Edexcel Jun 2014 (R) Paper 5 Q20 (with explained solutions) |

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Edexcel Jun 2014 (R) Paper 5 Q20

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SECTION B Answer ALL the questions. Write your answers in the spaces provided. 20 (a) When iron(II) sulfate is dissolved in water, the iron(II) ions are slowly oxidized to iron(III) ions by oxygen dissolved in the water. (i) Write the two ionic half-equations for this redox reaction in acid conditions. State symbols are not required. (ii) Hence write the overall ionic equation for the reaction. State symbols are not required. (b) 6.90 g of iron(II) sulfate crystals (FeSO4.7H2O) was dissolved in distilled water and the solution made up to 250 cm3 in a volumetric flask. After 24 hours, 25.0 cm3 portions of this solution were pipetted into a conical flask and titrated against acidified potassium manganate(VII) solution with a concentration of 0.0195 mol dm3. The mean titre was 24.90 cm3. (i) Write the ionic equation showing that 1 mol of manganate(VII) ions oxidizes 5 mol of iron(II) ions in acid conditions. State symbols are not required. (1) (1) (1) 12 *P42989rA01228*<br />
(ii) What is the colour of the solution in the conical flask at the end-point of the titration? (1)*(iii) Calculate the amount of Fe2+ ions in 250 cm3 of the solution after it had been left to stand for 24 hours. Hence calculate the percentage of the iron(II) ions that had been oxidized between the preparation of the solution and the titration. The molar mass of FeSO4.7H2O is 277.9 g mol1. (5) (iv) Suggest, with an explanation, the appropriate number of significant figures to give for the answer to (b)(iii). (1)*P42989rA01328* 13 Turn over<br />
(c) The most stable oxidation states in iron compounds are +2 and +3, but others do exist, for example in the ferrate(VI) ion, FeO4 conversion of ferrate(VI) to iron(III) is 2. The ionic half-equation for the FeO4 2(aq) + 8H+(aq) + 3e (cid:85) Fe3+(aq) + 4H2O(l) E (cid:57) = +2.20 V (i) Ferrate(VI) decomposes in neutral or acid solution, but is stable in alkali. Suggest why this is so. (1)(ii) Write the equation for the disproportionation of iron(III) into iron(II) and ferrate(VI) in aqueous solution. State symbols are not required. (iii) Use standard electrode potential values to determine the thermodynamic feasibility of this disproportionation. (2) (2) (Total for Question 20 = 15 marks) 14 *P42989rA01428*<br />

