Chemistry 9701/53 May/June 2019: question paper and mark scheme

Question

A student investigates the charge (z+) carried by aqueous manganese ions, Mnz+(aq). The electrochemical cell shown is set up for this investigation with the following two half-cells: ●● a standard copper(II) ion / copper half-cell (E o = +0.340 V) ●● a half-cell made from manganese and 0.500 mol dm–3 Mnz+(aq). (a) Label the items P and Q and state the concentration of the copper(II) ion solution in the copper half-cell. P …… copper manganese copper(II) ion 0.500 mol dm–3 solution aqueous Mnz+(aq) Q …… concentration of the copper(II) ion solution in the copper half-cell = ���������������������������������������� [1] (b) During the investigation the student plans to use solutions of Mnz+(aq) of lower concentration than 0.500 mol dm–3. (i) Calculate the volume of 0.500 mol dm–3 Mnz+(aq) needed to prepare 100.0 cm3 of 0.200 mol dm–3 Mnz+(aq). volume = …… cm3 [1] (ii) Describe how, using a 100 cm3 volumetric flask, the student should prepare exactly 100.0 cm3 of 0.200 mol dm–3 Mnz+(aq) using the volume of 0.500 mol dm–3 Mnz+(aq) calculated in (b)(i) and standard school or college apparatus. . …… . …… . …… . …… . …… [2] The cell potential of the electrochemical cell in (a) is measured. The 0.500 mol dm–3 Mnz+(aq) is then replaced by the 0.200 mol dm–3 solution and the cell potential is measured again. This is repeated for other lower concentrations of Mnz+(aq). All measurements are made at 25 °C. (iii) The results of the experiment are shown in the table. Complete column three of the table, calculating log[Mnz+] to two decimal places. Complete column four of the table, calculating E, the electrode potential of each manganese half‑cell, to three decimal places, using the equation shown. E (manganese half‑cell) = Ecell + 0.340 V electrode potential [Mnz+] cell potential, Ecell / V log[Mnz+] of each manganese / mol dm–3 half-cell, E / V 5.0 × 10–1 –1.529 –0.30 –1.189 2.0 × 10–1 –1.541 1.0 × 10–1 –1.550 7.5 × 10–2 –1.553 2.5 × 10–2 –1.567 8.0 × 10–3 –1.582 6.0 × 10–3 –1.590 4.0 × 10–3 –1.591 3.0 × 10–3 –1.594 5.0 × 10–4 –1.617 [2] (c) Plot a graph of electrode potential of manganese half‑cell (y‑axis) against log[Mnz+] (x-axis). Use a cross (×) to plot each data point. Draw a line of best fit. log[Mnz+] –3.0 –2.0 –1.0 0.0 –1.16 –1.17 –1.18 –1.19 –1.20 –1.21 electrode potential of manganese half-cell, E / V –1.22 –1.23 –1.24 –1.25 –1.26 –1.27 –1.28 [2] (d) (i) Circle the most anomalous point on your graph. [1] (ii) The student is careful to ensure that all solutions used are at the same temperature in all experiments. Suggest a possible explanation for the position of the anomalous point circled in (d)(i) relative to the line of best fit. . …… . …… . …… [1] (e) Your graph is a plot of E against log[Mnz+] and can be analysed using the Nernst equation at 25 °C. 0.059 E = Eo + log[Mnz+] z z is the value of the charge carried by the manganese ion E is the electrode potential / V E o is the standard electrode potential / V Use the Nernst equation and your graph to find the standard electrode potential, E o, of the manganese half-cell. E o = …… V [1] (f) (i) Determine the gradient of the graph. State the co‑ordinates of both points you used for your calculation. Record the value of the gradient to three significant figures. co‑ordinates 1 …… co-ordinates 2 …… gradient = …… [2] (ii) Use your answer to (f)(i) and the Nernst equation to calculate the value of z to three significant figures and give the formula of the manganese ion. Your calculation must show the use of the Nernst equation. (If you were unable to calculate an answer to (f)(i) you may use the value 0.0197. This is not the correct value.) z = …… formula of manganese ion = …… [2] (g) Lowering [Mnz+] causes the value of the electrode potential of the manganese half‑cell to become more negative. Suggest why this happens. . …… . …… [1] [Total: 16]

