Oxidation-Reduction (Redox) Reactions (College Board AP Chemistry)

Half reactions

• During redox reactions, reduction and oxidation occur a the same time
• However, it is convenient to consider them as separate process
• For example, let's consider the following net ionic equation

Sn²⁺ (aq) + 2Fe³⁺ (aq) → Sn⁴⁺ (aq) + 2Fe²⁺ (aq)

• Sn²⁺ oxidizes into Sn⁴⁺ by losing two electrons. Therefore, the oxidation half reaction can be written as:

Oxidation:

Sn²⁺ (aq)  → Sn⁴⁺ (aq) + 2e^–

• Fe³⁺ is reduced to Fe²⁺ by gaining one electron. Therefore the reduction half reaction can be written as:

Reduction:

 Fe³⁺ (aq) + 1e^– → Fe²⁺ (aq)

• The number of electrons gained in the reduction half reaction must be the equal to the number of electrons lost in the oxidation half reaction, so the iron half equation is multiplied times two

 2Fe³⁺ (aq) + 2e^– → 2Fe²⁺ (aq)

• • If this condition is met, the electrons cancel out when both half reactions are added
  • When both half reactions are added, the result must be the net ionic equation

Balancing chemical equations by the method of half reactions

• For balancing a redox reaction using the half reactions, follow these steps:

1. 1. Divide the equation into half reactions (one oxidation and one reduction)
   2. Balance each half reactions with the following rules:
      1. First, balance elements other than H and O
      2. Second, balance O atoms by adding H₂O
      3. Third, balance H atoms by adding H⁺
      4. Finally, balance charges by adding e^-
   3. Multiply half reactions by integers to make the number of electrons in both half reactions the same
   4. Add both half reactions, and cancel chemical species that appear on both sides
   5. Check if the atoms are balanced