Chemistry Past Paper November 2023 PDF

A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Worksheet 2 Papers 2023 – 18 Name ………………………………………………………………………………………… Date ……………………………… Details Electrochemistry – From 2023 - 2018 Papers 2023 Nov 20...

A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Worksheet 2 Papers 2023 – 18 Name ………………………………………………………………………………………… Date ……………………………… Details Electrochemistry – From 2023 - 2018 Papers 2023 Nov 2023 / 42, Q. 3 (a)–(e) – Electrochemistry 1 Some electrode potentials are shown in Table 3.1. Table 3.1 A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (a) (i) Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions. (ii) An electrochemical cell is set up using an Fe3+ / Fe2+ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit. positive electrode …………………………………………………………………………………… Electrons flow from the ….………………….. electrode to the ……………………... electrode. (b) The vanadium-containing species in the electrode reactions given in Table 3.1 are V, V2+, V3+, VO2+ and VO2+. (i) Identify one vanadium-containing species that does not react with Fe2+ ions under standard conditions. Use data from Table 3.1 to explain your answer. ………………………………………………………………………………………………………… ………………………………………………………………………………………………..…… A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (ii) Identify all the vanadium-containing species that will react with Fe2+ ions under standard conditions. ………………………………………………………………………………………………..…… (iii) Write an equation for one of the possible reactions identified in (ii). ………………………………………………………………………………………………..…… (c) Another electrochemical cell is set up using an Fe3+ / Fe2+ electrode and an alkaline ClO– / Cl– electrode. The concentration of Fe3+ is 1000 times greater than the concentration of Fe2+ in the Fe3+ / Fe2+ electrode. All other conditions are standard. (i) Use the Nernst equation to calculate the E value of the Fe3+ / Fe2+ electrode. Show your working. E =.............................. V (ii) Write an equation for the reaction that occurs in the cell, under these conditions. ………………………………………………………………………………………………..…… (d) Another electrochemical cell is set up using an Fe2+ / Fe electrode and an alkaline ClO– / Cl– electrode under standard conditions. Calculate the value of ΔGꝊ for the cell. ΔGꝊ =.............................. kJ mol–1  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (e) A solution of iron(II) sulfate, FeSO4(aq) is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640A is passed for 17.0 minutes. The mass of the cathode increases by 0.185 g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working. L =.............................. mol–1 Solution Nov 2023 / 42, Q. 3 (a)–(e) – Electrochemistry 1 (a) (i) hydrogen, delivery system, H+, platinum (1) 1 (a) (ii) iron …… hydrogen …… iron (1) 1 (b) (i) (for specified V2+, V3+ or VO2+) EꝊ is more positive than / above –0.44 AND more negative than / below 0.77 V (1) 1 (b) (ii) V and VO2+ (1) 1 (b) (iii) V + Fe2+ → V2+ + Fe OR VO2+ + 2H+ + Fe2+ → VO2+ + H2O + Fe3+ (1) 1 (c) (i) Nernst: E = 0.77 + (0.059 / z) log[ox] / [red] (1) 0.947 (1) 1 (c) (ii) 2Fe3+ + Cl– + 2OH– → 2Fe2+ + ClO– + H2O (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 1 (d) EꝊcell = 1.33 V (1) ∆GꝊ = –nEꝊcellF (1) –257 (1) 1 (e) 0.64 x 17 x 60 = 653 / 652.8 Coulombs (1) 652.8 ÷ 1.6 x 10–19 = 4.08 x 1021 (number of electrons) 4.08 x 1021 ÷ 2 = 2.04 x 1021 (number of atoms Fe) (1) 0.185 ÷ 55.8 = 3.31 x 10–3 (number of moles of Fe atoms) 2.04 x 1021 ÷ 3.31 x 10–3 = L = 6.153 x 1023 (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2023 / 42 June 2023 / 42, Q. 1 (c) – Electrochemistry (+ moles) 2 (c) Potassium sulfite, K2SO3, is used as a food additive. The concentration of sulfite ions, SO32–, can be determined by titration using aqueous acidified manganate(VII) ions, MnO4–. A 250cm3 solution contains 3.40g of impure K2SO3. 25.0cm3 of this solution requires 22.40cm3 of 0.0250moldm–3 acidified MnO4– to reach the end-point. All the SO32– ions are oxidised. None of the other species in the impure K2SO3 are oxidised. The reaction occurs as shown by the two half-equations. H2O + SO32– → SO42– + 2H+ + 2e– MnO4– + 8H+ + 5e– → Mn2+ + 4H2O (i) Give the ionic equation for the reaction between SO32– and acidified MnO4–. ………………………………………………………………………………………………..