JJC H2 Chem 2012 Prelim P3 Soln

Page 1 Mark Scheme for 2012 JC2 Preliminary Examination Paper 3 Minus [1m] overall for P3 for lack of 3 sf. Indicate on Cover page when penalised. General Comments for Q1 Many candidates probably left this question as their last attempted question, as they ran out of time and thus (i) did not manage complete answers or (ii) did not have enough time to calculate their answers or (iii) made a lot of careless mistakes. Marks for this question varies greatly from 4-5 to 17-20. 1. (a) (i) Anode: 2H2(g) ® 4H+(aq) + 4e- Cathode: O2(g) + 4H+(aq) + 4e- ® 2H2O(l) [1m] s.s not penalised [1m] s.s. not penalised · Small handful of candidates still make the usual errors: o Use of reversible arrows o Use of [O] and [R] o Wrong electrodes or no labelling of electrodes · A few wrote equations for alkaline medium · A few chose half eqn for H2O2 instead. (ii) · Requires a heavy storage tank of H2 gas on board a car / takes up more storage space / must be kept under high pressure · Difficult to refill H2 fuel due to lack of infrastructure · H2 gas is MORE explosive than fuels in internal combustion engines. · More expensive due to the expensive catalysts etc · Explosive at high temperatures [1m] for any relevant ans. Do not accept: “explosive” only Common Mistakes: · A few listed advantages instead · A small handful of candidates did not understand what to compare e.g. thinking that the hydrogen-oxygen fuel cell is used together with the internal combustion engine or to supply electricity to the internal combustion engine. (iii) Using 12 2 11 12 2 , PP P PTTT T æö æö= == ç÷ ç÷èøèø 53.29 ×10 373293 Pa= 54.19 ×10 [1m]: correct subst into formula. Accept 293.15 and 373.15 [1m]: correct ans, 3 s.f. · Many candidates solved using PV=nRT, which is a longer method. Common Mistake: · Wrong units for Temperature · Some gave a qualitative answer only, with no calculation.

Page 2 1. (b) (i) Reference electrode (cathode): Hg2Cl2(s) + 2e à 2Hg(l) + 2Cl-(aq) [1m] full arrow (ii) Measuring electrode (anode): H2(g) à 2H+ (aq) + 2e Overall: Hg2Cl2(s) + H2(g) à 2Hg(l) + 2Cl-(aq) + 2H+ (aq) [1m] full arrow Common Mistakes for (i) and (ii): · Did not seem to understand Qn or were simply confused with ‘Hg’ and ‘H’ e.g. o Mixed up ‘Hg’ with ‘H2’ or ‘Hg’ with ‘H’ ! o wrote reduction of chlorine to chloride only o wrote eqns for reaction between hydrogen and chlorine only · Did not notice formula of “Hg2Cl2”. · Did not check their equation for balancing of charges. (iii) At a higher pH, there is lower [H+], · equilibrium in 2H+ (aq) + 2e = H2(g) shifts t o the left (OR eqm shifts to favour oxidation of H2) and thus · E(H+/H2) becomes (more) negative. Thus Ecell becomes more positive. [1m] No penalty for “E,”. Just circle. Common Mistakes: · Thought “higher pH” = “higher [H+]” · Did not write reversible arrow for the eqm equation. · Thought that a reaction between H+ & Cl- will result in an eqm shift! · Writing “E value” to represent E(H+/H2). (c) (i) ( ) 22 0 2 0.0592 [ ] [ ]log 0.05920.41 0.28 2log[ ]2 0.13log[ ] 0.0592 cell H H ClEE nP H H +- + + æö=- ç÷ èø =- ´ =- +20.05920.41= 0.28 - log [H ]2 Solving, [H+] = 6.37 ´ 10-3 (or 6.38 ´ 10-3) mol dm-3 pH = -log(6.37 ´ 10-3) = 2.20 [1m] correct substitution & n=2 ecf for n from (b) ß need work from both half eqns [1m] 3sf, ecf [1m] 3sf, ecf · About half of the candidates (who attempted this part) are able to solve and calculate [H+], using whatever values substituted. Common Mistakes: · Substituted n = amount of acid used (or any other strange number) · Substituted their answer in (a)(iii) for PH2 · Substituted “1.01 x 105” for PH2