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Section B Question Number 20 (a)(i) Fe2+Fe3+ + e() Acceptable Answers Reject Mark O2 + 2H+ + 2e()H2O OR O2 + 4H+ + 4e()2H2O ALLOW Reversible arrows Equations in other direction Electrons subtracted on LHS of first equation Multiples Ignore state symbols even if incorrect O2 + 2H+ + 2Fe2+2Fe3+ + H2O OR O2 + 4H+ + 4Fe2+4Fe3+ + 2H2O ALLOW Multiples Reversible arrows Ignore state symbols even if incorrect No TE from 20(a)(i) Acceptable Answers Reject Question Number 20 (a)(ii) Mark Equation in the wrong direction, even with reversible sign Question Number 20 (b)(i) 5Fe2+ + MnO4 Acceptable Answers+ 8H+5Fe3+ + Mn2+ + 4H2O Ignore state symbols even if incorrect Reject Mark Question Number 20 (b)(ii) Acceptable Answers Reject Mark (Pale) pink Purple / mauve 6CH05/01R 1406<br />
Reject Mark Question Number *20 (b)(iii) Acceptable Answers= 24.90 x 0.0195 x 103 Amount of MnO4 = 4.8555 x 104 (mol)* (1) Amount of Fe2+ = answer * x 5 in 25 cm3 = 4.8555 x 104 x 5 = 2.42775 x 103 (mol) So in 250 cm3 = 2.42775 x 102 (mol) (1) (Mr (FeSO4.7H2O) = 277.9) ROUTE 1 (via moles) Amount of Fe2+ used to prepare the solution = 6.90 / 277.9 = 2.4829 x 102 (mol) (1) EITHER % of Fe2+ remaining at titration = 100 x 2.42775 x 102 / 2.4829 x 102 = 97.7785 (%) (1) % Oxidized = 10097.7785 = 2.221 (%) (1) OR Amount oxidized = 2.4829 x 10-22.42775 x 102 = 5.516 x 104 (mol) (1) % Oxidized = 5.516 x 104 x 100 / 2.4829 x 102 = 2.221 (%) (1) ROUTE 2 (via mass) mass from titration = 2.42775 x 102 x 277.9 = 6.7467 (g) (1) % of Fe2+ remaining at titration = 100 x 6.7467 / 6.9 = 97.7785 (%) (1) % Oxidized = 10097.7785 = 2.221 (%) (1) Ignore SF except 1 SF unless justified in b(iv) Correct answer no working scores 5 marks 90.22% obtained from failure to multiply by 10 scores 4 marks 6CH05/01R 1406<br />
Question Number 20 (b)(iv) Acceptable Answers 3 (significant figures) because all the data (except Ar(H)) is given to 3 SF OR 2 (significant figures) because the least precise data (Ar(H)) is 2 SF OR 2 (significant figures) because the data is to three figures. After processing only two figures are certain. OR 1 (significant figure) because of the subtraction of two similar numbers. Question Number 20 (c)(i) Alkali neutralizes the acid shifting the Acceptable Answers equilibrium to the left OR Alkali neutralizes the acid so E value for half cell becomes less (than +2.20 V) ALLOW Reacts with and removes for neutralizes IGNORE Just shifts equilibrium to the left Reject Mark Reject Mark Question Number 20 (c)(ii) Acceptable Answers 2 + 8H+ 4Fe3+ + 4H2O3Fe2+ + FeO4 OR Multiples Species (1) balance (1) Ignore state symbols even if incorrect Reject Mark 6CH05/01R 1406<br />
Question Number 20 (c)(iii) Acceptable Answers Required half cell value is Eo = (+)0.77 (1) Eo (Eo cell = (0.772.20 =) 1.43 V cell negative so disproportionation) not feasible (1) Reject Mar k 2 cell cell TE on calculated negative value of Eo No TE on positive value for Eo OR Correct application of anti-clockwise rule e.g. Fe3+(aq) + eFe2+(aq) Eo = +0.77 V 2(aq) + 8H+(aq) + 3eFe3+(aq) + 4H2O(l) FeO4 Eo =+2.20 V Equations in order of increasing Eo value and arrows shown (1) Anti-clockwise rule shows top reaction moves left and bottom reaction moves right so disproportionation not feasible (1) Total for Question 20 = 15 marks 6CH05/01R 1406<br />
Question Number Acceptable Answers 21 (a) (A transition metal) forms ions / oxidation states with partially filled / incomplete d orbital(s) / d sub-shell Reject Mark Question Number 21 (b)(i) Question Number 21 (b)(ii) Question Number 21 (b)(iii) Acceptable Answers Reject Mark Names without oxidation numbers. 2 (1) W = chromate(VI) (ion) / CrO4 X = chromium(III) hydroxide / Cr(OH)3 / Cr(OH)3(H2O)3 (1) Y = hexahydroxochromate(III) (ions) / [Cr(OH)6]3 /tetrahydroxochromate(III) (ions) / [Cr(OH)4] / [Cr(H2O)2(OH)4] (1) Z = chromium(II) (ions) / chromium(II) sulfate / Cr2+ / Cr2+(aq) / [Cr(H2O)6]2+ (1) ALLOW Name or formula of the compounds IGNORE Omission of square brackets around complexes Acceptable Answers A = ethanol / C2H5OH / ethanal / CH3CHO OR any primary or secondary alcohol or any aldehyde (1) B = zinc / Zn ALLOW magnesium / Mg (1) C = any acid (name or formula) (1) IGNORE Omission of (aq) with acid formula Concentration of acid Acceptable Answers 2 + H2O 2 + 2OH2CrO4 Cr2O7 OR Multiples Ignore state symbols even if incorrect Mark Reject CH3COH Alkali metals Tin / Sn H+ or H3O+ or acid Reject Mark 6CH05/01R 1406<br />
Question Number 21 (b)(iv) Question Number 21 (b)(v) Question Number 21 (c)(ii) Acceptable Answers (chromium(II) ions) oxidized by (oxygen in the) air ALLOW Just oxygen Question Number 21 (c)(i) (A ligand is a) molecule or (negative) Acceptable Answers ion with a (lone) pair (of electrons) ALLOW Species / Compound / group (1) Which forms a dative covalent bond with a (central) metal ion or atom (to form a complex) (1) ALLOW (if no other marked scored) Electron pair donor Acceptable Answers 3+ + 6NH3 3+ + 6H2O Cr(H2O)6Cr(NH3)6 ALLOW Cr(H2O)6Cr(NH3)4(H2O)2 Correct formula for ammine (1) Rest of the equation correct (1) 3+ + 4NH3 3+ + 4H2O Reject Mark Reject Mark Positive ion Reject Cr3+ and Cr3+(aq) Mark Reject Mark Acceptable Answers (NH4)2Cr2O7Cr2O3 + N2 + 4H2O Allow multiples (1) Chromium is reduced from (+)6 to (+)3 (1) Nitrogen is oxidized from 3 to 0 (1) Penalise use of changes / increases / decreases for oxidises or reduces once only Total for Question 21 = 17 marks 6CH05/01R 1406<br />

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