…… (ii) Calculate the percentage purity of the sample of K2SO3. Show your working. percentage purity of K2SO3 = ……………………..  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2023 / 42, Q. 1 (c) – Electrochemistry (+ moles) 2 (c) (i) 2MnO4– + 6H+ + 5SO32– → 2Mn2+ + 3H2O + 5SO42– (1) 2 (c) (ii) M1 M2 any two bullets or all four ⚫ moles MnO4– = 0.025 x 22.40 / 1000 = 5.6 x 10–4 ⚫ moles SO32– = 5.6 x 10–4 x 5 / 2 = 1.4 x 10–3 (in 25 cm3) ecf from (c)(i) and bullet 1 ⚫ moles SO32– = 1.4 x 10–2 (in 250 cm3) ecf bullet 2 ⚫ mass K2SO3 = 1.4 x 10–2 x 158.3 = 2.2162 g ecf bullet 3 OR moles K2SO3 (if 100% pure) = 3.40 ÷ 158.3 = 0.02148 M3 % purity = 100 x 2.2162 / 3.40 = 65.2 / 65.3 % ecf min 2sf OR % purity = 100 x 0.014 / 0.02148 = 65.2 / 65.3 % (3)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2023 / 42, Q. 9 (complete) – Electrochemistry 3 (a) Define standard cell potential, EꝊcell. ……………………………………………………………………………………………………………… ……………………………………………………………………………………………………..…… (b) An electrochemical cell is set up to measure EꝊcell of a cell consisting of an Fe3+ /Fe2+ half-cell and a Cl2 /Cl – half-cell. Draw a labelled diagram of this electrochemical cell. Include all necessary substances. It is not necessary to state conditions used. A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (c) The cell reaction for the electrochemical cell in (b) is shown. Cl2 + 2Fe2+ → 2Fe3+ + 2Cl– EꝊcell = +0.59V Calculate ΔGꝊ, in kJmol–1, for this cell reaction. ΔGꝊ = …………………….. kJmol–1 (d) Another experiment is set up using the same electrochemical cell. In this experiment the Fe2+ concentration is 0.15moldm–3. All other concentrations remain at their standard values. The Nernst equation is shown. E = EꝊ + (0.059 / z) log [oxidised species] [reduced species] (i) Use the Nernst equation to calculate the electrode potential, E, for the Fe3+ /Fe2+ half-cell in this experiment. [EꝊ : Fe3+ / Fe2+ = +0.77V] E = …………………….. V (ii) Use your answer to (d)(i) to calculate the Ecell for this electrochemical cell. Ecell = …………………….. V [Total: 8]  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2023 / 42, Q. 9 (complete) – Electrochemistry 3 (a) potential difference / voltage between the two half-cells / two electrodes (in a cell) under standard conditions (1) 3 (b) salt bridge voltmeter complete circuit [wires / salt bridge in / touching both solutions] Cl2 Cl– good delivery system (no arrow required) Pt Pt Fe2+ and Fe3+ Any three , any six , all nine (3) 3 (c) M1 (∆GꝊ) = –nFEcellꝊ OR –2 x 96500 x 0.59 M2 (∆GꝊ) = –2 x 96500 x 0.59 = 113870 J mol–1 (∆GꝊ) = –114 kJ mol–1 min 3sf ecf (2) 3 (d) (i) E = 0.77 + (0.059 / 1) log (1 / 0.15) use of z = 1 E = 0.82 (1) 3 (d) (ii) Ecell = 0.59 + 0.77 – (answer to (d)(i)) = 0.54 V ecf (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Papers 2022 Nov 2022 / 42 Nov 2022 / 42, Q. 3 (complete) – Electrochemistry 4 Data should be selected from Table 3.1 in order to answer some parts of this question. Table 3.1 (a) An electrochemical cell can be constructed from a Mg2+ / Mg half-cell and a MnO4– / Mn2+ half-cell. The standard cell potential of this cell can be calculated using the standard electrode potentials of the two half-cells. (i) Define standard electrode potential. Include details of the standard conditions used. ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ……..……………………………………………………………………………………………… A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (ii) Complete the diagram below to show an electrochemical cell constructed from a Mg2+ /Mg half‑cell and a MnO4– /Mn2+ half-cell. Label your diagram. (iii) Use a positive (+) sign and a negative (–) sign to identify the polarity of each of the two electrodes in your diagram. Use an arrow and the symbol ‘e’ to show the direction of electron flow in the external circuit. (iv) Calculate the standard cell potential, EꝊcell, of this cell. EꝊcell = …………………….. V (v) Construct an equation for the cell reaction. …………………………………………………………………………………………………….. (vi) Predict how the cell reaction will change, if at all, when the solution in the Mg2+ /Mg half-cell is diluted by the addition of a large volume of water. Explain your answer. ………………………………………………………………………………………………………… ………………………………………………………………………………………………………… ………………………………………………………………………………………………..