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2010 ‘A’ Levels Suggested Solutions Paper 1 1 2 3 4 5 B B B D B 6 7 8 9 10 C A A D A 11 12 13 14 15 B D C B C 16 17 18 19 20 B C C B C 21 22 23 24 25 D B B D D 26 27 28 29 30 A B C C D 31 32 33 34 35 A B A B C 36 37 38 39 40 B C A A C Paper 2 Question 1 (a) Overall eqn: 3C2O42- +2FeO42- + 16H+ à 6CO2 + 2Fe3+ + 8H2O Note: Qn stated that the ions are reduced to 2Fe3+ (and NOT Fe2+) under acidic conditions. You are tested on the concept of balancing redox half eqns, and then combining them to come up with the overall eqn.. Half eqns: C2O42- à 2CO2 + 2e (given) FeO42- + 8H+ + 3eà 6CO2 + Fe3+ + 4H2O Overall eqn: 3C2O42- +2FeO4- + 16H+ à 6CO2 + 2Fe3+ + 8H2O (b) If volume of FeO42- is 25.0 cm3 (measured using a pipette), burette reading would be: By proportion, A burette reading over 50cm3 would require careful refilling of the burette, making the titration difficult. If volume of C2O42- is 25.0 cm3 (measured using a pipette), burette reading would be: By proportion, A small burette reading of 8 cm3 would lead to high percentage error. (c) *When planning the procedure, quantities are to be provided as stated by the question, hence some calculation must be done to find out the mass / volumes of the reagents to be use. The general procedure is i. weighing of the chosen mass of sodium ethanedioate ii. dissolving the solid and preparation of the standard solution of sodium ethanedioate iii. carrying out titration (burette, pipette, conical flask, colour change at end point) (i) Calculation of mass of sodium ethanedioate to be weighed Calculation: Let 1 drop be 1 dm3 Amount of C2O42- used = 0.100 x 32 = 3.2 mol Amount of FeO42- required = 2/3 (3.2) = 2.13 mol Since 10 drops was used, [FeO42-] = 2.13/10 = 0.213 mol dm-3 . We let volume of FeO42- to be used = 25 cm3 Amount of FeO42- in 25 cm3 = 0.213 x 25/1000 = 5.33 x 10-3 mol Amount of C2O42- required to react = 5.33 x 10-3 x 3/2 = 8.00 x 10-3 mol For the volume of solution used to be 25 cm3, [C2O42-] = 8.00 x 10-3 / 25 x 1000 = 0.320 mol dm-3 Amount required to be dissolved in 250 cm3 volumetric flask = 0.320 x 250/1000 =0.0800 g Mass required to be dissolved in 250 cm3 = 0.0800 x [23.0(2) + 12.0(2) + 16.0(4)] = 10.72 g Procedure: 1. Using a weighing balance, weigh accurately between 10.5-10.9 g of Na2C2O4 in a clean and dry weighing bottle. (state a suitable range ±0.2g to weigh out) 2. Transfer the Na2C2O4 into a 100 cm3 beaker and weigh the weighing bottle with the residue. The difference in mass will give the mass of the Na2C2O4 that was used to prepare the solution. 3. Add 100 cm3 of H2SO4 (aq) to dissolve the solid. An excess of acid must be added to provide an acidic medium. (As seen in the balanced eqn in (a), acidic medium is required and we add H2SO4 to provide the acidic medium. H2SO4 is the best choice to acidify solutions as it will not be oxidized as easily as other acids, such as HCl) 4. Transfer the solution and the washings into a 250 cm3 volumetric flask and make up to mark with water. 5. Stopper and shake the volumetric flask to ensure thorough mixing. 6. Pipette 25.0 cm3 of the solution prepared into a 250 cm3 conical flask. 7. Top up the burette with FA1. 8. Titrate the solution against FA1 and stop when the first trace of orange is observed. (dark red FA1 will change to yellow (Fe3+) in this reaction. The end point is marked by the appearance of orange, due to 1 excess drop of FA1) 9. Repeat the titration until the readings are consistent (difference within 0.10 cm3). Note: Considering the colour change at the end point, if we have the C2O42- solution in the conical flask, the colour change is colourless to yellow and finally to orange at end point. If FA1 is in the conical flask instead, the colour change is red to orange to yellow at end point. The colour change from yellow to orange is more obvious than orange to yellow. That is why we choose to have the C2O42 solution in the conical flask. Of the three mineral acids (HNO3, HCl and H2SO4) available, only sulfuric acid is suitable for use because : • Ferrate (VI) ions would oxidise HCl to Cl2 (just as manganate (VII) ions do) • HNO3 itself is an oxidising agent and will interfere with the oxidising action of Ferrate (VI) ions.. Hence, the titration readings would not be accurate if HCl or HNO3 are used. (d) M ) mol 1000 2 M (y x ) mol n(FA1) = 3 1000 2 M 2 (y × ) ( yM ) 2yM 1000 × 1000 = 3 [FA1] = 3 = mol dm-3 x x 3x NB: Be very careful with your units when doing such questions. n(FA2) = (y x (e) FeO42- is a very strong oxidizing agent, hence it is corrosive/ may cause burns. Use gloves to avoid contact with the chemical. NB: SOME other reagents that must be handled with care: - SOCl2, PCl5 reacts vigorously with water, giving fumes of strong acid HCl. Hence reaction must be carried out in a fumehood. - LiAlH4 reacts with water, making it ineffective. Hence, reaction must be carried out in an anhydrous condition. - Organic compounds/solvents, are usually flammable, hence, reaction