Balance the redox reaction by half reaction method :
`Br_(2)+ H_(2)O_(2) to BrO_(3)^(-) + H_(2)O` (in acidic medium)

Explanation

(ii) `Br_(2) rarr 2BrO_(3)^(-)` (Oxidation half)
`" "H_(2)O_(2) rarr 2H_(2)O` (Reduction half)
Oxidation half
`Br_(2) rarr 2BrO_(3)^(-)+10e^(-)`
`Br_(2)+6H_(2)O rarr 2BrO_(3)^(-)+10e^(-)` (Balancing oxygen)
`Br_(2)+6H_(2)O rarr 2BrO_(3)^(-)+10e^(-)+12H^(+)` (Balancing hydrogen)
Reduction half
`H_(2)O_(2) rarr H_(2)O`
`H_(2)O_(2)+2e^(-) rarr 2H_(2)O` (Balancing charge and oxygen)
`H_(2)O_(2)+2e^(-)+2H^(+) rarr 2H_(2)O` (Balancing hydrogen)
Balancing total charge
(iii) `[Br_(2)+6H_(2)O rarr 2BrO_(3)^(-)+12H^(+)+10e] times 1`
`" "[H_(2)O_(2)+2H^(+)+2e^(-) rarr 2H_(2)O] times 5`
(iv) `Br_(2)+5H_(2)O_(2)+10H^(+)+10e^(-)+6H_(2)O rarr 2BrO_(3)^(-)+12H^(+)+10e^(-)+10H_(2)O`
Now, we get
`" "Br_(2)+5H_(2)O_(2) rarr 2BrO_(3)^(-)+4H_(2)O+2H^(+)`