VJC 2021 Prelim H2P4 Ans [FInal]

1 Ó VJC 2021 9729/04/PRELIM [Turn over Victoria Junior College 2021 H2 Chemistry Prelim Exam 9729/4 Suggested Answers 1 Determination the extent of oxidation of iron(II) solution exposed to air Read through the whole method before starting any practical work. Where appropriate, prepare a table for your results in the space provided. Show your working and appropriate significant figures in the final answer to each step of your calculations. It is essential to dissolve solid samples of iron(II) compounds in acids such as dilute sulfuric acid instead of water. This is to prevent excessive oxidation of iron(II) ions to iron(III) ions in the presence of oxygen over a period of time. FA 1 is 0.020 mol dm–3 acidified potassium manganate(VII), KMnO4. FA 2 is 1.0 mol dm–3 sulfuric acid, H2SO4. FA 3 is iron(II) salt dissolved in water and left standing for an extended period of time. Using the apparatus and chemicals provided, you are required to determine the extent of oxidation of iron(II) salt in FA 3. (a) Dilution of FA 3 1. Pipette 25.0 cm3 of FA 3 into a 100 cm3 volumetric flask. 2. Make up to the contents of the flask to the 100 cm3 mark with deionised water. Shake to obtain a homogeneous solution. 3. Label this solution diluted FA 3 which is to be used for (b) and (c). (b) Determination of amount of iron(II) ions originally present in FA 3. 1. Add 50 cm3 of diluted FA 3 using a measuring cylinder into a 100 cm3 beaker. 2. Add about 2 spatula of zinc powder into the beaker and stir. 3. Filter the mixture into a boiling tube. 4. Pipette 10.0 cm3 of the filtrate into a conical flask. 5. Add 10 cm3 of FA 2 into the conical flask using a measuring cylinder. 6. Titrate this solution against FA 1 from a burette until the first permanent pale pink colour is obtained. 7. Record your titration results, to an appropriate level of precision, in the space below. You are to perform this titration only once. Results: For both (b) and (c) table Correct headers with units; Tabulation of ALL burette readings to 2.d.p. Experiment 1 Final burette reading / cm3 19.00 Initial burette reading / cm3 0.00 Volume of FA 1 used / cm3 19.00 [2]

2 Ó VJC 2021 9729/04/PRELIM [Turn over (c) Determination of amount of iron(II) remaining in FA 3. 1. Pipette 10.0 cm3 of diluted FA 3 into a conical flask. 2. Add 10 cm3 of FA 2 into the conical flask using a measuring cylinder. 3. Titrate this solution against FA 1 from a burette until the first permanent pale pink colour is obtained. 4. Record your titration results, to an appropriate level of precision, in the space below. 5. Repeat the titration as many times as necessary to obtain consistent results. Results: Experiment 1 2 Final burette reading / cm3 13.00 13.00 Initial burette reading / cm3 0.00 0.00 Volume of FA 1 used / cm3 13.00 13.00 At least two consistent readings ±0.10 cm3 Accuracy [2m] [3] (d) Calculations (i) From your titration results in (c), calculate the average volume of FA 1 used. Average volume of FA 1 used = (13.00 + 13.00) ÷ 2 = 13.00 cm3 (2 d.p.) [1] (ii) Calculate the amount of iron(II) in 10.0 cm3 of diluted FA 3. MnO4–(aq) + 8H+(aq) + 5Fe2+(aq) ® Mn2+(aq) + 5Fe3+(aq) + 4H2O(l) Amount of MnO4– required = 13.00/1000 x 0.020 = 2.60 x 10–4 mol (3 s.f.) Amount of iron(II) in 10.0 cm3 of diluted FA 3 = 2.60 x 10–4 x 5 = 1.30 x 10–3 mol (3 s.f.) Amount of iron(II) in diluted FA 3 = 1.30 x 10–3 mol [1] (iii) Calculate the amount iron(II) in 25.0 cm3 of FA 3. Amount of iron(II) in 100 cm3 of diluted FA 3 = 1.30 x 10–3 x 𝟏𝟎𝟎𝟏𝟎.𝟎 = 1.30 x 10–2 mol Number of moles before dilution = Number of moles after dilution Amount of iron(II) in 25.0 of FA 3 = 1.30 x 10–2 mol (3 s.f.) Amount of iron(II) in FA 3 = 1.30 x 10–2 mol

