Oxidation, Reduction and Redox Equations - Balancing Redox Equations (A-Level Chemistry)

Balancing Redox Equations

Half-equations to Form Redox Equations

In a redox reaction, a reduction and an oxidation happen simultaneously. We can divide the reaction into two separate equations, one showing oxidation and the other showing reduction. These are called half-equations.

You should be able to form an overall redox equation by combining two half-equations.

Practice Question: Write the redox equation for when chlorine gas oxidises iron(II) ions to iron(III) ions. In the process, the chlorine is reduced to chloride ions

Method:

1. Identify half equations. First of all write the two half equations for the reduction and oxidation reactions that take place. The number of electrons lost or gained by an atom will be the same as its change in oxidation number.

Reduction half-equation: Cl2 + e- —> Cl-
(Oxidation state changes from 0 to -1)

Oxidation half-equation: Fe2+ —> Fe3+ + e-
(Oxidation state changes from +2 to +3)

2. Balance masses. Work on each half equation separately. Make sure that the number of atoms is the same at either side of the equation.

Reduction half-equation: Cl2 + e- —> 2Cl-

Oxidation half-equation: Fe2+ —> Fe3+ + e

3. Balance charges. Double check that the total charge at either side of the equation is the same.

Reduction half-equation: Cl2 + 2e- —> 2Cl-
(Charge at either side of the equation is -2)

Oxidation half-equation: Fe2+ —> Fe3+ + e-
(Charge at either side of the equation is +2)

4. Balance electrons. Make sure that both half-equations contain the same number of electrons.

For this to occur in this example, we need to double all the species in the oxidation half-equation.

Reduction half-equation: Cl2 + 2e- —> 2Cl-

Oxidation half-equation: 2Fe2+ —> 2Fe3+ + 2e

5. Combine and cancel. Finally you can combine the half-equations to make a full redox equation. Remember to cancel out the electrons on either side as they are the same on both sides.

Final redox equation: Cl2 + 2Fe2+ —> 2Cl- + 2Fe3

Practice Question: Potassium manganate (VII) is an important oxidising agent. When used as an oxidising agent in an acidic solution, the purple solution containing MnO4- ions gets reduced to a colourless solution containing Mn2+ ions. It oxidises iron(II) ions to iron(III) ions.

Method:

1. Identify half equations. First of all write the two half equations for the reduction and oxidation reactions that take place. Remember that the number of electrons gained or lost will equal the change in oxidation number.

Reduction half-equation: MnO4- + 5e- —> Mn+2
(Oxidation state changes from +7 to +2)

Oxidation half-equation: Fe2+ —> Fe3+ + e-
(Oxidation state changes from +2 to +3)

2. Balance masses. Work on each half equation separately. When in acidic conditions, H2O is added to balance oxygen atoms and H+ ions are added to balance hydrogens.

Reduction half-equation: 8H+ + MnO4- + 5e- —> Mn+2 + 4H2O

Oxidation half-equation: Fe2+ —> Fe3+ + e

3. Balance charges. Make sure that the charge at either side of the half equation is the same.

Reduction half-equation: 8H+ + MnO4- + 5e- —> Mn+2 + 4H2O
(Charge at both sides of the equation is +2)

Oxidation half-equation: Fe2+ —> Fe3+ + e-
(Charge at either side of the equation is +2)

4. Balance electrons. Ensure that both half-equations have the same number of equations.

For this to occur in this example, we need to double all the species in the oxidation half-equation.

Reduction half-equation: 8H+ + MnO4- + 5e- —> Mn+2 + 4H2O

Oxidation half-equation: 5Fe2+ —> 5Fe3+ + 5e

5. Combine and cancel. Finally you can combine the half-equations to make a full redox equation. Remember to cancel out the electrons on either side as they are the same on both sides, as well as anything else that is repeated.

Final redox equation: