Balancing Redox Equations

Understanding Redox

Redox (oxidation-reduction) reactions include all chemical reactions in which atoms have their oxidation states changed. Oxidation is the loss of electrons—or the increase in oxidation state—by a molecule, atom, or ion. Reduction is the gain of electrons—or the decrease in oxidation state—by a molecule, atom, or ion. To remember this, think that LEO the lion says GER (Loss of Electrons is Oxidation; Gain of Electrons is Reduction).

Describing the overall electrochemical reaction for a redox process requires balancing the component half-reactions for oxidation and reduction.

Simple Redox Reactions

Follow these rules to balance simple redox equations:

1. Write the oxidation and reduction half-reactions for the species that is reduced or oxidized.
2. Multiply the half-reactions by the appropriate number so that they have equal numbers of electrons.
3. Add the two equations to cancel out the electrons. The equation should be balanced.

An example is given below of the reaction of iron(III) sulfate with magnesium.

• Unbalanced reaction: Mg(s) + Fe₂(SO₄)₃(aq) → Fe(s) + MgSO4(aq)

This reaction is split into two half-reactions, one that involves oxidation and one that involves reduction.

• Reduction: Fe³⁺(aq) + 3e^- → Fe(s)
• Oxidation: Mg(s) → Mg²⁺(aq) + 2e^-

This pair of half-reactions can be balanced by ensuring that both have the same number of electrons. To do this, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2, so that each half-reaction has 6e^-.

• 2 Fe³⁺(aq) + 6e^- → 2 Fe(s)
• 3 Mg(s) → 3 Mg²⁺(aq) + 6e^-

Adding these two half reactions together gives the balanced equation:

• 2 Fe³⁺(aq) + 3 Mg(s) → 2 Fe(s) + 3 Mg²⁺(aq)

Notice that the sulfate ion (SO₄^2-) is ignored. This is because it does not take part in the reaction; it is a spectator ion.

Complex Redox Reactions

For reactions in aqueous solution, these reactions can be more complex, and involve adding H⁺, OH⁻, and H₂O in addition to electrons to compensate for oxidation changes.

Follow these steps when balancing acidic complex redox equations:

1. Write the oxidation and reduction half-reactions for the species including the element that is reduced or oxidized.
2. Balance both reactions for all elements except oxygen and hydrogen.
3. If the oxygen atoms are not balanced in either reaction, add water molecules to the side missing the oxygen. If the hydrogen atoms are not balanced, add hydrogen ions (H⁺).
4. Multiply the half-reactions by the appropriate number so that they have equal numbers of electrons.
5. Add the two equations to cancel out the electrons. The equation should be balanced.

If the reaction occurs in a base, proceed as if it is in an acid environment, but after step 4, add a hydroxide ion to both sides of the equation for each hydrogen ion added. Then, combine the hydroxide and hydrogen ions to form water. Next, cancel all of the water molecules that appear on both sides.

Both acidic and basic media conditions will now be explored more in depth.

Acidic Media

In acidic media, H⁺ ions and water are added to half-reactions to balance the overall reaction. For example, when manganese(II) reacts with sodium bismuthate:

• Unbalanced reaction: Mn²⁺(aq) + BiO₃^-(s) → Bi³⁺(aq) + MnO₄⁻ (aq)

Step 1: Write the oxidation and reduction half-reactions for the species including the element that is reduced or oxidized.

Reduction: $6H^+ + BiO_3^- \rightarrow Bi^{3+} + 3H_2O$ (Bi goes from a +5 to a +3)

Oxidation: $Mn^{2+} \rightarrow MnO_4^-$ (Mn goes from a +2 to a +5)

Step 2: Balance both reactions for all elements except oxygen and hydrogen.

In this case, both Bi and Mn are already balanced.

Step 3: If the oxygen atoms are not balanced in either reaction, add water molecules to the side missing the oxygen. If the hydrogen atoms are not balanced, add hydrogen ions (H⁺).

Reduction: $2e^- + 6H^+ + BiO_3^- \rightarrow Bi^{3+} + 3H_2O$

Oxidation: $4H_2O + Mn^{2+} \rightarrow MnO_4^- + 8H^+ + 5e^-$

Step 4: Multiply the half-reactions by the appropriate number so that they have equal numbers of electrons.

We need to multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2.

Reduction: $10e^- + 30H^+ + 5BiO_3^- \rightarrow 5Bi^{3+} + 15H_2O$

Oxidation: $8H_2O + 2Mn^{2+} \rightarrow 2MnO_4^- + 16H^+ + 10e^-$

Step 5: Add the two equations to cancel out the electrons. The equation should be balanced.

$10e^- + 30H^+ + 5BiO_3^- + 8H_2O + 2Mn^{2+} \rightarrow 5Bi^{3+} + 15H_2O + 2MnO_4^- + 16H^+ + 10e^-$

Cancel like species:

$14H^+ + 5BiO_3^- + 2Mn^{2+} \rightarrow 5Bi^{3+} + 7H_2O + 2MnO_4^-$

Check the balanced equation:

14 H on left and 14 H on right

5 Bi on left and 5 Bi on right

2 Mn on left and 2 Mn on right

15 O on left and 15 O on right

13+ charge on left and 13+ charge on right

Basic Media

In basic media, OH⁻ ions and water are added to half reactions to balance the overall reaction. For example, take the reaction between potassium permanganate and sodium sulfite:

Unbalanced reaction: MnO₄^- + SO₃^2- + H₂O → MnO₂ +SO₄^2- + OH^-

As in acidic media, the unbalanced reaction can be separated into its two half-reactions, each representing either reduction or oxidation.

• Reduction: 3 e⁻ + 2 H₂O + MnO₄⁻ → MnO₂ + 4 OH⁻
• Oxidation: 2 OH⁻ + SO₃²⁻ → SO₄²⁻ + H₂O + 2 e⁻

Balancing the number of electrons in the two half-reactions gives:

• 6 e⁻ + 4 H₂O + 2 MnO₄⁻ → 2 MnO₂ + 8 OH⁻
• 6 OH⁻ + 3 SO₃²⁻ → 3 SO₄²⁻ + 3 H₂O + 6 e⁻

Adding these two half-reactions together gives the balanced equation:

• 2 MnO₄^- + 3 SO₃^2- + H₂O → 2 MnO₂ + 3 SO₄^2- + 2 OH^-