2022 MI H2 Chem Prelim Paper 2 - Answer

Class Adm No Candidate Name: This question paper consists of 16 printed pages and 2 blank pages. 2022 Preliminary Exams Pre-University 3 H2 CHEMISTRY 9729/02 Paper 2 Structured Questions 14 Sep 2022 2 hours Candidates answer on the Question paper. Additional materials: Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Question 1 2 3 4 5 Total Marks 20 13 24 9 9 75

2 Answer all questions in the spaces provided. 1 Many household tools are made from stainless steel as it is highly resistant to corrosion and rust. Stainless steel contains iron, chromium, manganese, nickel, molybdenum, silicon and carbon which interacts with oxygen from water and air to form oxides. Of these elements, chromium has the highest resistance to corrosion. Despite the high resistance, rusting still occurs in a humid environment. A ‘Do -it-Yourself’ (DIY) kit has been launched in the market to remove rust from household tools and the set-up is shown in Figure 1.1. Figure 1.1 (a) (i) State if the iron sheet is the anode or the cathode. [1] anode (ii) Use of Data Booklet is relevant to this question. State and explain the observations found at the iron sheet electrode. [2] The iron sheet will becomes smaller. (;) Fe → Fe2+ + 2e Eº = –0.44 V 2H2O → O2 + 4H+ + 4e- Eº = +1.23 V (;) Since EºFe2+/Fe is more negative than Eº(O2/H2O), Fe is preferentially oxidized to Fe2+. (;) (iii) Suggest why carbonate ions from Na2CO3 electrolyte cannot be oxidised at the anode. [1] Carbon in carbonate is at the maximum oxidation state of +4 or the electrode potential involving carbonate ion is very positive. Rusty tool Na2CO3 (aq) Iron sheet

3 [Turn over (iv) A consumer saw the DIY kit and decided to purchase it and try out on his rusted dumbbell. He prepared the set up as shown in Figure 1.1 with a 20 A power source and switched it on for 6 hours. Determine the change in mass of the iron sheet. [2] Q = It = 20 x 6 x 60 x 60 = 432000 C Amount of electrons = 432000 / 96500 = 4.477 mol (;) Amount of Fe = 4.477 / 2 = 2.238 mol Mass of Fe = 2.238 x 55.8 = 125 g (;) (b) Fe2O3 is the main constituent of the rust found on household items. The standard electrode potentials involving the iron containing species can be found in Table 1.1. Table 1.1 Half-equations Eθ/ V 3Fe2O3 + H2O +2e ⇌ 2OH– + 2Fe3O4 –0.62 Fe2O3 + 3H2O + 6e ⇌ 2Fe + 6OH– –0.88 Fe3O4 + 4H2O + 8e ⇌ 3Fe + 8OH– –0.91 A layer was coated around the rusty tool after using the DIY kit. With reference to the standard electrode potentials given in Table 1.1, explain fully whether the layer coated is iron. [2] Using data from the question, 3Fe2O3 + H2O +2e 2OH– + 2Fe3O4 Eθ = –0.62V Fe2O3 + 3H2O + 6e 2Fe + 6OH– Eθ = –0.88V The layer is Fe3O4. Since EºFe2O3/Fe3O4 is more positive/less negative than EºFe2O3/Fe, Fe2O3 will be reduced to Fe3O4 and not to Fe. (;) Fe3O4 + 4H2O + 8e 3Fe + 8OH– Eθ = –0.91V It is not possible to regenerate pure iron as the reduction potential EºFe3O4/Fe is the least positive/most negative, thus Fe3O4 will not be further reduced. (;) Hence, the layer coated is Fe3O4. (c) The metal alloy that made up the main components of stainless steel are transition metals. (i) Define transition metal. [1]

4 A transition element is a d-block element which forms at least one simple ion , in compounds, with a partially filled d-subshell. (ii) State a physical property of transition metals which makes them a suitable material used for household tools compared to main group metals. [1] They have high densities (iii) Iron(III) ions catalyse the reaction between I- ions and S2O82- ions through homogeneous catalysis. Explain the term homogeneous catalyst. [1] The catalyst is in the same phase as the reactants and it speeds up the rate of reaction by lowering the activation energy without itself being chemically changed at the end of the reaction. . (iv) By considering relevant Eo values from the Data Booklet, and using balanced equations, determine whether chromium(III) ions are suitable catalyst for the reaction between I– ions and S2O82- ions. [2] No it is not suitable as the Ecell for one of the step is less than zero, making the reaction non-spontaneous. (;) 2Cr3+ + 2I- → I2 + 2Cr2+ Ecell = -0.41 – (+0.54) = -0.95 V(;) 2Cr2+ + S2O82- → 2Cr3+ + 2SO42- Ecell = +2.01 – (-0.41) = +2.42 V (v) Describe the structure and bonding in the element chromium. Draw a diagram to illustrate your answer. [2] (;) Cr has giant metallic structure held by strong electrostatic forces of attraction between (positive) chromium ions and ‘sea’ of delocalised electrons. (;) (d) Chromium(III) oxide has the same acid-base behaviour as aluminium oxide. (i) State the type of oxide chromium(III) oxide exists as in terms of its acid-base behaviour. [1] Amphoteric oxide

5 [Turn over (ii) By quoting relevant data from the Data Booklet, explain why c hromium(III) oxide and aluminium oxide have the same acid-base behaviour. [2] Cr3+ has an ionic radius of 0.062 nm while Al3+ has an ionic radius of 0.050 nm (;) while both have the charge of +3. As the charge density of the cations are similar(;), they have similar degree of covalent character, hence behave similarly. (iii) Use of Data Booklet is relevant to this question. Describe the observations when dilute NaOH is slowly added to the solution containing chromium(III) ions until in excess. [2] Grey-green precipitate is formed which dissolves in excess to give a dark green solution. [Total: 20] 2 (a) An aqueous solution of chlorine dioxide undergoes disproportionation in sodium hydroxide to form chlorate(V) ion, ClO3– , and hypochlorite ion, C lO–. The reaction is catalysed by OBr– ion. 4ClO2(aq) + 4OH–(aq) → 3ClO3–(aq) + ClO–(aq) + 2H2O(l) To study the kinetics of this reaction at 25 °C, initial rates were measured using various concentrations of ClO2, OH– and OBr–. experiment [ClO2]/ mol dm