Oxidation states and half-equations

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

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

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

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

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

This is called a Redox equation Half-equations 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- Oxidised Multiply by 3 to get 12e- Reduced This is called a Redox equation

Transition metals The transition metals have variable oxidation states 1) Vanadium in VO2+ +5 2) Chromium in Cr2O72- +6 Species Oxidation state of Mn Mn2+ +2 Mn(OH)3 +3 MnO2 +4 MnO42- +6 MnO4- +7

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

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

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

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

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: Cr2O72- + 6e- → 2Cr3+ Step 3: Cr2O72- + 6e- → 2Cr3+ + 7H2O Step 4: Cr2O72- + 6e- + 14H+ → 2Cr3+ + 7H2O +6 + 3e- → +3

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- → SO42- + 2e- Step 3: SO32- + H2O → SO42- + 2e- Step 4: SO32- + H2O → SO42- + 2e- + 2H+ +4 → +6 + 2e- This time the water is on the left to equal the oxygen's on the right

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