IJC_2008_H2 Chemistry_Prelim II_Paper3 answers

Innova JC Preliminary Examination H2 Chemistry Paper 2 Mark Scheme © INNOVA 2008 Answer Scheme 9746/03 1(a) Ecell betw Cu & VO3 – = +1.00 – (+0.34) = +0.66V >0 (energetically feasible) E cell betw Cu & VO2+ = +0.34 – (+0.34) = 0V The reaction is in equilibrium. Hence, products and reactants are in equal proportion. Cu can reduce VO2+ to V3+. Colour changes from: VO 3 – (yellow) to VO2+ (blue) to Blue green (VO2+ + V3+) Balanced equation: 2 VO3 – + 8H+ + Cu Æ 2VO2+ + 4H2O + Cu2+ 2VO 2+ + 4H+ + Cu Æ 2V3+ + 2H2O + Cu2+ (b)(i) CuSO4(s) dissolves in water to give a blue solution containing [Cu(H2O)6]2+ ions. Cu2+ + 6H2O Æ [Cu(H2O)6]2+ (ii) With excess NH3(aq), CuSO4(aq) gives a deep blue solution due to formation of soluble complex ion, [Cu(NH3)4]2+. [Cu(H2O)6]2+ + 4NH3 Ù [Cu(NH3)4]2+ + 6H2O Blue soln deep blue complex NH3 is a stronger ligand than water, so it can replace water from [Cu(H2O)6]2+ to form [Cu(NH3)4]2+. (c)(i) CH2OH [ ]1M (ii) CH3CH2OH + OH– Æ CH3CH2O– + H2O [ ]1M (iii) The 9nucleophile for compound G is OH–. Compound H: 9nucleophile is CH3CH2O–. (iv) Test: add NaOH(aq) to each sample & heat. Then add excess HNO3(aq), followed by AgNO3(aq). Obs: Compound F: white ppt of AgCl Bromobenzene: no white ppt. (d)(i) Reagents/conditions: PBr 3, heat Or HBr, reflux. (ii) Compound K: (CH 3)2CHCN Compound L: (CH3)2CHCOOCH2CH3

2 © INNOVA 2008 Answer Scheme 9746/03 (e)(i) Discharging (using it as a galvanic cell) Eqm (1) PbO 2(s) + 4H+(aq) + 2e Ù Pb2+(aq) + 2H2O(l) +1.47V Eqm (2) Pb2+(aq) + 2e Ù Pb(s) –0.13V Overall : PbO 2(s) + 4H+(aq) + Pb(s) Æ 2Pb2+(aq) + 2H2O(l) Ecell = E , (PbO2/Pb2+) – E , (Pb2+/Pb) = +1.47 – (–0.13) = +1.60 V > 0 (energetically feasible) (ii) SO 4 2-(aq) + Pb2+(aq) Æ PbSO4(s) The precipitation of PbSO4(s) reduced the [Pb2+(aq)] which caused eqm (1) to shift to the right, Ered (PbO2/Pb2+) is more positive than +1.47 V and eqm (2) to the left. Eoxid (Pb2+/Pb) is more negative than –0.13 V. As a result, the overall E cell increases to 2.0 V. 2 (a) K c = [] 3 2 − − ⎡⎤⎣⎦ ⎡⎤⎣⎦ I I I mol-1 dm3 (b) I - (aq) + I 2 (aq) I 3 - (aq) Initial amt 0.058 0.080 0 Change –0.0375 –0.0375 +0.0375 Eqm amt 0.0205 0.0425 +0.0375 Eqm conc 0.0205/0.5 0.0425 / 0.5 0.0375/0.5 Kc = [] 3 2 − − ⎡⎤⎣⎦ ⎡⎤⎣⎦ I I I = 0.0375 0.5 0.0205 0.0425 0.5 0.5 ⎡⎤ ⎢⎥⎣⎦ ⎡⎤ ⎡⎤ ⎢⎥ ⎢⎥⎣⎦ ⎣⎦ = 21.5 mol-1 dm3 (c) (i) Amount of iodine in toluene lay