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:

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

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

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

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.

$$\ce{NaBrO3 (aq) -> Br- (aq)}$$

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.

$$\ce{NaBrO3(aq) + 6e- -> Br- (aq)}$$

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-.

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

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.

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

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

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

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?

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    $\begingroup$ You have Na on the left, but none of it on the right. You'd better fix this one way or another (preferably by not having it on either side, but that does not really matter much, just be consistent). $\endgroup$ – Ivan Neretin Sep 7 '15 at 15:32

Now if you need to use $\ce{OH-}$ ions. Use the same scheme: $$\ce{BrO3- -> Br-}$$ There is no oxygen in the right part so we need to add $\ce{H2O}$ and/or $\ce{OH-}$ ions somewhere. Look closely at $\ce{H2O}$ and $\ce{OH-}$. Water have 2 hydrogen per oxygen atom and $\ce{OH-}$ have 1 hydrogen per oxygen atom. Problem here is that we can't just add element to one side. When we add $\ce{H2O}$ or $\ce{OH-}$ we change both oxygen and hydrogen amounts. We will need to solve system of equation for that. First equation will be hydrogen balance and second one will be oxygen balance. What we need to add to the left part and what we need to add to the right part? As you can see $\ce{OH-}$ contains the same amount of oxygen and less hydrogen compared to $\ce{H2O}$. The only way to balance those two equations is by adding $\ce{OH-}$ to the oxygen poor (right) part. It means that we will add water to the left part. Now lets make equations, $a$ will be the amount of water, $b$ is the amount of $\ce{OH-}$.

Hydrogen: $2a=b$

Hydrogen can be contained in $\ce{H2O}$ or $\ce{OH-}$ only, but one $\ce{H2O}$ molecule contains two times more then $\ce{OH-}$. It means we need to use 2 multiplier.

Oxygen: $3+a=b$

Because we already have 3 in left part. Solve this system. We get $a=3$ and $b=6$. $$\ce{BrO3- + 3H2O -> Br- + 6OH-}$$ Wow, this works! Now equalize charges. $$\ce{BrO3- + 3H2O +6e- -> Br- + 6OH-}$$ P.S. Adding water to the right side will make our system unsolvable in natural numbers. Take a look.

Hydrogen: $b=2a$

Oxygen: $3+b=a$

  • $\begingroup$ Thank you @Cosmodruid. System of equations is a great tool to use. $\endgroup$ – ChemiNoviceN0.NA Sep 10 '15 at 9:09

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:

$$\ce{BrO3- -> Br-}$$

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

$$\ce{BrO3- -> Br- + 3H2O}$$

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:

$$\ce{BrO3- + 6H+ -> Br- + 3H2O}$$

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.

$$\ce{BrO3- + 6H+ + 6e- -> Br- + 3H2O}$$

Half-reaction is ready.

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    $\begingroup$ I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance. $\endgroup$ – bon Sep 7 '15 at 17:00
  • $\begingroup$ This scheme is very useful for solving this acid redox problem. It works and the process is considerably faster than the textbook definition I used above. How would you apply this scheme to a redox reaction in a basic medium dealing with Hydroxide instead of Hydrogen ions? I've tried to apply this scheme to several other redox problems in basic solution but I have been unsuccessful. Thank you for the help you have given me. $\endgroup$ – ChemiNoviceN0.NA Sep 9 '15 at 9:14

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