The products formed when an aqueous solution of `NaBr` is electrolysed in a cell having inert electrodes are `:`

Options

A. Na and `Br_(2)`
B. Na and `O_(2)`
C. `H_(2),Br_(2)andNaOH`
D. `H_(2)andO_(2)`

Answer

c

Explanation

`NaBr hArr NA^(+)+Br^(-)`
`H_(2) O hArr H^(+) +OH^(-)`
At cathode `2H^(+)+2e^(-) to H_(2)`
At node ` 2Br^(-)+OH^(-) to NaOH`
As discharge potential of `H^(+)` ions is lower than that of `Na^(+)` ions of `OH^(-)` ions, `H_(2)` is liberated at cathode and `Br_(2)` liberated at anode.