Define a reducing agent as a substance that reduces another substance and is itself oxidised

Chemical Reactions – Redox (IGCSE 0620)

Objective

By the end of this unit students will be able to:

• Define oxidation, reduction, oxidising agents and reducing agents (using the three IGCSE conventions).
• Write oxidation numbers in Roman‑numeral form in all equations, half‑reactions and answer boxes.
• Apply the oxidation‑number rules to assign oxidation numbers in ions, molecules and complex ions.
• Identify the oxidising and reducing agents in any redox reaction.
• Predict colour‑change tests that signal oxidation or reduction.
• Balance redox equations by the half‑reaction method (acidic or basic media).

Key Definitions (IGCSE wording)

Term	Definition (Cambridge syllabus)
Oxidation	Any of the following three equivalent statements: Loss of electrons (oxidation number increases). Gain of oxygen. Loss of hydrogen.
Reduction	Any of the following three equivalent statements: Gain of electrons (oxidation number decreases). Gain of hydrogen. Loss of oxygen.
Oxidising agent	A substance that causes another substance to be oxidised; it is itself reduced.
Reducing agent	A substance that causes another substance to be reduced; it is itself oxidised.
Redox reaction	A chemical change in which oxidation and reduction occur simultaneously.

Examples of the three definitions

• Oxidation as loss of electrons: Zn → Zn²⁺ + 2e⁻
• Oxidation as gain of oxygen: 2 Fe + 3 O₂ → Fe₂O₃
• Oxidation as loss of hydrogen: CH₄ + 2 O₂ → CO₂ + 2 H₂O

Roman‑Numeral Notation

In the Cambridge syllabus oxidation numbers are always written as Roman numerals. Examples:

• Fe³⁺ → Fe III
• Cu²⁺ → Cu II
• Mn⁷⁺ → Mn VII

All oxidation numbers in equations, half‑reactions and answer boxes must be written as Roman numerals.

Exercise: Convert the following to Roman‑numeral form: Cr³⁺, Co²⁺, Pb⁴⁺.

Oxidation‑Number Rules (the “Roman Numerals” system)

Rule	Explanation / Example
1. Elements in their elemental form have oxidation number 0.	H₂, O₂, Fe(s) → 0
2. A mono‑atomic ion has an oxidation number equal to its charge.	Na⁺ → +I, Cl⁻ → –I
3. Hydrogen is usually +I (except when bonded to metals, where it is –I).	H₂O → H +I, NaH → H –I
4. Oxygen is usually –II (exceptions: peroxides –I, superoxides –½, OF₂ where O is +II).	H₂O₂ → O –I, OF₂ → O +II
5. The sum of oxidation numbers in a neutral compound is 0; in an ion it equals the ion’s charge.	SO₄²⁻ → S +VI, each O –II (4 × –II = –VIII); +VI + (–VIII) = –II
6. Halogens are –I unless combined with a more electronegative element.	Cl⁻ → –I, ClO₃⁻ → Cl +V
7. Carbon’s oxidation number varies; assign using rule 5.	CO₂ → C +IV, CH₄ → C –IV

Worked Example – Determining oxidation numbers in NO₃⁻

1. Write the known oxidation numbers:
   • Oxygen = –II (rule 4).
2. Let the oxidation number of nitrogen be x. The ion carries a –1 charge (rule 5):
   x + 3(–II) = –I
3. Solve for x:
   x – VI = –I → x = +V
4. Therefore, in nitrate ion, N +V and each O –II.

How a Reducing Agent Works

A reducing agent donates electrons to another species. By losing electrons its oxidation number increases, so the reducing agent is oxidised.

Example – Zinc metal as a reducing agent:

$$\text{Zn} \;\longrightarrow\; \text{Zn}^{2+} + 2e^-$$

Zn changes from 0 to +II (Zn II); the two electrons are taken up by the oxidising agent.

Identifying Reducing and Oxidising Agents

1. Assign oxidation numbers to every element in the reactants and products (use the rules above).
2. The species whose oxidation number **increases** is the **reducing agent**.
3. The species whose oxidation number **decreases** is the **oxidising agent**.
4. Confirm that the total electrons lost by the reducing agent equal the total electrons gained by the oxidising agent.

