TMJC 2025 H2 Chem P4 Answer

CANDIDATE NAME Answers CIVICS GROUP H2 CHEMISTRY 9729/04 Paper 4 Practical 3 September 2025 2 hours 30 minutes Candidates answer on the Question Paper. This document consists of 19 printed pages and 3 blank pages. READ THESE INSTRUCTIONS FIRST Write your name and Civics Group in the spaces at the top of the page. Give details of the practical shift and laboratory where appropriate, in the boxes provided. 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 in the spaces provided on the question paper. The use of an approved calculator is expected, where appropriate. You may lose marks if you do not show your working or if you do not use appropriate units. Qualitative Analysis Notes are printed on pages 21 and 22. The number of marks is given in brackets [ ] at the end of each question or part question. Shift Laboratory For Examiner’s Use 1 / 10 2 / 19 3 / 17 4 / 9 Total / 55 TAMPINES MERIDIAN JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION

2 Tampines Meridian Junior College 2025 JC2 Preliminary Examinations H2 Chemistry Answer all questions in the spaces provided. 1 Qualitative analysis tests involving solid iron(III) chloride FA 1 is a solution containing iron(III) chloride, FeCl3.nH2O FA 2 is a solution containing an unknown L– ligand Carry out the following tests. Record your observations in Table 1.1. Unless otherwise stated , the volumes given in Table 1.1 are approximate and should be estimated rather than measured. Test and identify any gases evolved. If there is no observable change, write no observable change. Table 1.1 Test Observations Explanation (teacher’s notes) (a) (i) Add about 6 cm depth of FA1 into a test tube. Add aqueous silver nitrate in excess. Filter the mixture. Separate the filtrate into four test tubes for (iii), (iv), (v) and (vi). white ppt formed in yellow/orange solution White residue/ppt obtained Yellow/orange filtrate/solution obtained Cl– is present to form AgCl ppt [Fe(H2O)6]3+ complex is present (ii) Place the filter funnel containing the residue on a new test tube, add aqueous ammonia over the residue and collect the filtrate. To this filtrate, add nitric acid dropwise until no further changes. The white ppt was soluble/dissolves in aqueous ammonia to form a colourless/pale yellow/orange solution. With nitric acid, white ppt formed. White ppt remains insoluble in excess nitric acid. Cl– is present. [Ag(NH3)2]+ is formed. It should be a colourless solution but the ppt is covered with traces of Fe3+, hence the solution appears pale yellow. Reversal of equilibrium. HNO3 neutralise the NH3 and hence AgCl is formed again. (iii) To the first test tube containing the filtrate from (i), add 2 cm3 of hydrogen peroxide. Observe the mixture until no further changes are seen. Brown/orange solution formed. Effervescence observed. Gas evolved relights a glowing splint. O2 (g) is evolved. Brown solution turns back to yellow/orange on standing O2 evolved. FA 2 acts as a reducing agent; Fe3+ acts as an oxidising agent.

3 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry [Turn Over (iv) To the second test tube containing the filtrate from (i), add an equal volume of aqueous sodium hydroxide. Then add 2 cm3 of hydrogen peroxide. Observe the mixture until no further changes are seen. Red-brown ppt formed (More) effervescence observed. Gas evolved relights a glowing splint. O2(g) is evolved. Fe(OH)3 is produced. More O2 released as H2O2 exists as HO2– and have a greater reducing ability (gets oxidised more easily by remaining Fe3+). Solid Fe(OH)3 does not undergo reduction (v) To the third test tube containing the filtrate from (i), add 1 cm3 of FA 2. (Blood) red/black solution formed. [Fe(H2O)6]3+ undergoes a ligand exchange reaction to form [Fe(SCN)]2+. (vi) To the last test tube containing the filtrate from (i), add half a spatula of iron filings. Then add an equal volume of hydrochloric acid. Dispose the reaction mixture once observations are completed to prevent gas build up. No observable change (when Fe is added) Effervescence observed Gas evolved “pops” with a lighted splint. H2(g) is evolved Yellow/orange solution turns light green/colourless Comproportionation reaction occurred. Fe is oxidised to Fe 2+, while Fe3+ is reduced to Fe 2+, as shown by the colour change (Fe 2+(aq) is green) [5] (b) (i) With the aid of the following equilibrium equations, explain the observations obtained in Test (ii) when nitric acid was added to the filtrate. AgCl(s) Ag+(aq) + Cl–(aq) -- Equilibrium 1 Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) -- Equilibrium 2 ……………………………………………………………………………………………….. [1]     The white solid residue is AgC l. When NH 3(aq) is added, the colourless and soluble silver complex [Ag(NH3)2]+ is formed in the filtrate. When HNO3 is added, the H+ from the acid neutralises NH 3(aq), resulting in equilibrium 2 to shift left, hence reproducing Ag+ and the white solid AgCl in equilibrium 1 again.

4 Tampines Meridian Junior College 2025 JC2 Preliminary Examinations H2 Chemistry (ii) In the presence of a strong base, hydrogen peroxide reacts with OH– to form HO2–. electrode reaction EƟ/V Fe3+ + e– ⇌ Fe2+ +0.77 Fe(OH)3 + e– ⇌ Fe(OH)2 + OH– –0.56 O2 + 2H+ + 2e– ⇌ H2O2 +0.68 H2O2 + 2H+ + 2e– ⇌ 2H2O +1.77 O2 + H2O + 2e– ⇌ HO2– + OH– –0.08 With reference to the electrode potential given above, calculate the overall E Ɵ values for the reactions in Tests (iii) and (iv) respectively. Hence, account for the difference in the rate of effervescence produced when hydrogen peroxide was added in Tests (iii) and (iv). ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… …………….………………………………………………………………………….……… [2] (iii) Based on your observation in Test (v), state the type of reaction that has occurred. ………………………………………………………………..…………………………...…. [1] (iv) In Test (vi), a comproportionation reaction occurred, where two species of the same element undergoes a redox reaction to form the same product. Write an ionic equation to represent the reaction. ………………………………………………………………..…………………………...…. [1] [Total: 10] EƟ for (iii) = 0.77 – 0.68 = +0.09 V EƟ for (iv) = 0.77 – (–0.08) = +0.85 V (Fe(OH)3(s) does not undergo reduction. If Fe(OH)3 values are used, the resulting Eo values will be negative, which goes against any observable reaction.) The effervescence is more vigorous for step (iv) The overall EƟ value is more positive for step (i v) when hydrogen peroxide reacts with Fe3+ in alkaline medium, representing a more spontaneous reaction. OR As the EƟ (O2/ HO2–) is less positive than that of E Ɵ(H2O2/H2O), HO2– is a stronger reducing agent than H2O2 Ligand exchange reaction Fe + 2Fe3+ → 3Fe2+

5 Tampines Meridian Junior College 2025 JC2 Preliminary Examination H2 Chemistry [Turn Over 2 Determination of water of crystallisation in a hydrated iron(III) salt Iron(III) chloride solutions are often used as a coagulant to remove impurities in sewage and industrial waste. FA 3 is hydrated iron(III) chloride with the formula FeCl3.nH2O. The addition of excess zinc to a solution of FA 3 converts the Fe3+ ions to Fe2+ ions. The amount of Fe2+ ions can then be determined quantitatively by titration against a standard solution of potassium manganate(VII), KMnO4. The r