TJC_H2_CHEM_P3_Qn_&_ans Prelim

1 2016 TJC H2 Chemistry Preliminary Exam [Turn Over CHEMISTRY 9647/03 Paper 3 Free Response 13th September 2016 2 hours Candidates answer on separate paper. Additional materials: Answer paper Graph Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, Civics Group, Centre number and Index number in the spaces provided on the cover page and on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, glue or correction fluid. Answer any four questions. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for good English and clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 15 printed pages.

2 2016 TJC H2 Chemistry Preliminary Exam [Turn Over Answer any four questions 1 Ethanedioic acid is a substance found in many plant foods. Cabbage is among the plant foods with high ethanedioic acid content. However, its anion, C2O42- can bind to iron to form iron (II) ethanedioate, which renders much of the iron in cabbage unusable by the body. (a) 50.0 cm3 sample of iron(II) ethanediaote, FeC2O4 was extracted from 300 g of cabbage, diluted in water and the solution made up to 250 cm 3. A 25.0 cm3 portion of this solution was acidified and required 26.90 cm 3 of 0.0100 mol dm -3 potassium manganate(VII) for oxidation of iron(II) to iron(III) and ethanedioate ions to carbon dioxide. (i) State the change in oxidation number for manganese and carbon in the reaction. [1] Mn: From +7 to +2 [Or decrease by 5] C: From +3 to +4 [Or increase by 1] (ii) Write down all the relevant ion-electron half equations and hence the overall redox equation for the reaction between potassium manganate(VII) and iron(II) ethanedioate. [2] MnO4- + 8H+ + 5e-  Mn2+ + 4H2O  C2O42-  2CO2 + 2e- Fe2+  Fe3+ + e-  3MnO4- + 24H+ + 5C2O42- + 5Fe2+  3Mn2+ + 5Fe3+ + 12H2O + 10CO2 (iii) Calculate the