Free until 14 Nov · 80% off for the first 100 · Pricing

Chemistry 9701/42 May/June 2023: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_s23_qp_429701_s23_ms_429701/42/M/J/23

Paper
Paper 4 A Level Structured Questions
Total marks
100
Time
2 hours
Questions
9
Code on the paper
9701/42/M/J/23
9701_s23_ms_42.pdf

9701/42 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 113 marks

Question

(a) Group 2 carbonates decompose when heated to form the metal oxide and carbon dioxide. (i) Suggest a mechanism for the decomposition of the carbonate ion by adding two curly arrows in Fig. 1.1. O C CO2 + O2– –O O– Fig. 1.1 [1] (ii) Describe the variation in the thermal stability of Group 2 carbonates. Explain your answer. …… [3] (b) (i) Define lattice energy. …… [2] o (ii) The lattice energy of the Group 2 carbonates, ΔH latt (MCO3), becomes less exothermic down the group. o The lattice energy of the Group 2 oxides, ΔH latt (MO), also becomes less exothermic down the group. o o ΔH latt (MCO3) and ΔH latt (MO) change by different amounts going down the group. Suggest how the standard enthalpy change of the decomposition reaction for Group 2 carbonates changes down the group. Explain your reasoning in terms of the relative sizes of the anions and the relative changes in lattice energy down the group. …… [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 250 cm3 solution contains 3.40 g of impure K2SO3. • 25.0 cm3 of this solution requires 22.40 cm3 of 0.0250 mol dm–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–. …… [1] (ii) Calculate the percentage purity of the sample of K2SO3. Show your working. percentage purity of K2SO3 = …… [3] (d) Potassium disulfite, K2S2O5, is another food additive. The disulfite ion, S2O52–, has the displayed formula shown in Fig. 1.2. O O– α S S O O O– Fig. 1.2 Deduce the geometry (shape) around the S(α) atom in S2O52–. geometry around S(α) …… [1] [Total: 13]

Mark scheme

1(a)(i) 1 ARROW 1 starts from O– to the C–O bond AND ARROW 2 starts at the C-O bond to other O– ion 1(a)(ii) M1 increases (down the group) 3 M2 (cat)ionic radius / ion size increases (down the group) OR charge density of M2+ decreases M3 less polarisation / distortion of anion / carbonate ion / CO3(2)– 1(b)(i) M1 energy released when one mole of a ionic solid / compound is formed 2 M2 from gas (phase) ion(s) / gaseous ion(s) (under standard conditions) 1(b)(ii) (ΔHdecomp / it) becomes more positive/less negative 2 (down the group) size /(ionic) radii of oxide ion is smaller (than carbonate ion) ORA so ΔHlatt of oxides becomes ORA less exothermic faster OR less negative faster OR changes more OR changes faster Any two [1], all three [2] 1(c)(i) 2MnO4– + 6H+ + 5SO32– → 2Mn2+ + 3H2O + 5SO42– 1 1(c)(ii) M1 M2 3 any two bullets [1] or all four [2] moles MnO4– = 0.025 × 22.40 / 1000 = 5.6 × 10–4 moles SO32– = 5.6 × 10–4 × 5 / 2 = 1.4 × 10–3 (in 25 cm3) ecf from (c)(i) and bullet 1 moles SO32– = 1.4 × 10–2 (in 250 cm3) ecf bullet 2 mass K2SO3 = 1.4 × 10–2 × 158.3 = 2.2162 g ecf bullet 3 OR moles K2SO3 (if 100% pure) = 3.40 ÷ 158.3 = 0.02148 M3 % purity = 100 × 2.2162 / 3.40 = 65.2 / 65.3 % ecf min 2sf OR % purity = 100 × 0.014 / 0.02148 = 65.2 / 65.3 % 1(d) S(α) tetrahedral 1

