Redox Equations

This lesson covers: 

1. Half-equations show oxidation or reduction
2. Balancing half-equations
3. Combining half-equations into full redox equations
4. Disproportionation reactions

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 (MnO^4-) 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: MnO₄^-_(aq) ➔ Mn²⁺_(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: MnO₄^-_(aq) ➔ Mn²⁺_(aq) + 4H₂O_(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: MnO₄^-_(aq) + 8H⁺_(aq) ➔ Mn²⁺_(aq) + 4H₂O_(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: MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- ➔ Mn²⁺_(aq) + 4H₂O_(l)

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

5Fe²⁺_(aq) ➔ 5Fe³⁺_(aq) + 5e^-

Step 4: Combine half-equations into full redox equation

MnO₄^-_(aq) + 8H⁺_(aq) + 5e^- + 5Fe²⁺_(aq) ➔ Mn²⁺_(aq) + 4H₂O_(l) + 5Fe³⁺_(aq) + 5e^-

Step 5: Cancel equal electrons on both sides

MnO₄^-_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) ➔ Mn²⁺_(aq) + 4H₂O_(l) + 5Fe³⁺_(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 MnO₄^- to +2 in Mn²⁺, a change of -5.
• Fe is being oxidised from +2 in Fe²⁺ to +3 in Fe³⁺, a change of +1, but there are 5 Fe²⁺ 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

In disproportionation reactions, a single element is both oxidised and reduced.

Example:

Cl₂ + 2OH⁻ ➔ ClO⁻ + Cl⁻ + H₂O

Oxidation number of Cl:

• Increases from 0 (in Cl₂) to +1 (in ClO^-) so Cl is oxidised.
• Decreases from 0 (in Cl₂) to -1 (in Cl^-) so Cl is reduced.