Redox IEdexcel A-Level Chemistry: Topic test

1. 1
   Hydrogen peroxide, H₂O₂, is stored in brown bottles because it slowly decomposes. Sodium hydride, NaH, is a white solid used as a reducing agent in organic synthesis.

   (a)

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

   [1 mark]

   • A−2
   • B−1
   • C0
   • D+1

   (b)

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

   [1 mark]

   • A−1
   • B0
   • C+1
   • D+2

   (c)

   Hydrogen peroxide decomposes according to the equation 2H₂O₂ → 2H₂O + O₂. Use oxidation numbers to explain why this is a disproportionation reaction.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Chromium forms compounds in several oxidation states. Orange potassium dichromate(VI), K₂Cr₂O₇, is used as an oxidising agent in acidified solution.

   (a)

   What is the oxidation number of chromium in the dichromate(VI) ion, Cr₂O₇²⁻?

   [1 mark]

   • A+3
   • B+4
   • C+6
   • D+7

   (b)

   What is the formula of chromium(III) sulfate?

   [1 mark]

   • ACrSO₄
   • BCr(SO₄)₃
   • CCr₃(SO₄)₂
   • DCr₂(SO₄)₃

   (c)

   In acidified solution, dichromate(VI) ions are reduced to chromium(III) ions. State the change in oxidation number of each chromium atom and explain, in terms of electrons, why dichromate(VI) ions act as an oxidising agent.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Acidified potassium iodate(V), KIO₃, reacts with aqueous potassium iodide. Iodine forms and the solution turns brown.

   (a)

   Deduce the oxidation number of iodine in the iodate(V) ion, IO₃⁻. Write the half-equation for the reduction of iodate(V) ions to iodine in acidic solution.

   [3 marks]

   (b)

   Write the half-equation for the oxidation of iodide ions to iodine. Use it, together with your answer to part (a), to construct the overall ionic equation for the reaction. State which species is the reducing agent and explain your answer.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A technician studies two reactions of copper compounds. In the first, red copper(I) oxide, Cu₂O, is warmed with dilute sulfuric acid. A brown solid, copper metal, forms and the solution turns blue as aqueous copper(II) sulfate forms. In the second, aqueous copper(II) sulfate is added to aqueous potassium iodide. A white precipitate of copper(I) iodide, CuI, forms in a brown solution of iodine, I₂.

   (a)

   Use oxidation numbers and electron transfer to explain why the reaction of copper(I) oxide with dilute sulfuric acid is a disproportionation reaction. Write the two half-equations for copper species and the overall ionic equation for the reaction of Cu⁺ ions.

   [6 marks]

   (b)

   The ionic equation for the second reaction is 2Cu²⁺ + 4I⁻ → 2CuI + I₂. Use oxidation numbers to identify the oxidising agent and the reducing agent. Write the two half-equations. Explain why this is not a disproportionation reaction.

   [6 marks]

   Total for question 4: 12 marks

5. 5
   Manganese(IV) oxide, MnO₂, reacts with warm concentrated hydrochloric acid to form manganese(II) chloride, chlorine and water: MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O.

   (a)

   Which species is the oxidising agent in the reaction of manganese(IV) oxide with concentrated hydrochloric acid?

   [1 mark]

   • AMnO₂
   • BHCl
   • CMnCl₂
   • DCl₂

   (b)

   Which is the correct half-equation for the reduction of manganese(IV) oxide in acidic solution?

   [1 mark]

   • AMnO₂ + 4H⁺ → Mn²⁺ + 2H₂O
   • BMnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O
   • CMnO₂ + 2H⁺ + 2e⁻ → Mn²⁺ + H₂O
   • DMnO₂ + 4H⁺ + 4e⁻ → Mn²⁺ + 2H₂O

   (c)

   Write the half-equation for the oxidation of chloride ions to chlorine. State the change in the oxidation number of chlorine.

   [2 marks]

   Total for question 5: 4 marks

6. 6
   Vanadium forms compounds in several oxidation states. Vanadium(V) oxide, V₂O₅, is the catalyst in the Contact process. It oxidises sulfur dioxide to sulfur trioxide while being reduced to V₂O₄, and it is then re-oxidised by oxygen.

   (a)

   What is the oxidation number of vanadium in the VO₂⁺ ion?

   [1 mark]

   • A+2
   • B+3
   • C+5
   • D+6

   (b)

   What is the formula of vanadium(III) chloride?

   [1 mark]

   • AV₃Cl
   • BVCl
   • CV₂Cl₃
   • DVCl₃

   (c)

   The reaction in the Contact process is V₂O₅ + SO₂ → V₂O₄ + SO₃. Use oxidation numbers to identify the reducing agent.

   [2 marks]

   Total for question 6: 4 marks

7. 7
   Calcium hydride, CaH₂, is an ionic solid containing Ca²⁺ and H⁻ ions. It is used to dry organic solvents because it reacts with water: CaH₂(s) + 2H₂O(l) → Ca(OH)₂(aq) + 2H₂(g).

   (a)

   Deduce the oxidation numbers of hydrogen in CaH₂, in H₂O and in H₂. Explain why the hydride ion behaves as a reducing agent in this reaction.

   [3 marks]

   (b)

   Write the half-equation for the oxidation of hydride ions and the half-equation for the reduction of water to hydrogen and hydroxide ions. Use them to construct the ionic equation for the reaction. Explain why the reaction is not disproportionation.

   [4 marks]

   Total for question 7: 7 marks

8. 8
   Green potassium manganate(VI), K₂MnO₄, is stable in alkaline solution. When the solution is acidified, the green colour is replaced by the purple of manganate(VII) ions, MnO₄⁻, and a brown precipitate of manganese(IV) oxide, MnO₂, forms. When hydrogen peroxide is added to the purple solution, the purple colour disappears and oxygen gas is given off, leaving a solution that contains manganese(II) ions.

   (a)

   Use oxidation numbers to explain why the acidification of manganate(VI) ions is a disproportionation reaction. Use half-equations to construct the ionic equation for the reaction.

   [6 marks]

   (b)

   Use oxidation numbers to explain why hydrogen peroxide acts as a reducing agent when added to the purple manganate(VII) solution. Construct the ionic equation for the reaction, using half-equations.

   [6 marks]

   Total for question 8: 12 marks

End of questions