Redox Reactions MCQs – 40 Practice Questions with Answers and Explanations

Master redox chemistry with 40 free practice questions covering oxidation and reduction, oxidation numbers, half-equations and redox titrations. Every question has a worked answer explanation, and the questions build in difficulty from GCSE and IGCSE core ideas to A-Level, IB and AP stretch material.

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Section 1: Oxidation & Reduction Basics

Q1. What does the mnemonic OIL RIG tell us about oxidation?

  • (A) Oxidation is the loss of electrons
  • (B) Oxidation is the gain of electrons
  • (C) Oxidation is the gain of protons
  • (D) Oxidation is the loss of oxygen
Show Answer

Answer: A – OIL stands for Oxidation Is Loss (of electrons), and RIG stands for Reduction Is Gain (of electrons).

Q2. In the reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), which species is oxidised?

  • (A) Zn
  • (B) Cu2+
  • (C) SO42−
  • (D) Cu
Show Answer

Answer: A – Each zinc atom loses two electrons: Zn → Zn2+ + 2e−. Loss of electrons is oxidation; the Cu2+ ions gain those electrons and are reduced.

Q3. Magnesium reacts with dilute hydrochloric acid: Mg + 2HCl → MgCl2 + H2. The oxidising agent in this reaction is:

  • (A) Mg
  • (B) Cl−
  • (C) H+ from the acid
  • (D) H2
Show Answer

Answer: C – The H+ ions gain electrons (2H+ + 2e− → H2), so they oxidise the magnesium. An oxidising agent is itself reduced.

Q4. Which statement about rusting is correct?

  • (A) Iron gains electrons as it rusts
  • (B) Rusting is a reduction of iron
  • (C) Rusting is a neutralisation reaction
  • (D) Iron is oxidised, forming hydrated iron(III) oxide
Show Answer

Answer: D – Rusting is oxidation: Fe → Fe3+ + 3e−, and the product is hydrated iron(III) oxide, Fe2O3.xH2O. Both oxygen and water are needed.

Q5. In terms of oxygen transfer, a substance that gains oxygen in a reaction is:

  • (A) oxidised
  • (B) reduced
  • (C) neutralised
  • (D) decomposed
Show Answer

Answer: A – Under the older oxygen-based definition, oxidation is gain of oxygen and reduction is loss of oxygen. This matches the electron definition in reactions such as Mg burning in oxygen.

Q6. A reducing agent is a substance that:

  • (A) donates electrons to another species and is itself oxidised
  • (B) gains electrons itself and is itself reduced
  • (C) is always a metal
  • (D) always contains oxygen
Show Answer

Answer: A – A reducing agent gives electrons away (causing reduction of the other species), so the reducing agent itself loses electrons and is oxidised.

Q7. Chlorine displaces bromine from potassium bromide: Cl2 + 2KBr → 2KCl + Br2. Which species is reduced?

  • (A) Br−
  • (B) K+
  • (C) Cl2
  • (D) Br2
Show Answer

Answer: C – Chlorine gains electrons: Cl2 + 2e− → 2Cl−. Gain of electrons is reduction. The bromide ions lose electrons and are oxidised.

Q8. Copper(II) oxide is reduced by carbon: 2CuO + C → 2Cu + CO2. The oxidising agent is:

  • (A) C
  • (B) CO2
  • (C) Cu
  • (D) CuO
Show Answer

Answer: D – CuO provides the oxygen and its Cu2+ ions gain electrons to form Cu, so CuO oxidises the carbon and is itself reduced. Carbon is the reducing agent.

Q9. In the reaction 2Mg + O2 → 2MgO, each magnesium atom:

  • (A) gains two electrons
  • (B) gains one electron
  • (C) loses one electron
  • (D) loses two electrons
Show Answer

Answer: D – Mg → Mg2+ + 2e−. Each magnesium atom loses two electrons, which is why magnesium is oxidised.

Q10. Which of these reactions is NOT a redox reaction?

  • (A) Zn + CuSO4 → ZnSO4 + Cu
  • (B) CH4 + 2O2 → CO2 + 2H2O
  • (C) NaOH + HCl → NaCl + H2O
  • (D) 2Mg + O2 → 2MgO
Show Answer

Answer: C – Neutralisation only transfers an H+ ion from the acid to OH−; no oxidation number changes. In the other three, electrons are transferred.

Section 2: Oxidation Numbers

Q11. What is the oxidation number of sulfur in H2SO4?

  • (A) +4
  • (B) +6
  • (C) −2
  • (D) +2
Show Answer

Answer: B – Working: 2(+1) + S + 4(−2) = 0, so S = 0 − 2 + 8 = +6.

Q12. What is the oxidation number of nitrogen in the ammonium ion, NH4+?