……  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (b) A molten magnesium salt is electrolysed for 15.0 minutes by a constant current. 4.75 × 1022 magnesium atoms are produced at the cathode. Calculate the value of the current used. current =.............................. A [Total: 11] Solution Nov 2022 / 42, Q. 3 (Complete) – Electrochemistry 4 (a) (i) the voltage produced by a half-cell compared with a standard hydrogen electrode (1) 1 mol dm–3, 298 K, 1 atm (1) 4 (a) (ii) Mg wire and Pt wire (1) voltmeter, salt bridge, complete circuit (1) solutes Mg2+ and MnO4–, Mn2+, H+ (1) 4 (a) (iii) Mg is minus, Pt is plus arrow points towards MnO4– / Mn2+ half-cell (1) 4 (a) (iv) 3.90 V (1) 4 (a) (v) 5Mg + 2MnO4– + 16H+ → 5Mg2+ + 2Mn2+ + 8H2O (1) 4 (a) (vi) no change and dilution will make Mg2+ / Mg potential even more negative (1) 4 (b) either: 4.75 x 1022 x 2 x 1.60 x 10–19 = 15 200 C OR 2 x 96 500 x (4.75 x 1022) / 6.02 x 1023 = 15 228 C (1) 15 200 / (15 x 60) = 16.9 A OR 15 228 / (15 x 60) = 16.9 A (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2022 / 42, Q. 5 (complete) – Electrochemistry 5 (a) Complete Table 5.1 to predict the substance liberated at each electrode during electrolysis of the indicated electrolyte with inert electrodes. Table 5.1 substance liberated substance liberated electrolyte at the anode at the cathode PbBr2(l) concentrated NaCl(aq) Cu(NO3)2(aq) (b) An electrolytic cell is set up to determine a value for the Avogadro constant, L. The electrolyte is dilute sulfuric acid and both electrodes are copper. When a current of 0.600A is passed through the acid for 30.0 minutes, the anode decreases in mass by 0.350g. (i) State the relationship between the Faraday constant, F, and the Avogadro constant, L. …………………………………………………………………………………………………..… (ii) Use the experimental information in (b) and data from the table on page 23 to calculate a value for the Avogadro constant, L. Show all working. Avogadro constant, L =..............................  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2022 / 42, Q. 5 (Complete) – Electrochemistry 5 (a) Two for one mark, four for two marks, six for three marks 5 (b) (i) F = Le OR F is directly proportional to L (1) 5 (b) (ii) number of Cu2+ formed = 0.35 / 63.5 = 5.51 x 10–3 Q = I x t = 0.60 x 30 x 60 = 1080 C number of electrons = 1080 / 1.6 x 10–19 = 6.75 x 1021 ecf number of Cu2+ ions = 6.75 x 1021 / 2 = 3.375 x 1021 ecf number of Cu2+ ions per mole (L) = 3.375 x 1021 / 5.51 x 10–3 = 6.12 x 1023 ecf min 2sf all five points for four marks ALLOW valid alternate calculations for L  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Specimen 2022 Sp. 2022, P.4, Q. 7 (c) – Electrochemistry 6 (c) A solution containing a mixture of Sn2+(aq) and Sn4+(aq) is added to a solution containing a mixture of Fe2+(aq) and Fe3+(aq). Table 7.2 lists electrode potentials for some electrode reactions of these ions. Table 7.2 electrode reaction EꝊ / V Fe2+ + 2e– ⇌ Fe –0.44 Fe3+ + 3e– ⇌ Fe –0.04 Fe3+ + e– ⇌ Fe2+ +0.77 Sn2+ + 2e– ⇌ Sn –0.14 Sn4+ + 2e– ⇌ Sn2+ +0.15 EꝊ data from the table can be used to predict the reaction that takes place when the two solutions are mixed. (i) Write an equation for this reaction. …………………………………………………………………………………………………... (ii) Calculate EꝊcell for this reaction. ……………...………………………………………………………………………………………. …………………………………………………………………………………………………... Solution Sp. 2022, P.4, Q. 7 (c) – Electrochemistry Question --- 6 (c) (i) Sn2+ + 2Fe3+ ⎯→ Sn4+ + 2Fe2+ ; (1) 6 (c) (ii) EꝊcell = 0.62 (V) (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Papers 2021 Nov 2021 / 42 Nov 2021 / 42, Q. 3 (a), (b), (c), (d) – Electrochemistry 7 Pure water is a very poor conductor of electricity. However, when hydrogen chloride gas is dissolved in water, ions are formed and a current flows during electrolysis. HCl(g) + aq → H+(aq) + Cl–(aq) The overall change after electrolysis is that hydrogen chloride gas is converted into hydrogen and chlorine. 