Ionic half-equations show electron transfer

Half-equations illustrate the electrons lost or gained during redox reactions.

For example, the half-equation for the oxidation of sodium is:

Na ➔ Na^+^ + e^-^

By combining oxidation and reduction half-equations, you can form the complete redox equation.

Worked example 1 - Writing a redox equation

Write the redox equation for the reaction between magnesium and oxygen to produce magnesium oxide.

Step 1: Write oxygen reduction half-equation

O₂ + 4e^- ➔ 2O^2-

Step 2: Write magnesium oxidation half-equation

Mg ➔ Mg²⁺ + 2e^-

Step 3: Double magnesium half-equation to make the number of electrons the same in both half-equations

2Mg ➔ 2Mg²⁺ + 4e^-

Step 4: Combine half-equations into full redox equation

O₂ + 4e^- + 2Mg ➔ 2MgO + 4e^-

Step 5: Cancel equal electrons on both sides

O₂ + 2Mg ➔ 2MgO

Worked example 2 - Writing a redox equation

Write the redox equation for the reaction between aluminium and chlorine to produce aluminium chloride.

Step 1: Write aluminium oxidation half-equation

Al ➔ Al³⁺ + 3e^-

Step 2: Write chlorine reduction half-equation

Cl₂ + 2e^- ➔ 2Cl^-

Step 3: Double aluminium half-equation and triple chlorine half-equation to make the number of electrons the same in both half equations

2Al ➔ 2Al³⁺ + 6e^-

3Cl₂ + 6e^- ➔ 6Cl^-

Step 4: Combine half-equations into full redox equation

2Al + 3Cl₂ + 6e^- ➔ 2AlCl₃ + 6e^-

Step 5: Cancel equal electrons on both sides

2Al + 3Cl₂ ➔ 2AlCl₃

Worked example 3 - Writing a redox equation

Acidified manganate(VII) ions (MnO₄^-) can be reduced to Mn²⁺ by Fe²⁺ ions. Write the full redox equation for this reaction.

Step 1: Write iron oxidation half-equation

Fe²⁺_(aq) ➔ Fe³⁺_(aq) + e^-

Step 2: Write manganate reduction half-equation

• Manganate is being reduced. Start by writing this down: MnO_4_^-^(aq) ➔ Mn^2+^(aq)
• To balance the oxygens, add water to the side with fewer oxygen atoms. In this case, add 4H₂O to the right-hand side of the equation: MnO_4_^-^(aq) ➔ Mn^2+^(aq) + 4H_2_O_(l)_
• To balance the hydrogens, add H⁺ to the side with fewer hydrogen atoms. In this case, add 8H⁺ to the left-hand side of the equation: MnO_4_^-^(aq) + 8H^+^(aq) ➔ Mn^2+^(aq) + 4H_2_O_(l)_
• Finally, to balance the charges, add e^- to the side with the higher positive charge. In this case, add 5e^- to the left-hand side of the equation: MnO_4_^-^(aq) + 8H^+^(aq) + 5e^-^ ➔ Mn^2+^(aq) + 4H_2_O_(l)_

Step 3: Multiply iron half-equation by 5 to make the number of electrons the same in both half-equations

5Fe^2+^(aq) ➔ 5Fe^3+^(aq) + 5e^-^

Step 4: Combine half-equations into full redox equation

MnO_4_^-^(aq) + 8H^+^(aq) + 5e^-^ + 5Fe^2+^(aq) ➔ Mn^2+^(aq) + 4H_2_O_(l)_ + 5Fe^3+^(aq) + 5e^-^

Step 5: Cancel equal electrons on both sides

MnO_4_^-^(aq) + 8H^+^(aq) + 5Fe^2+^(aq) ➔ Mn^2+^(aq) + 4H_2_O_(l)_ + 5Fe^3+^(aq)

To check if the numbers of electrons involved in the equation are correct, we can look at the changes in oxidation numbers of the elements being oxidised and reduced:

• Mn is being reduced from +7 in MnO_4_^-^ to +2 in Mn^2+^, a change of -5.
• Fe is being oxidised from +2 in Fe^2+^ to +3 in Fe^3+^, a change of +1, but there are 5 Fe^2+^ ions, so the total change is +5. Since these oxidation number changes are equal in magnitude but opposite in sign, and they balance the 5 electrons being transferred, we can confirm that the numbers of electrons involved agree with the balanced equation.

Disproportionation is simultaneous oxidation and reduction

Disproportionation is a unique type of redox reaction where a single element is both oxidised and reduced.

For example, in the reaction:

Cl_2_ + 2OH^−^ ➔ ClO^−^ + Cl^−^ + H_2_O

The oxidation number of Cl:

• Increases from 0 (in Cl_2_) to +1 (in ClO^-^) so Cl is oxidised.
• Decreases from 0 (in Cl_2_) to -1 (in Cl^-^) so Cl is reduced.