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Edexcel Jun 2015 Paper 5 Q22
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![22 This question is about vanadium and its ions. (a) Consider the data below. Electrode system V2+(aq)|V(s) V3+(aq), V2+(aq)|Pt Standard electrode potential E(cid:57)/ V 1.18 0.26 (i) Draw a labelled diagram showing how to set up a cell, using the two electrode systems in the table above, in order to measure E (cid:57) Include standard conditions in your labelling. cell. (ii) Write an equation for the reaction in this cell. State symbols are not required. (b) (i) Complete the table below with the missing standard electrode potentials. Use the table starting on page 14 of your Data Booklet. (3) (2) (1) Electrode system [VO2+(aq) + 2H+(aq)], [V3+(aq) + H2O(l)]|Pt [VO2 +(aq) + 2H+(aq)], [VO2+(aq) + H2O(l)]|Pt I2(aq),2I(aq)|Pt [2H+(aq) + O2(g)], [H2O2(aq)]|Pt Standard electrode potential E(cid:57)/ V +0.54 +0.68 10 *P45073RA01028*<br />
(ii) The colours of the different oxidation states of vanadium are shown below. Oxidation state +5 +4 +3 +2 Colour yellow blue green violet For each of the following experiments, A and B, calculate the E (cid:57) value for the proposed reaction. Use your answers to predict whether or not a reaction occurs in each case. Give the formula of the vanadium product formed where a reaction occurs and give one observation you would make in each experiment. (6) Experiment A: Hydrogen peroxide is added to an aqueous solution containing VO2 + ions.Experiment B: An aqueous solution of potassium iodide is added to an aqueous solution containing VO2+ ions.*P45073RA01128* 11 Turn over<br />
(c) An experiment was carried out to determine the percentage purity of a sample of ammonium vanadate(V), NH4VO3. An impure sample of ammonium vanadate(V) with mass 0.150 g was dissolved + ions. Excess zinc in dilute sulfuric acid. This produced a solution containing VO2 powder was added to the solution, and this reduced the VO2 + ions to V2+ ions. The solution containing V2+ ions was titrated with potassium manganate(VII) of concentration 0.0200 mol dm3. The manganate(VII) ions oxidized the V2+ back to VO2 +. The volume of potassium manganate(VII) required was 35.50 cm3. (i) The manganate(VII) ions react as shown: MnO4+ 8H+ + 5e (cid:314) Mn2+ + 4H2O Show, by writing the appropriate half equation or otherwise, that 5 mol V2+ . react with 3 mol MnO4 (ii) Calculate the number of moles of manganate(VII) ions used in the titration. (1) (1) 12 *P45073RA01228*<br />
(iii) Calculate the number of moles of VO2 + in the original solution, and hence the percentage purity of the sample of NH4VO3. Give your answer to three significant figures. Molar mass of NH4VO3 = 116.9 g mol1. (3) (Total for Question 22 = 17 marks) *P45073RA01328* 13 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/l_smalljpgJune_2015_QP_-_Unit_5_Edexcel_Chemistry_A-level_22-0.jpg)
![Acceptable Answers Question Number 22(a)(i) Beaker with V electrode in solution containing V2+(aq) AND beaker containing V2+(aq) and V3+(aq) with Pt electrode N.B. Both solution levels must be shown (1) Labelled salt bridge AND connections to voltmeter ALLOW Suitable name or formula of salt for label ALLOW Salts eg NaCl in salt bridge (1) Ion concentrations = 1 mol dm-3 ALLOW M for mol dm-3 Concentrations given in one beaker only (1) Beaker positions may be reversed Ignore references to temperature and pressure Mark Reject Salt bridge neither dipping into nor touching solution unless penalised in MP1 Salt bridge containing an alkali/acid 1 mole of V2+ and 1 mole of V3+ 6CH0501 1506<br />
Acceptable Answers Question Number 22(a)(ii) First mark Reject Mark e- included 2V3+ + V3V2+ Balanced equation, either direction ALLOW Eqm sign forIGNORE State symbol even if incorrect (1) Second mark Correct direction ALLOW If balancing is incorrect or e- included in equation (1) Question Number 22b(i) Acceptable Answers Reject ([VO2+(aq) + 2H+(aq)], [V3+(aq)+ H2O(l)] Pt) +(aq) + 2H+(aq)], ([VO2 [VO2+(aq)+ H2O(l)] Pt) Sign and value needed +0.34 +1.00 Mark 6CH0501 1506<br />
Question Number 22(b)(ii) Question Number 22c(i) Question Number 22(c)(ii) Acceptable Answers Reject Mark Mark Violet Stays violet A: (+)0.32 (V) (1) VO2+ (may be shown as a product in an overall equation) (1) EITHER Bubbles / effervescence (of colourless gas) OR Colour changes (from yellow) to blue TE on negative Ecell for stays yellow ALLOW (from yellow) to green if justified by partial reduction (1) B: 0.2(0) (V) (1) no change / stays blue (1) If B=+0.2 or other positive value allow colour change from blue to green or brown. EITHER Consistent use of rule that reaction occurs when Ecell is positive OR Consistent use of rule that no reaction occurs when Ecell negative ALLOW If implied but not stated specifically (1) Acceptable Answers Reject + + 4H+ + 3e Reverse equation unless used to deduce final correct equation. V2+ + 2H2OVO2 OR Ox number of V increases by 3, ox number of Mn decreases by 5 ALLOW Balanced full equation 5V2+ + 3MnO4 5VO2 + +3Mn2+ + + 2H2O - + 4H+Acceptable Answers (35.50 x 0.0200/1000) = 7.1(0)x 10-4 / 0.00071 Reject Mark 6CH0501 1506<br />
Acceptable Answers Question Number 22(c)(iii) In final answer 92.2 scores 3 marks 33.2 scores 2 marks (ratio inverted) 55.3 scores 2 marks (ratio 1:1) METHOD 1 Mol V2+ reacting = 7.10x 10-4 x 5/3 =1.18333 x 10-3 = mol VO2 TE on answer to (c)(ii) (1) Mass NH4VO3 = (1.183 x 10-3 x 116.9) = 0.1382927 g TE from 4.26 x 10-3 =0.497994 (1) % purity = (0.1382927x 100/ 0.150) = (92.19333) = 92.2% TE from 0.497994 = 33.2% (1) METHOD 2 If 100% pure, moles of NH4VO3 =0.150/116.9=1.283 x 10-3 (1) Mol V2+ reacting = 7.10x 10-4 x 5/3 =1.18333 x 10-3 = mol VO2 TE on answer to (c)(ii) (1) % purity = =1.18333 x 10-3 x 100/1.283 x 10-3 = 92.2% (1) ALLOW TE at each step provided that each number used is to at least 2sf Mark Reject x 3/5 = 4.26x 10-4 Total for Question 22 = 17 marks 6CH0501 1506<br />](https://learnah.org/wp-content/uploads/2018/08/June_2015_MS_-_Unit_5_Edexcel_Chemistry_A-level_22-0.jpg)
