Answer
$2BrO_{3}^{-}(aq) + 3N_{2}H_{4} (g) → 2Br^{-} (aq) + 3N_{2} (g) + 6H_{2}O(l)$
Oxidizing agent: $2BrO_{3}^{-}
Reducing agent: $N_{2}H_{4}$
Work Step by Step
We add water to right side to balance the equation as it is in acidic medium. Oxidation state of Br is changed from +5 to -1, hence it is reduced and state of N is changed from -2 to 0, hence it is oxidized.
