Oxidation number and its changes

This page covers one Form 5 content standard from Redox Equilibrium: oxidation number and its changes. The previous standard defined oxidation and reduction. Oxidation number turns those ideas into a precise, countable tool, so that you can settle what is oxidised and what is reduced in any equation, including reactions where nothing obvious gains or loses oxygen, hydrogen or electrons.

What an oxidation number is

The oxidation number (or oxidation state) of an atom is a signed value that describes how many electrons an atom appears to have gained or lost compared with the free element. It is written with the sign first, for example +2, -1, +7. You assign it using a fixed order of rules.

The rules, in priority order:

1. The oxidation number of an atom in a free (uncombined) element is 0, for example Fe, O₂, Cl₂, S₈ are all 0.
2. For a monatomic ion, the oxidation number equals the charge, Na⁺ is +1, Cl⁻ is -1, Al³⁺ is +3, S²⁻ is -2.
3. In compounds, some elements have almost-fixed values: Group 1 metals are +1, Group 2 metals are +2, fluorine is -1, hydrogen is usually +1 (but -1 in metal hydrides such as NaH), and oxygen is usually -2 (but -1 in peroxides such as H₂O₂).
4. The sum of all oxidation numbers in a neutral compound is 0, and in a polyatomic ion is equal to the charge on the ion.

Rule 4 is the workhorse: once the fixed values are placed, you solve for the unknown atom like a simple equation.

How changes in oxidation number show redox

Compare each element’s oxidation number before and after the reaction:

• An increase in oxidation number (more positive) is oxidation.
• A decrease in oxidation number (more negative) is reduction.

The oxidising agent contains the element whose oxidation number decreases; the reducing agent contains the element whose oxidation number increases. This is exactly consistent with the electron definition, losing electrons pushes the value up (oxidation), gaining electrons pushes it down (reduction).

Naming with Roman numerals (Stock nomenclature)

The Roman numeral in a name such as iron(II) sulfate or copper(II) oxide is the oxidation number of that metal. So iron(II) means iron with oxidation number +2, and iron(III) means +3. Manganese in MnO₄⁻ has oxidation number +7, which is why potassium manganate(VII) carries the (VII). Getting comfortable reading names this way saves time in the exam.

Worked example

Question. Determine the oxidation number of manganese in the manganate(VII) ion, MnO₄⁻. Then, in the reaction MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, state which element is oxidised and which is reduced, and name the oxidising agent.

Step 1, assign oxidation number of Mn in MnO₄⁻. Oxygen is -2 (rule 3), and there are four oxygens, giving 4 x (-2) = -8. Let the oxidation number of Mn be x. The ion has an overall charge of -1, so by rule 4: x + (-8) = -1, therefore x = +7. Manganese is +7 in MnO₄⁻.

Step 2, find Mn after the reaction. In Mn²⁺, a monatomic ion, the oxidation number equals the charge: +2 (rule 2).

Step 3, find the change for Mn. Manganese goes from +7 to +2, a decrease of 5. A decrease is reduction, so manganese is reduced.

Step 4, find the change for Fe. Iron goes from +2 (in Fe²⁺) to +3 (in Fe³⁺), an increase of 1. An increase is oxidation, so iron is oxidised.

Step 5, name the oxidising agent. The oxidising agent contains the element that is reduced (manganese), so MnO₄⁻ (the manganate(VII) ion) is the oxidising agent, and Fe²⁺ is the reducing agent. Note the electron bookkeeping balances: manganese drops by 5 for one ion, and iron rises by 1 for each of five iron ions, 5 total, so electrons lost equal electrons gained.

Answer. Mn is +7 in MnO₄⁻ and is reduced to +2; iron is oxidised from +2 to +3. The manganate(VII) ion is the oxidising agent.

Practice question

Determine the oxidation number of chromium in the dichromate(VI) ion, Cr₂O₇²⁻.

Answer. Oxygen is -2, and there are seven oxygens: 7 x (-2) = -14. Let each chromium be x; there are two of them, so 2x. The ion charge is -2, so 2x + (-14) = -2, giving 2x = +12 and x = +6. Each chromium has oxidation number +6 (which is why the ion is named dichromate(VI)).

Exam tip

Write the oxidation number above every atom you are asked about, and always include the sign, a value written as “2” instead of “+2” can be marked wrong. When identifying redox, quote the change explicitly: “sulfur is oxidised, its oxidation number increases from -2 to +4”. For polyatomic ions, do not forget that the total must equal the charge on the ion, not zero. And remember the two exceptions that catch people out: hydrogen is -1 in metal hydrides, and oxygen is -1 in peroxides. These small habits protect easy marks throughout SPM Chemistry.

Where this fits in the chapter

Oxidation number is the tool you will reuse in every later standard: it confirms which electrode reaction is oxidation in electrolysis and in cells, and it explains why a more reactive metal displaces a less reactive one.

• Up to the chapter hub: Redox Equilibrium.
• Sideways: the chapter revision notes and the key terms list for precise definitions.

Our online 1-to-1 SPM Chemistry lessons, in English and from RM50 per hour, spend time on the exact reactions the exam favours, manganate(VII) and dichromate(VI) redox especially, so that assigning oxidation numbers under time pressure becomes routine rather than a gamble.