2HCl(g) → H2(g) + Cl2(g) When a current of 3.10 A is passed through the solution for Y minutes, 351 cm3 of chlorine are produced at the anode, measured under room conditions. (a) Calculate the number of chlorine molecules produced during the electrolysis. number of chlorine molecules = …………………….. (b) Calculate the total number of electrons transferred to produce this number of chlorine molecules. total number of electrons = …………………….. (c) Calculate the quantity of charge, in coulombs, of the total number of electrons calculated in (b). quantity of charge = …………………….. C  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (d) Calculate the time, Y, in minutes, for which the current flows. Y = …………………….. minutes Solution Nov 2021 / 42, Q. 3 (a), (b), (c), (d) – Electrochemistry Question --- 7 (a) 0.351 / 24 = 0.015 (mol) (1) 0.015 x 6.02 x 1023 = 9.0 × 1021 / 8.8 × 1021 (1) 7 (b) 1.76 x 1022 / 1.8 x 1022 Working : 8.8 x 1021 x 2 = 1.76 x 1022 Details : The mole ratio Cl2 : e– = 1 : 2 (given by the equation 2Cl– ⎯→ Cl2 + 2e–) So, the moles of chlorine gas calculated in the previous part is multiplied by 2  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 7 (c) 2817 / 2816 / 2820 / 2800 C Working : 1.76 x 1022 x 1.6 x 10–19 = 2.816 x 103 = 2816 C Details : The number of electrons calculated in the previous part is to be multiplied with charge on a single electron … which is given on the third page of data booklet i.e. –1.6 x 10–19. The important point here is that the negative sign in power is retained but the negative sign of the coefficient is not to be used in such calculations. 7 (d) 15 / 15.1 / 15.05 / 15.15 minutes Working : t = 2816 / 3.10 = 908 s , 908 / 60 = 15.1 minutes Details : Rearranging Q = It  t = Q / I So, putting the value of Q calculated in the last part (2816 C) in the numerator and putting the value of I given in the question (3.10) in the denominator we get the answer 908 s Since he requires the answer in minutes asked in this part, so we have to convert this time in seconds into minutes by dividing it with 60.  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Nov 2021 / 42, Q. 6 (d) – Electrochemistry 8 (d) Oxygen can oxidise [Co(NH3)6]2+ to [Co(NH3)6]3+ under standard conditions in alkaline solutions. [Co(NH3)6]3+ + e– ⇌ [Co(NH3)6]2+ EꝊ = +0.10 V (i) Use this information and the Data Booklet to calculate the value for this oxidation of [Co(NH3)6]2+. ……………………………………………………………………………………………..………. ……………………………………………………………………………………………..………. EꝊcell = ……………………. V (ii) Write an ionic equation for this oxidation of [Co(NH3)6]2+. …...………………………………………………………………………………………..…… (iii) Predict, by selecting suitable data from the Data Booklet, whether oxygen can oxidise Co2+(aq) in acidic solution, in the absence of ammonia. Explain your answer. ……………………………………………………………………………………………..………. ……...……………………………………………………………………………………..…… Solution Nov 2021 / 42, Q. 6 (d) – Electrochemistry Question --- 8 (d) (i) EꝊcell = +0.30 (1) 8 (d) (ii) 4[Co(NH3)6]2+ + O2 + 2H2O ⇌ 4[Co(NH3)6]3+ + 4OH– (1) 8 (d) (iii) M1 No, because1.82 V and 1.23 V (1) M2 EꝊcell = –0.59 / 1.23 lower than 1.82 / 1.82 greater than 1.2 (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2021 / 42 June 2021 / 42, Q. 1 (e) – Electrochemistry 9 (e) The [Cr2(O2CCH3)4(H2O)2] complex reacts with aqueous acid to form Cr2+(aq) ions. Cr2+(aq) ions react with O2(aq) under acidic conditions. Cr3+(aq) ions are formed. Use the Data Booklet to answer the following questions. (i) Construct an ionic equation for the reaction of Cr2+(aq) with O2(aq) under acidic conditions. ……...…….……………………………………………………………………………………..... (ii) Calculate EꝊcell cell for the reaction in (e)(i). EꝊcell = ……………………. V Solution June 2021 / 42, Q. 1 (e) – Electrochemistry Question --- 9 (e) (i) 4Cr2+ + O2 + 4H+ → 4Cr3+ + 2H2O OR 2Cr2+ + O2 + 2H+ → 2Cr3+ + H2O2 (2) M1: correct species M2: balancing 9 (e) (ii) EꝊcell = 1.23 – (–0.41) = (+)1.64V OR EꝊcell = 0.68 – (–0.41) = (+)1.09 V (1) value linked to (e)(i)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2021 / 42, Q. 2 (c)(i)(ii) – Electrochemistry 10 (c) (i) Barium ethanedioate, BaC2O4, decomposes on heating to produce barium oxide and a mixture of two different gases. Construct an equation for the decomposition of barium ethanedioate. …..…….……………………………………………………………………………………..... (c) (ii) An impure sample of BaC2O4, of mass 0.500 g, is added to 50.0 cm3 of 0.0200 mol dm–3 acidified MnO4–(aq), an excess. A redox reaction takes place and all the BaC2O4 reacts. The resulting solution, containing unreacted acidified MnO4–, is titrated with 0.0500 mol dm– 3 Fe2+(aq). The end-point is reached when 30.40 cm3 of 0.0500 mol dm–3 Fe2+(aq) has been added. C2O42– ⇌ 2CO2 + 2e– MnO4– + 8H+ + 5e– ⇌ Mn2+ + 4H2O Fe2+ ⇌ Fe3+ + e– Calculate the percentage by mass of BaC2O4 in the 0.500 g impure sample. Show your working. [Mr: BaC2O4, 225.3] percentage by mass of BaC2O4 = ……………….