Question 212 marks

Question

(a) State two typical chemical properties of a transition element. 1 …… 2 …… [1] (b) Aqueous solutions of cobalt(II) salts contain the complex ion [Co(H2O)6]2+. (i) Define complex ion. …… [1] (ii) Samples of [Co(H2O)6]2+ are reacted separately with an excess of aqueous ammonia, with an excess of concentrated HCl and with an excess of aqueous sodium hydroxide, as shown in Fig. 2.1. excess excess NH3(aq) NaOH(aq) A [Co(H2O)6]2+(aq) C excess concentrated HCl B Fig. 2.1 Complete Table 2.1 about the reactions shown by [Co(H2O)6]2+. Table 2.1 reagent added to formula of cobalt colour and state of type of reaction [Co(H2O)6]2+(aq) species formed cobalt species formed an excess of NH3(aq) A= an excess of B= concentrated HCl an excess of NaOH(aq) C= [4] (c) The ethanedioate ion, C2O42–, can act as a bidentate ligand. (i) Explain what is meant by a bidentate ligand. …… [2] (ii) The complex [Co(H2O)2(C2O4)BrCl ]− exists as stereoisomers. Complete the three-dimensional diagrams in Fig. 2.2 to show four stereoisomers of [Co(H2O)2(C2O4)BrCl ]−. The C2O42– ligand is represented using O O. O O Co Co O O O O Co Co O O Fig. 2.2 [3] (iii) State the oxidation state of cobalt in this complex and a type of stereoisomerism shown. oxidation state of cobalt …… type of stereoisomerism …… [1] [Total: 12]

Mark scheme

2(a) Any two from: 1 they have variable / multiple oxidation states OWTTE they behave as catalysts they form complex ions / complexes they form coloured compounds/ions 2(b)(i) (is a molecule or ion formed by a central) metal atom / metal ion bonded / surrounded by (one or more) ligands 1 2(b)(ii) 4 formula of colour / state of reagent added type of reaction species formed species formed brown solution an excess of ligand [Co(NH3)6]2+ ALLOW yellow- NH3(aq) exchange brown solution an excess of ligand [CoCl4]2– blue solution concentrated HCl exchange Co(OH)2 blue ppt. NaOH(aq) OR precipitation ALLOW pink ppt Co(OH)2(H2O)4 formula and colour / state: any two [1], any four [2], all six [3] type of reaction: all three correct [1] 2(c)(i) M1 (a species) that donates / uses two lone pairs 2 M2 to form a two dative / coordinate bond to a metal atom / metal ion 2(c)(ii) Any four from: 3 Any two [1], any three [2], all four [3] 2(c)(iii) oxidation state of cobalt (+)3 1 AND type of stereoisomerism cis-trans OR geometric(al) OR optical

Question 312 marks

Question

(a) Complete Table 3.1 by placing one tick (3) in each row to indicate the sign of each type of energy change under standard conditions. Table 3.1 can be either energy change always positive always negative negative or positive bond energy enthalpy change of atomisation enthalpy change of formation [1] (b) Define standard enthalpy change of atomisation. …… [2] (c) Table 3.2 shows some energy changes. Table 3.2 energy change value / kJ mol–1 standard enthalpy change of atomisation of silver +285 first ionisation energy of silver +731 second ionisation energy of silver +2074 bond energy of O=O +496 bond energy of O–O +150 first electron affinity of oxygen –141 second electron affinity of oxygen +798 first ionisation energy of oxygen +1314 standard enthalpy change of formation of silver oxide, Ag2O(s) –31 o Calculate the lattice energy, ΔH latt , of Ag2O(s) using relevant data from Table 3.2. It may be helpful to draw a labelled energy cycle. Show your working. o ΔH latt of Ag2O(s) = …… kJ mol–1 [3] (d) Suggest the trend in the magnitude of the lattice energies of the silver compounds Ag2S, Ag2O and Ag2Se. Explain your answer. least exothermic most exothermic [2] (e) Silver sulfite, Ag2SO3(s), is sparingly soluble in water. (i) Give an expression for the solubility product, Ksp, of Ag2SO3. Ksp = [1] (ii) Calculate the equilibrium concentration of Ag+ in a saturated solution of Ag2SO3 at 298 K. [Ksp: Ag2SO3, 1.50 × 10–14 mol3 dm–9 at 298 K] [Ag+] = …… mol dm–3 [1] o (f) The standard enthalpy change of solution, ΔH sol , of AgNO3(s) in water is +22.6 kJ mol–1. Suggest how the feasibility of dissolving AgNO3(s) in water changes with temperature. Explain your answer. …… [2] [Total: 12]

