# 7.3 Oxidation and Reduction (Part 2)

F.5 Chemistry · Topic VII Redox Reactions, Chemical Cells and Electrolysis · Regular Course Lesson 1 (Part 2)

How to read the Electrochemical Series, and what chlorine, nitric acid, concentrated sulphuric acid and sulphite do as oxidizing and reducing agents.

## Key points

- **Electrochemical Series (E.C.S.) ／ 電化序**: rows of **oxidizing agent + ne^−^ ⇌ reducing agent** (oxidizing agents left, reducing agents right), Li^+^/Li at the top, F~2~/F^−^ at the bottom. **Oxidizing power increases down** the series (strongest F~2~); **reducing power increases up** (strongest Li).
- **Predicting a reaction**: the half equation **lower** in the series goes **as written**, the **higher** one runs **in reverse**: Cl~2~ + 2e^−^ → 2Cl^−^ with Fe^2+^ → Fe^3+^ + e^−^ gives Cl~2~ + 2Fe^2+^ → 2Cl^−^ + 2Fe^3+^.
- **Chemical cell**: Mg/Cu cell, Mg + Cu^2+^ → Mg^2+^ + Cu. The metal higher in the series loses electrons, so they flow from **Mg to Cu** in the external circuit.
- **Chlorine** is a powerful oxidizing agent (Cl 0 → −1): it **displaces ／ 置換** Br~2~ from KBr(aq) and I~2~ from KI(aq). Oxidizing power: I~2~ < Br~2~ < Cl~2~.
- **Disproportionation ／ 歧化反應**: one species is oxidized **and** reduced. Cl~2~ + **cold dilute** NaOH → NaCl + NaOCl + H~2~O (Cl → −1 and +1); 3Cl~2~ + **hot conc.** NaOH → 5NaCl + NaClO~3~ + 3H~2~O (Cl → −1 and +5).
- **Nitric acid**, a strong oxidizing agent. **Dilute**: NO~3~^−^ + 4H^+^ + 3e^−^ → NO + 2H~2~O (N +5 → +2); colourless NO turns brown in air, 2NO + O~2~ → 2NO~2~. **Concentrated**: NO~3~^−^ + 2H^+^ + e^−^ → NO~2~ + H~2~O (N +5 → +4), brown fumes at once.
- Conc. HNO~3~ oxidizes most metals, **even those below H in the E.C.S.** such as Cu (not Au or Pt), and C → CO~2~, Fe^2+^ → Fe^3+^, SO~3~^2−^ → SO~4~^2−^. Very dilute HNO~3~ acts as an ordinary acid (oxidizing agent H^+^, H~2~ given off).
- **Conc. H~2~SO~4~** (98 %), especially hot, is reduced to **SO~2~** (S +6 → +4): metals (not Au, Pt) → sulphates + SO~2~; C → CO~2~, S → SO~2~. Dilute H~2~SO~4~ attacks only metals above H, with H^+^ as the oxidizing agent.
- Conc. H~2~SO~4~ on halides: NaCl → HCl (steamy fumes); NaBr → HBr, which reduces it to SO~2~ and becomes Br~2~; NaI → HI, which reduces it to **H~2~S** and becomes I~2~. Reducing power: HCl < HBr < HI.
- **Sulphite ion ／ 亞硫酸根離子** (SO~2~ + H~2~O ⇌ H~2~SO~3~ ⇌ 2H^+^ + SO~3~^2−^) is a strong reducing agent: SO~3~^2−^ + H~2~O → SO~4~^2−^ + 2H^+^ + 2e^−^ (S +4 → +6).

## Must know

| Reaction | What you see | Equation |
|---|---|---|
| Cl~2~(aq) + KBr(aq) | colourless → yellowish brown; orange-red in heptane | Cl~2~ + 2Br^−^ → 2Cl^−^ + Br~2~ |
| Cl~2~(aq) + KI(aq) | colourless → brown; purple in heptane | Cl~2~ + 2I^−^ → 2Cl^−^ + I~2~ |
| Cu + warm dilute HNO~3~ | solution turns blue; brown fumes at the mouth of the tube | 3Cu + 2NO~3~^−^ + 8H^+^ → 3Cu^2+^ + 2NO + 4H~2~O |
| Mg + conc. HNO~3~ | brown fumes of NO~2~ | Mg + 4HNO~3~ → Mg(NO~3~)~2~ + 2NO~2~ + 2H~2~O |
| Cu + hot conc. H~2~SO~4~ | colourless SO~2~ given off | Cu + 2H~2~SO~4~ → CuSO~4~ + SO~2~ + 2H~2~O |
| NaI + conc. H~2~SO~4~ | purple fumes of I~2~ condensing to a black solid; H~2~S | 8HI + H~2~SO~4~ → 4I~2~ + H~2~S + 4H~2~O |
| SO~2~ / SO~3~^2−^ + acidified KMnO~4~ | purple → colourless | 5SO~3~^2−^ + 2MnO~4~^−^ + 6H^+^ → 5SO~4~^2−^ + 2Mn^2+^ + 3H~2~O |
| SO~2~ / SO~3~^2−^ + acidified K~2~Cr~2~O~7~ | orange → green | 3SO~3~^2−^ + Cr~2~O~7~^2−^ + 8H^+^ → 3SO~4~^2−^ + 2Cr^3+^ + 4H~2~O |
| SO~2~ / SO~3~^2−^ + Br~2~(aq) | yellow-brown → colourless | SO~3~^2−^ + H~2~O + Br~2~ → SO~4~^2−^ + 2H^+^ + 2Br^−^ |
| SO~2~ / SO~3~^2−^ + Fe^3+^(aq) | yellow → pale green | SO~3~^2−^ + H~2~O + 2Fe^3+^ → SO~4~^2−^ + 2H^+^ + 2Fe^2+^ |

