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Electrolysis of KBr(aq) gives Br2 at anode but of KF(aq) does not give F2. Give reason for disparity in behaviour.
Hi Belani,
The activities of the anion based on EOP follows the order:
F- > OH- > Cl- > Br- > I-
At anode least active anion is deactivated.
In aq. solution OH- is present.
So in the electrolysis of KBr(aq), Br- will be deactivated. While in the electrolysis of KF(aq), OH-will be deactivated.
askiitian.expert- chandra sekhar
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