1.

Electrolysis of KBr (aq) gives Br_(2) at anode but of KF (aq) does not give F_(2) . Give reason for disparity in behaviour.

Answer»

Solution :`L_(F_(2)//F^(-))^(@)`has the HIGHEST reduction potential. THEREFORE, `F_(2)`is not liberated. Reduction potential of `O_(2)`is higher than that of `Br_(2)` . Thus, ELECTROLYSIS of KBr (aq) gives `Br_(2)`and not `O_(2)`


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