Page 3 1. (c) (ii) Ka 3[ ] (6.37 10 ) 0.104c +- ´== 22H = 3.90 ´ 10-4 mol dm-3 OR Ka 3 3 [ ] (6.37 10 ) [ ] 0.104 6.37 10c +- +- ´== - -´ 22H H = 4.15 ´ 10-4 mol dm-3 (or 4.16 ´ 10-4 mol dm-3) If students get [H+] = 6.38 ´ 10-3 mol dm-3, Ka 3 3 [ ] (6.38 10 ) [ ] 0.104 6.38 10c +- +- ´== - -´ 22H H = 4.17 ´ 10-4 mol dm-3 Ka 3[ ] (6.38 10 ) 0.104c +- ´== 22H = 3.91 ´ 10-4 mol dm-3 [1m] for Ka, 3sf, ecf · Most candidates earned the ecf marks. · Most candidates (who got this part correct) solved for Ka using [] aK c + = 2H · A few students got all previous calculations correct but were stuck at K a. Perhaps they did not notice that mandelic acid is a monobasic acid. (iii) Mandelic acid, being the stronger acid, has a larger Ka value. The electron withdrawing O atom of the - OH group disperses the negative charge and stabilises CH(OH)COO- , making CH(OH)COO- more stable than CH2COO- . [1m] Not awarded if mandelic acid is deduced as weak acid (contradiction) [1m] Anion must be clear Accept: - “electron withdrawing -OH group” - “electronegative” Common Mistakes: · Electron donating -OH group · Weaker acid has a larger Ka · ‘stabilising negative charge’ · ‘mandelic anion’ or ‘anion’ or ‘-COO- anion’ · Using ‘it’ to describe the acid, the anion, the -OH group (basically everything!) and thus making answers vague/unclear/misleading. (iv) · Add acidified K2Cr2O7 to each of the unknowns and heat in a hot water bath. (Accept “hot acidified K2Cr2O7) · Mandelic acid turns orange K 2Cr2O7 green but not phenylacetic acid. [1m] Circle if ‘heat under reflux’ but no penalty [1m] Common Mistakes: · Using KMnO4 · Using KMnO4 then Brady’s reagent · Using KMnO4 (or K2Cr2O7), followed by Tollens’ reagent. · Using PCl5

Page 4 1. (c) (v) CH(OH)COOH CH2OH CH(OH)CNCHO OR the less recommended way: CH(OH)COOH CH2OH CHOH Cl CHOH CN [1m] for each correct intermediate structure 2(ü): 1m ; 3(ü): 2m , only awarded if corresponding structures are correct Award marks proportionately, for any other long synthesis method suggested. 1m max for any haywire synthesis method but correct cyano-intermediate & reagent and conditions for hydrolysis to get the end product -1m if answers are correct but not clearly presented · Have to remind students that Free Radical Substitution is not a recommended way of synthesis. It gives a mixture of products. Common Mistakes: · Some only listed the sequence of reagents & conditions required w/o giving intermediates for ‘synthetic route’ · Suggested step to convert -OH group to -CN group ß(MANY STUDENTS) · Suggested step to convert -OH group to -COOH group ß(MANY STUDENTS) · Using “KMnO4/K2Cr2O7 and heat under reflux” to oxidise alcohol to aldehyde · Confused on when to use HCN and NaCN (for nuclephilic sub & addition) · Listed a mixture of two acids for hydrolysis e.g. ‘H2SO4, HCl, heat under reflux’ · Did not use ‘dilute’ or ‘aq’ for acid hydrolysis (i.e. simply “H2SO4, heat”) · Used ‘conc’ acid for hydrolysis. Acidified K2Cr2O7(aq), distil (ü) HCN, trace amt of NaCN/base (ü) dilute H2SO4, heat or H2SO4(aq), heat or dilute HCl, heat or HCl(aq), heat (ü) Limited Cl2, uv light (ü) NaCN in ethanol, heat under reflux (ü) dilute H2SO4, heat or H2SO4(aq), heat or dilute HCl, heat or HCl(aq), heat (ü)

General Comments for Q2 Part (a) was generally well attempted and the average score is 7 out of 12. Students generally lost most marks in (a)(i)(ii)(vi). Part (b) was poorly attempted and the average score is 3 out of 8. Students generally scored all marks in (b)(iii)(v). For details, please see the comments under individual subparts. 2. (a) (i) Menthone is expected to be more volatile. (or Menthol is expected to be less volatile) Less energy is required to overcome the * weaker permanent dipole- permanent dipole interaction **between menthone molecules than the *stronger hydrogen bonds **between menthol molecules. Reject: · J