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Assigning and 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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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. For a one-atom ion it equals its charge. In a covalent molecule, it is not a measured charge on the atom. For example, sulfur is +6 in neutral H₂SO₄; that does not mean the molecule contains a free S⁶⁺ ion. In a redox reaction:

  • oxidation state increases → oxidation
  • oxidation state decreases → reduction

2. Key Ideas

Core ruleExample
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, including every atom, = 0 for a neutral compound.H₂SO₄ total = 0
Sum, including every atom, = overall charge for an ion.SO₄²⁻ total = -2
Three-step calculation method
  1. Assign the known values and multiply each by the number of atoms in the formula.
  2. Set their sum equal to the overall charge: 0 for a neutral compound, or the stated charge for an ion.
  3. Solve for the unknown and include its sign, for example + 5 rather than 5.

3. Detailed Explanations

Oxidation-state calculation and redox methodA decision flow. An element has oxidation state zero. A one-atom ion has oxidation state equal to its charge. For compounds and polyatomic ions, assign known values, set their sum equal to zero or the ion charge, and solve. An increase is oxidation and a decrease is reduction.Assign first, then compareWhat type of particle?Use the formula and overall chargeElementOxidation state = 0Cl₂: Cl = 0One-atom ionOxidation state = chargeFe³⁺: Fe = +3Compound orpolyatomic ionAssign values, then addsum = overall chargeShow the sum and solveNO₃⁻: x + 3(−2) = −1, so x = +5Compare the same elementI: −1 → 0increase = oxidationCl: 0 → −1decrease = reductionBoth changes occur in the same redox reaction.
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. The nitrate calculation and iodide/chlorine changes are separate examples of these two stages.

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.

When the unknown element occurs more than once: in Al₂O₃, let x be the oxidation state of one aluminium atom. Then 2x + 3(-2) = 0, so 2x = +6 and x = +3. The + 6 is the total aluminium contribution, not the oxidation state of each atom.

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.
Scope note

Oxygen’s usual −2 value is not universal. In hydrogen peroxide, H₂O₂, hydrogen is +1 and the neutral total gives 2(+1) + 2x = 0, so each oxygen is −1. Use any supplied rule or oxidation state 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 the four oxygen atoms in SO₄²⁻).
  • 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

Show the change as well as the answer

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

6. Worked Examples

Modelled example 1

NO₂

Core

Problem

Find the oxidation state of nitrogen in NO₂.
Study the worked solution

  1. 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.

  2. Use the neutral total

    Method

    Set the sum equal to zero.

    Reason

    NO₂ is a neutral compound.

    Working

    x + 2(-2) = 0.

  3. Solve with the sign

    Method

    Rearrange for nitrogen.

    Reason

    Nitrogen must balance the total oxygen contribution.

    Working

    x-4 = 0, so x = +4.

Common misconception 2

NO₃⁻

Find and correct the mistake

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

View solution step by step

  1. 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.

  2. 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.

Challenge 3

KMnO₄

Minimal support

Multiple-rule transfer

Find the oxidation state of manganese in KMnO₄.

Combine two known values

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

  1. 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.

  2. 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.

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

Try independently: A learner uses x + 3(-2) = 0 for sulfur in the sulfite ion, SO₃²⁻, and gets +6. Correct the total and find sulfur’s oxidation state. Then compare it with sulfur in neutral SO₂.

Show answer and reasoning

For sulfite, the total must be −2: x + 3(-2) = -2, so x = +4. For neutral sulfur dioxide, x + 2(-2) = 0, so sulfur is also +4. The two formulas have different overall charges and atom counts, but sulfur has the same oxidation state. Oxidation state is not the formula’s overall charge.

Try independently: Find nitrogen’s oxidation state in N₂O₅. A learner obtains +10. Explain the missing step.

Show answer and reasoning

2x + 5(-2) = 0, so 2x = +10 and x = +5. The +10 is the combined contribution of two nitrogen atoms. Divide by two to find the oxidation state of each nitrogen atom.

Practise and check

Practise and check

Test the core rules, neutral compounds, polyatomic ions and oxidation-state changes in redox reactions.

Open the Redox Chemistry topic check
Syllabus and review details

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