……  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2021 / 42, Q. 2 (c)(i)(ii) – Electrochemistry Question --- 10 (c) (i) BaC2O4 → BaO + CO2 + CO OR BaC2O4 → BaO + 2CO2 + ½O2 (1) 10 (c) (ii) M1: [a] initial moles MnO4– = 0.0200 x 0.050 = 1.00 x 10–3 [b] moles Fe2+ = 0.050 x 0.0304 = 1.52 x 10–3 M2: [a] moles MnO4– unreacted = 1.52 x 10–3 / 5 = 3.04 x 10–4 [b] moles MnO4– reacted = 1.00 x 10–3 – 3.04 x 10–4 = 6.96 x 10–4 M3: moles C2O42– reacted = 6.96 x 10–4 x 5/2 = 1.74 x 10–3 M4: mass of BaC2O4 = 225.3 x 1.74 x 10–3 = 0.392g % Purity of BaC2O4 = 100 x 0.392/0.50 = 78.4  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2021 / 42, Q. 3 (a) (i), (ii) – Electrochemistry 11 (a) (i) Define the term standard electrode potential. …..…….…………………………………………………………………………………….......... …..…….…………………………………………………………………………………….......... …..…….……………………………………………………………………………………..... Three redox systems, A, B and C, are shown. The ligand 1,2-diaminoethane, H2NCH2CH2NH2, is represented by en. A [Ru(H2O)6]3+ + e– ⇌ [Ru(H2O)6]2+ B [Ru(NH3)6]3+ + e– ⇌ [Ru(NH3)6]2+ C [Ru(en)3]3+ + e– ⇌ [Ru(en)3]2+ Two electrochemical cells are set up to compare the standard electrode potentials, EꝊ, of three half- cells. The diagrams show the relative potential of each electrode. (ii) Use this information to complete the table by adding the labels A, B and C to deduce the order of EꝊ for the three half-cells. EꝊ redox system most negative least negative  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2021 / 42, Q. 3 (a) (i), (ii) – Electrochemistry Question --- 11 (a) (i) M1: voltage of an electrode / a half-cell compared to / connected to (standard) hydrogen electrode / half-cell M2: (at concentration of) 1 mol dm–3 AND (pressure of) 1 atm / 101 kPa (or in Pa) AND 298 K / 25oC (2) 11 (a) (ii) EꝊ redox system most negative B C least negative A (1)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2021 / 42, Q. 3 (b) – Electrochemistry 12 (b) (i) An electrochemical cell consists of a Br2/Br– half-cell and a Ag+/Ag half-cell, under standard conditions. Use the Data Booklet to calculate the EꝊcell. Deduce the direction of electron flow in the wire through the voltmeter between these two half-cells. EꝊcell = ……………………. V direction of electron flow from ………………………….. to ……..…………………….. (ii) Water is added to the Ag+/Ag half-cell in (b)(i). Suggest the effect of this addition on the Ecell. Place a tick () in the appropriate box. less positive no change more positive Explain your answer. …..…….…………………………………………………………………………………….......... …..…….…………………………………………………………………………………….......... …..…….…………………………………………………………………………………….... Solution June 2021 / 42, Q. 3 (b) – Electrochemistry Question --- 12 (b) (i) EꝊcell = 1.07 – 0.80 = (+)0.27 V AND direction of electron flow = Ag+ / Ag to Br2 / Br– (1) 12 (b) (ii) M1: EꝊcell third box ticked M2: [Ag+] decreases AND so (Ag+ / Ag) equilibrium shifts towards the left OR [Ag+] decreases AND E for (Ag+ / Ag) becomes less positive / more negative (2)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Papers 2020 Nov 2020 / 42 Nov. 2020 / 42, Q. 4 – Electrochemistry 13 (a) Identify the substances liberated at the anode and at the cathode during the electrolysis of saturated KCl (aq). at the anode ………….………………………………………………………………………………... at the cathode …….…………………………………………………………………………..……….. (b) When dilute sulfuric acid is electrolysed, oxygen is liberated at the anode. Dilute sulfuric acid is electrolysed for 15.0 minutes using a current of 0.750 A. Calculate the volume of oxygen that is liberated under room conditions. volume of oxygen =.............................. cm3 (c) The halogens chlorine, bromine and iodine differ in their strengths as oxidising agents. These strengths are indicated by the EꝊ values for these halogens. (i) Give the EꝊ values for chlorine, bromine and iodine acting as oxidising agents. ….………………………………….