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Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant. |
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Answer» Solution :The oxidising POWER of halogens DECREASES in the order `F_(2)gtCl_(2)gtBr_(2)gtl_(2).F_(2)` being the strongest OXIDANT can oxidise `Cl^(-),Br^(-)` and `l^(-)` ions. `Cl_(2)` oxidises `Br^(-)` and `l^(-)` ions, whereas `Br_(2)` can oxidise only `l^(-)` ions. `l_(2)` is unable to oxidise none of these. The reactions are as follows: Oxidising reactions of `F_(2)` : `{:(F_(2)(g)+2Cl^(-)(aq)to2F^(-)(aq)+Cl_(2)(g)),(F_(2)(g)+2Br^(-)(aq)to2F^(-)(aq)+Br_(2)(l)),(F_(2)(g)+2l^(-)(aq)to2F^(-)(aq)+l_(2)(s)):}}` Oxidising reactions of `Cl_(2)` : `{:(Cl_(2)(g)+2Br^(-)(aq)to2Cl^(-)(aq)+Br_(2)(l)),(Cl_(2)(g)+2l^(-)(aq)to2Cl^(-)(aq)+l_(2)(g)):}}` Oxidising reaction of `l_(2)` : `Br_(2)(l)+2l^(-)(aq)to2Br^(-)(aq)+l_(2)(s)}` Thus, fluorine is the BEST oxidant. The reducing power of hydrohalic acids decreases in the order `HlgtHBrgtHClgtHF.HI` and HBr reduce `H_(2)SO_(4)` to `SO_(2)` while HCl and HF are unable to do so. `2HBr+H_(2)SO_(4)toSO_(2)+2H_(2)O+Br_(2)` `2HI+H_(2)SO_(4)toSO_(2)+2H_(2)O+l_(2)` HCl reduces `MnO_(2)` to `Mn^(2+)` but HF is unable to do so. This indicates that HCl is a stronger reducing agentd than HF. `MnO_(2)+4HCl toMnCl_(2)+Cl_(2)+2H_(2)O` `MnO_(2)+4HFto` No reaction Thus, HI is the best reductant among hydrohalic compounds. |
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