Oxidation numbers and redoxEdexcel International A Level Chemistry: Subtopic test
10 questions, 27 marks
Edexcel International A Level Chemistry
Oxidation numbers and redox
Total 27 marks
Name
Class
Date
- 1Sulfur forms compounds and ions in which its oxidation number ranges from −2 to +6. Examples include hydrogen sulfide, H₂S, sulfur dioxide, SO₂, the sulfate ion, SO₄²⁻, and the sulfite ion, SO₃²⁻. In the contact process sulfur dioxide is oxidised to sulfur trioxide by oxygen.(a)What is the oxidation number of sulfur in the sulfate ion, SO₄²⁻?[1 mark]
- A+4
- B+8
- C+6
- D−2
(b)What is the systematic name of sodium sulfite, Na₂SO₃?[1 mark]- ASodium sulfate(VI)
- BSodium sulfate(II)
- CSodium sulfate(III)
- DSodium sulfate(IV)
(c)In the reaction 2SO₂ + O₂ → 2SO₃, state which element is oxidised and which is reduced. Give the oxidation numbers to support your answer.[2 marks]Total for question 1: 4 marks
- 2Hydrides and peroxides contain elements in unusual oxidation states. Sodium hydride, NaH, reacts with water: NaH + H₂O → NaOH + H₂. Hydrogen peroxide, H₂O₂, decomposes when a catalyst is added: 2H₂O₂ → 2H₂O + O₂.(a)What is the oxidation number of hydrogen in sodium hydride, NaH?[1 mark]
- A+1
- B−1
- C0
- D+2
(b)Which statement about the decomposition of hydrogen peroxide is correct?[1 mark]- AIt is a disproportionation, because oxygen is both oxidised (−1 to 0) and reduced (−1 to −2)
- BOxygen is oxidised only
- COxygen is reduced only
- DIt is not a redox reaction because there is no change of oxidation number
(c)Deduce the changes in oxidation number of hydrogen in the reaction NaH + H₂O → NaOH + H₂, and state which species is the reducing agent.[2 marks]Total for question 2: 4 marks
- 3Acidified potassium manganate(VII) solution is purple. When it is added to a solution containing iron(II) ions, the purple colour disappears and iron(III) ions and manganese(II) ions form.(a)Write ionic half-equations for (i) the reduction of MnO₄⁻ to Mn²⁺ in acid solution and (ii) the oxidation of Fe²⁺ to Fe³⁺.[3 marks](b)Combine your half-equations to give the overall ionic equation. Hence name the oxidising agent and state the change in oxidation number of manganese.[4 marks]
Total for question 3: 7 marks
- 4A student studies these four reactions.
Reaction 1: Cl₂(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H₂O(l), with cold dilute sodium hydroxide.
Reaction 2: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
Reaction 3: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l).
Reaction 4: 2Cu⁺(aq) → Cu(s) + Cu²⁺(aq).(a)Use oxidation numbers to explain why Reaction 1 is a disproportionation reaction, and write the ionic equation for it.[6 marks](b)Classify Reactions 2, 3 and 4 as redox, disproportionation or neither, using oxidation numbers to justify each answer.[6 marks]Total for question 4: 12 marks
End of questions
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).