HCI Prelim H2 CHEM P2 ans

7 Paper 2 1 (a) (i) Cu(s) + 2Fe3+(aq) → Cu2+(aq) + 2Fe2+(aq) [1] (ii) Positive. More aqueous ions produced and thus more ways that energy can be distributed in the system through the motion of ions. [1] 1 (b) M1 [1/2] voltmeter and salt bridge [1/2] a single water bath or hot plate (not separate) M2 Correct Fe3+/Fe2+ half cell: [1/2] electrode + [1/2] solutions (with correct concentration) M3 Correct Cu2+/Cu half cell: [1/2] electrode + [1/2] solution (with correct concentration) M4 [1/2] 5.56 g of solid [1/2] details for weighing – electronic balance, weighing bottle/ beaker, reweigh or tare and rinse out all solid M5 Preparation of acidified FeSO4 solution: [1/2] adding acid before topping up to 100 cm3 mark [1/2] 100 cm3 volumetric flask M6 Preparation of FeSO4 solution:  Transfer solid to beaker  Dissolve solid in beaker  Transfer solution to volumetric flask

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 2  Transfer washings to volumetric flask  Top up volumetric flask  Shake [1/2] any 3 – 5 points [1] all 6 points M7 [1/2] measure volumes of solutions used for both half -cells (minimum volume of electrolyte in each half-cell: 20 cm3) [1/2] use appropriate apparatus e.g. measuring cylinder for measurement of solutions for both half-cells M8 [1/2] measure temperature using thermometer or set temperature of thermostatically controlled water bath [1/2] range of temperature used at least 30 ºC – should have at least 5 readi ngs with at least 5 ºC intervals (lowest temperature should not be below 20 ºC) M9 Sketch of graph: [1/2] Correct choice of axes – E vs T (ignore units) [1/2] Linear graph with positive gradient Allow a plot of ∆G vs T. In this case, a linear graph with negative gradient would be obtained. Candidate needs to explain clearly how ∆G can be obtained. M10 [1] explain how entropy change could be obtained from the graph – e.g. Gradient = ∆S/ nF (for E vs T graph) Gradient = ∆S (for ∆G vs T graph) Sample procedure 1. Using an electronic weighing balance, weigh accurately 5.56g of FeSO4.7H2O in a weighing bottle. 2. Transfer this solid into a 100 cm3 beaker. Reweigh the weighing bottle to account for any residual solid. Record the mass of the solid used. 3. Dissolve the solid using about 20 cm3 of 0.50 mol dm3 H2SO4(aq). 4. Transfer this solution into a 100 cm3 volumetric flask. Rinse the beaker thoroughly and transfer all washings into the flask. Top up to the mark with deionised water/H2SO4(aq) and shake well to obtain a homogeneous solution. Label this solution as FA1.

8 5. Transfer 15 cm3 of FA1 and 15 cm3 of 0.20 mol dm-3 FeCl3 into a 100 cm3 beaker using separate burettes. 6. Transfer 30 cm3 of 0.10 mol dm-3 CuSO4 into a 100 cm3 beaker using a burette. 7. Set up the apparatus as shown in the diagram above. 8. Ensure that the water bath is maintained at 25oC by measuring the temperature with a thermometer. 9. Record the cell potential using a voltmeter. 10. Record the cell potential at different temperatures: 30oC, 35oC, 40oC, 45oC, 50oC, 55oC. Plot a graph of E against T. ∆S = gradient of the graph x F x n 2 (a) (i) ½m each:  O atom has more protons (or higher nuclear charge than C atom)  shielding effect is similar (or almost the same)  O has higher effective nuclear charge  its outer electrons are pulled (or attracted) more closely to the nucleus (ii) 1m: S atom has 1 more occupied quantum shell than O and C. 2 (b) (i) 1m: blood red (ii) ½m each:  [Fe(CN)6]3– has (much) higher Kstab than [Fe(H2O)5(SCN)]2+  SCN– (and H2O) unable to replace CN – as the ligands attached to Fe3+ i.e. no ligand exchange (iii) 1m: [Fe(edta)]– has (much) higher Kstab than [Fe(edta)]2–. So [Fe(edta)]– is less likely to be reduced to [Fe(edta)]2–. OR reduction equilibrium [Fe(edta)]– + e– ⇌ [Fe(edta)]2– lies more to the left. E T