Common Reducing Agents (IGCSE syllabus)

Reducing Agent	Typical Oxidation‑Number Change	IGCSE‑style Example Reaction
Hydrogen gas, H₂	0 → +I (H I)	H₂ + Cl₂ → 2 HCl
Carbon (coke), C	0 → +IV (C IV) in CO₂	C + O₂ → CO₂
Metals (e.g., Zn, Fe, Mg)	0 → positive ion (Zn II, Fe II/III, Mg II)	Zn + CuSO₄ → ZnSO₄ + Cu
Hydrogen sulphide, H₂S	S –II → 0 (S⁰ in S₈)	H₂S + Cl₂ → 2 HCl + S
Carbon monoxide, CO	C +II → +IV (in CO₂)	CO + ½ O₂ → CO₂
Aluminium, Al	0 → +III (Al III)	2 Al + 3 CuCl₂ → 2 AlCl₃ + 3 Cu

Common Oxidising Agents (IGCSE syllabus)

Oxidising Agent	Typical Oxidation‑Number Change	IGCSE‑style Example Reaction
Chlorine gas, Cl₂	0 → –I (Cl I)	Cl₂ + 2 NaBr → 2 NaCl + Br₂
Potassium permanganate, KMnO₄ (acidic medium)	Mn VII → Mn II	5 Fe²⁺ + MnO₄⁻ + 8 H⁺ → 5 Fe³⁺ + Mn²⁺ + 4 H₂O
Hydrogen peroxide, H₂O₂ (acidic)	O –I → –II (in H₂O)	2 Fe²⁺ + H₂O₂ + 2 H⁺ → 2 Fe³⁺ + 2 H₂O
Nitric acid, HNO₃ (conc.)	N V → N IV (NO₂) or N III (NO)	3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O
Oxygen gas, O₂	0 → –II (O II)	2 Fe + O₂ → 2 FeO

Colour‑Change Tests for Redox

Colour change gives a quick visual clue that a redox process has taken place. Recognise the following Cambridge‑style tests:

Test reagent (oxidising agent)	Colour before	Colour after reduction	Indicates
Potassium permanganate, KMnO₄ (acidic)	Purple	Colourless (Mn²⁺) or faint pink (MnO₂)	Presence of a reducing substance (e.g., Fe²⁺, H₂S)
Iodine solution, I₂ (in KI)	Brown‑violet	Colourless (I⁻)	Reducing agents such as sulfite, thiosulfate, Fe²⁺
Silver nitrate, AgNO₃ (acidic)	Colourless	White precipitate of AgCl if Cl₂ is reduced to Cl⁻	Detection of halogen oxidation/reduction

Balancing Redox Equations – Half‑Reaction Method (IGCSE)

1. Write the unbalanced skeletal equation.
2. Separate into oxidation and reduction half‑reactions.
3. Balance each half‑reaction:
   • Balance all atoms except H and O.
   • Balance O by adding H₂O.
   • Balance H by adding H⁺ (acidic) or OH⁻ (basic).
   • Balance charge by adding electrons (e⁻).
4. Equalise the number of electrons transferred in the two half‑reactions (multiply if necessary).
5. Add the half‑reactions** together and cancel species that appear on both sides (including e⁻, H₂O, H⁺/OH⁻).
6. Check** that atoms and overall charge are balanced.

Worked Example – Balancing in acidic solution:

$$\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$$

1. Oxidation half‑reaction (Fe²⁺ → Fe³⁺) $$\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$$
2. Reduction half‑reaction (MnO₄⁻ → Mn²⁺)
   1. Balance O: add 4 H₂O on the right. $$\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
   2. Balance H: add 8 H⁺ on the left. $$8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
   3. Balance charge: left side +7, right side +2 → add 5 e⁻ to the left. $$5e^- + 8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
3. Equalise electrons (multiply the oxidation half‑reaction by 5): $$5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^-$$
4. Add and cancel electrons: $$5\text{Fe}^{2+} + 8\text{H}^+ + \text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}$$
5. Check: atoms and total charge (–1) are balanced on both sides.

Sample Redox Reaction Walk‑through

Reaction: Zn (s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)

1. Assign oxidation numbers (using the rules):
   • Zn: 0 → +II (Zn II)
   • Cu in CuSO₄: +II → 0 (Cu 0)
   • S in SO₄²⁻: +VI (unchanged)
   • O in SO₄²⁻: –II (unchanged)
2. Identify the agents:
   • Zn’s oxidation number increases → **reducing agent**.
   • Cu²⁺’s oxidation number decreases → **oxidising agent**.
3. Write half‑reactions:
   • Oxidation (Zn): $$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$$
   • Reduction (Cu²⁺): $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$$
4. Combine (electrons cancel automatically) → the overall equation is already balanced.