Edexcel Jan 2015 (IAL) Paper 5 Q21 (with explained solutions) |

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Edexcel Jan 2015 (IAL) Paper 5 Q21

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SECTION B Answer ALL the questions. Write your answers in the spaces provided. 21 Brass is an alloy of copper and zinc, often with traces of other metals. The copper content of brass can be determined by dissolving the metal in concentrated nitric acid and measuring, by titration, the concentration of the copper(II) ions formed. (a) When concentrated nitric acid reacts with copper, the copper dissolves and one of the products is dinitrogen tetroxide, N2O4. (i) Use the data on page 15 of the Data Booklet to write the ionic half-equations for this reaction of copper with concentrated nitric acid. State symbols are not required. (ii) Write the overall equation for the reaction of copper with concentrated nitric acid and calculate E(cid:57) cell for the reaction. State symbols are not required. (iii) State one observation that you would expect to make when copper dissolves in concentrated nitric acid. (2) (2) (1)*P45045A01128* 11 Turn over<br />
(b) 1.35 g of a sample of rivet brass was dissolved in concentrated nitric acid. The resulting mixture was boiled and then allowed to cool before being transferred to a volumetric flask. The solution was made up to 250 cm3 with distilled water and mixed thoroughly. Excess potassium iodide solution was added to 25.0 cm3 samples of this solution, and the liberated iodine determined by titration with a solution of sodium thiosulfate of concentration 0.0505 mol dm3. The mean titre was 26.35 cm3. (i) Write the ionic equation for the reaction of the copper(II) ions with iodide ions to form copper(I) iodide and iodine. State symbols are not required. (1) (1) (1) (ii) Write the ionic equation for the reaction of iodine with thiosulfate ions. State symbols are not required. (iii) Use the equations in (b)(i) and (b)(ii) to show that the amount of copper(II) ions is equal to the amount of thiosulfate ions.12 *P45045A01228*<br />
(iv) Calculate the percentage by mass of copper in the sample of rivet brass. (4) *P45045A01328* 13 Turn over<br />
(c) (i) The reaction mixture in (b) was boiled before being transferred to a volumetric flask. This removed dissolved nitrogen oxides which would otherwise oxidize the iodide ions. Explain the effect that omitting this step would have on the value obtained for the percentage of copper. (2)(ii) Any nitrogen oxides that remain after boiling can be removed by the addition of urea. When this was done, the mean titre changed by 0.25 cm3. By considering the uncertainties in the various measurements, explain whether the addition of urea is worthwhile. (2)(d) Both copper and zinc are d-block elements, but only copper is a transition metal. (i) Explain the term d-block element. (1)14 *P45045A01428*<br />
(ii) Explain why copper is classed as a transition metal but zinc is not. (1)*(iii) Explain why the complexes of copper(II) ions are coloured. (4)(iv) Although zinc is not a transition metal, zinc(II) ions form complexes. Explain why these complexes are colourless. (1)(Total for Question 21 = 23 marks) *P45045A01528* 15 Turn over<br />