The last four rows are the tests for SO~2~: sulphite is the reducing agent (S +4 → +6) and the coloured oxidizing agent changes colour.

## Common mistakes

- ✗ "Bromine is a stronger oxidizing agent than chlorine." ✓ Oxidizing power increases **down** the E.C.S.: Cl~2~ sits below Br~2~, so Cl~2~ displaces Br~2~ from bromide, never the reverse.
- ✗ "Copper reacts with dilute HCl or dilute H~2~SO~4~, acids attack metals." ✓ Cu is below H in the E.C.S.: no reaction with non-oxidizing acids. It reacts only with oxidizing acids: dilute or conc. HNO~3~ (NO / NO~2~) and hot conc. H~2~SO~4~ (SO~2~).
- ✗ "In Cl~2~ + 2NaOH → NaCl + NaOCl + H~2~O, NaOH is the reducing agent." ✓ Na, O and H keep their O.N.; chlorine goes 0 → −1 **and** 0 → +1, so Cl~2~ is both the oxidizing and the reducing agent (disproportionation).
- ✗ "Brown fumes with dilute HNO~3~, so it is reduced to NO~2~." ✓ Dilute HNO~3~ is reduced to **colourless NO** (N +5 → +2); the brown NO~2~ forms when NO meets air. Only concentrated HNO~3~ gives NO~2~ directly (+5 → +4).
- ✗ "SO~2~ turns acidified KMnO~4~ colourless, so SO~2~ is an oxidizing agent." ✓ SO~2~ / SO~3~^2−^ is the **reducing** agent: S rises from +4 to +6 and it is oxidized to SO~4~^2−^; MnO~4~^−^ is the one reduced.

## Quick check

1. Using the E.C.S., predict whether Br~2~(aq) reacts with KI(aq) and with KCl(aq). Write the ionic equation for any reaction.
2. Chlorine is passed into hot concentrated sodium hydroxide solution. Give the oxidation numbers of chlorine in the two chlorine-containing products and name the type of reaction.
3. A metal X warmed with acid Y gives a colourless solution and a brown gas. Choose X from copper or zinc and Y from concentrated nitric acid or concentrated sulphuric acid, and explain.
4. Write the ionic equation for copper with warm dilute nitric acid, and explain why brown fumes appear at the mouth of the test tube.
5. Sodium sulphite solution is added to acidified potassium dichromate solution. State the colour change, write the ionic equation, and give the change in oxidation number of sulphur.

## Answers

1. Br~2~ + 2I^−^ → 2Br^−^ + I~2~ (solution turns brown). No reaction with KCl: Br~2~ is a weaker oxidizing agent than Cl~2~.
2. −1 in NaCl and +5 in NaClO~3~; disproportionation (Cl~2~ is both oxidized and reduced).
3. Zinc with concentrated nitric acid: Zn^2+^(aq) is colourless (Cu^2+^ would be blue) and conc. HNO~3~ gives brown NO~2~ (conc. H~2~SO~4~ gives colourless SO~2~).
4. 3Cu + 2NO~3~^−^ + 8H^+^ → 3Cu^2+^ + 2NO + 4H~2~O. The colourless NO reacts with oxygen in air: 2NO + O~2~ → 2NO~2~ (brown).
5. Orange → green. 3SO~3~^2−^ + Cr~2~O~7~^2−^ + 8H^+^ → 3SO~4~^2−^ + 2Cr^3+^ + 4H~2~O. S: +4 → +6 (oxidized).

## Did you know?

Gold and platinum shrug off concentrated nitric acid, but a 1 : 3 mixture of concentrated nitric and hydrochloric acids, aqua regia ("royal water"), dissolves them both.