Writing half equations
The reduction half equation shows how the oxidant is reduced
The oxidation half equation shows how the reductant is oxidised
The oxidation half equation shows how the reductant is oxidised
Half equations must always be balanced
Step 1 - balance the atoms of the element undergoing the change in oxidation number
Step 2 - balance oxygen atoms by adding water
Step 3 - balance hydrogen atoms by adding H+ ion
Step 4 - balance the charge by adding electrons the to most positive side
Step 1 - balance the atoms of the element undergoing the change in oxidation number
Step 2 - balance oxygen atoms by adding water
Step 3 - balance hydrogen atoms by adding H+ ion
Step 4 - balance the charge by adding electrons the to most positive side
SO2 -> SO42-
1. The S atoms are already balanced 2. There are 2 oxygen atoms on the left and 4 on the right so we add 2 water molecules to the left SO2 + 2H2O -> SO42- 3. There are 4 hydrogen atoms on the left and 0 on the right so we add 4 H+ ions to the right SO2 + 2H2O -> SO42- + 4H+ 4. There is an overall charge of 0 on the left and 2+ on the right so we add 2 electrons to the right (+2 -2 = 0) SO2 + 2H2O -> SO42- + 4H+ + 2e- This is an oxidation reaction as the oxidation number of S is increasing from +4 in SO2 to +6 in SO42- |
IO3- -> I2
1. We must balance the I atoms 2IO3- -> I2 2. To balance the oxygen atoms we need to add 6 water molecules to the right 2IO3- -> I2 + 6H2O 3. There are 12 hydrogen atoms on the right so we must add 12 H+ ions to the left 2IO3- + 12H+ -> I2 + 6H2O 4. The overall charge on the left is 10+ and 0 on the right so we add 10 electrons to the left to balance out the charges 2IO3- + 12H+ + 10e- -> I2 + 6H2O This is a reduction reaction as the oxidation number of the I decreases from +5 in IO3- to 0 in I2. |
Observations linked to species
Not only do you need to be able to write the half equation and determine if it is an oxidation or reduction reaction, you also need to record the relevant colour and state (aq, s, l, or g). This means you need to be familiar with your redox couples from the previous page.
Combining half equations
Fully worked example - a typical question
Potassium dichromate with iron (II) ions
To 1mL of potassium dichromate solution in a test tube, add 1mL of hydrochloric acid (this acidifies the potassium dichromate). Add a solution containing iron (II) ions dropwise, shaking until the colour changes.
Oxidation: Fe2+ -> Fe3+ + e-
Reduction: Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O
To work out the full equation, you must balance the electrons - To do this you will need to multiply the first equation by 6. We don't want any electrons in our full equation.
6Fe2+ -> 6Fe3+ + 6e-
Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O
Now we can cancel out the electrons
6Fe2+ -> 6Fe3+
Cr2O72- + 14H+ -> 2Cr3+ + 7H2O
Now we can combine the two equations
Full: Cr2O72- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+
Name the species that is oxidised and justify your choice:
Fe2+ is the species that is oxidised. The oxidation number of Fe in Fe2+ is increasing from +2 to +3 in Fe3+
Name the species that is reduced and justify your choice:
Cr2O72- is the species that is reduced. The oxidation number of Cr in Cr2O72- is decreasing from +6 to +3 in Cr3+
Observations linked to species:
When the orange dichromate ions, Cr2O72- (aq) in solution are added to the pale green iron (II) ions, Fe2+ (aq) in solution, a solution forms which contains orange-brown iron (III) ions, Fe3+ (aq) and blue-green chromium (III) ions, Cr3+ (aq).
Potassium dichromate with iron (II) ions
To 1mL of potassium dichromate solution in a test tube, add 1mL of hydrochloric acid (this acidifies the potassium dichromate). Add a solution containing iron (II) ions dropwise, shaking until the colour changes.
Oxidation: Fe2+ -> Fe3+ + e-
Reduction: Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O
To work out the full equation, you must balance the electrons - To do this you will need to multiply the first equation by 6. We don't want any electrons in our full equation.
6Fe2+ -> 6Fe3+ + 6e-
Cr2O72- + 14H+ + 6e- -> 2Cr3+ + 7H2O
Now we can cancel out the electrons
6Fe2+ -> 6Fe3+
Cr2O72- + 14H+ -> 2Cr3+ + 7H2O
Now we can combine the two equations
Full: Cr2O72- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+
Name the species that is oxidised and justify your choice:
Fe2+ is the species that is oxidised. The oxidation number of Fe in Fe2+ is increasing from +2 to +3 in Fe3+
Name the species that is reduced and justify your choice:
Cr2O72- is the species that is reduced. The oxidation number of Cr in Cr2O72- is decreasing from +6 to +3 in Cr3+
Observations linked to species:
When the orange dichromate ions, Cr2O72- (aq) in solution are added to the pale green iron (II) ions, Fe2+ (aq) in solution, a solution forms which contains orange-brown iron (III) ions, Fe3+ (aq) and blue-green chromium (III) ions, Cr3+ (aq).
Add 1mL of unacidified potassium permanganate to a solution containing iron (II) ions
Oxidation half equation |
Fe2+ -> Fe3+ + e- (x3) |
Reduction half equation |
MnO4- + 4H+ + 3e- -> MnO2 + 2H2O |
Full ionic equation |
3Fe2+ -> 3Fe3+ + 3e- MnO4- + 4H+ + 3e- -> MnO2 + 2H2O 3Fe2+ + MnO4- + 4H+ -> 3Fe3+ + MnO2 + 2H2O |
Name the species which is oxidised and justify |
Fe2+ as the oxidation number of the Fe increases from +2 in Fe2+ to +3 in Fe3+ |
Name the species which is reduced and justify |
MnO4- as the oxidation number of Mn decreases from +7 in MnO4- to +4 in MnO2 |
Observations linked to species |
When the purple permanganate ions, MnO4- (aq) in solution are added to the pale green iron (II) ions Fe2+ (aq) in solution, a solution forms which contains orange-brown iron (III) ions, Fe3+ (aq) and a black-brown solid, MnO2 (s) |
Add 1mL of acidified potassium permanganate to a solution containing 1mL of sulfite ions
Oxidation half equation |
SO32- + H2O -> SO42- + 2H+ + 2e- |
Reduction half equation |
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O |
Full ionic equation |
To balance the electrons, we can multiple the oxidation half equation by 5 and the reduction half equation by 2 5SO32- + 5H2O -> 5SO42- + 10H+ + 10e- 2MnO4- + 16H+ + 10e- -> 2Mn2+ + 8H2O Not only do the electrons cancel out, but we can balance the H+ ions and the H2O. There are 10H+ on the right and 16 on the left - we can cancel out the 10 and write 6H+ on the left There are 5H2O's on the left and 8 on the right - we can cancel out the 5 and write 3H2O on the right 5SO32- + 2MnO4- + 6H+ -> 5SO42- + 2Mn2+ + 3H2O |
Name the species which is oxidised and justify |
SO32-, as the oxidation number of the S increases from +4 in SO32- to +6 in SO42- |
Name the species which is reduced and justify |
MnO4- as the oxidation number of Mn decreases from +7 in MnO4- to +2 in Mn2+ |
Observations linked to species |
When the purple permanganate ions, MnO4- (aq) in solution are added to the colourless sulfite ions, SO32- (aq) in solution, a solution forms which contains colourless SO42- (aq) ions and colourless Mn2+ ions (aq) |