Mark scheme

3(a) 1 always always can be either energy change positive negative negative or positive bond energy enthalpy change of atomisation enthalpy change of formation 3(b) M1 (enthalpy change) when one mole of gaseous atoms is produced IGNORE energy released 2 M2 from its element(s) in its standard state / standard conditions / 298 K AND 1 atm 3(c) M1 use of correct six numbers only –31 / 285 / 731 / –141 / 798 / 496 3 M2 2 × used correctly with Ag (2 × 285 (570) and 2 × 731 (1462)) AND 0.5 with O=O (496 (248)) M3 correct signs and evaluation –31 = (2 × 285) + (2 × 731) + (–141) + (798) + x + (0.5 × 496) x = –2968 kJ mol–1 3(d) Ag2Se Ag2S Ag2O OWTTE 2 least exothermic most exothermic charge density of anion decreases down the group ORA / radius/size of anion increases down the group / Se2– largest radius / O2– smallest radius / O has smallest ionic radius less attraction between the ions / ionic bond gets weaker (with Ag2Se) ORA Any two [1], all three [2] 3(e)(i) (Ksp = ) [Ag+]2[SO32–] [1] 1 3(e)(ii) x = 3√1.5 × 10–14 / 4 = 1.55 × 10–5 (mol dm-3) 1 [Ag+] = 1.55 × 10–5 × 2 = 3.11 × 10–5 (mol dm-3) min 2sf ecf (e)(i) 3(f) feasibility / it increases as temperature increases ORA 2 ΔS is positive / ΔS is >0 / entropy change is positive (and ΔH is positive) OR –TΔS becomes more negative / TΔS becomes more positive as ΔG becomes / is negative / ΔG < 0 Any two [1], all three [2]

Question 410 marks

Question

(a) In aqueous solution, iron(III) ions react with iodide ions, as shown. 2Fe3+ + 2I– 2Fe2+ + I2 A series of experiments is carried out using different concentrations of Fe3+ and I–, as shown in Table 4.1. Table 4.1 experiment [Fe3+] / mol dm–3 [I–] / mol dm–3 initial rate / mol dm–3 s–1 1 0.0400 0.0200 2.64 × 10–4 2 0.1200 0.0200 7.92 × 10–4 3 0.0800 0.0400 2.11 × 10–3 (i) Explain what is meant by overall order of reaction. …… [1] (ii) Use the data in Table 4.1 to deduce the order of reaction with respect to Fe3+ and with respect to I–. Explain your reasoning. …… [2] (iii) Use your answer to (a)(ii) to construct the rate equation for this reaction. rate = [1] (iv) Use your answer to (a)(iii) and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = …… units …… [2] (v) Describe qualitatively the effect of an increase in temperature on the rate constant and on the rate of this reaction. …… [1] (b) In aqueous solution, iodide ions react with acidified hydrogen peroxide, as shown. 2I– + H2O2 + 2H+ I2 + 2H2O The initial rate of reaction is found to be first order with respect to I–, first order with respect to H2O2 and zero order with respect to H+. Fig. 4.1 shows a possible four-step mechanism for this reaction. step 1 H2O2 + I– IO– + H2O step 2 H+ + IO– HIO step 3 HIO + I– I2 + OH– step 4 OH– + H+ H2O Fig. 4.1 (i) Suggest which of the steps, 1, 2, 3 or 4, in this mechanism is the rate-determining step. Explain your answer. …… [1] (ii) Identify a step in Fig. 4.1 that involves a redox reaction. Explain your answer in terms of oxidation numbers. …… [1] (iii) Suggest the role of HIO in this mechanism. Explain your reasoning. …… [1] [Total: 10]

Mark scheme

4(a)(i) the sum/total of the power / exponent to which a concentration of a reactant is raised in the rate equation / law 1 4(a)(ii) M1 (expt 1 and 2) when [Fe3+] × 3, rate × 3 so first order w.r.t. Fe3+ 2 M2 (expt 1 and 3) when [Fe3+] × 2, [I–] × 2, rate × 8 so second order w.r.t. I– OR (expt 2 and 3) when [Fe3+] ÷ 1.5, [I–] × 2, rate × 2.7 so second order w.r.t. I– 4(a)(iii) (rate =) k[Fe3+][I–]2 [1] ecf (a)(ii) 1 4(a)(iv) k = rate / [Fe3+][I–]2 = (2.64 × 10–4) / (0.04 × 0.022) 2 k = 16.5 min 2sf ecf units = mol-2 dm6 s–1 ecf 4(a)(v) (k and rate of reaction) both increase 1 4(b)(i) step 1 as this has one mole of H2O2 and one mole of I– 1 OR step 1 as correct stoichiometry / coefficients of 1 H2O2 and 1 I– OR step 1 as number of moles of each reactant are consistent with rate equation / their orders 4(b)(ii) step 1 I oxid no –1 → +1 AND O oxid no –1 → –2 1 OR step 3 I oxid no –1 → 0 AND I oxid no +1 → 0 4(b)(iii) intermediate AND formed (in step 2) and used up (in step 3) 1 ALLOW oxidising agent (in step 3) AND oxidises I– (to I2)

