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Justify giving reaction that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodicacid in the best reductant. |
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Answer» Solution :The standard redox potentials for (or standard reductionpotentials) of the redox couples formed by halgoens are - A redox couple consists of reduced from and an oxidised form .Larger the value of `E^(0)`for a redox couple ,greater is the tendency of its oxidised from to get reduced and smaller is the tendency of its reduced from to ger oxidised .The reverse is true when the value of `E^(0)` for a redox couple is small .Conbining this idea with standard ELECTRODE potentials of the redox couples given we can infer that the tendency of oxidised forms (i.e.,`F_(2),CI_(2),Br_(2)andI_(2))`to get reduced or the strength of oxidising power of the oxidised forms follows the order: `F_(2)ltCI_(2)ltBr_(2)ltI_(2)` and the tendency of reduced froms (i.,e`F^(-)CI^(-)Br^(-)andI^(-)`to get oxidised or the strength of reducing power of the reduced forms follows the order : `I^(-)ltBr^(-)ltCI^(-)ltF^(-)`As the oxidising power of `F_(2)` is highest among the halongs ,it is capable of oxidising other halides to the corresponding halogensn No other halogen except `F_(2)` has this ability. `F_(2)(g)2CI^(-)(aq)to2F^(-)(aq)+CI_(2)(g)` `F_(2)(g)2Br^(-)(aq)to2F^(-)(aq)+Br_(2)(l)` `F_(2)(g)+2I^(-)(aq)to2I^(-)(aq) to2F^(-)(aq)+I_(2)(s)` `CI_(2)(g)+2Br^(-)(aq)to2CI^(-)(aq)+br_(2)(l)` `CI_(2)(g)+2I^(-)(aq)to+2CI^(-)(aq)+I_(2)(s)` `Br_(2)(g)+3I^(-)(aq)to+2Br^(-)(aq)+I_(2)(s)` Therefore ,`F_(2)` has the strongest oxidising power among the halogens . The oxidation of a hydrohalic acid to get oxidised or the reducing power of a hydrohalic acid is high when the halide ion of the hydrohalic acid exhibits greater tendency of getting oxidised .As the tendency of halide ions to get oxidised follows the order `I^(-)ltBr^(-)ltCI^(-)ltF^(-)`, the reducing powet of hydrohallic ACIDS will follow the order `HIltHBrltHCIltHF`. This is confirmed from the following reactions : HI or HBr can reduce `H_(2)SO_(4)`to `SO_(2)`,but HCI or HIE cannot reduce . `2HI+H_(2)SO_(4)toI_(2)+SO_(2)+2H_(2)O` `2HBr+H_(2)SO_(4)toBr_(2)+SO_(2)+2H_(2)O` `I(-)` can reduce `Cu^(2+)` to `Cu^(+)` but `Br^(-)` cannot `2Cu^(2+)(aq)+4I^(-)(aq)toCU_()I_(2)+I_(2)(s)` `Cu^(2+)(aq)+2br^(-)(aq) to No reaction` Therefore ,we can conclude that Hi is the stongest reducing AGENT among the hydrohalic acids. 
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