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CIE Jun 2013 v3 Paper 4 Q1
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![Examiners For Use Section A Answer all the questions in the spaces provided. (a) What is meant by the term standard electrode potential, SEP?[2] (b) Draw a fully labelled diagram of the apparatus you could use to measure the SEP of the Fe3+ / Fe2+ electrode. [5] (c) The reaction between Fe3+ ions and I ions is an equilibrium reaction. 2Fe3+(aq) + 2I(aq) 2Fe2+(aq) + I2(aq) (i) Use the Data Booklet to calculate the for this reaction.(ii) Hence state, with a reason, whether there will be more products or more reactants at equilibrium.(iii) Write the expression for Kc for this reaction, and state its units. Kc = unitsUCLES 2013 9701/43/M/J/13<br />
Examiners For Use An experiment was carried out using solutions of Fe3+(aq) and I(aq) of equal concentrations. 100 cm3 of each solution were mixed together, and allowed to reach equilibrium. The concentrations at equilibrium of Fe3+(aq) and I2(aq) were as follows. [Fe3+(aq)] = 2.0104 mol dm3 [I2(aq)] = 1.0102 mol dm3 (iv) Use these data, together with the equation given in (c), to calculate the concentrations of Fe2+(aq) and I(aq) at equilibrium. [Fe2+(aq)] =mol dm3 (v) Calculate the Kc for this reaction. [I(aq)] =mol dm3 Kc =[8] [Total: 15]UCLES 2013 9701/43/M/J/13 [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2013_v3_QP_-_Paper_4_CIE_Chemistry_A-level_1-0.jpg)
![Page 2 Mark Scheme GCE A LEVELMay/June 2013 Syllabus 9701 Paper 43 (a) The potential of an electrode compared to that of a standard hydrogen electrode (SHE) or the EMF of a cell composed of the test electrode and the SHE all measurement concentrations of 1 mol dm3 and 298 K / 1 atm pressure [1] [1] [2] (b) H2 and good delivery system [1] Fe2+/Fe3+ solution labelled [1] platinum electrodes (both) [1] salt bridge and voltmeter [1] or H2SO4 [1] (acid is not sufficient) [5] H+ or HCl (c) (i) E = 0.770.54 = 0.23 (V) (ii) Since E is positive/ E >0 [1] So more products / the equilibrium will be over to the right / forward reaction is favoured ecf from (c)(i) [1] (iii) Kc = [Fe2+]2[I2] / [Fe3+]2[I]2 units are mol1 dm3 ecf on expression (iv) ([Fe2+] must always be twice [I2], so) [Fe2+] = 0.02 (mol dm3) ([I] must always be equal to [Fe3+], so) [I] = 2104 (mol dm3) (v) Kc = {(0.02)20.01} / {(2 x 104)2(2104)2} correct expression (allow ecf from incorrect expression in (c)(iii)) (allow ecf from (c)(iv)) = (4106) / (1.6101.5) = 2.5109 (mol1 dm3) [1] [1] [1] [1] [1] [1] [8] [Total: 15]Cambridge International Examinations 2013<br />](https://learnah.org/wp-content/uploads/2018/08/June_2013_v3_MS_-_Paper_4_CIE_Chemistry_A-level_1.jpg)