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# Worked example: Balancing a simple redox equation

AP.Chem:
TRA‑2 (EU)
,
TRA‑2.C (LO)
,
TRA‑2.C.1 (EK)
A redox equation can be balanced using the following stepwise procedure: (1) Divide the equation into two half-reactions. (2) Balance each half-reaction for mass and charge. (3) Equalize the number of electrons transferred in each half-reaction. (4) Add the half-reactions together. In this video, we'll use these steps to balance the redox equation for the reaction between aluminum metal and hydrogen ions. Created by Sal Khan.

## Want to join the conversation?

• How would I balance Br- +ClO3- >>> Br2 + Cl-
• We first need to identify which atoms are being reduced and which are being oxidized. The bromine begins with an oxidation state of -1 and proceeds to 0. This is an increase which means it is being oxidized. Automatically this means the chlorate is being reduced, but we can prove this too with oxidation states. The oxygens will have a constant oxidation state of -2 which means only the chlorine is changing. It starts with +5 and proceeds to -1. A decrease in the oxidation number means a reduction.

Next we separate the reaction into two half-reactions; an oxidation and reduction reaction.

Oxidation: Br^(-) → Br2
Reduction: ClO3^(-) → Cl^(-)

We then balance the half-reactions similarly to a normal chemical equation except here in a redox reaction we have to balance the mass AND the charge. Starting with the oxidation reaction, to balance the mass we simply add a '2' coefficient in front of the bromide.

2Br^(-) → Br2

Afterwards we balance the charge by adding electrons to one side of the equation. We can do this since in a redox reaction we have transfers of electrons. Initially the left side has a net charge of -2 and the right is 0. So to balance this we need to add two electrons to the right so both sides are -2. That balances the oxidation half-reaction.

2Br^(-) → Br2 + 2e^(-)

For the reduction half-reaction the chlorines are already balanced, but not the oxygens. We solve this by adding water molecules to the right side of the equation. We can do this because most redox reactions are assumed to be in a water solution.

ClO3^(-) → Cl^(-) + 3H2O

So while this balances the oxygens, it creates a new issue where hydrogen is introduced and is not balanced now. We solve this by adding hydrogen cations to the left side of the equation.

ClO3^(-) + 6H^(+) → Cl^(-) + 3H2O

Now that the masses are balanced, we balance the charge. The left side has a net charge of +5 while the right side is -1. Therefore we add six electron to the left side. This balances the reduction half-reaction.

ClO3^(-) + 6H^(+) + 6e^(-) → Cl^(-) + 3H2O

The last step is to add the two half-reactions into a single reaction again to get the overall redox reaction. In the overall reaction we want the electrons to cancel each other out from the two reactions since we can't have electrons in the final equation since they're not actually chemicals. The issue is that the oxidation reaction has two electrons, but the reduction reaction has six. We need the number of electrons to be the same so they cancel completely. We need the least common multiple of 2 and 6, which is 6. This means we need to multiple the entire oxidation by a certain number to have six elections; which is three here.

3x(2Br^(-) → Br2 + 2e^(-)) = 6Br^(-) → 3Br2 + 6e^(-)

Now we can finally add the two equations together and see that the six electrons from both equations cancel out.

* 6Br^(-) + ClO3^(-) + 6H^(+) + 6e^(-) → 3Br2 + Cl^(-) + 3H2O + 6e^(-)

* 6Br^(-) + ClO3^(-) + 6H^(+) → 3Br2 + Cl^(-) + 3H2O

And there's the final balanced redox reaction. Specifically thought it's balanced in an acidic solution. It is possible to balance it in a basic solution is needed.

Hope that helps.
• Do you always balance the number of atoms before the charges?
(1 vote)
• You don't absolutely have to balance mass before change, but it is generally easier to do so in redox reactions.
• How I can balance Mg+Fe(2)O(3)>>>Fe+MgO
(1 vote)
• I won't just give you the answer, but it's pretty straightforward if you balance them in a certain order.
Balance the elements in this order: Iron, oxygen, then magnesium.

Hope that helps.
• Near the end of the video the 6e- is added to 2Al(s) and 6H+(aq) why? same goes for the stuff added together after that.
(1 vote)
• Balancing redox reactions are slightly different from regular reactions in that not only do we have to balance mass, but we also have to balance charge. We do this primarily by adding electrons as an intermediate step which we represent with the symbol e^(-).

Once we've added electrons to each half reaction to balance the charge, we have to multiply each reaction to get the electrons to their LCM. The reason we need to is that when we add the half reactions back into a full reaction the electrons will cancel out. And we need to do this because electrons can't show up in the overall reaction's equation because they're not truly reactants/products. Instead what's really happening is just that electrons (six in this case) are moving from the aluminum atoms to the hydrogen atoms.

Hope that helps.
(1 vote)
• Why does H2 have no charge?
I would have thought the charge is +2 because each H has a charge of +1.
(1 vote)
• Hydrogen has the possibility of taking on several different charges as an ion including +1 or -1. But two neutral atoms of hydrogen will form a neutral molecule of diatomic hydrogen. Each atom has 1 proton and 1 electron, which is neutral, so two of those together would still be neutral.

Hope that helps.
(1 vote)