Edexcel Jun 2016 (IAL) Paper 5 Q (with explained solutions) |

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Edexcel Jun 2016 (IAL) Paper 5 Q

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20 Manganese exists in different oxidation states which can be interconverted, using suitable oxidizing and reducing agents. to complete the table below in which two E (cid:57) values are missing. (a) Use relevant standard electrode potential values, from page 17 of the Data Booklet, D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A Half-equation MnO4 (aq) + e (cid:85) MnO4 2(aq) MnO4 2(aq) + 2H2O(l) + 2e (cid:85) MnO2(s) + 4OH(aq) Mn3+(aq) + e (cid:85) Mn2+(aq) MnO4 (aq) + 8H+(aq) + 5e (cid:85) Mn2+(aq) + 4H2O(l) BiO3 (aq) + 6H+(aq) + 2e (cid:85) Bi3+(aq) + 3H2O(l) Fe3+(aq) + e (cid:85) Fe2+(aq) (1) E (cid:57)/ V +1.49 +1.51 +1.60 +0.77 (b) The apparatus below is used to measure the standard electrode potential of Mn3+(aq) + e (cid:85) Mn2+(aq) hydrogen gas platinum electrode solution containing H+(aq) 16 *P46665A01632*<br />
Identify, by name or formula, the substances needed in the salt bridge and the right-hand half-cell to measure the standard electrode potential. The measurement is made at 298 K and 1 atm. State the other essential condition for solution C. (4) A Salt bridge containing a solution ofB Electrode made ofC Solution containingEssential conditionA E R A S I H T N I E T I R W T O N O D (c) Manganate(VII) ions, MnO4 , can be made by oxidizing manganese(II) ions with bismuthate(V) ions, BiO3 , in acid solution. (i) Use the information in (a) to write the overall ionic equation for this reaction. State symbols are not required. A E R A S I H T N I E T I R W T O N O D (ii) Hence calculate E (cid:57) cell for this reaction. Include a sign and units in your answer. (2) (1) A E R A S I H T N I E T I R W T O N O D *P46665A01732* 17 Turn over<br />
*(d) Iron(II) ions in solution are oxidized by the air to iron(III) ions. The amounts of iron(II) and iron(III) ions in solution can be found by using titrations with acidified potassium manganate(VII) which only reacts with the iron(II) ions. MnO4+ 5Fe2+ + 8H+ (cid:314) Mn2+ + 5Fe3+ + 4H2O25.0cm3 portions of a solution A, containing a mixture of iron(II) and iron(III) ions, were acidified and titrated with potassium manganate(VII) solution of concentration 0.0195mol dm3.The mean titre was 16.80cm3.About 150 cm3 of solution A was reacted with excess zinc, which reduced the iron(III) ions to iron(II) ions. The excess zinc was filtered off.25.0cm3 portions of this reduced solution, which contained iron(II) ions but no iron(III) ions were acidified and titrated with potassium manganate(VII) solution of concentration 0.0195 mol dm3.The mean titre was 18.20cm3. Calculate the mass of iron(II) ions and the mass of iron(III) ions in 500cm3 of the original solution A. (5) (Total for Question 20 = 13 marks) 18 *P46665A01832* D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A D O N O T W R I T E I N T H I S A R E A<br />

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Acceptable Answers MnO4(aq) + eMnO42(aq) 2(aq) + 2H2O(l) + 2eMnO4 MnO2(s) + 4OH(aq) Reject Mark +0.56 Missing + (1) +0.59 Mark (4) Both correct for the mark Acceptable Answers Reject KI NaI Incorrect unit eg mol dm3 If name and formula are given, both must be correct A (Salt bridge containing a solution of) potassium nitrate / KNO3 ALLOW potassium chloride / KCl / sodium chloride / NaCl /sodium nitrate / NaNO3 (1) B (Electrode made of) platinum /Pt (1) C (Solution containing) manganese(II) and manganese(III) ions /Mn2+ and Mn3+ ions ALLOW Soluble salts of manganese(II) and manganese(III) ions (1) (Essential condition) stand alone mark 1 mol dm-3 ALLOW this if written in C ALLOW 1 molar / 1M / equal concentrations of Mn2+ and Mn3+ / manganese(II) and manganese(III) ions (1) IGNORE any temperature or pressure Question Number 20(a) Question Number 20(b)<br />
Acceptable Answers Reject Mark (2) 2Mn2+ + 5BiO3 All correct formulae on both sides - + 14H+2MnO4 - + 5Bi3+ + 7H2O ALLOW(1) Balancing correct formulae Conditional on all formulae correct ALLOW multiples IGNORE state symbols, even if incorrect IGNORE other equations as working before final equation (1) IGNORE electrons left in if they have been crossed through Note: Balanced equation with uncancelled electrons or uncancelled H+ ions / H2O scores (1) Acceptable Answers cell = 1.601.51 =) +0.09 V (Eo Sign, value and unit required TE on incorrect starting oxidation state of manganese Reject +0.1 V Mark (1) cell = 1.601.49 =) +0.11 V For Mn2+ to Mn3+ (Eo For MnO2 to MnO4 (Eo For MnO4 (Eo 2- to MnO4- 2- cell = 1.600.59 =) +1.01 V cell = 1.600.56 =) +1.04 V Question Number 20(c)(i) Question Number 20(c)(ii)<br />
Acceptable Answers Question Number 20*(d) Reject Mark (5) NOTE This calculation involves 8 individual mathematical operations (see 1 to 8 below), 4 on each titration and an additional subtraction. The first mark is awarded for the first operation finding the number of moles of Fe2+ ions from one of the titrations. The second mark is awarded after three further operations and then each subsequent mark for every other operation. The subtractions scores 1 mark on its own. One possible suggested solution is as follows For the original solution of A 1. Moles of MnO4 which reacts with 25 cm3 of original solution A = 3.276 x 10-4 (mol) (1) 2. Moles of Fe2+ in original solution AND 3. Moles of Fe2+ in 500 cm3 of original solution A AND 4. Mass of Fe2+ in 500 cm3 of original solution A = Answer to 3 x 55.8 = 1.828 (g) = Answer to 2 x 20 = 3.276 x 10-2 (mol) = Answer to 1 x 5 = 1.638 x 10-3 (mol) = 16.80 x 0.0195 1000 (1) ALLOW The three operations ( x 5, x 20 and x 55.8) in any order. Ar Fe = 56 (instead of 55.8)<br />
(5) For the fully reduced solution of A 5. Moles of MnO4 which reacts with 25 cm3 of reduced solution = 18.20 x 0.0195 1000 = 3.549 x 10-4 (mol) AND 6. Moles of Fe2+ in the fully reduced solution A = Answer to 1 x 5 = 1.7745 x 10-3 (mol) (1) 7. Moles of Fe2+ in 500 cm3 of the fully reduced solution A = Answer to 6 x 20 = 3.549 x 10-2 (mol) AND 8. Mass of Fe2+ in 500 cm3 of reduced solution Y = Answer to 7 x 55.8 = 1.980342 (g) Mass of Fe3+ in original solution = Answer to 8 - Answer to 7 = 0.15288 (g) ALLOW The three operations ( x 5, x 20 and x 55.8) in any order. Ar Fe = 56 (instead of 55.8) ALLOW other methods IGNORE SF except 1SF (1) (1) (Total for Question 20 = 13 marks)<br />

Question 20 21 22 23

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