  • (A) −3
  • (B) +3
  • (C) −4
  • (D) +5
Show Answer

Answer: A – Working: N + 4(+1) = +1 (the charge on the ion), so N = +1 − 4 = −3.

Q13. What is the oxidation number of chromium in the dichromate ion, Cr2O72−?

  • (A) +3
  • (B) +12
  • (C) +7
  • (D) +6
Show Answer

Answer: D – Working: 2Cr + 7(−2) = −2, so 2Cr = +12 and Cr = +6. The trap answer +12 forgets there are two chromium atoms.

Q14. What is the oxidation number of manganese in the manganate(VII) ion, MnO4−?

  • (A) +4
  • (B) +6
  • (C) +2
  • (D) +7
Show Answer

Answer: D – Working: Mn + 4(−2) = −1, so Mn = −1 + 8 = +7, which is why the ion is named manganate(VII).

Q15. What is the oxidation number of oxygen in hydrogen peroxide, H2O2?

  • (A) −2
  • (B) −1
  • (C) 0
  • (D) +1
Show Answer

Answer: B – Peroxides are the exception to the usual −2 rule: 2(+1) + 2(O) = 0 gives O = −1.

Q16. What is the oxidation number of hydrogen in sodium hydride, NaH?

  • (A) −1
  • (B) +1
  • (C) 0
  • (D) +2
Show Answer

Answer: A – In metal hydrides hydrogen is the exception to its usual +1 rule: Na is +1, so H must be −1 for the compound to be neutral.

Q17. What is the oxidation number of chlorine in the perchlorate ion, ClO4−?

  • (A) +5
  • (B) +7
  • (C) +3
  • (D) +1
Show Answer

Answer: B – Working: Cl + 4(−2) = −1, so Cl = −1 + 8 = +7, the highest oxidation state of chlorine.

Q18. Which of these changes is a reduction?

  • (A) Fe2+ → Fe3+
  • (B) Cu → Cu2+
  • (C) Fe3+ → Fe2+
  • (D) S2− → S
Show Answer

Answer: C – In reduction the oxidation number decreases: iron goes from +3 to +2. In each of the other changes the oxidation number increases, which is oxidation.

Q19. Which reaction is an example of disproportionation?

  • (A) Cl2 + 2OH− → Cl− + ClO− + H2O
  • (B) Zn + 2HCl → ZnCl2 + H2
  • (C) Cl2 + 2Br− → 2Cl− + Br2
  • (D) CaCO3 → CaO + CO2
Show Answer

Answer: A – Chlorine starts at oxidation number 0 and is simultaneously reduced to −1 (in Cl−) and oxidised to +1 (in ClO−). One species acting as both oxidising and reducing agent is disproportionation.

Q20. In the reaction H2S → S + 2H+ + 2e−, the sulfur is:

  • (A) reduced, because its oxidation number decreases
  • (B) neither oxidised nor reduced
  • (C) both oxidised and reduced
  • (D) oxidised, because its oxidation number rises from −2 to 0
Show Answer

Answer: D – Sulfur goes from −2 in H2S to 0 in S. An increase in oxidation number (with loss of electrons) is oxidation.


Section 3: Half-Equations & Balancing Redox

Q21. Which half-equation correctly shows the oxidation of Fe2+ to Fe3+?

  • (A) Fe2+ + e− → Fe3+
  • (B) Fe3+ + e− → Fe2+
  • (C) Fe2+ → Fe3+ + e−
  • (D) Fe2+ → Fe3+ + 2e−
Show Answer

Answer: C – Oxidation is loss of electrons, so the electron appears on the right. Charge balances: +2 on the left, +3 − 1 = +2 on the right.

Q22. In acid solution, manganate(VII) is reduced as follows: MnO4− + 8H+ + xe− → Mn2+ + 4H2O. What is the value of x?

  • (A) 3
  • (B) 5
  • (C) 2
  • (D) 4
Show Answer

Answer: B – Balance charge: the right side totals +2, so −1 + 8 − x = +2, giving x = 5. This also matches manganese falling from +7 to +2, a gain of 5 electrons.

Q23. In acid solution, dichromate is reduced: Cr2O72− + 14H+ + ne− → 2Cr3+ + 7H2O. What is the value of n?

  • (A) 3
  • (B) 4
  • (C) 5
  • (D) 6
Show Answer

Answer: D – Balance charge: −2 + 14 − n must equal 2(+3) = +6, so n = 6. Each chromium falls from +6 to +3, gaining 3 electrons, and there are two chromium atoms.

Q24. Which half-equation correctly shows the oxidation of chloride ions?

