Electrolysis of KBr(aq) gives Br2 at anode but KF(aq) does not give F2

1.	Electrolysis of KBr(aq) gives Br2 at anode but KF(aq) does not give F2. Give reason.
Answer» Oxidation takes place at anode. Now higher the oxidation Potential, easier to oxidize. Oxidation potential of Br^-, H₂O, F^- are in the following order. Br^- > H₂O > F^- Therefore in aq. Solution of KBr. Br^- ions are oxidized to Br₂ in preference to H₂O. On the other hand, in aq. Solution of KF, H₂O is oxidized in preference to F^-. Thus in this case oxidation of H₂O at anode gives O₂ and no F² is produced.

Electrolysis of KBr(aq) gives Br2 at anode but KF(aq) does not give F2. Give reason.

Answer»

Oxidation takes place at anode. Now higher the oxidation Potential, easier to oxidize. Oxidation potential of Br^-, H₂O, F^- are in the following order.

Br^- > H₂O > F^-

Therefore in aq. Solution of KBr. Br^- ions are oxidized to Br₂ in preference to H₂O. On the other hand, in aq. Solution of KF, H₂O is oxidized in preference to F^-. Thus in this case oxidation of H₂O at anode gives O₂ and no F² is produced.

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Electrolysis of KBr(aq) gives Br2 at anode but KF(aq) does not give F2. Give reason.

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