Edexcel Jan 2015 (IAL) Paper 6 Q3 (with explained solutions) |

Skip to content

Edexcel Jan 2015 (IAL) Paper 6 Q3

Answers available below

3 This is an experiment to determine the oxidation number of vanadium in a purple solution, T, of a vanadium compound. Preparation of solution T Solution T was formed when 25.00 cm3 of a 0.100 mol dm3 solution of sodium vanadate(V), NaVO3, was reduced by heating with excess zinc and dilute sulfuric acid. When the reduction was complete, the yellow NaVO3 solution had turned purple. Titration of solution T The mixture was filtered through glass wool, directly into 50.00 cm3 of 0.0200 mol dm3 potassium manganate(VII), KMnO4, solution. Further potassium manganate(VII) solution of the same concentration was added from a burette to this reaction mixture, which was kept at a temperature of about 80 oC. The end point is reached when all the vanadium ions had been oxidized back into vanadate(V) ions by the manganate(VII) ions. The end point occurred when a further 25.00 cm3 had been added. (a) (i) Draw a diagram of the apparatus for carrying out the titration, while keeping the titration mixture at about 80 oC. (2) *P45046A0616*<br />
(ii) What is removed from the reaction mixture by filtering through glass wool? (1)(iii) Suggest why the mixture is filtered directly into potassium manganate(VII) solution before carrying out the rest of the titration. (1)(iv) Explain why an indicator is not required for this titration. (1)(b) (i) Calculate the number of moles of vanadate(V) ions, VO3 0.100 mol dm3 solution of sodium vanadate(V), NaVO3. , in 25.00 cm3 of a (ii) Calculate the total volume of potassium manganate(VII) solution. Hence the total number of moles of potassium manganate(VII) used to oxidize the purple vanadium solution, T. (1) (2) *P45046A0716* Turn over<br />
(iii) Complete the half equation for the reduction of manganate(VII) ions to manganese(II) ions. MnO4+ H+ + e Mn2+ + H2O (iv) By considering either the number of electrons transferred or by using the changes in oxidation numbers, calculate the oxidation number of vanadium in the purple solution, T. You must show your working. (c) In acidic solution, the vanadate ions, VO3are changed into VO2 +. Write an ionic equation for this reaction. State symbols are not required. (1) (3) (1) *P45046A0816*<br />
(d) Some standard electrode potentials of tin and vanadium are given below. Sn2+(aq)|Sn(s) V2+(aq)|V(s) V3+(aq),V2+(aq)|Pt [VO2+(aq) + 2H+(aq)], [V3+(aq) + H2O(l)]|Pt +(aq) + 2H+(aq)], [VO2+(aq) + H2O(l)]|Pt [VO2 0.14 V 1.18 V 0.26 V +0.34 V +1.00 V Use these values to predict the lowest oxidation number of vanadium that can be produced from VO2 + using tin as the reducing agent. Explain your reasoning. (2)(Total for Question 3 = 15 marks) *P45046A0916* Turn over<br />
BLANK PAGE 10 *P45046A01016*<br />

Show answer

Question Number 3(a)(i) Acceptable Answers Reject Mark Burette and (conical) flask / beaker and Either Heated water bath / direct heat Can be shown by heat/arrow (1) Thermometer in flask / water (1) OR Heating mantle / hot plate (1) With thermostatic control (1) ALLOW 1 mark for heating separate from the titration Acceptable Answers Reject (The excess / unreacted) zinc / Zn((s)) (is removed) Allow Insoluble zinc Insoluble reactant Zinc Zinc solid / left over Insoluble impurities Insoluble reactant alone Mark WCH0601 1501 Question Number 3(a)(ii)<br />
Question Number 3(a)(iii) Question Number 3(a)(iv) Question Number 3(b)(i) Reject Mark Acceptable Answers To prevent it / T being oxidized by air / oxygen ALLOW To prevent oxidation OR The vanadium(II) is easily oxidized by air / oxygen OR (As a) large volume / volume greater than 50 cm3 of potassium manganate(VII) is required Acceptable Answers Reject EITHER Because the potassium manganate(VII) is self-indicating ALLOW Reaction is self-indicating ALLOW Potassium manganate (VII) changes colour during the reaction/ at the end point. OR at the end point a pink / purple solution forms (from a blue/yellow/green solution) ALLOW Modified pink e.g. yellow pink because of the yellow vanadate(V) Acceptable Answers 25 x 0.10 = 2.5 x 10-3/0.0025 (mol) 1000 Mark Just obvious / clear colour change without potassium manganate(VII) Reject Mark WCH0601 1501<br />
Question Number 3(b)(ii) Question Number 3(b)(iii) Reject Mark Acceptable Answers Total volume of 0.02 mol dm-3 solution of potassium manganate(VII) = 25 + 50 = 75 cm3 (1) (75) x 0.02 = 1.5 x 10-3/0.0015 (mol) 1000 (1) ALLOW for 1 mark 25 x 0.02 = 5 x 10-4/0.0005 (mol) 1000 OR 50 x 0.02 = 1 x 10-3/0.001 (mol) 1000 ALLOW Internal TE for incorrect volume in first calculation for second mark. Acceptable Answers MnO4+ 8H+ + 5e Mn2+ + 4H2O Reject Mark WCH0601 1501<br />
Question Number 3(b)(iv) Question Number 3(c) Mark Acceptable Answers Reject METHOD 1 2.5 x 10-3 mol of vanadium ions lose 5 x 1.5 x 10-3 = 7.5 x 10-3 mol electrons (1) Therefore 1 mol of vanadium ions lose 3 mol of electrons (1) As final oxidation state is +5 the oxidation state of vanadium in the purple solution is +2 (1) METHOD 2 Ratio of Mn:V = 0.0015:0.0025 = 3:5 (1) Oxidation number of Mn changes by 5 so oxidation number of vanadium must change by 3 (1) As final oxidation state is +5 the oxidation state of vanadium in the purple solution is +2 (1) METHOD 3 First Mark +2 (with no working) Second Mark Working backwards from this: Any mention of transfer of 3 electrons OR V5+ + 3e(-) V2+ OR V2+ - 3e(-) V5+ IGNORE Mn(VII) +3e(-) Mn(II) Third Mark 7.5 x 10-3 mols of electrons / change in oxidation number removed from 2.5 x 10-3 V3+ OR 7.5 x 10-3 mols of manganate(VII) ion react with 2.5 x 10-3 mols V3+ ALLOW TE from (b)(ii) and (iii) Acceptable Answers VO3+ 2H+ VO2 + + H2O Reject Mark WCH0601 1501<br />
Question Number 3(d) Reject Mark Acceptable Answers First Mark V3+ / V(III) / (V)+3 (1) Second Mark Any of the following calculations : V(IV)/VO2+ to V(III)/ V3+ = +0.48 (V) OR V(III)/V3+ to V(II)/ V2+ =-0.12 (V) OR ALLOW V(V)/VO2 OR Sn2+(aq)|Sn(s) has a more negative electrode potential than the last two vanadium potentials (so vanadium reduced to V3+) Accept reverse argument OR By the anticlockwise rule if shown with appropriate arrows (1) + to V(IV)/VO2+ =+0.66 (V) Total for Question 3 = 15 marks WCH0601 1501<br />

Question 1 2 3 4

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.