Use a Roman numeral to indicate the oxidation number of an element in a compound

Cambridge IGCSE Chemistry 0620 – Redox (6.4)

Learning Objective

Use Roman numerals to indicate the oxidation number of an element in a compound and apply this knowledge to identify, write and balance redox reactions (AO1‑AO3).

1. Oxidation‑Number Rules (Core 1)

• Elements in their elemental form have oxidation number 0 (e.g. N₂, O₂, Fe).
• For a mono‑atomic ion the oxidation number equals the ionic charge (e.g. Na⁺ = +1, Cl⁻ = –1).
• Oxygen is usually –2, except in peroxides (–1) and in OF₂ (+2).
• Hydrogen is +1 when bonded to non‑metals and –1 when bonded to metals.
• Fluorine is always –1.
• The sum of oxidation numbers in a neutral compound is 0; in a poly‑atomic ion it equals the ion’s overall charge (e.g. in SO₄²⁻ the total is –2).

2. Writing Oxidation Numbers with Roman Numerals (Core 1)

1. Identify the element whose oxidation state you need.
2. Apply the rules above to calculate its oxidation number.
3. Write the element symbol followed by the oxidation number in Roman numerals, enclosed in parentheses.

Examples

• FeCl₃ → Fe(III)Cl₃ (Fe = +3)
• CuSO₄ → Cu(II)SO₄ (Cu = +2)
• MnO₂ → Mn(IV)O₂ (Mn = +4)
• K₂Cr₂O₇ → K₂Cr(VI)₂O₇ (Cr = +6)
• Zn + CuSO₄ → ZnSO₄ + Cu ( Zn(0) → Zn(II), Cu(II) → Cu(0) ) – a metal‑displacement redox example.

3. Definitions of Oxidation and Reduction

• Historical (oxygen‑transfer) definition – Oxidation = gain of oxygen; Reduction = loss of oxygen.
• Modern (electron‑transfer) definition (Supplement 6) – Oxidation = loss of electrons; Reduction = gain of electrons.

Both definitions are equivalent for the reactions studied at IGCSE level because a loss of electrons is always accompanied by a gain of oxygen (or loss of hydrogen).

4. Identifying Redox Reactions (Core 2 & 4)

A reaction is redox if any of the following occur:

• Oxygen is transferred from one species to another.
• There is a change in oxidation number for any element.
• A more reactive metal displaces a less reactive metal (metal‑metal displacement).

Example – 2 Mg + O₂ → 2 MgO (oxygen transferred to magnesium).

5. Determining Which Species Is Oxidised and Which Is Reduced (Core 5)

1. Assign oxidation numbers to every atom in the reactants and products.
2. The element whose oxidation number increases is oxidised (loss of electrons).
3. The element whose oxidation number decreases is reduced (gain of electrons).

Worked example

2 Mg + O₂ → 2 MgO

Mg: 0 → +2 → oxidised O: 0 → –2 → reduced.

6. Oxidising and Reducing Agents (Supplement 7)

• Oxidising agent – the species that is reduced (it gains electrons).
• Reducing agent – the species that is oxidised (it loses electrons).

Common agents (with oxidation numbers in Roman numerals)

7. Writing Half‑Equations (Core 3)

1. Identify the species that is oxidised and the one that is reduced.
2. Write separate equations showing only the element that changes oxidation number.
3. Balance all atoms except O and H.
4. Balance O by adding H₂O (aqueous solution).
5. Balance H by adding H⁺ (acidic medium) or OH⁻ (basic medium).
6. Balance charge by adding electrons (e⁻) to the more positive side.

Acidic example – Oxidation of Fe²⁺ to Fe³⁺

Fe²⁺ → Fe³⁺ + e⁻

Basic example – Reduction of MnO₄⁻ to MnO₂

MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

8. Balancing Redox Equations – Ion‑Electron (Half‑Reaction) Method (Core 3)

1. Write the unbalanced skeletal equation.
2. Separate into oxidation and reduction half‑equations (as in section 7).
3. Balance each half‑equation for atoms and charge.
4. Equalise the number of electrons lost and gained (multiply half‑equations as required).
5. Add the half‑equations, cancel species that appear on both sides, and write the final balanced overall equation.

Worked Example – Acidic Medium

Balancing the reaction between potassium permanganate and iron(II) sulphate in acidic solution:

Unbalanced:  KMnO₄ + FeSO₄ + H₂SO₄ → K₂SO₄ + Fe₂(SO₄)₃ + MnSO₄ + H₂O

• Reduction half‑equation (Mn(VII) → Mn(II)):

  MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

• Oxidation half‑equation (Fe(II) → Fe(III)):

  Fe²⁺ → Fe³⁺ + e⁻

Multiply the oxidation half‑equation by 5, add, and simplify:

2 KMnO₄ + 5 FeSO₄ + 8 H₂SO₄ → K₂SO₄ + 5 Fe₂(SO₄)₃ + 2 MnSO₄ + 8 H₂O

Worked Example – Basic Medium

Balancing the reaction of hypochlorite ion with iodide ion in alkaline solution:

ClO⁻ + I⁻ → Cl⁻ + I₂

• Reduction (ClO⁻ → Cl⁻):

  ClO⁻ + H₂O + 2e⁻ → Cl⁻ + 2OH⁻

• Oxidation (I⁻ → I₂):

  2I⁻ → I₂ + 2e⁻

Electrons are already equal, so adding the half‑equations gives:

ClO⁻ + I⁻ + H₂O → Cl⁻ + I₂ + OH⁻

9. Colour‑Change Tests for Oxidation States (Supplement 8)

10. Practical Context (Supplement 9)

• Industrial – KMnO₄ as a disinfectant and oxidising agent in water treatment.
• Laboratory – Redox titrations (e.g. Fe²⁺ ↔ KMnO₄) rely on sharp colour changes at the equivalence point.
• Everyday – Rusting of iron (Fe → Fe₂O₃) is a slow redox process involving oxidation by atmospheric O₂.

11. Common Oxidation States of Transition Metals (Roman Numerals)

12. Practice Questions

1. Write the oxidation state of iron in Fe₂O₃ using Roman numerals.
2. Write the oxidation state of copper in CuSO₄ using Roman numerals.
3. Write the oxidation state of manganese in the permanganate ion MnO₄⁻ using Roman numerals.
4. Write the oxidation state of chromium in dichromate ion Cr₂O₇²⁻ using Roman numerals.
5. Identify the oxidising and reducing agents in the reaction:
   

   2 FeSO₄ + H₂SO₄ + KMnO₄ → K₂SO₄ + Fe₂(SO₄)₃ + MnSO₄ + H₂O

6. Balance the following redox reaction in acidic solution:
   

   Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O

7. Balance the reaction of hypochlorite with iodide in basic solution:
   

   ClO⁻ + I⁻ → Cl⁻ + I₂

Answers

1. Fe(III)
2. Cu(II)
3. Mn(VII)
4. Cr(VI)
5. Oxidising agent: KMnO₄ (Mn(VII) reduced to Mn(II)).
   

   Reducing agent: FeSO₄ (Fe(II) oxidised to Fe(III)).

6. Balanced (acidic): 3 Cu + 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O
7. Balanced (basic): ClO⁻ + I⁻ + H₂O → Cl⁻ + I₂ + OH⁻

13. Suggested Diagram (for teacher use)