Question 514 marks

Question

(a) Methylbenzene can undergo different reactions, as shown in Fig. 5.1. H2 / Pt Br2 / UV light reaction 1 reaction 2 reaction 3 Br2 / FeBr3 + Fig. 5.1 (i) Draw structures in Fig. 5.1 for the possible organic products of the three reactions shown. [3] (ii) Complete Table 5.1. Table 5.1 type of reaction mechanism reaction 1 reaction 2 [2] (b) When methylbenzene reacts with an electrophile, a substitution reaction occurs. No addition reaction takes place under these conditions. Explain why no addition reaction takes place. …… [1] (c) The reaction of methylbenzene with thionyl bromide, SOBr2, in the presence of an iron(III) bromide catalyst, FeBr3, is shown in Fig. 5.2. O S FeBr3 + SOBr2 Br + HBr Fig. 5.2 The mechanism of this reaction is similar to that of the bromination of benzene. The first step of the mechanism generates the SOBr+ electrophile, as shown. SOBr2 + FeBr3 SOBr+ + FeBr4– (i) The reaction of methylbenzene with SOBr+ ions is shown in Fig. 5.3. Complete the mechanism in Fig. 5.3. Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. intermediate O S+ Br O S Br + …… Fig. 5.3 [3] (ii) The reaction shown in Fig. 5.2 produces a small amount of a by-product, P, with the molecular formula C14H14OS. Suggest a structure for by-product P. [1] (d) Acyl bromides, RCOBr, can be synthesised by the reaction of a carboxylic acid and SOBr2. This is a similar reaction to the synthesis of acyl chlorides using SOCl 2. (i) Give an equation for the reaction between ethanoic acid and SOBr2. …… [1] (ii) Suggest the relative ease of hydrolysis of acyl bromides, RCOBr, acyl chlorides, RCOCl, and alkyl chlorides, RCl. Explain your answer. …… > …… > …… easiest to hydrolyse hardest to hydrolyse [3] [Total: 14]

Mark scheme

5(a)(i) 3 M1 M2 M3 BOTH 5(a)(ii) M1 reaction 1: hydrogenation / reduction 2 M2 mechanism 2: (free) radical substitution 5(b) the substitution product is stabilised by delocalization of π-electrons / by π-electrons in the ring 1 OR the addition product is not stabilised by delocalisation (of π-)electrons ALLOW addition product will remove π-electron delocalised system 5(c)(i) 3 M1 curly arrow from inside the hexagon to S atom M2 structure of the intermediate M3 curly arrow from C-H bond into the ring AND formation/loss of H+ 5(c)(ii) 1 5(d)(i) CH3COOH + SOBr2 → CH3COBr + HBr + SO2 1 5(d)(ii) M1 acyl bromides > acyl chlorides > alkyl chlorides 3 easiest hardest M2 M3 Any two from: Acyl bromides easiest in acyl bromides C–Br is much weaker ORA AND due to less orbital overlap / Br having a larger atomic radii Acyl halides easier than alkyl chlorides carbon in C–Cl / C–Br / C–X bond is more δ+/electron deficient OR C–Cl / C–Br bond is weaker (than C-Cl in alkyl chlorides) AND attached to an oxygen atom / two electronegative atoms / electron withdrawing C=O group ORA alkyl chlorides hardest in alkyl chlorides C–Cl bond strengthened AND by positive inductive effect / electron donating effect of alkyl / R group

Question 68 marks

Question

(a) Perindopril is a drug used to treat heart disease. perindopril O O H O OH N O N Fig. 6.1 (i) State the number of chiral carbon atoms present in one molecule of perindopril. …… [1] (ii) Suggest one benefit and one disadvantage of producing a drug such as perindopril as a single pure optical isomer. benefit …… disadvantage …… [2] (b) (i) Name all the functional groups in perindopril. …… [2] (ii) A sample of perindopril is hydrolysed with hot aqueous acid. Draw the structures of the three organic products of the complete acid hydrolysis of perindopril. [3] [Total: 8]

