Bob Neilson, TCU

Balance the following reaction that occurs in basic solution. Write complete, balanced equations for the oxidation and reduction half-reactions and the net ionic equation.

N₂H₄ + BrO₃^- ==> NO + Br^-

(1) Check oxidation numbers to determine what is oxidized and what is reduced.

Bromine goes from +5 in BrO₃^- to -1 in Br^-. Thus, BrO₃^- is being reduced.
Nitrogen goes from -2 in N₂H₄ to +2 in NO. Thus, N₂H₄ is being oxidized.

So, the two unbalanced half reactions are:

Reduction: BrO₃^- ==> Br^-

Oxidation: N₂H₄ ==> NO

(2) Balance atoms other than H and O: N₂H₄ ==> 2 NO

(3) Balance O with H₂O, and then

(4) Balance H with H⁺

2 H₂O + N₂H₄ ==> 2 NO + 8 H⁺

6 H⁺ + BrO₃^- ==> Br^- + 3 H₂O

(Note: the atoms are balanced at this point but the charges are not!)

(5) Balance charge with electrons

2 H₂O + N₂H₄ ==> 2 NO + 8 H⁺ + 8 e^-

6 e^- + 6 H⁺ + BrO₃^- ==> Br^- + 3 H₂O

(6) Since the reaction is occuring in basic solution, the hydrogen ions (H⁺) must be neutralized by adding equal numbers of OH^- ions to both sides of the equations.

8 OH^- + 2 H₂O + N₂H₄ ==> 2 NO + 8 H⁺ + 8 e^- + 8 OH^-

6 OH^- + 6 e^- + 6 H⁺ + BrO₃^- ==> Br^- + 3 H₂O + 6 OH^-

Now simplify by combining H⁺ with OH^- and substracting H₂O where possible

Reduction: 8 OH^- + N₂H₄ ==> 2 NO + 8 e^- + 6 H₂O

Oxidation: 3 H₂O + 6 e^- + BrO₃^- ==> Br^- + 6 OH^-

(7) You should now have the complete, balanced half-reactions! Rewrite them and check the balance of all atoms and charges.

(8) Multiply the balanced half-reactions by appropriate coefficients so as to make the number of electrons cancel. In this case, multiply the reduction by 3 and the oxidation by 4 for a total of 24 electrons on each side.

3 x [8 OH^- + N₂H₄ ==> 2 NO + 8 e^- + 6 H₂O]
i.e., 24 OH^- + 3 N₂H₄ ==> 6 NO + 24 e^- + 18 H₂O

4 x [3 H₂O + 6 e^- + BrO₃^- ==> Br^- + 6 OH^-]
i.e., 12 H₂O + 24 e^- + 4 BrO₃^- ==> 4 Br^- + 24 OH^-

(9) Add the half-reactions together, being careful not to omit anything.

24 OH^- + 3 N₂H₄ + 12 H₂O + 24 e^- + 4 BrO₃^- ==>
4 Br^- + 24 OH^- + 6 NO + 24 e^- + 18 H₂O

(10) Cancel species like H₂O, OH^-, or H⁺ that may appear on both sides. In this case, substract 24 e^-, 24 OH^- and 12 H₂O from each side.

3 N₂H₄ + 4 BrO₃^- ==> 4 Br^- + 6 NO + 6 H₂O

(11) If necessary add spectator ions to get the balanced molecular eqaution. That is not possible in this case because we started with an unbalanced ionic equation. Then, check the final balance -- all atoms and total charges must be equal on both sides of the equations.

3 N₂H₄ + 4 BrO₃^- ==> 4 Br^- + 6 NO + 6 H₂O