20   9 This question is about silver compounds. (a) The diagram shows a Born–Haber cycle for the formation of silver(I) oxide, Ag2O. All quantities are measured in kJ mol–1. 2Ag+(g) + Ag2O(s) O2–(g) +798 O(g) 2Ag(s) + ½O2(g) ΔatH d(Ag(s)) × 2 +731 × 2 –141.1 –31 +249.2 –2969 (i) Complete the diagram by adding appropriate species and state symbols to the empty boxes. (2) (ii) Explain why the value for the first electron affinity of oxygen is negative and the value for the second electron affinity is positive. 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Turn over 21   (iii) Calculate a value for the standard enthalpy change of atomisation of silver, ΔatH d, using the Born–Haber cycle. (3) 22   (b) Another silver compound is silver chloride, AgCl. Values for its lattice energy can be found by experiment or by theoretical calculation. Compound Experimental lattice energy / kJ mol–1 Theoretical lattice energy / kJ mol–1 Silver chloride –905 –833 (i) Give two assumptions used in the model to calculate the theoretical lattice energy. 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(ii) Explain the difference in the two values for the lattice energy of silver chloride by considering the possible bonding models. 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(Total for Question 9 = 13 marks) Turn over 23   10 Manganese compounds can be used to determine the amounts of dissolved molecular oxygen in water samples. (a) Draw the dot‑and‑cross diagram for an oxygen molecule, O2 . Show outer shell electrons only. (1) (b) The solubility of oxygen in water under standard conditions is 1.22 × 10–3 mol dm–3. Comment on this value by considering the type and strength of the intermolecular forces in • pure water • pure oxygen • a mixture of water and oxygen. Detailed descriptions of the forces involved are not required. (4) .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... .................................................................................................................................................................................................................................................................................... 24   (c) The amount of dissolved oxygen in a sample of river water was found using the process outlined. • excess alkaline manganese(II) sulfate, MnSO4 , was added to a 150 cm3 sample of river water • the Mn2+ ions reacted with the dissolved oxygen forming a precipitate of manganese(IV) oxide hydroxide 2Mn2+(aq) + O2(aq) + 4OH–(aq) → 2MnO(OH)2(s) • the precipitate was then dissolved using excess sulfuric acid, forming Mn4+(aq) ions MnO(OH)2(s) + 4H+(aq) → Mn4+(aq) + 3H2O(l) • excess potassium iodide solution was then added, forming iodine Mn4+(aq) + 2I–(aq) → Mn2+(aq) + I2(aq) • the liberated iodine was then titrated with sodium thiosulfate solution, Na2S2O3(aq), of concentration 0.00518 mol dm–3 I2(aq) + 2S2O3 2–(aq) → 2I–(aq) + S4O6 2–(aq) • the mean volume of the titre of Na2S2O3(aq) was 34.20 cm3. (i) Calculate the concentration of dissolved oxygen in the sample of river water, in g dm–3. (5) 25   You may use this space to continue your answer to 10(c)(i). (ii) The concentration of oxygen in water is often expressed in parts per million (ppm), where 1 ppm equals 1 g of solute dissolved in 1 × 106 g of solvent. Calculate the concentration of the oxygen in the sample of river water in ppm. Assume the density of the river water is 1.00 g cm–3. (1) Turn over 26   (d) Some data is shown for electrode systems involving the Mn3+(aq) ion. Half‑cell Electrode system E d / V A MnO2(s) + 4H+(aq) + e–  Mn3+(aq) + 2H2O(l) +0.95 B Mn3+(aq) + e–  Mn2+(aq) +1.51 Explain why Mn3+ ions are unstable in aqueous solution. Include an equation and the type of reaction that occurs. 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(Total for Question 10 = 15 marks) TOTAL FOR PAPER = 90 MARKS 27   BLANK PAGE 28  