Mark scheme

6(a)(i) 5 / five 1 6(a)(ii) M1 benefit: 2 higher biological efficiency / activity (of the drug) OR less side effects OR smaller dose required (as drug more potent) M2 disadvantage: the need to separate (a racemic mixture into a single stereoisomer) OR lower yield (of biologically active molecule/product) OR need a chiral catalyst / enzyme (in the synthesis so expensive) 6(b)(i) carboxylic acid/carboxyl, ester, amide, amine 2 Any two [1], all four [2] 6(b)(ii) 3 3 × [1] for each structure

Question 78 marks

Question

(a) Explain why phenol is brominated much more easily than benzene is brominated. …… [3] (b) Iodine monobromide, I–Br, reacts with benzene in the presence of an Al Br3 catalyst. Predict whether the organic product will be bromobenzene or iodobenzene. Explain your answer. …… [1] (c) Fig. 7.1 shows some reactions of phenol. Na(s) excess Br2(aq) Q OH R reaction 1 reaction 2 O2N OH N N OH Fig. 7.1 (i) Give an equation for the reaction of phenol with Na(s). …… [1] (ii) Draw the structure of the organic product, R, formed when phenol reacts with an excess of Br2(aq). [1] (iii) State the reagents and conditions for reaction 1 and reaction 2 in Fig. 7.1. reaction 1 …… reaction 2 …… [2] [Total: 8]

Mark scheme

7(a) M1 p-orbital / lone pair from O atom is overlaps / delocalised into the ring 3 M2 greater π electron density around the ring OR makes the ring more electron rich / positions 2,4,6 more electron rich M3 polarises electrophiles more easily 7(b) iodobenzene AND as Br is more electronegative (than I / Iδ+/I+ in the electrophile) 1 7(c)(i) C6H6O / C6H5OH + Na → C6H5O–Na+ + ½H2 1 7(c)(ii) 1 7(c)(iii) M1 reaction 1: aqueous HNO3 / dilute HNO3 2 M2 reaction 2: alkaline/ NaOH AND C6H5N2+ (Cl–) OR alkaline / NaOH AND benzene / phenyl diazonium ion / salt

Question 815 marks

Question

(a) Describe the difference in reactivity between chloroethane and chlorobenzene with OH–(aq). Explain your answer. …… [2] (b) Compound T, C5H9O2Cl, is a useful synthetic intermediate. Fig. 8.1 shows some reactions of T. T W X O excess NaOH(aq) + Cl O heat step 1 Y O excess HCl (aq) OH + NC O heat step 2 Z, C6H13NO2 Fig. 8.1 (i) Give the systematic name for T. …… [1] (ii) Draw the structures of W, X, Y and Z in Fig. 8.1. [4] (iii) State the reagents and conditions for steps 1 and 2 in Fig. 8.1. step 1 …… step 2 …… [2] (c) The proton (1H) NMR spectrum of compound T, C5H9O2Cl, in CDCl 3 is shown in Fig. 8.2. T O Cl O δ 1.2 δ 3.9 δ 3.7 δ 2.8 5 4 3 2 1 0 chemical shift δ / ppm Fig. 8.2 Table 8.1 chemical shift range environment of proton example δ / ppm alkane –CH3, –CH2–, >CH– 0.9–1.7 alkyl next to C=O CH3–C=O, –CH2–C=O, >CH–C=O 2.2–3.0 alkyl next to aromatic ring CH3–Ar, –CH2–Ar, >CH–Ar 2.3–3.0 alkyl next to electronegative atom CH3–O, –CH2–O, –CH2–Cl 3.2–4.0 attached to alkene =CHR 4.5–6.0 attached to aromatic ring H–Ar 6.0–9.0 aldehyde HCOR 9.3–10.5 alcohol ROH 0.5–6.0 phenol Ar–OH 4.5–7.0 carboxylic acid RCOOH 9.0–13.0 alkyl amine R–NH– 1.0–5.0 aryl amine Ar–NH2 3.0–6.0 amide RCONHR 5.0–12.0 (i) Suggest why CDCl 3 is used as a solvent instead of CHCl 3 for the proton (1H) NMR spectrum. …… [1] (ii) Complete Table 8.2 for the proton (1H) NMR spectrum of T. Table 8.2 chemical number of 1H atoms environment of proton splitting pattern shift δ / ppm responsible for the peak 1.2 2.8 3.7 3.9 [4] (iii) Explain the splitting pattern of the peak at δ 3.9 ppm. …… [1] [Total: 15] o

