ChemistryISC Class 11 · Redox Reactions

Oxidation Numbers, Oxidising and Reducing Agents and Types of Redox Reactions

Q: Is a fractional oxidation number real?

It is an average over atoms of the same element in different environments, as in Fe₃O₄ or Na₂S₄O₆.

Q: Can a reaction have oxidation without reduction?

No. Electrons lost by one species must be gained by another.

Q: Is hydrogen peroxide an oxidising or a reducing agent?

Both, depending on the partner: oxygen is −1, between 0 and −2, so it can be reduced to H₂O or oxidised to O₂.

Updated 17 October 2026 · Fact-checked

Oxidation is an increase in oxidation number (loss of electrons) and reduction is a decrease (gain of electrons). The oxidising agent is reduced and the reducing agent is oxidised. Redox reactions are classified as combination, decomposition, displacement, disproportionation (and comproportionation).

Moving right on the oxidation number line is oxidation, moving left is reduction; in disproportionation one species Cl₂ does both.

Illustration drawn for QuestionBuddy.

Understand Oxidation Numbers and Types of Redox Reactions

Rules for oxidation numbers. (1) Atoms of an element in its free state: 0. (2) A monatomic ion: its charge. (3) Fluorine is −1 in all compounds. (4) Oxygen is −2 in most compounds, −1 in peroxides (H₂O₂, Na₂O₂), −½ in superoxides (KO₂) and +2 in OF₂. (5) Hydrogen is +1 with non-metals and −1 with metals (NaH, CaH₂). (6) The sum of the oxidation numbers of all atoms is zero in a neutral compound and equals the charge in an ion. (7) Alkali metals are +1 and alkaline earth metals +2 in compounds.

Calculation. Write the algebraic sum equal to the charge. In Na₂S₄O₆: 2(+1) + 4x + 6(−2) = 0 gives x = +2.5, an average; the actual structure has sulphur atoms in different states. In Fe₃O₄: 3x − 8 = 0, so x = +8/3 (average).

Redox language. Oxidation: loss of electrons or increase in oxidation number (also addition of oxygen or removal of hydrogen). Reduction: gain of electrons or decrease in oxidation number. The substance that gets oxidised is the reducing agent (reductant); the substance that gets reduced is the oxidising agent (oxidant). The reactions always occur together.

Types of redox reactions. *Combination*: A + B → AB (C + O₂ → CO₂). *Decomposition*: 2KClO₃ → 2KCl + 3O₂. *Displacement*: Zn + CuSO₄ → ZnSO₄ + Cu (metal or non-metal displacement; the activity series decides). *Disproportionation*: one species is simultaneously oxidised and reduced, for example 2H₂O₂ → 2H₂O + O₂ and Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. *Comproportionation* is the reverse: two states of an element form an intermediate state (NH₄NO₂ → N₂ + 2H₂O).

Test. A reaction is a redox reaction only if at least one element changes its oxidation number; acid-base and precipitation reactions are not redox.

Oxidation Numbers and Types of Redox Reactions: formulas and key facts

Sum rule: $Σ (o x i d a t i o nn u mb er s) = n e t c ha r g e$; Zero for a neutral compound.
Oxidation and reduction: Oxidation: oxidation number increases | Reduction: oxidation number decreases; Electrons are lost or gained.
Agents: Oxidising agent: is reduced | Reducing agent: is oxidised; They change by the same total number of electrons.
Special oxygen states: $P er o x i d e : - 1∣ S u p er o x i d e : - ½∣ O F_{2} : + 2$; Normal oxides: −2.

Oxidation Numbers and Types of Redox Reactions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Rust

Suppose iron reacts with oxygen and water to form rust: 4Fe + 3O₂ + 2xH₂O → 2Fe₂O₃·xH₂O.

Find the changes in oxidation number and the total electrons transferred.

Try it first, then show the solutionSolution

Fe goes from 0 to +3 (oxidation); O goes from 0 to −2 (reduction).
Electrons lost: 4 Fe × 3 = 12; electrons gained: 6 O atoms × 2 = 12.
Iron is the reducing agent and oxygen is the oxidising agent.

Answer: Fe: 0 → +3; O: 0 → −2; 12 electrons transferred.

