AJC H2 Chem 2012 Prelim P3 Soln

2 2012 H2 Chemistry Preliminary Examination Paper 3 Solutions 1 (a) (i) 4FeCr2O4 + 8Na2CO3 + 7O2 2Fe2O3 + 8Na2CrO4 + 8CO2 a = 8, b = 7, c = 2, d = 8, e = 8 1 Any dilute acid e.g. dilute H2SO4 or dilute HCl (b) voltmeter electron flow V salt H2 at 25 °C & 1 atm Pt electrode bridge platinised Pt electrode 1.00 mol dm–3 Cr2O7 2– (aq), 1.00 mol dm–3 H+ (aq) 1.00 mol dm–3 Cr3+ (aq), 1.00 mol dm–3 H+ (aq) correct components of each half–cell direction of electron flow (c) Zn2+ + 2e Zn –0.76 V Cr3+ + e Cr2+ –0.41 V Zn reduces Cr3+ to blue Cr2+(aq); E ocell = +0.35 V > 0, 2H+ + 2e H2 0.00 V The acid present in the solution oxidises blue Cr2+ back to green Cr3+; Eocell = + 0.41 V, producing H2 gas. (i) Mole ratio of CrOx n– : Cr2O7 2– = 7.5 x 10–3 : 2.5 x 10–3 = 3 : 1 Thus, 3 mol of CrOxn– disproportionate to 1 mol of Cr2O72– and 1 mol of Cr3+ Let O.N. of Cr be y. y – 3 = 2(6) – 2y y = 5 O.N. of Cr in the ion CrOxn– = +5 (e) (i) Fe3+ + e Fe2+ +0.77 V Cr3++ e Cr2+ –0.41 V Eocell = +1.18 V Fe3+ + Cr2+ Fe2+ + Cr3+

2 (ii) When cell discharges, iron half –cell becomes less positive and the chromium half–cell becomes more positive. Passage of current would stop. Membrane a llows anions to pass through to maintain electrical neutrality in the two half– cells. (not allowing the positive ions to pass through, otherwise self–discharge or short circuit occurs) [2] (iii) Eocell = Eored – Eoox 1.59 = +0.40 – Eoanode Eoanode = –1.19 V Anode: Zn + 4OH– Zn(OH)42– + 2e (or Zn + 2OH– Zn(OH)2 + 2e) (iv) Q = 0.8 x 1.9/65.4 x 2 x 96500 = 4486 C Current = 4486 / (30 x 24 x 60 x 60) = 1.73 x 10–3 A (v) ICB is expected to discharge faster as it involves simple electron transfer reactions, while energy is needed to break covalent bonds in O2 (and H2O) in the zinc–air battery. (vi) Zn–air battery has higher energy density as it uses air as the oxidising agent, unlike the heavier oxidising agent, MnO 2, used in the alkaline battery. Thus, more zinc can be packed within a cell of similar weight. 2 (a) (i) It is the amount of energy evolved when 1 mo le of solid ionic compound is formed from its constituent gaseous ions at 298 K and 1 atm. (ii) I: As the cationic radius increases down the group, inter–ionic distance increases OR state that L.E. a and cationic radius (r+) increases down the group. Thus, the strength of electrostatic forces of attraction between the M 2+ and SO42– ions decreases. Hence, the magnitude of L.E. decreases. II: DHhydration of the cation a + + r q As the cationic radius increases down the group, OR the charge density of the cation decreases. Thus, the strength of ion– dipole interactions formed between M 2+ and water molecules decreases and hence DHhydration becomes less exothermic. (iii) DHsolution = – L.E. + DHhydration of M2+ + DHhydration of anion For Group II sulfates,