Electrolysis of kbr(aq) gives br2 at anode but kf(aq) does not give f2. give reason.
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Higher is the oxidation potential, easier to oxidize a specie. The oxidation potential of Br- is higher than water whereas the oxidation potential of F- is lower than water.
Therefore, in aq. solution of KBr, Br- ions are oxidized to Br2 in preference to H2O. Whereas, in aq. solution of KF, H2O is oxidized in preference to F-. Thus, oxidation of H2O at anode gives O2 and no F2 is produced
Therefore, in aq. solution of KBr, Br- ions are oxidized to Br2 in preference to H2O. Whereas, in aq. solution of KF, H2O is oxidized in preference to F-. Thus, oxidation of H2O at anode gives O2 and no F2 is produced
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