3 Ó VJC 2021 9729/04/PRELIM [Turn over [1] (iv) Using your titration results in (b), calculate the amount iron(II) originally present in 25.0 cm3 of FA 3. Amount of MnO4– required = 19.00/1000 x 0.020 = 3.80 x 10–4 mol Amount of iron(II) originally present in 10.0 cm3 of diluted FA 3 = 3.80 x 10–4 x 5 = 1.90 x 10–3 mol Amount of iron(II) in 25.0 of FA 3 = Amount of iron(II) originally present in 100 cm3 of diluted FA 3 = 1.90 x 10–3 x 𝟏𝟎𝟎𝟏𝟎.𝟎 = 1.90 x 10–2 mol (3 s.f.) Amount of iron(II) originally present in FA 3 = 1.90 x 10–2 mol [1] (v) Calculate the percentage of iron(II) in FA 3 oxidised upon standing for an extended period of time. Amount of iron(II) oxidised = 1.90 x 10–2 – 1.30 x 10–2 = 6.00 x 10–3 mol Percentage of iron(II) oxidised = 𝟔.𝟎𝟎𝐱𝟏𝟎!𝟑𝟏.𝟗𝟎𝐱𝟏𝟎!𝟐 x 100 % = 31.6 % (3 s.f.) Percentage of iron(II) in FA 3 oxidised = 31.6 % [2] (e) Evaluations (i) State one assumption made for the addition of zinc powder in (b). Explain how the titre volume would be affected if the assumption is not valid. The assumption was that all of the Fe3+ was reduced to Fe2+ by the zinc powder. If some Fe3+ remained, less FA 1 will be needed to oxidise the Fe2+ present. Thus, titre volume in (b) would be lower. [2] (ii) With the help of relevant calculations, show that the iron(II) is more stable in air under acidic conditions compared to alkaline conditions. Fe3+ + e– ⇌ Fe2+ Eo = +0.77 V Fe(OH)3 + e– ⇌ Fe(OH)2 + OH– Eo = –0.56 V O2 + 4H+ + 4e– ⇌ 2H2O Eo = +1.23 V O2 + 2H2O + 4e– ⇌ 4OH– Eo = +0.40 V

4 Ó VJC 2021 9729/04/PRELIM [Turn over In acidic conditions, Eocell = Eored – Eoox = +1.23 – (+0.77) = 0.46 V In alkaline conditions, Eocell = Eored – Eoox = +0.40 – (–0.56) = 0.96 V for correct calculation of Eocell values. Since Eocell in acidic medium is less positive than that in alkaline medium, iron(II) undergoes oxidation less readily in acidic conditions i.e. iron(II) is more stable in acidic medium. [2] [Total: 15] 2 Determination of the activation energy of the reaction between ammonium peroxodisulfate and potassium iodide. Equation 1 represents the reaction between peroxodisulfate ions and iodide ions. equation 1 S2O82–(aq) + 2I–(aq) ® I2(aq) + 2SO42– The reaction is first order with respect to both the concentration of peroxodisulfate ions and the concentration of iodide ions. rate = k[S2O82–][I–] where k = rate constant When starch is added to the reaction mixture, a blue colour is immediately seen due to the formation of an iodine–starch complex. If a small amount of sodium thiosulfate, Na2S2O3, is also present in the reaction mixture, the formation of the blue colour is delayed. The Na2S2O3 reacts with I2 as shown in equation 2. equation 2 I2(s) + 2S2O32–(aq) ® S4O62–(aq) + 2I–(aq) The activation energy, Ea, for the reaction can be obtained using the Arrhenius Equation: k = A𝑒–"#$% where k is the rate constant, A is the pre-exponential constant R is molar gas constant and T is the reaction temperature in Kelvin, The activation energy, Ea, of the reaction can be determined by performing a series of experiments where the reaction mixture is kept constant, but different reaction temperatures are used, and the same end-point is timed. Plotting ln k against 1/TK will give a best-fit straight line where the gradient of the line is – Ea/R. FA 4 is 0.200 mol dm–3 potassium iodide, KI. FA 5 is 0.00500 mol dm–3 sodium thiosulfate, Na2S2O3.

5 Ó VJC 2021 9729/04/PRELIM [Turn over FA 6 is 0.100 mol dm–3 ammonium peroxodisulfate, (NH4)2S2O8. Solution S is starch solution. You are to perform a series of five experiments, at different temperatures, in which the concentrations of FA 4, FA 5 and FA 6 are kept constant. The time taken for the dark blue colour to form allows the reaction rate to be determined. (a) Procedure You will attempt five experiments. In each ex