# 7.3 Oxidation and Reduction

F.5 Chemistry · Topic VII Redox Reactions, Chemical Cells and Electrolysis · Regular Course Lesson 1 (Part 1)

Four ways to say "oxidized", how to assign oxidation numbers, the agents and colour changes you must recognize, and how to balance a redox equation.

## Key points

- **Oxidation ／ 氧化**: a species **gains oxygen**, **loses hydrogen** or **loses electrons**. **Reduction ／ 還原**: it **loses oxygen**, **gains hydrogen** or **gains electrons**. The two always take place together (**redox** = **RED**uction + **OX**idation).
- A **redox reaction** involves gain or loss of electrons, so it covers reactions with no oxygen or hydrogen at all: Al → Al^3+^ + 3e^−^ (oxidation), Fe^3+^ + 3e^−^ → Fe (reduction).
- A **reducing agent ／ 還原劑** causes reduction: it loses electrons and is itself **oxidized**. An **oxidizing agent ／ 氧化劑** causes oxidation: it gains electrons and is itself **reduced**.
- The **oxidation number ／ 氧化數 (O.N.)** of an element in a compound is the charge an atom of the element would have if it existed as an ion. **Oxidation: O.N. increases. Reduction: O.N. decreases.** A reaction with no change in O.N. is not redox.
- **O.N. rules**: an element is 0 (N~2~, S, all metals); a monoatomic ion equals its charge; Group I +1, Group II +2; H +1 (−1 in metal hydrides, e.g. NaH); F −1; Cl, Br, I mostly −1; O −2 (−1 in peroxides). The O.N. sum is 0 in a compound and equals the charge in a polyatomic ion.
- **Worked O.N.**: S in Na~2~SO~4~: $x + (-2)\times4 = -2$, so +6. N in NH~4~^+^: $x + (+1)\times4 = +1$, so −3. Pb in [Pb(OH)~4~]^2−^: $x - 8 + 4 = -2$, so +2.
- **Stock system**: a Roman numeral gives the O.N. of the metal or central element: Cu~2~O copper(I) oxide, FeCl~3~ iron(III) chloride, MnO~4~^−^ manganate(VII), Cr~2~O~7~^2−^ dichromate(VI), NO~2~^−^ nitrate(III), SO~3~^2−^ sulphate(IV); HNO~2~ is nitric(III) acid.
- **Half equation method**: write the two half equations; balance atoms, then O with H~2~O and H with H^+^ (acidic solution), then charge with e^−^; multiply so electrons lost = electrons gained; add and cancel the electrons.
- **Oxidation number method**: assign O.N., find the electrons lost or gained per formula unit, insert coefficients so gained = lost, balance the other atoms, then H^+^ (acidic) for charge and H~2~O for O.

## Must know

| Agent | Colour change | Half equation |
|---|---|---|
| Acidified KMnO~4~(aq) (oxidizing) | purple → colourless | MnO~4~^−^ + 8H^+^ + 5e^−^ → Mn^2+^ + 4H~2~O |
| Acidified K~2~Cr~2~O~7~(aq) (oxidizing) | orange → green | Cr~2~O~7~^2−^ + 14H^+^ + 6e^−^ → 2Cr^3+^ + 7H~2~O |
| Dilute HNO~3~ (oxidizing) | NO~3~^−^ → NO | NO~3~^−^ + 4H^+^ + 3e^−^ → NO + 2H~2~O |
| Conc. HNO~3~ (oxidizing) | NO~3~^−^ → NO~2~ | NO~3~^−^ + 2H^+^ + e^−^ → NO~2~ + H~2~O |
| Conc. H~2~SO~4~ (oxidizing) | H~2~SO~4~ → SO~2~ | H~2~SO~4~ + 2H^+^ + 2e^−^ → SO~2~ + 2H~2~O |
| Cl~2~, Br~2~(aq) (oxidizing) | Br~2~(aq): yellow-brown → colourless | Br~2~ + 2e^−^ → 2Br^−^ |
| Fe^3+^(aq) (oxidizing) | yellow → pale green | Fe^3+^ + e^−^ → Fe^2+^ |
| Fe^2+^(aq) (reducing) | pale green → yellow | Fe^2+^ → Fe^3+^ + e^−^ |
| SO~2~, SO~3~^2−^(aq) (reducing) | → SO~4~^2−^ | SO~3~^2−^ + H~2~O → SO~4~^2−^ + 2H^+^ + 2e^−^ |
| Br^−^, I^−^(aq) (reducing) | colourless → yellow-brown; colourless → brown | 2I^−^ → I~2~ + 2e^−^ |