Question

A student plans to prepare propanone from propan‑2‑ol and test the product. Reagents provided to the student and some of their hazards are shown in the table. reagent hazard propan-2-ol flammable concentrated sulfuric acid corrosive potassium dichromate(VI) oxidising distilled water non-hazardous (a) (i) The full equation for the reaction between propan‑2‑ol and acidified potassium dichromate(VI) is shown. 3CH3CH(OH)CH3 + K2Cr2O7 + 4H2SO4 3(CH3)2CO + K2SO4 + Cr2(SO4)3 + 7H2O Calculate the minimum mass of potassium dichromate(VI) that is needed for complete oxidation of 5.00 g of propan‑2‑ol to propanone. Give your answer to three significant figures. [Ar: K, 39.1; Cr, 52.0; O, 16.0; C, 12.0; H, 1.0] mass K2Cr2O7 = …… g [2] (ii) The student is provided with a set of instructions to prepare the propanone. step 1 Add concentrated sulfuric acid to 5.0 g of propan‑2‑ol in a round‑bottomed flask, a few drops at a time. step 2 Dissolve the mass of potassium dichromate(VI) calculated in (a)(i) in a few cm3 of distilled water. Add this aqueous potassium dichromate(VI) slowly to the mixture in the step 3 round-bottomed flask. step 4 Heat the mixture under reflux. step 5 Separate the propanone from the reaction mixture using distillation. The student is also provided with the boiling points of propan‑2‑ol and propanone. compound boiling point / °C propan-2-ol 82.5 propanone 56.5 Complete the diagram to show how the propanone is separated from the reaction mixture in step 5. Label your diagram fully including the location of propan-2-ol and propanone after distillation has taken place. There is no need to include clamps. clamp heated collection water-bath flask round-bottomed reaction flask mixture [3] (iii) The reaction mixture needs heating for reflux to take place. Explain why a water-bath is used to heat the mixture. . …… . …… . …… [1] (iv) The propanone separated from the mixture in step 5 contains sulfuric acid as an impurity which needs to be removed. Name a reagent that could be added to remove the sulfuric acid and explain how the student would ensure that all of the acid is no longer present. reagent …… explanation …… . …… [2] (b) (i) When propanone reacts with a solution of 2,4-DNPH an insoluble compound, X, is produced according to the following equation. 2,4-DNPH CH3 N C H2N NH HN CH3 NO2 NO2 CH3COCH3 + + H2O NO2 NO2 The melting point of X can be used to confirm the identity of the carbonyl compound that has reacted with 2,4‑DNPH. To do this, solid X must be separated from the mixture. This can be done using method A or method B. Buchner funnel suction pump Buchner flask method A: gravity filtration using a filter funnel, filter paper and a conical flask method B: filtration under reduced pressure using a Buchner funnel and Buchner flask, filter paper and a suction pump to reduce the pressure in the Buchner flask Suggest one major advantage of using method B rather than method A. . …… . …… [1] (ii) The student places a washed sample of X in a drying oven for an hour. The student records the mass of X. The student wants to ensure that X is completely dry. Describe what the student should do to ensure that X is completely dry. . …… . …… [1] (iii) 5.00 cm3 of propanone reacts with an excess of 2,4‑DNPH. The mass of dry X produced is 11.84 g. Calculate the percentage yield of X in this reaction. Mr X = 238 Mr CH3COCH3 = 58 density CH3COCH3 = 0.789 g cm–3 percentage yield of X = …… % [3] (iv) Explain why a 100% yield in the preparation of a pure sample of X is not possible. . …… . …… . …… [1] [Total: 14]