………………………………………………………….. A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (ii) Deduce which of chlorine, bromine and iodine will react with a solution of Sn2+(aq) under standard conditions. Explain your answer. Include a relevant equation in your explanation. …..…………………………………………………………………………………………………. …..…………………………………………………………………………………………………. …..………………………………………………………………………………………..…… (iii) An excess of chlorine is added to a solution of acidified Mn2+(aq) under standard conditions. Give the formula of the product of this reaction that contains manganese. …….…………………………………………………………………………………………… (d) An electrochemical cell can be made by connecting an Fe3+ / Fe2+ half-cell to an S2O82– / SO42– half-cell under standard conditions. (i) Calculate the standard cell potential of this electrochemical cell. EꝊcell = ……………………... V (ii) State the material that should be used as the electrode in each half-cell. in the Fe3+ / Fe2+ half-cell …….…………………………………………………………………. in the S2O82– / SO42– half-cell ………………………………………..………………………….. (iii) Describe one change to each half‑cell that would increase the value of the cell potential. The temperature should remain at 298 K. Fe3+ / Fe2+ half-cell ………………..……………………………………………………………... …………………………………………..…………………………………………………………. S2O82– / SO42– half-cell ……………………………………….………………………………….. …………………………………………………………………..…………………………………. [Total: 12]  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution Nov 2020 / 42, Q. 4 – Electrochemistry Question --- 13 (a) chlorine AND hydrogen 13 (b) 15 x 60 x 0.75 = 675 C –3 – 675 / 96 500 = 7.0 × 10 moles e 7.0 x 10–3 x 0.25 gives 1.75 x 10–3 moles O2 1.75 x 10–3 x 24000 = 42 (41.969) cm3 O2 OR 15 × 60 x 0.75 = 675 C 675 / 1.60 x 10 = 4.22 x 10 e = 7.01 x 10–3 moles e– –19 21 – gives 1.75 x 10–3 moles O2 = 42 (42.047) cm3 13 (c) (i) 1.36 1.07 0.54 13 (c) (ii) all of them (all EꝊ values) greater than 0.15 / EꝊ cell greater than zero e.g. Sn2+ + X2 ⎯→ Sn4+ + 2X 13 (c) (iii) MnO2 13 (d) (i) 1.24 V 13 (d) (ii) platinum, platinum 13 (d) (iii) increase [Fe2+] or decrease [Fe3+] increase [S2O82–] or decrease [SO42–]  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2020 / 42 June 2020 / 42, Q. 2(f) – Electrochemistry 14 (f) The half-equation for the reduction of iodate(V) ions is shown. IO3– + 6H+ + 5e– → ½ I2 + 3H2O EꝊ = +1.19 V Use data from the Data Booklet to predict whether a reaction is feasible when aqueous solutions of acidified iodate(V) ions and bromide ions are mixed. Explain your answer. ……..……………………………………………………………………………………………………. ……….……………………………………………………………………………………...……… Solution June 2020 / 42, Q. 2(f) – Electrochemistry Question --- 14 (f) It is feasible as the Ecell will be positive/+0.12 V 1  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2020 / 42, Q. 8 – Electrochemistry 15 (a) (i) Define the term standard cell potential. …..…………………………………………………………………………………………………. ……..………………………………………………………………………………………………. ……….………………………………………………………………………………...……… An electrochemical cell is set up to measure the standard electrode potential of a cell, EꝊcell, made of a Co3+ / Co2+ half-cell and a Cl2 / Cl– half-cell. (ii) Complete the table with the substance used to make the electrode in each of these half‑cells. half-cell electrode Co3+ / Co2+ Cl2 / Cl (iii) Use data from the Data Booklet to calculate the EꝊcell. EꝊcell = …………………….. V (iv) Write the equation for the overall cell reaction. ……….………………………………………………………………………………...………  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (b) A fuel cell is an electrochemical cell that can be used to generate electrical energy. A methanol-oxygen fuel cell can be used as an alternative to a hydrogen-oxygen fuel cell. When the cell operates, the carbon atoms in the methanol molecules are converted into carbon dioxide. CH3OH + H2O → CO2 + 6H+ + 6e– Calculate the volume of CO2, in cm3, formed when a current of 2.5 A is delivered by the cell for 30 minutes. Assume the cell is operated at room conditions. volume of CO2 = …………………….. cm3 [Total: 7] Solution June 2020 / 42, Q. 8 – Electrochemistry Question --- 15 (a) (i) M1 potential difference between two half-cells/two electrodes in a cell M2 under conditions of 1 atm., 298 K, (all) solutions being 1 mol dm–3 15 (a) (ii) both platinum 15 (a) (iii) EꝊcell = 1.82 – 1.36 = = (+)0.46 V 15 (a) (iv) 2Co3+ + 2Cl– ⎯→ Cl2 + 2Co2+ 15 (b) M1 Q = 2.5 x 30 x 60 C = 4500 C AND 