Element Oxidation state Exceptions

H +1 Metal hydrides, i.e. NaH where H is -1

Group 1: Li, Na, K +1Group 2: Mg, Ca, Ba +2Group 3: Al +3

F -1

Cl -1Unless it is with F or O, i.e. ClF3 where Cl is +3

O -2 Many exceptions

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

Oxidation is the loss of electrons

Reduction is the gain of electrons

O Oxidation I Is L Loss of electrons R Reduction I Is G Gain of electrons

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

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

Example 3: Combine the following half-equations

Al → Al3+ + 3e-

O2 + 4e- → 2O2-

4Al → 4Al3+ + 12e-

3O2 + 12e- → 6O2-

We can combine the equations by cancelling the electrons

4Al + 3O2 → 4Al3+ + 6O2-

Multiply by 4 to get 12e-

Multiply by 3 to get 12e-

OxidisedReduced

This is called a Redox equation

The transition metals have variable oxidation states

1) Vanadium in VO2+ +5

2) Chromium in Cr2O72- +6

Species Oxidation state of MnMn2+ +2

Mn(OH)3 +3

MnO2 +4

MnO42- +6

MnO4- +7

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

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

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

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

OxidisedReduced

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: Cr2O72- + 6e- → 2Cr3+

Step 3: Cr2O72- + 6e- → 2Cr3+ + 7H2O

Step 4: Cr2O72- + 6e- + 14H+ → 2Cr3+ + 7H2O

+6 + 3e- → +3

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- → SO4

2-

Step 2: SO32- → SO4

2- + 2e-

Step 3: SO32- + H2O → SO4

2- + 2e-

Step 4: SO32- + H2O → SO4

2- + 2e- + 2H+

+4 → +6 + 2e-

This time the water is on the left to equal the oxygen's on the right

Combining the two half-equations (overall equation):

Cr2O72- + 6e- + 14H+ → 2Cr3+ + 7H2O

SO32- + H2O → SO4

2- + 2e- + 2H+

Cr2O72- + 6e- + 14H+ → 2Cr3+ + 7H2O

3SO32- + 3H2O → 3SO4

2- + 6e- + 6H+

This is the overall equation

Cr2O72- + 8H+ + 3SO3

2- → 2Cr3+ + 4H2O + 3SO42-

x 3 to get 6e-

OxidisedReduced