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

Explanation

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