Mark scheme

8(a) chlorobenzene is less reactive than chloroethane OWTTE 2 p-orbital / lone pair on Cl will overlap / delocalise into the ring due to partial double C-Cl bond OWTTE OR C-Cl bond strengthened (more) Any two [1], all three [2] 8(b)(i) ethyl 3-chloropropanoate 1 8(b)(ii) W 4 O X OH HO ONa Y O Z O HO H2N OH O O 4 × [1] for each structure 8(b)(iii) M1 step 1 NaCN / KCN in ethanol (heat) 2 M2 step 2 H2 + Ni / H2 + Pt ALLOW LiAlH4 8(c)(i) (because solvent / D / CDCl3) doesn’t give a signal / peak / absorption ORA 1 8(c)(ii) 4 chemical splitting number of 1H atoms environment of proton shift δ / ppm pattern responsible for the peak 1.2 alkyl / alkane / (R-)CH(3) triplet 3 2.8 alkyl next to C=O / triplet 2 CH(2)C=O 3.7 alkyl next to triplet 2 electronegative atom / CH(2)-Cl 3.9 alkyl next to quartet / 2 electronegative atom / quadruplet CH(2)-O Any three [1], any six [2], any nine [3], all twelve [4] 8(c)(iii) (δ = 3.9) three H on neighbouring / adjacent C / it’s next to a CH3 1

Question 98 marks

Question

(a) Define standard cell potential, E cell . …… [1] o (b) An electrochemical cell is set up to measure E cell of a cell consisting of an Fe3+ / Fe2+ half-cell and a Cl 2 / Cl – half-cell. Draw a labelled diagram of this electrochemical cell. Include all necessary substances. It is not necessary to state conditions used. [3] (c) The cell reaction for the electrochemical cell in (b) is shown. o Cl 2 + 2Fe2+ 2Fe3+ + 2Cl – E cell = +0.59 V Calculate ΔG o , in kJ mol–1, for this cell reaction. ΔG o = …… kJ mol–1 [2] (d) Another experiment is set up using the same electrochemical cell. In this experiment the Fe2+ concentration is 0.15 mol dm–3. All other concentrations remain at their standard values. The Nernst equation is shown. [oxidised species] E = E o + (0.059 / z) log [reduced species] (i) Use the Nernst equation to calculate the electrode potential, E, for the Fe3+ / Fe2+ half-cell in this experiment. [E o : Fe3+ / Fe2+ = +0.77 V] E = …… V [1] (ii) Use your answer to (d)(i) to calculate Ecell for this electrochemical cell. Ecell = …… V [1] [Total: 8] Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 24 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.4 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge

Mark scheme

9(a) potential difference / voltage between the two half-cells / two electrodes (in a cell) under standard conditions 1 9(b) 3 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 [1], any six [2], all nine [3] 9(c) M1 (ΔGo ) = – nFEcello OR –2 × 96500 × 0.59 2 M2 ΔGo = –2 × 96500 × 0.59 = –113870 J mol-1 ΔGo = –114 kJ mol-1 min 3sf ecf 9(d)(i) E = 0.77 + (0.059 / 1) log (1/0.15) use of z = 1 1 E = 0.82 V 9(d)(ii) Ecell = 0.59 + 0.77 – (answer to (d)(i)) = 0.54 V ecf 1

Checked your answers against these? Let AI mark the whole paper, on your own working.

Mark my 9701/42

How this paper is marked

Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.

Check your answers against it

Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.

9701/42 in other series

Find another Chemistry paper

Every AS & A Level Chemistry mark scheme

Question papers and mark schemes © Cambridge University Press & Assessment. CIE Buddy is independent and not connected to Cambridge.

9701/42 questions

Where can I find the 9701/42 May/June 2023 mark scheme?

On this page, free: the mark scheme for AS & A Level Chemistry 9701/42, May/June 2023 (9701_s23_ms_42), and the question paper (9701_s23_qp_42), to read or download, with each question's marking points written out underneath.

How many marks is 9701/42 out of?

Chemistry 9701/42 for May/June 2023 is out of 100 marks, with 2 hours to answer it, across 9 questions.

Can I get my 9701/42 answers marked?

Yes. Upload your answers to 9701/42/M/J/23 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.