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Electrolysis of KBr(aq) gives Br 2 at anode but of KF(aq) does not give F 2 . 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
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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