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), NaOH`
D. `H_(2)` and `O_(2)`

Answer

c

Explanation

`NaBr overset("ionises")rarrNa^(+)(aq)+Br^(-)(aq)`
since the electrode potential of `H_(2)O(-0.83 V)` is higher than that of `Na^(+)` (-2.71) therefore `H_(2)O` is reduced in preference to `Na^(+)` ions at anode
`2 br^(-)(aq)rarr Br_(2)(g)+2E^(-)`
thus `H_(2)` is liberated at the cathode `Br_(2)` at the anode and NaOH is left in the solution