96500 OR 579000 seen moles of CO2 = 4500 / 579000 = 7.8 x 10–3 or 7.77 x 10–3 M2 volume of CO2 = 7.77 x 10–3 x 24000 = 187 cm3  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Papers 2019 Nov. 2019 / 42 Nov 2019 / 42, Q. 1 (Whole) – Electrochemistry 16 An electrochemical cell is constructed using two half-cells. a Br2/Br– half-cell an Mn3+/Mn2+ half-cell (a) State the material used for the electrode in each half-cell. Br2/Br– half-cell.......................................................................................................................... Mn3+/Mn2+ half-cell..................................................................................................................... (b) The cell is operated at 298 K. The Br2/Br– half-cell has standard concentrations. The Mn3+/Mn2+ half-cell has [Mn3+] = 0.500 mol dm–3 and [Mn2+] = 0.100 mol dm–3. (i) Use the Nernst equation to calculate the electrode potential, E, of the Mn3+/Mn2+ half‑cell under these conditions. E =.............................. V (ii) Calculate the Ecell under these conditions. Ecell =.............................. V (iii) Write an equation for the overall cell reaction that occurs. ………………………………………..………………….…………………………………….. A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (c) An aqueous solution of copper(II) sulfate is electrolysed using copper electrodes. A current of 1.50 A is passed for 3.00 hours. 5.09 g of copper is deposited on the cathode. The charge on one electron is –1.60 × 10–19 C. The relative atomic mass of copper is 63.5. Use these data to calculate an experimentally determined value for the Avogadro constant, L. Give your answer to three significant figures. L =.............................. mol–1 (d) Explain why magnesium metal cannot be obtained by the electrolysis of dilute aqueous magnesium sulfate. Your answer should include data from the Data Booklet. …………………………………………………………………………………………………………... …………………………………………………………………………………………………………... …………………………………………………………………………………………………………... …………………………………………………………………………………………………..…… [Total: 13]  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution Nov 2019 / 42, Q. 1 (Whole) – Electrochemistry Question --- 16 (a) Platinum and platinum 16 (b) (i) M1: Nernst quoted correctly E = EꝊ + 0.0590 / zlog [ox] / [red] or E = 1.49 + 0.0590log5 M1: (+)1.53 V minimum 2 sig. fig. 16 (b) (ii) +/– 0.46 minimum 2 sig. fig. 16 (b) (iii) M1: Mn3+ + 2Br– ⎯→ Mn2+ + Br2 M2: 2Mn3+ + 2Br– ⎯→ Mn2+ + Br2 16 (c) M1: 16200 C M2: 1.0125 x 1023 electrons (use of 1.60 x 10–19) M3: 0.0802 moles of copper (use of 5.09 and 63.5) M4: 0.1603 moles electrons M5: L = 6.32 x 1023 (correct answer other approaches acceptable including: M1: 16200 C M2: 1.0125 x 1023 electrons (use of 1.60 × 10–19) M3: 5.0625 x 1022 copper atoms M4: 0.0802 moles of copper (use of 5.09 and 63.5) M5: L = 6.32 x 1023 (correct answer ) 16 (d) M1: Mg2+ + 2e– ⇌ Mg EꝊ = –2.38 and 2H+ + 2e- ⇌ H2 EꝊ = 0.00 M2: hydrogen produced instead / hydrogen easier to reduce / hydrogen preferentially reduced / hydrogen has more positive EꝊ  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Nov 2019 / 42, Q. 6(d) – Electrochemistry 17 (d) When chlorine gas is bubbled into FeSO4(aq) the colour of the solution changes from pale green to yellow. Use data from the Data Booklet to explain this observation. Include an equation in your answer. Reference to electron movement between orbitals is not needed. ……..……………………………………………………………………………………………………. ………..…………………………………………………………………………………………………. ………….…………………………………………………………………………………………… Solution Nov 2019 / 42, Q. 6(d) – Electrochemistry Question --- 17 (d) M1: EꝊ values 1.36 and 0.77 quoted M2: 2FeSO4 + Cl2 ⎯→ Fe2(SO4)2Cl2 or 2Fe2+ + Cl2 ⎯→ 2Fe3+ + 2Cl–  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2019 / 42 June 2019 / 42, Q. 1 (d), (e) – Electrochemistry 18 (d) Define the term standard electrode potential, EꝊ. ……..……………………………………………………………………………………………………. ………..…………………………………………………………………………………………………. ………….…………………………………………………………………………………………… (e) (i) Complete and label the diagram to show how the standard electrode potential, EꝊ, of Ag+(aq) / Ag(s) could be measured under standard conditions. (ii) Use the Data Booklet to label the diagram in (e)(i) to show which is the positive electrode, the direction of electron flow in the external circuit when a current flows.  