question_answer 25)

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

Answer:

\[E_{{{F}_{2}}/{{F}^{-}}}^{{}^\circ }=++2.87\,\text{V};\,\,\,E_{C{{l}_{2}}/C{{l}^{-}}}^{{}^\circ }=(+1.36V);\] \[E_{B{{r}_{2}}/Br}^{{}^\circ }=++1.09\,\text{V;}\,\,\,\,\,\,\text{E}_{{{I}_{2}}/{{I}^{-}}}^{{}^\circ }=0.54V\] 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. \[C{{u}^{2+}}+4{{I}^{-}}\to C{{u}_{2}}{{I}_{2}}+{{I}_{2}}\]