Assigning & Calculating Oxidation States

Calculate oxidation states in elements, ions and compounds, then use increases and decreases to identify oxidation and reduction in K324 / 6092 redox reactions.

SEC G3 Pure Chemistry 2027

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Learning objectives

define redox in terms of electron transfer and changes in oxidation state
identify redox reactions in terms of oxygen/hydrogen gain/loss, electron gain/loss and changes in oxidation state

Oxidation states provide a consistent way to track redox when oxygen or hydrogen is not transferred. Assign the values, compare the same element before and after, then state whether the value increases or decreases.

1. Definition

The oxidation state (also called oxidation number) is a bookkeeping value assigned to an atom. In a redox reaction:

oxidation state increases → oxidation
oxidation state decreases → reduction

2. Key Ideas

Core rule	Example
An element has oxidation state 0.	Na(s): 0, Cl₂(g): 0
A one-atom ion has oxidation state equal to its charge.	Fe³⁺: +3, Cl⁻: -1
Group 1 metals are +1 in compounds.	K in KMnO₄: +1
Group 2 metals are +2 in compounds.	Ca in CaCl₂: +2
Oxygen is usually -2.	O in SO₂: -2
Hydrogen is usually +1.	H in H₂O: +1
Sum of oxidation states = 0 for a neutral compound.	H₂SO₄ total = 0
Sum of oxidation states = ion charge for an ion.	SO₄²⁻ total = -2

Write the usual oxidation states for the atoms you know.
Set their sum equal to the overall charge: 0 for a neutral compound, or the stated charge for an ion.
Solve for the unknown and include its sign, for example + 5 rather than 5.

3. Detailed Explanations

Start with the type of particle, apply the correct total, and show the sum equation. To identify redox, compare the oxidation state of the same element before and after the reaction.

Oxidation state is a symbolic model; it is not a colour or another direct observation. A macroscopic colour change may be evidence that a reaction occurred, while formulae and oxidation-state changes show which species was oxidised or reduced.

A. Neutral compounds: sum equals 0

Example: Find the oxidation state of sulfur in H₂SO₄.

2(H) + 1(S) + 4(O) = 0; 2(+1) + x + 4(-2) = 0; 2 + x - 8 = 0; x = +6

So sulfur is +6 in H₂SO₄.

B. Ions: sum equals the ion charge

Example: Find the oxidation state of sulfur in SO₄²⁻.

x + 4(-2) = -2; x - 8 = -2; x = +6

C. Using changes to identify redox

Consider the displacement reaction:

2KI(aq) + Cl₂(aq) → 2KCl(aq) + I₂(aq)

Iodine changes from -1 in I⁻ to 0 in I₂: the oxidation state increases, so iodide ions are oxidised.
Chlorine changes from 0 in Cl₂ to -1 in Cl⁻: the oxidation state decreases, so chlorine is reduced.

The usual values above cover the core method used here. If an exam question supplies an unfamiliar rule or oxidation state, use the information given rather than assuming the usual value.

4. Common Mistakes

Forgetting to multiply by the number of atoms (e.g., writing “O is -2” but not doing 4(-2) for O₄).
Setting every sum equal to zero: a polyatomic ion must add to its overall ion charge.
Omitting the sign: write + 6, not 6.
Mixing oxidation state with ionic charge in covalent molecules: oxidation state is a bookkeeping value, not a measured charge on each atom.
Comparing different elements instead of tracking the same element before and after a reaction.

5. Exam Tips

When asked “what is oxidised/reduced”, use oxidation states: write the number before and after, then state “increases/decreases”.

Show the sum equation so the multiplier and overall charge are visible.
State both values and the direction of change: “iodine increases from -1 to 0, so it is oxidised.”
Link the result to redox definitions and agents, then practise the tests for oxidising and reducing agents.

6. Worked Examples

Problem

Find the oxidation state of nitrogen in

NO₂.

Study the worked solution

Assign the known value
Method
Give each oxygen its usual oxidation state of -2.
Reason
No exception is indicated, and there are two oxygen atoms.
Working
Oxygen contribution = 2(-2) = -4.
Use the neutral total
Method
Set the sum equal to zero.
Reason
NO₂ is a neutral compound.
Working
x + 2(-2) = 0.
Solve with the sign
Method
Rearrange for nitrogen.
Reason
Nitrogen must balance the total oxygen contribution.
Working
x-4 = 0, so x = +4.

Learner working

A student starts the nitrate calculation with

x + 3(-2) = 0. Identify the error and find nitrogen’s oxidation state in

NO₃⁻.

Use the ion's overall charge

Required sum

−10

Nitrogen oxidation-state number

View solution step by step

Correct the total
Method
Set the sum equal to -1, not zero.
Reason
A polyatomic ion’s oxidation states add to its overall charge.
Working
x + 3(-2) = -1.
Solve and report
Method
Solve for nitrogen and include its sign.
Reason
The three oxygens contribute -6, so nitrogen must contribute + 5 to leave -1 overall.
Working
x-6 = -1; x = +5.

Multiple-rule transfer

Find the oxidation state of manganese in

KMnO₄.

Combine two known values

K oxidation-state number

Four O atoms' total

Mn oxidation-state number

Hints

Hint 1: two usual values

Use K

= +1 and each O

= -2.

Hint 2: neutral formula

The written compound has no overall charge, so all contributions sum to zero.

View solution step by step

Write known contributions
Method
Assign potassium and oxygen first.
Reason
Group 1 metals are +1 and oxygen is usually -2.
Working
1(+1) + x + 4(-2) = 0.
Solve for manganese
Method
Balance the neutral total.
Reason
+ 1 and -8 leave -7, so manganese must contribute + 7.
Working
1 + x-8 = 0; x = +7.

Next step

Try Identify oxidation and reduction by numbers independently.

7. Mind Stretchers

Mind stretcher 1: Identify oxidation and reduction by numbersExtension

Question: In 2KI(aq) + Cl₂(aq) → 2KCl(aq) + I₂(aq), show (with oxidation states) what is oxidised and what is reduced.

Show Answer

I: -1 in KI to 0 in I₂ → increases → oxidised.
Cl: 0 in Cl₂ to -1 in KCl → decreases → reduced.

Mind stretcher 2: Correct the totalExtension

Question: A student writes x + 3(-2) = 0 for nitrogen in NO₃⁻. Identify the error and calculate the correct oxidation state.