  • (A) 2Cl− → Cl2 + 2e−
  • (B) Cl2 + 2e− → 2Cl−
  • (C) 2Cl− + 2e− → Cl2
  • (D) Cl− → Cl2 + e−
Show Answer

Answer: A – The chloride ions lose electrons, so electrons are products. Atoms and charge both balance: −2 on each side. Option B is the reverse process, reduction.

Q25. The half-equations Zn → Zn2+ + 2e− and Cu2+ + 2e− → Cu combine to give the overall ionic equation:

  • (A) Zn + Cu2+ → Zn2+ + Cu
  • (B) Zn2+ + Cu → Zn + Cu2+
  • (C) Zn + Cu → Zn2+ + Cu2+
  • (D) 2Zn + Cu2+ → 2Zn2+ + Cu
Show Answer

Answer: A – Both half-equations involve 2 electrons, so they add directly and the electrons cancel. Zinc is the more reactive metal, so the reaction goes in this direction.

Q26. Iodine reacts with thiosulfate ions: I2 + 2S2O32− → 2I− + S4O62−. The half-equation for the reduction of iodine is:

  • (A) 2I− → I2 + 2e−
  • (B) I2 → 2I− + 2e−
  • (C) I2 + 2e− → 2I−
  • (D) I2 + e− → I−
Show Answer

Answer: C – Iodine falls from oxidation number 0 to −1, gaining one electron per iodine atom, so I2 gains 2 electrons in total. This reaction is the basis of iodine-thiosulfate titrations.

Q27. In the balanced half-equation Cr2O72− + aH+ + 6e− → 2Cr3+ + bH2O, what is the value of a?

  • (A) 7
  • (B) 6
  • (C) 8
  • (D) 14
Show Answer

Answer: D – The 7 oxygen atoms form 7H2O (so b = 7), which needs 14 hydrogen atoms, so a = 14 H+ ions.

Q28. Which half-equation is correctly balanced for both atoms and charge?

  • (A) Al3+ + 2e− → Al
  • (B) Zn2+ + 2e− → Zn
  • (C) Fe3+ + 3e− → Fe2+
  • (D) Cu2+ + e− → Cu
Show Answer

Answer: B – Charge: +2 − 2 = 0 on the left, 0 on the right. In A the charge is +1 on the left; in C it is 0 against +2; in D it is +1 against 0.

Q29. During the electrolysis of molten lead(II) bromide, the cathode half-equation is Pb2+ + 2e− → Pb. This process is:

  • (A) reduction, because the lead ions gain electrons
  • (B) oxidation, because the lead ions gain electrons
  • (C) neutralisation
  • (D) disproportionation
Show Answer

Answer: A – Reduction is gain of electrons, and it always happens at the cathode (the negative electrode in electrolysis) because that is where positive ions pick up electrons.

Q30. Using the half-equations MnO4− + 8H+ + 5e− → Mn2+ + 4H2O and Fe2+ → Fe3+ + e−, how many moles of Fe2+ react with 1 mole of MnO4−?

  • (A) 1
  • (B) 3
  • (C) 5
  • (D) 8
Show Answer

Answer: C – One MnO4− ion accepts 5 electrons, and each Fe2+ ion supplies 1 electron, so the ratio MnO4− : Fe2+ is 1 : 5. This ratio is used in manganate(VII) titrations.

Section 4: Redox Titrations & Applications (A-Level / AP stretch)

Q31. A 25.0 cm3 sample of an Fe2+ solution is titrated with 0.0200 mol/dm3 acidified KMnO4. The titre is 20.0 cm3. Using the ratio MnO4− : Fe2+ = 1 : 5, what is the concentration of the Fe2+ solution?

  • (A) 0.0160 mol/dm3
  • (B) 0.0400 mol/dm3
  • (C) 0.0800 mol/dm3
  • (D) 0.400 mol/dm3
Show Answer

Answer: C – Working: moles MnO4− = 0.0200 × (20.0/1000) = 4.00 × 10−4 mol. Moles Fe2+ = 5 × 4.00 × 10−4 = 2.00 × 10−3 mol. Concentration = 2.00 × 10−3 / (25.0/1000) = 0.0800 mol/dm3. Option A forgets the 1 : 5 ratio.

Q32. Why is no indicator needed in a manganate(VII) titration?

  • (A) The Mn2+ formed is deep pink
  • (B) MnO4− is intensely purple and is decolourised as it reacts, so it is self-indicating
  • (C) Fe2+ changes colour sharply at the end-point
  • (D) The acid used acts as an indicator
Show Answer

Answer: B – MnO4− is deep purple but its product Mn2+ is almost colourless, so each drop is decolourised until the Fe2+ runs out; the first permanent pale pink colour marks the end-point.