KMnO~4~ and K~2~Cr~2~O~7~ are acidified with dilute H~2~SO~4~; dilute HNO~3~ is not used because it is itself an oxidizing agent.

## Common mistakes

- ✗ "The oxidizing agent is oxidized." ✓ The oxidizing agent is **reduced**: it gains electrons and its O.N. decreases. The reducing agent is the one oxidized.
- ✗ O is always −2 and H is always +1. ✓ O is −1 in peroxides (H~2~O~2~); H is −1 in metal hydrides (NaH, MgH~2~). An element on its own (Cl~2~, O~2~, S, Fe) is 0.
- ✗ Any reaction with acids, oxides or gases is redox. ✓ Check O.N. first: neutralization (NH~3~ + HNO~3~ → NH~4~NO~3~), precipitation and acid + carbonate change no O.N., so they are not redox.
- ✗ Acidify KMnO~4~ with HCl or HNO~3~. ✓ Dilute H~2~SO~4~ only: HCl may be oxidized to Cl~2~ and HNO~3~ is itself an oxidizing agent.
- ✗ MnO~4~^−^ + 8H^+^ + Fe^2+^ → Mn^2+^ + Fe^3+^ + 4H~2~O, "atoms balance, done". ✓ Check charge too: +9 on the left, +5 on the right. MnO~4~^−^ gains 5e^−^ and each Fe^2+^ loses 1e^−^, so 5Fe^2+^ → 5Fe^3+^ (+17 on both sides).

## Quick check

1. Give the oxidation number of S in H~2~S~2~O~7~ and of N in NH~4~^+^.
2. In 2S~2~O~3~^2−^ + I~2~ → S~4~O~6~^2−^ + 2I^−^, which species is the oxidizing agent, and what is the oxidation number of S in S~2~O~3~^2−^?
3. Iron(II) sulphate solution is added to acidified potassium permanganate solution. State the colour change of each solution.
4. Balance: Cr~2~O~7~^2−^ + SO~3~^2−^ → Cr^3+^ + SO~4~^2−^ in acidic solution.
5. Given MnO~4~^−^ + 8H^+^ + 5e^−^ → Mn^2+^ + 4H~2~O and C~2~O~4~^2−^ → 2CO~2~ + 2e^−^, how many moles of MnO~4~^−^ react with one mole of C~2~O~4~^2−^?

## Answers

1. S: +6 ($2(+1) + 2x + 7(-2) = 0$). N: −3.
2. I~2~ (its O.N. falls from 0 to −1, so it gains electrons). S in S~2~O~3~^2−^: +2.
3. Permanganate: purple → colourless (very pale pink). Iron(II): pale green → yellow.
4. Cr~2~O~7~^2−^ + 8H^+^ + 3SO~3~^2−^ → 2Cr^3+^ + 4H~2~O + 3SO~4~^2−^.
5. 0.4 mol (2 mol MnO~4~^−^ take 10e^−^ from 5 mol C~2~O~4~^2−^, so 2 : 5).

## Did you know?

A cut apple turns brown because an enzyme lets oxygen oxidize compounds in the fruit; lemon juice slows this because its vitamin C is a reducing agent and is oxidized first.