Question

Balance the following redox reactions in acidic solutions:

$Br{O}_3^{-}+N_2{H}_4⟶{Br}^{-}+N_2$

• A. $2Br{O}_3^{-}+{3N}_2{H}_4⟶{2Br}^{-}+{3N}_2+6H_{2}O$
• B. $3Br{O}_3^{-}+{3N}_2{H}_4⟶{2Br}^{-}+{3N}_2+4H_{2}O$
• C. $3Br{O}_3^{-}+{3N}_2{H}_4⟶{2Br}^{-}+{3N}_2+6H_{2}O$
• D. None of these

Answer 1

The correct option is A $2Br{O}_3^{-}+{3N}_2{H}_4⟶{2Br}^{-}+{3N}_2+6H_{2}O$

The unbalanced redox reaction is as follows:
$Br{O}_3^{-}+N_2{H}_4⟶{Br}^{-}+N_2$
All atoms other than H and O are balanced.
The oxidation number of Br changes from +5 to -1. The net change in the oxidation number is 6.
The oxidation number of N changes from -2 to 0. The net change in the oxidation number is 2. Total change in the oxidation number for 2 nitrogen atoms is 4.
To balance the increase in oxidation number with decrease in oxidation number, multiply $Br{O}_3^{-}$ and ${Br}^{-}$ with 2 and multiply $N_2{H}_4$ and $N_2$ with 3.
$2Br{O}_3^{-}+3N_2{H}_4⟶2{Br}^{-}+3N_2$
To balance O atoms, add 6 water molecules on RHS.
$2Br{O}_3^{-}+{3N}_2{H}_4⟶{2Br}^{-}+{3N}_2+6H_{2}O$
The H atoms are balanced. This is the balanced chemical equation.