1. Oxidation states and half-equations

2. Common oxidation states
Element
Oxidation state
Exceptions H +1
Metal hydrides, i.e. NaH where H is -1
Group 1: Li, Na, K
Group 2: Mg, Ca, Ba +2
Group 3: Al +3 F -1 Cl
Unless it is with F or O, i.e. ClF3 where Cl is +3 O -2
Many exceptions

3. Oxidation states Work out the oxidation state of the following:
1) Iron in Fe2O3 +3
2) Nitrogen in KNO3 +5
3) Sulfur in SO2 +4
4) Chlorine in ClO3- +5
5) Phosphorus in PCl4+ +5

4. Oxidation and Reduction
Oxidation is the loss of electrons
Reduction is the gain of electrons
O Oxidation
I Is
L Loss of electrons
R Reduction
G Gain of electrons

5. Half-equations Example 1: Reaction between magnesium and chlorine:
Mg (s) + Cl2 (g) → MgCl2 (s)
Write the two half-equations and identify which species is oxidised and which is reduced
Mg → Mg2+ + 2e-
Cl2 + 2e- → 2Cl-
The number of electrons in the two half-equations must be the same
Oxidised
Reduced

6. Half-equations Example 2: Balance the equation below:
2Al (s) + 3Cl2 (g) → 2AlCl3 (s)
Write the two half-equations and identify which species is oxidised and which is reduced
2Al → 2Al3+ + 6e-
3Cl2 + 6e- → 6Cl-
The number of electrons in the two half-equations must be the same
Oxidised
Reduced

7. This is called a Redox equation
Half-equations
Example 3: Combine the following half-equations
Al → Al3+ + 3e-
O2 + 4e- → 2O2-
4Al → 4Al e-
3O2 + 12e- → 6O2-
We can combine the equations by cancelling the electrons
4Al + 3O2 → 4Al3+ + 6O2-
Multiply by 4 to get 12e-
Oxidised
Multiply by 3 to get 12e-
Reduced
This is called a Redox equation

8. Transition metals The transition metals have variable oxidation states
1) Vanadium in VO2+ +5
2) Chromium in Cr2O
Species
Oxidation state of Mn
Mn2+ +2
Mn(OH)3 +3
MnO2 +4
MnO42- +6
MnO4- +7

9. Half-equations Writing half-equations involving transition metals:
Step 1: Write the two ions
Step 2: Using the oxidation states write the electrons
Step 3: Add water (H2O) to the right side to equal the oxygen atoms on the left
Step 4: Add hydrogen ions (H+) to the left side to equal the hydrogen atoms on the right

10. Half-equations
Example 4: Manganese(VII) ions (MnO4-) can be reduced to Manganese(II) ions Mn2+ by iron(II) ions (Fe2+) which are oxidised to iron(III) ions (Fe3+)
Work out the two half-equations and then write an overall equation for the reaction
Step 1: MnO4- → Mn2+
Step 2: MnO4- + 5e- → Mn2+
Step 3: MnO4- + 5e- → Mn2+ + 4H2O
Step 4: MnO4- + 5e- + 8H+ → Mn2+ + 4H2O
+7 + 5e- → +2

11. Number of electrons in the two half-equations must be the same
Example 4: Manganese(VII) ions (MnO4-) can be reduced to Manganese(II) ions Mn2+ by iron(II) ions (Fe2+) which are oxidised to iron(III) ions (Fe3+)
Step 1: Fe2+ → Fe3+
Step 2: Fe2+ → Fe3+ + e-
Combining the two half-equations (overall equation):
MnO4- + 5e- + 8H+ → Mn2+ + 4H2O
Fe2+ → Fe3+ + e-
Multiply by 5 to get 5e-
Number of electrons in the two half-equations must be the same

12. MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
Half-equations
MnO4- + 5e- + 8H+ → Mn2+ + 4H2O
5Fe2+ → 5Fe3+ + 5e-
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
This is the overall equation
It is also called a redox equation
Reduced
Oxidised

13. Half-equations
Example 5: Dichromate(VI) ions (Cr2O72-) can be reduced to chromium(III) ions (Cr3+) by sulfite ions (SO32-) which are oxidised to sulfate ions (SO42-)
Work out the two half-equations and then write an overall equation for the reaction
Step 1: Cr2O72- → 2Cr3+
Step 2: Cr2O e- → 2Cr3+
Step 3: Cr2O e- → 2Cr3+ + 7H2O
Step 4: Cr2O e- + 14H+ → 2Cr3+ + 7H2O
+6 + 3e- → +3

14. This time the water is on the left to equal the oxygen's on the right
Half-equations
Example 5: Dichromate(VI) ions (Cr2O72-) can be reduced to chromium(III) ions (Cr3+) by sulfite ions (SO32-) which are oxidised to sulfate ions (SO42-)
Step 1: SO32- → SO42-
Step 2: SO32- → SO e-
Step 3: SO32- + H2O → SO e-
Step 4: SO32- + H2O → SO e- + 2H+
+4 → e-
This time the water is on the left to equal the oxygen's on the right

15. Cr2O72- + 8H+ + 3SO32- → 2Cr3+ + 4H2O + 3SO42-
Half-equations
Combining the two half-equations (overall equation):
Cr2O e- + 14H+ → 2Cr3+ + 7H2O
SO32- + H2O → SO e- + 2H+
3SO H2O → 3SO e- + 6H+
This is the overall equation
Cr2O H+ + 3SO32- → 2Cr3+ + 4H2O + 3SO42-
x 3 to get 6e-
Reduced
Oxidised