The oxidation number method · Chemistry Class 11–12 | StemPy Lab
Course 7.2·Lesson 1 of 3
The oxidation number method
A redox equation is balanced when the electrons lost equal the electrons gained. Follow the oxidation numbers, find what changes, match the changes, then balance the rest. Worked for permanganate with iron(II), copper with nitric acid and zinc with very dilute nitric acid.
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Two friends and a bag of marbles
Two friends play a game with marbles. Every marble that one friend loses, the other friend gains. Nobody can win a marble that nobody lost.
A redox reaction is the same game with electrons. The substance that is oxidised loses electrons. The substance that is reduced gains exactly the same number. So the first rule of balancing is simple: electrons lost equal electrons gained.
You met oxidation numbers earlier in this chapter. They let us count electrons without drawing every bond. Now we use them to balance equations that are too hard to balance by looking at them.
The five steps
Write the skeleton equation. Put the correct formulas of the reactants and products, with no coefficients yet.
Give oxidation numbers to the atoms that might change.
Find the change. A rise in oxidation number is oxidation. A fall is reduction. Write the change for one atom of each kind.
Match the changes. Multiply so that the total rise equals the total fall.
Balance the rest by looking. First the atoms that did not change, then charge, then hydrogen, then oxygen. In acid, use H+ and water.
A blue box on the left holds the permanganate ion, labelled oxidising agent. Manganese goes from plus 7 to plus 2 and gains 5 electrons. On the right, five circles each hold an iron(II) ion, labelled reducing agent. Each iron goes from plus 2 to plus 3 and loses 1 electron. An orange arrow from the iron ions to the box is labelled 5 electrons. Below, the text says 5 ions times 1 electron equals 5 electrons, so one permanganate ion reacts with five iron(II) ions.
Worked example 1: permanganate and iron(II) in acid
Purple permanganate turns almost colourless when it meets iron(II) ions in acid. The skeleton equation is:
MnO4−+Fe2+→Mn2++Fe3+
Oxidation numbers. Manganese goes from +7 to +2. Iron goes from +2 to +3.
The change. Each manganese atom gains 5 electrons. Each iron ion loses 1 electron.
Match. One permanganate ion takes 5 electrons, and each iron(II) ion gives only 1. So five iron ions must react with each permanganate ion:
MnO4−+5Fe2+→Mn2++5Fe3+
Charge. The left side has −1+10=+9. The right side has +2+15=+17. The reaction is in acid, so we add 8H+ to the left. Now both sides carry +17.
Hydrogen and oxygen. Eight H+ on the left need four water molecules on the right. Check oxygen: four atoms on each side.
MnO4−+8H++5Fe2+→Mn2++4H2O+5Fe3+
Practise with new numbers
Worked example 2: copper and dilute nitric acid
Copper dissolves in dilute nitric acid, and a colourless gas, nitric oxide, comes off. The skeleton equation is:
Cu+HNO3→Cu(NO3)2+NO+H2O
Oxidation numbers. Copper goes from 0 to +2, a rise of 2. Nitrogen goes from +5 in nitric acid to +2 in NO, a fall of 3.
Match. The smallest number that both 2 and 3 divide into is 6. So three copper atoms lose 6 electrons, and two nitrogen atoms gain 6:
3Cu+HNO3→3Cu(NO3)2+2NO+H2O
Balance the nitrogen. Not all the nitrate is reduced. Six nitrate ions stay in the copper(II) nitrate, and two nitrogen atoms go into NO. So eight HNO3 are needed.
Hydrogen and oxygen. Eight hydrogen atoms on the left give four water molecules on the right. Oxygen: 24 atoms on the left, and 18+2+4=24 on the right.
3Cu+8HNO3→3Cu(NO3)2+2NO+4H2O
Nitric acid does two jobs here. Two of the eight molecules are the oxidising agent. The other six only supply nitrate ions to the salt.
Try it in Python
The program finds, for three reactions, how many atoms of each agent react together. It uses the smallest number of electrons that both changes divide into.
Common mistakes
Counting all the nitric acid as the oxidising agent. In the copper reaction only two of the eight molecules are reduced.
Using the change for an ion instead of an atom. Match the change for one atom, then count how many such atoms the formula holds.
Fixing charge with water. Water carries no charge. Use H+ or OH− for that.
Changing a subscript. Only coefficients change. A subscript makes a different substance.
Balancing charge before the electrons match. Match the electrons first. Charge and atoms come after.
Remember these
Electrons lost equal electrons gained: the total rise in oxidation number equals the total fall.
Match the changes with the smallest numbers that both divide into.
Atoms that did not change, then charge, then hydrogen, then oxygen.
In acid, add H+ for charge and water for hydrogen and oxygen.
Check atoms and charge at the end.
Check yourself
Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.
One oxidising atom gains 3 electrons. One reducing atom loses 2. What is the smallest number of electrons moved, and how many atoms of each agent react?
Why are only two of the eight nitric acid molecules reduced when copper dissolves?
Zinc changes to Zn2+ when it dissolves in an acid. Which way does its oxidation number move, and by how much?
After the electrons are matched, an acid equation has a charge of +9 on the left and +17 on the right. What do you add, and how many?
Eight H+ are on the left. How many water molecules appear on the right, and why?
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