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 OR edta ligand stabilises +3 oxidation state of Fe relative to +2 state.  reduction potential for Fe(III)/Fe(II) with edta as ligands or for [Fe(edta)]–/[Fe(edta)]2– would be lower than +0.77 V. 2 (c) (i) ½m each:  yellow solution: [CuCl4]2–  blue solution: [Cu(H2O)6]2+  pale blue ppt: Cu(OH)2 or [Cu(OH)2(H2O)4]  dark blue solution: [Cu(NH3)4(H2O)2]2+ (ii) 1m each: Test II [CuCl4]2– + 6H2O  [Cu(H2O)6]2+ + 4Cl– Test III [Cu(H2O)6]2+ + 4NH3  [Cu(NH3)4(H2O)2]2+ + 4H2O Accept  2 (d) (i) ½m each:  Eo(H2O/H2) = –0.83 V and Eo(Al3+/Al) = –1.66 V or Eo(H+/H2) = 0 V and Eo(Al3+/Al) = –1.66 V  Eo(H2O/H2) is higher or less negative, so H2O would be preferentially reduced to H2, Al3+ would not be reduced to give Al. or Eo(H+/H2) is higher or less negative, so H+ would be preferentially reduced to H2, Al3+ would not be reduced to give Al. (ii) 1m: 2AlO2–  Al2O3 + ½O2 + 2e– or 4AlO2–  2Al2O3 + O2 + 4e– (iii) 1m: calculates no. of moles of electrons 1m: calculates time in hours Amount of Al deposited = 1000×103 27 = 3.704  104 mol Amount of electrons passed = 3  3.704104 = 1.111  105 mol Q = It = 1.111105  96500 = 1.072 1010 C t = 1.072×1010 104 = 1.072  106 s = 298 h (or 297.8 h) 2 (e) 1m: concludes +6 1m: calculates 0.00225 mol XeF2 and 0.00150 mol Cr(III) 1m: explains how to conclude +6, e.g. 1 mol Cr(III) loses 3 mol electrons Sample working:

9 Amount of XeF2 = 15.2 1000  25 169 = 0.002249 mol XeF2 reduced to Xe, oxidation state of Xe changes from +2 to 0 Amount of electrons transferred = 2  0.002249 = 0.004497 mol Amount of Cr(III) = 10 1000  0.150 = 0.00150 mol 0.004497 0.00150 = 3 mol of electrons lost from 1 mol of Cr(III)  oxidation state of Cr increases by 3 units  oxidation state of Cr is +6 after reaction 3 (a) The carbon atom of the reactant (but-2-ene) in the CC bond is sp2 hybridised [1/2] and the bond angle around the carbon atom is 120o [1/2]. The carbon atom of the product (butane) in the CC bond is sp3 hybridised [1/2], and the bond angle around the carbon atom is 109.5o. (accept 109o) [1/2] (b) (i) Steam, 300C, 70 atm (or 60 atm), (conc) H3PO4 [1] OR conc H2SO4 followed by warming with water. A : [1] (ii) Both but-2-ene and alcohol are simple covalent compound s /simple, discrete molecules /have simple molecular structure. [1/2] For but -2-ene, there are weak dispersion forces between its molecules. For alcohol A (butan-2-ol) with polar OH bond, there are strong intermolecular hydrogen bonding between its molecules. [1/2] More energy is required to overcome the stronger intermolecular hydrogen bonding in alcohol A than the weak dispersion forces in but- 2-ene which is therefore in gaseous state. [1/2] OR the energy provided by room temperature is insufficient to overcome the strong hydrogen bonds between alcohol A molecules,

2016 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 Hence alcohol A exists as a liquid at room temperature and but-2-ene exists as gas. [1/2] (iii) B: [1] C: [1] 3 (c) (i) Ratio of 1-bromobutane : 2-bromobutane = 1 : 3 [1] In 1-bromobutane: there are 6 possible primary (1 o) H atom s for substitution and in 2 -bromobutane: there are only 4 secondary (2 o) H atoms for substitution [1/2] Relative rate of substitution suggests that the mole ratio of 1-bromobutane : 2-bromobutane = 1 x 6 : 4.5 x 4 = 6 : 18 [1/2] for working (ii) Type of mechanism: Free radical substitution [1] uv light Initiation: BrBr 2 Br [1/2] Propagation: [1]  CH3CH2CH2CH3 + Br