Q33. Dichromate ions oxidise iron(II) ions: Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O. How many moles of Fe2+ are oxidised by 0.00200 mol of Cr2O72−?

  • (A) 0.00200 mol
  • (B) 0.00600 mol
  • (C) 0.0120 mol
  • (D) 0.000333 mol
Show Answer

Answer: C – Working: the ratio Cr2O72− : Fe2+ is 1 : 6, so moles Fe2+ = 6 × 0.00200 = 0.0120 mol. Option D inverts the ratio.

Q34. Which titration needs a separate indicator to detect the end-point?

  • (A) Acidified KMnO4 titrated against Fe2+
  • (B) Acidified K2Cr2O7 titrated against Fe2+
  • (C) Acidified KMnO4 titrated against ethanedioate
  • (D) Acidified KMnO4 titrated against hydrogen peroxide
Show Answer

Answer: B – Manganate(VII) is self-indicating (purple to colourless), so A, C and D need no indicator. The orange to green colour change of dichromate is not sharp enough, so an indicator such as sodium diphenylamine sulfonate is used.

Q35. Standard electrode potentials: Cl2/Cl− = +1.36 V, Br2/Br− = +1.09 V, I2/I− = +0.54 V. Which is the strongest oxidising agent?

  • (A) I2
  • (B) Br−
  • (C) Cl−
  • (D) Cl2
Show Answer

Answer: D – The more positive the E value, the more readily the oxidised species gains electrons. Cl2 has the most positive electrode potential, so it is the strongest oxidising agent and can oxidise both Br− and I−.

Q36. Chlorine reacts with hot concentrated alkali to form a chlorate(V) salt: 3Cl2 + 6OH− → 5Cl− + ClO3− + 3H2O. This is a disproportionation because chlorine is:

  • (A) only oxidised
  • (B) only reduced
  • (C) neither oxidised nor reduced
  • (D) both oxidised (0 to +5 in ClO3−) and reduced (0 to −1 in Cl−)
Show Answer

Answer: D – Working: in ClO3−, Cl + 3(−2) = −1, so Cl = +5. The same element, starting at 0, rises to +5 and falls to −1, which is disproportionation. (In cold dilute alkali the products are Cl− and ClO− instead.)

Q37. What is the oxidation number of chlorine in the chlorate(V) ion, ClO3−?

  • (A) +1
  • (B) +3
  • (C) +5
  • (D) +7
Show Answer

Answer: C – Working: Cl + 3(−2) = −1, so Cl = +5, matching the (V) in the name chlorate(V).

Q38. In an iodine-thiosulfate titration, a 25.0 cm3 sample of iodine solution reacts with exactly 20.0 cm3 of 0.100 mol/dm3 sodium thiosulfate. Given I2 + 2S2O32− → 2I− + S4O62−, what is the concentration of the iodine solution?

  • (A) 0.0800 mol/dm3
  • (B) 0.0400 mol/dm3
  • (C) 0.0200 mol/dm3
  • (D) 0.100 mol/dm3
Show Answer

Answer: B – Working: moles S2O32− = 0.100 × (20.0/1000) = 2.00 × 10−3 mol. Moles I2 = half of this = 1.00 × 10−3 mol. Concentration = 1.00 × 10−3 / (25.0/1000) = 0.0400 mol/dm3. Option A forgets the 2 : 1 ratio.

Q39. Potassium manganate(VII) titrations are acidified with dilute sulfuric acid and NOT hydrochloric acid, because:

  • (A) hydrochloric acid is too weak to provide H+ ions
  • (B) MnO4− would oxidise Cl− ions to Cl2, using up extra manganate and giving a wrong titre
  • (C) sulfuric acid is a stronger oxidising agent than manganate
  • (D) hydrochloric acid reacts with the glass of the burette
Show Answer

Answer: B – MnO4− (E = +1.51 V) is a strong enough oxidising agent to oxidise Cl− (E for Cl2/Cl− = +1.36 V) to chlorine, so HCl cannot be used; dilute H2SO4 provides H+ without interfering.

Q40. Given E°(Fe3+/Fe2+) = +0.77 V and E°(I2/I−) = +0.54 V, which reaction takes place when the solutions are mixed?

  • (A) I2 oxidises Fe2+ to Fe3+
  • (B) Fe3+ oxidises I− to I2
  • (C) No reaction occurs in either direction
  • (D) Fe2+ reduces Fe3+ to Fe
Show Answer

Answer: B – The couple with the more positive E provides the oxidising agent: Fe3+ (+0.77 V) oxidises I−, giving 2Fe3+ + 2I− → 2Fe2+ + I2, with E cell = +0.77 − 0.54 = +0.23 V, so the reaction is feasible. This is why Fe3+ liberates iodine from iodide solutions.

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