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2019 / 42, Q. 1 (d), (e) – Electrochemistry Question --- 18 (d) The potential difference when a half-cell is connected to a (standard) hydrogen electrode under standard conditions OR the potential difference / voltage / EMF between a hydrogen electrode and another half-cell under standard conditions 18 (e) (i) salt bridge ⚫ voltmeter / V ⚫ Ag ⚫ Ag+ (or soluble silver salt) ⚫ Pt ⚫ H2 (and delivery correct) + H+ (or named strong acid) ⚫ 1 atm. pressure ⚫ 1 mol dm–1 (and 298 K) ⚫ mark as ⚫  ⚫  ⚫  ⚫  18 (e) (ii) Ag electrode labelled and arrow (in the external circuit moving towards this electrode)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Papers 2018 Nov 2018 /42 Nov. 2018/42, Q. 5 (c) – Electrochemistry 19 (c) Acidified manganate(VII) ions, MnO4–, can also be used to analyse solutions containing nitrite ions, NO2–, by titration. In acidic solution, NO2– ions exist as HNO2. (i) Use the Data Booklet to construct an ionic equation for this reaction. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……..…………………………………………………………………………………………… (ii) Use EꝊ values to calculate the EꝊcell for this reaction. EꝊcell = …………………….. V [Total: 3] Solution Nov. 2018/42, Q. 5 (c) – Electrochemistry Question --- 19 (c) (i) 5NO2– + 2MnO4– + 6H+ ⎯→ 2Mn2+ + 5NO3– + 3H2O OR 5HNO2 + 2MnO4– + H+ ⎯→ 2Mn2+ + 5NO3– + 3H2O all species correct balanced 19 (c) (ii) EꝊcell = 1.52 – 0.94 = 0.58 (V)  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2018/42 June 2018/42, Q. 3 (a), (b), (c) – Electrochemistry 20 (a) Complete the table by predicting the identity of the substance liberated at each electrode during electrolysis with inert electrodes. substance liberated substance liberated electrolyte at the anode at the cathode NaOH(aq) dilute CuCl2(aq) concentrated MgCl2(aq) (b) (i) The electrolysis of molten ZnBr2 is a redox process. Identify the ion that is oxidised and the ion that is reduced. Use ionic half-equations to explain your answer. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……..…………………………………………………………………………………………… (ii) Describe one visual observation that would be made during this electrolysis. ……..…………………………………………………………………………………………… A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 (c) Dilute sulfuric acid is electrolysed for 50.0 minutes using inert electrodes and a current of 1.20 A. A different gas is collected above each electrode. The volumes of the two gases are measured under room conditions. Calculate the maximum volume of gas that could be collected at the cathode. volume = …………………….. cm3 [Total: 10]  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 Solution June 2018/42, Q. 3 (a), (b), (c) – Electrochemistry Question --- 20 (a) anode cathode NaOH (aq) oxygen / O2 hydrogen / H2 dilute CuCl2 (aq) oxygen / O2 copper / Cu conc MgCl2 (aq) chlorine / Cl2 hydrogen / H2 20 (b) (i) 2Br– → Br2 + 2e– or 2Br– – 2e– → Br2 Zn2+ + 2e– → Zn Zinc / Zn2+ reduced and Br– / bromide oxidised 20 (b) (ii) liquid / molten metal or orange-brown / reddish brown vapour / gas (at anode) or amount of melt / electrolyte decreases 20 (c) 50 × 60 × 1.2 or 3600 C (calculation of number of Coulombs) 3600 / 96 500 or 0.0373 F (calculation of number of Faradays) 0.0373 F / 2 or 0.01865 / 0.0187 mol H2 (use of stoichiometry) 0.01865 × 24 000 = 448–449 (Use of 24 000 & answer to 3sf) 2 points = 1 mark 3 points = 2 marks 4 points = 3 marks  A2/L WS. Electroch. – P.4 2023-18 Saad Hameed MSc Chemistry  : 0300-4291902 June 2018/42, Q. 5 (d) – Electrochemistry 21 (d) Co2+(aq) can be oxidised to Co3+(aq). (i) Use the Data Booklet to suggest a suitable oxidising agent for this reaction. ……..…………………………………………………………………………………………… (ii) Calculate the EꝊcell of this reaction. EꝊcell = …………………….. V (iii) Write an equation for the reaction between Co2+ and the oxidising agent you chose in (d)(i). ……..…………………………………………………………………………………………… Solution June 2018/42, Q. 5 (d) – Electrochemistry Question --- 21 (d) (i) F2 or S2O82– 21 (d) (ii) +1.05 or +0.19 21 (d) (iii) 2Co2+ + F2 → 2Co3+ + 2F– or 2Co + S2O8 → 2Co + 2SO42– 2+ 2– 3+ 21 (d) (iii) 2Co2+ + F2 → 2Co3+ + 2F– or 2Co2+ + S2O82– → 2Co3+ + 2SO42– …………………………………………… With Best Wishes [email protected]