MI Prelim H2 Chem P3 Suggested answers for exchange

Class Adm No Candidate Name: 2025 Preliminary Examination Pre-University 3 H2 CHEMISTRY 9729/03 Paper 3 Free Response 15 Sep 2025 2 hours Candidates answer on separate paper. Additional materials: Data Booklet Suggested answers H

2 Section A Answer all the questions in this section. 1 Copper and iron are transition metals that form a wide array of complexes due to their ability to exhibit multiple oxidation states. Copper and iron complexes showcase diverse chemical and biological roles, ranging from catalysis to electrochemical processes. (a) (i) State two physical properties of transition metals which differ from that of main group elements. [2] They have high melting and boiling points. They have higher densities. (ii) When copper(I) oxide solid, Cu2O was added to sulfuric acid, H2SO4(aq), and warmed, a blue solution and a pink solid were observed. Identify the blue solution and the pink solid and write a balanced chemical equation for the reaction. [2] Blue solution: CuSO4 - Pink solid: Cu [1] Cu2O + H2SO4 → Cu + CuSO4 + H2O [1] (iii) Ethylenediaminetetraacetic acid (EDTA) is a compound used in chelation therapy. Suggest why [Cu(EDTA)]2− and [Cu(NH3)4(H2O)2]2+ complexes have different colours. [1] Presence of ligands in the solution cause the degenerate 3d orbitals to split into two different energy levels. Different ligands split the d-orbitals to different extents. (b) A sample of copper(II) sulfate solution was added to an excess o f aqueous potassium iodide to make a 250 cm3 solution. 2CuSO4 + 4KI ⟶ 2CuI + I2 + 2K2SO4 The amount of iodine produced can be found by titrating a sample of this solution with sodium thiosulfate, Na2S2O3, solution. 25.0 cm3 of the iodine-containing solution required 20.00 cm3 of 0.10 mol dm−3 sodium thiosulfate solution for complete reaction. I2 + 2S2O32– ⟶ S4O62– + 2I– Calculate the amount of copper(II) sulfate present in the original sample. [2] Comparing mole ratio, 2 mol S2O32– : 1 mol I2 2 mol Cu2+ : 1 mol I2 Hence, 1 mol Cu2+ : 1 mol S2O32– Amount of Cu2+ in 25 cm3 = (20/1000) x 0.10 = 2.00 x10-3 mol ; Amount of Cu2+ in the original solution = (250/25) x 2.00 x10-3 = 2.00 x 10-2 mol ;

3 [Turn over (c) Chromium is also a transition metal. Potassium dichromate(VI), K2Cr2O7, is commonly used as an oxidising agent in organic chemistry reactions. For example, a solution of K2Cr2O7 can be used to oxidise 1-butanol to butanoic acid. (i) State the oxidation number of the underlined carbon in 1-butanol, CH3CH2CH2CH2OH. [1] -1 (ii) By writing the oxidation and reduction half equations, construct an ionic equation for the reaction between potassium dichromate(VI) and 1-butanol in an acidic solution. [3] [R]: Cr2O72–+ 14H+ + 6e– → 2Cr3+ + 7H2O x2 [O]: CH3CH2CH2CH2OH + H2O → CH3CH2CH2COOH + 4H+ + 4e– x3 Overall Equation 3CH3CH2CH2CH2OH + 2Cr2O72– + 16H+ → 3CH3CH2COOH + 4Cr3+ + 11H2O A student proposed a reaction mechanism for the oxidation of 1-butanol and suggested that steps 1 and 2 of the mechanism are as follows: Step 1: Protonation of 1-butanol to form a better leaving group. Step 2: Breaking of the C−O bond to form a carbocation. (iii) Draw the structure of the protonated intermediate in step 1. [1] CH3CH2CH2CH2OH2+ (iv) It was found that the proposed mechanism was not feasible as the carbocation formed in step 2 was unstable. Suggest why the carbocation formed was unstable. [1] The carbocation in step 2 is unstable as it is a primary carbocation. There is only one electron donating group present hence the positive charge is less dispersed, resulting in an unstable carbocation. (d) A team of scientists is designing a power source for field sensors. The aim is to develop a sustainable electrochemical cell using abundant and recyclable materials. After consideration, the team narrows their choice of materials to zinc, copper, iron, and silver electrodes. (i) Propose a combination of two materials that will give rise to the most efficient electrochemical cell with the highest voltage. Explain your answer and justify by means of a calculation. [3] Zn – Ag electrochemical cell 1m Best oxidising agent: Ag⁺ (most positive E° value, +0.80 V) Best reducing agent: Zn (less positive E° value, –0.76 V) 1m E°cell = E°(cathode) – E°(anode) = +0.80 – (–0.76) = +1.56 V 1m

4 (ii) Draw a fully labelled diagram of the electrochemical cell proposed in (d)(i), indicating clearly the direction of electron flow. [3] 1m metal electrodes 1m electrolyte solution + salt bridge 1m direction of electron flow (iii) State the observations observed over time as the electrochemical cell is allowed to run. [1] The Zn electrode decreases in size (dissolves) and the Ag electrode thickens / increase in size as Ag metal deposits on it. (iv) Suggest one potential challenge the team of scientists might face should they proceed with large scale productions of the electrochemical cells proposed in (d)(i). [1] High production costs as Ag is expensive Or Environmental concerns – Ag / Ag+ is toxic [Total: 21]

5 [Turn over 2 The cracking of decane, C10H22, produces octane and ethene, as shown in the equation below. C10H22 → C8H18 + C2H4 Octane is a key chemical component of gasoline, while ethene is a common starting material for the synthesis of many two-carbon containing organic compounds. The straight-chain isomer of octane is known as n-octane and exists in the liquid state at room temperature. (a) Highly branched alkanes are favo ured in gasoline, because the branched structure makes them more resistant to ‘knocking’ in combustion engines compared to n-octane. This resistance to ‘knocking’ leads to smoother engine operation. (i) Draw the skeletal structure and state the IUPAC name of the most highly branched isomer of n-octane. [2] 2,2,3,3-tetramethyl butane or tetramethylbutane (ii) State and explain if the branched isomer in (a)(i) will have a higher or lower boiling point compared to n-octane. [2] The branched isomer will have a lower boiling point as it is more spherical, with lesser surface area for contact. Hence, less extensive instantaneous dipole–induced dipole forces of attractions exists between the branched molecules, therefore less energy is required to overcome the forces of attraction. (iii) Using bond energy values from the Data Booklet , calculate the enthalpy change of combustion of n-octane. [3] C8H18 + 25/2 O2 → 8CO2 + 9H2O ∆Hc of octane = bonds broken – bonds form = 7 (C-C) + 18 (C-H) + 25/2 (O=O) – [16 (C=O) + 18(O-H)] = 7(350) + 18(410) + 25/2(496) – [16(805) + 18(460)] = 16030 – 21160