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Mark Reject Alternative nitrate(V) reductions Mark Reject uncancelled electrons Section B Question Number 21(a)(i) Acceptable Answersonly+4H+ + 2e() Penalise omission of charge on NO3 once in (a)(i) and (a)(ii) Penalise an incorrect coefficient in (a)(i) and (a)(ii) once only Cu2+ + 2e()Cu (Eo = +0.34 V) (1) 2NO3N2O4 + 2H2O (Eo = +0.80 V) ALLOW multiples equations reversed reversible / double-headed arrows 2 NO2 for N2O4 (1) IGNORE Eo at this point State symbols even if incorrect Question Number 21(a)(ii) Cu + 2NO3 Acceptable Answers+ 4H+Cu2+ + N2O4 + 2H2O ALLOW multiples reversible / double-headed arrows 2 NO2 for N2O4 (1) No TE for equation from incorrect half- equations Eo TE for Eo half-equations IGNORE State symbols even if incorrect cell (= +0.800.34) = (+)0.46 (V) (1) cell value on incorrect selection of<br />
Reject Colourless gas bubbles Mark Reject Cu(NO3)2 +2KICuI + I2 + 2KNO3 Mark Reject 2Na2S2O3 + I2 Na2S4O6 + 2KI Mark OR Green solution ALLOW (pale) yellow fumes / gas OR effervescence / bubbling / fizzing OR blue solution IGNORE modifiers of blue IGNORE References to copper dissolving Acceptable Answer In (b)(i) and (b)(ii) penalise (correct) non-ionic equations once. Cu2+ + 2ICuI + I2 OR 2Cu2+ + 4ICu2I2 + I2 ALLOW Cu2+ + ICu+ + I2 OR 2Cu2+ + 2I2Cu+ + I2 OR Multiples IGNORE State symbols even if incorrect 22I + S4O6 Acceptable Answer I2 + 2S2O3 OR Multiples 2 Question Number 21(a)(iii) Brown fumes / gas Acceptable Answer Question Number 21(b)(i) Question Number 21(b)(ii)<br />
Reject Mark Just re-writing the equations. Question Number 21(b)(iii) 2 Acceptable Answer 2 mol Cu2+ forms 1 mol I2 which reacts with 2 mol S2O3 OR Multiples in this explanation OR Any clear explanation in words No TE on incorrect equations in (b)(i) and (b)ii) Question Number 21(b)(iv) mol S2O3 Acceptable Answer Reject Mark 2 in 25 cm3 = 0.0505 x 26.35 /1000 = 1.330675 x 103 ans* (1) mol Cu2+ in 250 cm3 = mol Cu in sample = 10 x ans* (1) = 1.330675 x 102 ans** mass Cu = ans** x 63.5 = 1.330675 x 102 x 63.5 (1) = 0.84498 (g) ans*** % copper in rivet brass = 100 x ans***/ 1.35 = 62.591/ 62.6 % (1) Correct answer with no working scores 4 If incorrect ratio used then max 3 Answers >100% max 3 IGNORE SF except one Do not penalise correct intermediate rounding Question Number 21(c)(i) More iodine would be formed (1) Acceptable Answer (Titre / volume of thiosulfate would be larger) so (calculated) % copper would be higher (1) Second mark dependent on first Reject Mark<br />
Question Number 21(c)(ii) MP1 and MP2 are stand alone Acceptable Answer Reject Mark 1.9% Total apparatus error greater than effect of urea Marking Point 1 Percentage difference in the titres is (approximately) 100 x 0.25/26.35 = 0.94877 / 0.95% (1) Marking Point 2 This MP should only be awarded if the candidate appreciates that the addition of urea improves experimental accuracy. The percentage error in the burette reading is ()100 x 0.1/26.35 = ()0.3795% and so change is a significant improvement OR Difference in titres is greater than uncertainty / error in burette reading OR Calculation any other specific apparatus uncertainty and use of urea has a significant effect OR Error without urea is significant when compared with the typical apparatus uncertainty (so the addition of urea improves accuracy) (1) Acceptable Answer (When the electronic structure is built up according to the aufbau rules) the last electron goes into the (3)d-subshell / one of the d-orbitals / the d-orbitals Mark Reject Just electrons present in (3)d-subshell outer / valence electrons are in d- subshell shell for subshell Question Number 21(d)(i)<br />
Mark Mark Acceptable Answer copper forms (one or more stable) ions having partially filled (3)d orbitals / subshell (but zinc does not) OR Zinc does not form an ion with a partially filled 3(d) orbital/subshell (but copper does) Reject 3d shell Just zinc only forms an ion with a full 3d subshell Reject Orbital / shell is split d-d splitting emitted Acceptable Answer Penalise use of orbital (singular) once only in (d)(iii) and (d)(iv) (3)d orbitals / (3)d subshell split (by the attached ligands) (1) Electrons are promoted (from lower to higher energy d orbital(s) / levels) OR Electrons move from lower to higher energy d orbital(s) / levels ALLOW dd transitions occur (1) Absorbing energy /photons of a certain frequency (in the visible region) ALLOW Absorbing light (1) Reflected / transmitted / remaining light is coloured / yellow / in the visible region ALLOW Complementary colour seen Reflected / transmitted / remaining light / frequency is seen (1) No mention of (3)d then max 3 IGNORE reference to electrons relaxing / dropping to the ground state Question Number 21(d)(ii) Question Number 21(d)*(iii)<br />
Question Number 21(d)(iv) Acceptable Answer (3)d subshell / (all) (3)d orbitals of zinc(II) are full (so electron transitions are not possible) Ignore No unpaired electrons Mark Reject (3)d orbital full Full 3d subshell is not split Total for Question 21 = 23 marks<br />

Question 21 22 23 24

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