{{O}_{4}}\to 2{{H}_{2}}O+S{{O}_{2}}+{{I}_{2}}$ $2HBr+{{H}_{2}}S{{O}_{4}}\to 2{{H}_{2}}O+S{{O}_{2}}+B{{r}_{2}}$ HI reduces cupric ion to cuprous ion but $HBr$is unable to do so, thus HI is strongest reducing agent. $C{{u}^{2+}}+4{{I}^{-}}\to C{{u}_{2}}{{I}_{2}}+{{I}_{2}}$