Structured Questions
Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.
Question 1: Oxidation States & Redox Equations
6 marksNitrogen forms a range of oxides and ions with different oxidation states.
(a) Calculate the oxidation state of nitrogen in each of the following species: NO₃⁻, NH₄⁺, and N₂O. [3]
(b) When copper metal reacts with dilute nitric acid, copper(II) nitrate, nitrogen monoxide gas (NO), and water are formed. Write a balanced overall ionic redox equation for this reaction. [3]
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(a)
- NO₃⁻: +5 [1]
- NH₄⁺: -3 [1]
- N₂O: +1 [1]
(b)
- Oxidation: Cu → Cu²⁺ + 2e⁻ [1]
- Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O [1]
- Overall equation: 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O [1]
Question 2: Manganate(VII) Titration Calculations
9 marksIn a laboratory experiment, a student dissolved an iron tablet containing iron(II) sulfate in dilute sulfuric acid. The solution was titrated against 0.0200 mol dm⁻³ potassium manganate(VII) solution.
(a) Write the half-equation for the reduction of manganate(VII) ions (MnO₄⁻) to manganese(II) ions (Mn²⁺) in acidic conditions. [2]
(b) Explain, in terms of electron transfer, why Fe²⁺ acts as a reducing agent in this reaction. [2]
(c) The student found that 25.0 cm³ of the iron(II) solution required 22.50 cm³ of the MnO₄⁻ solution for complete reaction. The overall redox equation is: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O. Calculate the concentration, in mol dm⁻³, of the iron(II) ions in the solution. [5]
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(a)
- MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O [2] (1 mark for correct species and balancing atoms; 1 mark for correct number of electrons)
(b)
- Fe²⁺ loses an electron to form Fe³⁺ [1]
- A reducing agent is an electron donor / is oxidised during the reaction [1]
(c)
- Moles of \(\text{MnO}_4^- = 0.0200 \times \frac{22.50}{1000} = 4.50 \times 10^{-4}\text{ mol}\) [1]
- Ratio of Fe²⁺ to MnO₄⁻ is 5:1 [1]
- Moles of Fe²⁺ = 5 × 4.50 × 10⁻⁴ = 2.25 × 10⁻³ mol [1]
- Concentration of \(\text{Fe}^{2+} = \frac{2.25 \times 10^{-3}}{25.0 / 1000}\) [1]
- Concentration = 0.0900 mol dm⁻³ [1]
Question 3: Disproportionation of Chlorine
5 marksChlorine gas reacts with cold, dilute aqueous sodium hydroxide in a redox reaction.
(a) Write a balanced chemical equation for this reaction. [2]
(b) State the oxidation states of chlorine in the reactants and products, and explain why this reaction is described as disproportionation. [3]
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(a)
- Cl₂ + 2NaOH → NaCl + NaClO + H₂O [2] (1 mark for correct products, 1 mark for balanced equation)
(b)
- Chlorine in Cl₂ has an oxidation state of 0 [1]
- Chlorine in NaCl is -1 (reduced) and in NaClO is +1 (oxidised) [1]
- It is disproportionation because the same element (chlorine) is simultaneously oxidised and reduced in the reaction [1]