Memory hook: Electrons lost always equal electrons gained.

Tricks to remember Oxidation Numbers and Types of Redox Reactions

Short memory aids. Say each one aloud once, then use it on a question the same day.

OIL RIG
Oxidation Is Loss, Reduction Is Gain of electrons.
Use it when: Naming oxidation and reduction.
The reducing agent gets oxidised
The agent does the opposite to itself.
Use it when: Identifying agents.

Oxidation Numbers and Types of Redox Reactions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Oxidation number of chromium · 1 mark · MCQBoard level

The oxidation number of chromium in K₂Cr₂O₇ is:
(a) +3
(b) +4
(c) +6
(d) +7

Try it first, then show the solutionSolution

2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6.

Answer: (c) +6

Example 2: Calculating oxidation numbers · 3 marks · Short answerBoard level

Find the oxidation number of (i) S in H₂SO₄, (ii) S in Na₂S₄O₆ (average) and (iii) Mn in KMnO₄.

Try it first, then show the solutionSolution

(i) 2(+1) + x + 4(−2) = 0, so x = +6. (1 mark)
(ii) 2(+1) + 4x + 6(−2) = 0, so 4x = 10 and x = +2.5. (1 mark)
(iii) (+1) + x + 4(−2) = 0, so x = +7. (1 mark)

Answer: (i) +6; (ii) +2.5; (iii) +7.

Example 3: Disproportionation · 4 marks · Short answerHarder board level

Identify the oxidising and reducing agents in (a) Zn + CuSO₄ → ZnSO₄ + Cu, and show that (b) 2H₂O₂ → 2H₂O + O₂ and (c) Cl₂ + 2NaOH → NaCl + NaClO + H₂O are disproportionation reactions.

Try it first, then show the solutionSolution

(a) Zn goes 0 → +2 (reducing agent); Cu goes +2 → 0 (oxidising agent). (1 mark)
(b) O in H₂O₂ is −1; it becomes −2 in H₂O (reduced) and 0 in O₂ (oxidised): the same species is both. (1.5 marks)
(c) Cl goes 0 → −1 in NaCl and 0 → +1 in NaClO. Cl₂ is both oxidised and reduced. (1.5 marks)

Answer: (a) Zn is the reductant; CuSO₄ the oxidant. (b), (c): the same species is oxidised and reduced, so they are disproportionations.

Example 4: Unusual oxidation states · 4 marks · Short answerHarder board level

Find the oxidation numbers of S in the peroxodisulphate ion S₂O₈²⁻ (which has one peroxide O–O link), of Cr in CrO₅ (one O²⁻ and four peroxide-type oxygens), and of O in KO₂ and in OF₂.

Try it first, then show the solutionSolution

S₂O₈²⁻: 2x + 6(−2) + 2(−1) = −2, so 2x = 12 and S = +6 (not +7). (1 mark)
CrO₅: x + 1(−2) + 4(−1) = 0, so Cr = +6, not +10. (1 mark)
KO₂: +1 + 2x = 0, so O = −½ (superoxide). (1 mark)
OF₂: x + 2(−1) = 0, so O = +2. (1 mark)

Answer: S +6; Cr +6; O −½ in KO₂; O +2 in OF₂.

Example 5: Classifying reactions · 5 marks · Long answerHOTS / competency

For each, give the oxidation-number changes and the type: (i) 2KClO₃ → 2KCl + 3O₂; (ii) 4P + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂; (iii) NH₄NO₂ → N₂ + 2H₂O; (iv) 3NO₂ + H₂O → 2HNO₃ + NO.

Try it first, then show the solutionSolution

(i) Cl: +5 → −1 (reduced); O: −2 → 0 (oxidised): decomposition, intramolecular redox. (1 mark)
(ii) P: 0 → −3 in PH₃ and 0 → +1 in NaH₂PO₂: disproportionation. (1.5 marks)
(iii) N: −3 (in NH₄⁺) and +3 (in NO₂⁻) both go to 0: comproportionation. (1.5 marks)
(iv) N: +4 → +5 in HNO₃ and +4 → +2 in NO: disproportionation. (1 mark)

Answer: (i) decomposition; (ii) and (iv) disproportionation; (iii) comproportionation.

