AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/4/4-9)
Unit 4 · Topic 4.9
4.9 Oxidation-Reduction (Redox) Reactions
Oxidation is losing electrons and reduction is gaining them, and they always happen together. To balance a redox equation, split it into two half-reactions, balance atoms and charge in each, make the electrons lost equal the electrons gained, then add the halves back together.
Key terms
- oxidation
- reduction
- half-reaction
- electron transfer
- oxidation number
Half-reactions
A half-reaction shows just one side of the electron transfer, with electrons written in. In Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), the two halves are:
Oxidation: Cu → Cu²⁺ + 2e⁻ (electrons are products). Reduction: Ag⁺ + e⁻ → Ag (electrons are reactants).
Every electron lost in oxidation must be gained in reduction. Electrons never appear in the final, overall equation.
Balancing in acidic solution
Most AP problems take place in acidic solution. Follow these steps. (If a problem says the solution is basic, balance it as if it were acidic first, then add enough OH⁻ to both sides to turn each H⁺ into H₂O, and cancel water.)
- Split the reaction into an oxidation half and a reduction half, using oxidation numbers to tell which is which.
- Balance all atoms other than O and H.
- Balance O by adding H₂O to the side that needs oxygen.
- Balance H by adding H⁺ to the side that needs hydrogen.
- Balance charge by adding electrons (e⁻) to the more positive side.
- Multiply one or both half-reactions so the electrons lost equal the electrons gained.
- Add the halves, cancel anything that appears on both sides (electrons, H₂O, H⁺), and check atoms and charge.
Counting electrons with oxidation numbers
You can check the number of electrons in a half-reaction with oxidation numbers. Electrons transferred = change in oxidation number × number of atoms that change.
In Cr₂O₇²⁻ → 2Cr³⁺, each Cr goes from +6 to +3, a change of 3, and there are 2 Cr atoms, so 6 electrons are gained. In MnO₄⁻ → Mn²⁺, Mn goes from +7 to +2, so 5 electrons are gained. If your half-reaction has a different number of electrons, recheck it.
Why balance this way
Many redox equations look impossible to balance by inspection, because both atoms and charge have to work out. Half-reactions break the job into two easy pieces. They also show exactly how many electrons are transferred, which you'll need later for electrochemistry (Unit 9): the number of moles of electrons, n, appears in ΔG° = −nFE°.
The final check is the most important step. Count each element and add up the charges on both sides. If either fails, a half-reaction is off.
Useful half-reactions to recognize
| Half-reaction | Electrons |
|---|---|
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | 5 gained |
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | 6 gained |
| Fe²⁺ → Fe³⁺ + e⁻ | 1 lost |
| 2I⁻ → I₂ + 2e⁻ | 2 lost |
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Equalizing electrons
Balance: Cu(s) + Ag⁺(aq) → Cu²⁺(aq) + Ag(s).
Show the solutionHide the solution
- Step 1: Oxidation: Cu → Cu²⁺ + 2e⁻.
- Step 2: Reduction: Ag⁺ + e⁻ → Ag.
- Step 3: Copper loses 2 electrons and each silver ion gains only 1, so multiply the reduction by 2: 2Ag⁺ + 2e⁻ → 2Ag.
- Step 4: Add and cancel the 2e⁻: Cu + 2Ag⁺ → Cu²⁺ + 2Ag. Check charge: +2 on each side.
Answer: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
- Example 2
Balancing in acid
Balance in acidic solution: Cr₂O₇²⁻(aq) + I⁻(aq) → Cr³⁺(aq) + I₂(s).
Show the solutionHide the solution
- Step 1: Reduction (Cr goes from +6 to +3): Cr₂O₇²⁻ → 2Cr³⁺. Add 7H₂O to the right for oxygen, then 14H⁺ to the left for hydrogen: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O.
- Step 2: Charge: left = −2 + 14 = +12; right = +6. Add 6e⁻ to the left: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
- Step 3: Oxidation: 2I⁻ → I₂ + 2e⁻.
- Step 4: Multiply the oxidation by 3 to make 6 electrons: 6I⁻ → 3I₂ + 6e⁻.
- Step 5: Add and cancel electrons: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O.
- Step 6: Check: 2 Cr, 7 O, 14 H, 6 I on each side. Charge: −2 + 14 − 6 = +6 on the left; +6 on the right.
Answer: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6I⁻(aq) → 2Cr³⁺(aq) + 3I₂(s) + 7H₂O(l)
- Example 3
Atoms balance but charge doesn't (classic trap)
A student balances MnO₄⁻ + Fe²⁺ + H⁺ → Mn²⁺ + Fe³⁺ + H₂O as MnO₄⁻ + Fe²⁺ + 8H⁺ → Mn²⁺ + Fe³⁺ + 4H₂O. Check it and fix it.
Show the solutionHide the solution
- Step 1: Atoms: 1 Mn, 1 Fe, 4 O and 8 H on each side, so atoms balance.
- Step 2: Charge: left = −1 + 2 + 8 = +9; right = +2 + 3 = +5. Not balanced, so electrons lost don't equal electrons gained.
- Step 3: Half-reactions: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, and Fe²⁺ → Fe³⁺ + e⁻. Multiply the iron half by 5.
- Step 4: Sum: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. Charge: −1 + 10 + 8 = +17 on the left; +2 + 15 = +17 on the right.
Answer: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Common mistakes
- Writing electrons on the wrong side: oxidation half-reactions produce electrons; reduction half-reactions consume them.
- Checking atoms but not charge.
- Balancing O by adding O₂ or O²⁻. In acidic solution, use H₂O and H⁺.
- Leaving electrons in the final equation.
On the exam
- Free-response questions often give two half-reactions and ask for the balanced overall equation, or ask how many moles of electrons are transferred. Show the multiplication step.
- The number of electrons transferred comes back in Unit 9 for cell potential and electrolysis calculations, so practice getting it right here.
Connected topics
Videos
Check yourself
4 questions on 4.9 Oxidation-Reduction (Redox) Reactions. Pick an answer to see if you got it, and why.
What is the oxidation number of manganese in the permanganate ion, MnO₄⁻?
A copper wire placed in AgNO₃(aq) becomes coated with silver metal, and the solution turns blue as Cu²⁺ ions form. Which of the following is the correctly balanced net ionic equation?
In acidic solution, permanganate ions oxidize iron(II) ions. The two half-reactions are:
Reduction: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)
Oxidation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Half-reactions
In the balanced overall equation with the lowest whole-number coefficients, how many Fe²⁺ ions react with each MnO₄⁻ ion?
In which of the following species does nitrogen have an oxidation number of +3?
0 of 4 answered