Redox Reaction Equation with elusive oxygen I've been balancing redox equations for the last couple of weeks and have become quite accustomed to the procedure however I've come across an equation that eludes me every which way when it comes to balancing oxygen. I come to the next to last step to balancing my redox equation but am unable to write my net ionic equation.


  Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed.


I am asked to balance this reaction in an acidic solution.

I first write out my equation:

$$\ce{N2H4(l) + NaBrO3(aq) -&gt; N2(g) + Br- (aq)}$$

I break my equation up into two half equations, one being my oxidant equation and the other my reductant. I balance them both:

$$\ce{N2H4(l) -&gt; N2(g) + 4H+ (aq) + 4e-}$$

The other half of the equation is where I get stuck with balancing the oxygen. I end up balancing the hydrogen on both sides to be 10 but I come up short of one oxygen on my product side.

$$\ce{NaBrO3 (aq) -&gt; Br- (aq)}$$

I use the algebraic method to assign an oxidation number to $\ce{Br}$ on reactant side of half equation.

$\ce{Br + Na + 3O = 0}$

$\ce{Br + 1 + 3(-2) = 0}$

$\ce{Br - 5 = 0}$

$\ce{Br = 5^{+}}$

$\ce{Br}$ on the product side has an oxidation number of -1 equal to its net charge

In order to balance the oxidation charge of both sides, I add $\ce{6e-}$ to the reactant side. This then brings the oxidation charge to -1 on both sides of the half reaction.

$$\ce{NaBrO3(aq) + 6e- -&gt; Br- (aq)}$$

I then compensate net charge of the acidic reaction by adding 5 hydrogen ions to the reactant side to bring the net charge on both sides to 1-.

$$\ce{NaBrO3(aq) + 6e- + 5H+ (aq) -&gt; Br- (aq)}$$

I then must add water to the product side in order to balance Hydrogen and oxygen atoms in the half reaction. I have 5 hydrogen for reactants so I must add 2.5 water molecules on the product side to match hydrogen.

$$\ce{NaBrO3(aq) + 6e- + 5H+(aq)-&gt;Br- (aq) + $2.5$H2O(l)}$$ 

Staying customary to the law of combining volumes and Avogadro's hypothesis I multiply the half equation by 2 to get a whole number for water

$$\ce{2NaBrO3(aq) + 12e- + 10H+(aq) -&gt; 2Br- (aq) + 5H2O(l)}$$ 

The oxygen atoms here elude me. I can't figure out how to balance them. I figured I took a wrong turn on the way but my troubleshooting comes to no avail thus far. Can anyone fill me in on this little secret?
First of all you need to write equations with real ions and molecules that do exist in solution. Your oxidation agent here is $\ce{BrO3-}$. In neutral media you can use $\ce{H+}$, $\ce{OH-}$ and $\ce{H2O}$. Your half-reaction can be made from this scheme:

$$\ce{BrO3- -&gt; Br-}$$

There is no oxygen in the right part so we need to add $\ce{H2O}$ there, we need 3 molecules.

$$\ce{BrO3- -&gt; Br- + 3H2O}$$

Now there are no H atoms in the left part and 6 H atoms in the right part. Fix this by adding 6 protons, $\ce{H+}$, in the left part:

$$\ce{BrO3- + 6H+ -&gt; Br- + 3H2O}$$

Charge of left part is +5 and charge of right part is -1. Now equalize charges in left and right parts by adding 6 electrons to the left part.

$$\ce{BrO3- + 6H+ + 6e- -&gt; Br- + 3H2O}$$

Half-reaction is ready.