Where students lose marks in Oxidation Numbers and Types of Redox Reactions

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong oxidation number of oxygen
Using −2 for the peroxide O₂²⁻.
Do this: In peroxides oxygen is −1; in superoxides −½.
Usually costs: 1 to 2 marks.
Mistake: Mixing up agents
Calling the substance that is oxidised the oxidising agent.
Do this: The oxidising agent is the substance reduced.
Usually costs: 1 mark.
Mistake: Fractional states left unexplained
Writing +2.5 for S in Na₂S₄O₆ without saying it is an average.
Do this: Say "average oxidation number".
Usually costs: 1 mark.

Exam tips

Write the sum equation for each oxidation number.
List the change in oxidation number for each element in a redox question.
Name the oxidant and the reductant in words.
For disproportionation, show one species going both up and down.

Practice questions on Oxidation Numbers and Types of Redox Reactions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In a redox reaction, the substance that loses electrons is called:
(a) the oxidising agent
(b) the reducing agent
(c) a spectator
(d) a catalyst
Hint
It is itself oxidised.
Answer
(b) the reducing agent
Q2 · 2 marks · Short answerBoard level
Find the oxidation number of Cl in HClO₄ and of S in SO₄²⁻.
Hint
Use the sum rule.
Answer
HClO₄: 1 + x − 8 = 0, so Cl = +7. SO₄²⁻: x − 8 = −2, so S = +6.
Q3 · 3 marks · Short answerHarder board level
Which of the following are redox reactions? (a) NaOH + HCl → NaCl + H₂O (b) CaCO₃ → CaO + CO₂ (c) 2Mg + O₂ → 2MgO (d) AgNO₃ + NaCl → AgCl + NaNO₃
Hint
Look for changes in oxidation number.
Answer
Only (c) is a redox reaction (Mg 0 → +2, O 0 → −2). The others involve no change in oxidation numbers.
Q4 · 3 marks · Short answerHarder board level
Find the oxidation number of carbon in CH₃OH, HCHO, HCOOH and CO₂ and arrange them in the order of increasing oxidation state.
Hint
Use H = +1, O = −2.
Answer
CH₃OH: −2; HCHO: 0; HCOOH: +2; CO₂: +4. The order of increasing oxidation state is CH₃OH < HCHO < HCOOH < CO₂.
Q5 · 3 marks · Short answerHarder board level
Explain why fluorine shows no positive oxidation state while chlorine can reach +7.
Hint
Think about electronegativity and d orbitals.
Answer
Fluorine is the most electronegative element and has no d orbitals, so it is always −1 in compounds. Chlorine has vacant 3d orbitals and can expand its octet, reaching +7 as in HClO₄.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a factory studies the corrosion 4Fe + 3O₂ → 2Fe₂O₃ (simplified).
(i) State the oxidation number of Fe at the start and the end. (1 mark)
(ii) Name the oxidising agent. (1 mark)
(iii) Find the number of electrons transferred for 4 mol of Fe. (1 mark)
(iv) State what happens to the oxidation number of oxygen. (1 mark)
Hint
Electrons lost by Fe equal electrons gained by O.
Answer
(i) 0 and +3. (ii) Oxygen (O₂). (iii) 4 × 3 = 12 mol of electrons. (iv) It decreases from 0 to −2 (reduction).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define oxidation in terms of electrons.Show answer
Loss of electrons.
2.Oxidation number of O in H₂O₂?Show answer
−1.
3.What is disproportionation?Show answer
A reaction in which the same species is both oxidised and reduced.
4.Oxidation number of an element in its free state?Show answer
Zero.

Oxidation Numbers and Types of Redox Reactions: frequently asked questions

Is a fractional oxidation number real?

It is an average over atoms of the same element in different environments, as in Fe₃O₄ or Na₂S₄O₆.

Can a reaction have oxidation without reduction?

No. Electrons lost by one species must be gained by another.

Is hydrogen peroxide an oxidising or a reducing agent?

Both, depending on the partner: oxygen is −1, between 0 and −2, so it can be reduced to H₂O or oxidised to O₂.