TJC_H2_CHEM_P2_Qn_and_ans Prelim

1 2016 TJC H2 Chemistry Preliminary Exam [Turn over a s CANDIDATE NAME CIVICS GROUP / CENTER NUMBER S INDEX NUMBER CHEMISTRY 9647/02 Paper 2 Structured Questions 29th August 2016 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Civics Group, centre number, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. 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 21 printed pages. For Examiner’s Use 1 / 12 2 / 15 3 / 12 4 / 9 5 / 24 Paper 1 / 40 Paper 3 / 80 Total PRELIMINARY EXAMINATIONS HIGHER 2

2 2016 TJC H2 Chemistry Preliminary Exam [Turn over Answer all the questions 1 Planning (P) Potentiometric titration is a technique similar to direct titration of a redox reaction and can be used to determine the concentration of a reducing ion such as Fe 2+ without the use of an indicator. It involves measurement of E cell at intervals as an oxidising titrant is added. The titration is carried out on one side of a cell whose other half is a reference electrode which can be a standard hydrogen electrode or a calomel electrode. Consider the titration of Fe 2+ with standard Ce 4+, the course of which is monitored potentiometrically as shown in Figure 1 below. The reference electrode used is the calomel electrode shown in Figure 2. To follow the course of the reaction, a pair of electrodes is inserted into the reaction mixture. At the calomel reference electrode, Reference half reaction is: Hg2Cl2(s) + 2e 2Hg(l) + 2Cl- E = + 0.24 V At the suitable electrode, there are two reactions that will come to equilibrium: Fe3+ + e Fe2+ E = + 0.77 V Ce4+ + e Ce3+ E = + 1.70 V As the Ce4+ titrant is added, Fe2+ is oxidised to Fe3+ and the cell potential is controlled by the ratio of reduced and oxidised iron according to the Nerst equation which causes the potential to rise as more iron becomes oxidised. Ecell =  E (Fe3+/Fe2+) - 0.05916 log ][Fe ][Fe 3 2    – 0.24 ………….(1) Figure 2: A calomel electrode Figure 1: Set-u