Electrolysis of KBr(aq) gives Br2 at anode but KF(aq)does not give F2. Give a reason.
Oxidation takes place at anode. Now higher the oxidation Potential, easier to oxidize. Oxidation potential of Br−Br−, H2OH2O, F−F− are in the following order. Br−>H2O>F−Br−>H2O>F− Therefore in the Aqueous Solution of KBr. Br−Br− ions are oxidized to Br2Br2 in preference to H2OH2O . On the other hand, in the aqueous solution of KFKF, H2OH2O is oxidized in preference to F−F− . Thus in this case oxidation of H2OH2O at anode gives O2O2 and no F2F2 is produced.