is to be carried out in a fumehood, and away from naked flame. - Organic compounds that contain benzene rings are cancer causing, hence reaction is to be carried out in fumehood, and use gloves. Question 2 (a) NaBr + H2SO4 à NaHSO4 + HBr (Group VII chemistry) CH3(CH2)2CH2OH + HBr à CH3(CH2)2CH2Br + H2O (b) 25 × 0.81 = 0.274 mol 74 n(NaBr) = n(HBr) = 35 / 103 = 0.340 mol (Note: use Mr data from table) n(butan-1-ol) = Since butan-1-ol reacts with HBr in a 1:1 mole ratio, NaBr is in excess. (c) (i) Concentrated H2SO4 reacts very exothermically when diluted with water due to the highly exothermic enthalpy change of hydration by water. (ii) Inorganic by-product : Br2(g) + SO2(g) NaBr + H2SO4 à NaHSO4 + HBr 2HBr(g) + H2SO4(l) Br2(g) + SO2(g) + 2H2O(l) Organic by-product: But-1-ene CH3CH2CH2CH2OH à CH3CH2CH2C=CH2 + H2O It was expected that candidates would consider the well-known reactions of concentrated H2SO4 with NaBr or with HBr to produce Br2 and SO2. Many candidates did this while fewer were able to write a correct equation for the reaction. Correct answers for the identity of the organic by-product were less common. The dehydration of alcohols by concentrated H2SO4 is well known when asked directly but many candidates did not bring such knowledge to this question. A significant number of those who did deduce what would happen did not identify the alkene as but-1-ene, giving the ambiguous answer ‘butene’ or ‘C4H8’ which Examiners did not accept. (d) Heating for 45 min is required as the reaction has a high activation energy and is slow, because strong covalent bonds need to be broken in the reactants. (This is the reason to why heating is required for most of the organic reactions as organic reactions involves breaking (and forming) covalent bonds.) (e) Butan-1-ol and water Their b.p. are very close to that of 1-bromobutane. (f) Lower 1-bromobutane (1.35 g cm-3) is more dense than water (1.00 g cm-3). (g)(i) Butan-1-ol (g)(ii) CH3(CH2)2CH2OH + H+ à CH3(CH2)2CH2OH2+ This reaction is unusual as you’ve been taught that alcohols are neutral. But note that the use of conc HCl is to remove the organic impurity that cannot be separate by Step 5. The ONLY organic reactant used is butan-1-ol making it the only possible candidate as the organic impure that the question is looking for. (refer to (e)) Butan-1-ol is immiscible with water due to the 4C alkyl chain that makes it non polar. Hence, something must be happening between the acid and the butan-1-ol to cause it to be soluble in the water layer. The only viable option is the protonation of the alcohol occurred in order for the resulting ion to form ion-dipole with water. (g)(iii) It is able to form strong ion-dipole attraction with the water molecules. (h) Conc. HCl 2HCl + Na2CO3 à 2NaCl + H2O + CO2 (i) Water Traces of water that was used in step 5 and 6 could be still present in the product. Anhydrous calcium chloride is a drying agent that can be used to remove water. i.e. ahydrous calcium chloride is added to the reaction mixture, and it will absorb any traces of water present. The calcium chloride is then removed by filtration. (j) 99 – 103 oC (note the range) Note Bpt of 1-bromobutane is 102 OC. ‘General rule’ is -3 +1 (as taught by Mr Lee during Prelims review of P2 Q2) Question 3 (a) NO is a radical hence it would lead to ozone depletion NO is readily oxidized to NO2, a brown toxic gas that causes respiratory illness. (b) initial rate / Nm-2s-1 (ρNO)2 / N2 m-4 graph cuts through the origin. If ρNO is choosen to be the x-axis instead, a curve will be obtained, the relation between rate and ρNO cannot be deduced immediately without showing some calculation. (c) The order of reaction with respect to ρNO is 2. The graph is linear, hence rate increases linearly with (ρNO)2. (d) First order. When the partial pressure of oxygen is halved, the gradient of the graphs is halved, i.e. the rate of the reaction is halved. (e)(i) Rate = k(ρNO)2(ρO2) Must be expressed in terms of partial pressure, not [ ]! (e)(ii) N-2 m4 s-1 Question 4 (ai) Magnitude of the LE decrease from MgSO4 to BaSO4. (aii) Ionic radius of Mg2+: 0.065 nm Ionic radius of Ca2+: 0.099 nm Ionic radius of Sr2+: 0.113 nm Ionic radius of Ba2+: 0.135 nm |q q | Since | LE |∝ + − where q is the charge on the ion, and r is the ionic radius r+ + r− The ionic radius increases from magnesium ion to barium ion, hence, |LE| decreases. (b)(i) (b)(ii) The charge density decreases from magnesium ion to barium ion due to the increase in size of the ions. Hence the ion-dipole ion dipole formed with water decreases in strength. (c)(i) Nucleophilic addition (c)(ii) Hydrolysis OH (d) H C CH3 (e) CH3CH2CH2COO− CH3CH2CH2COOH Question 5 (a)(i) When proteins are denatured, the R–group interactions are disrupted, causing only secondary, tertiary and quaternary structures to be destroyed permanently, hence losing its biological activity. The primary structure remains intact. (a)(ii) Collagen may be denatured by addition of acid to it. This will disrupt the ionic bonding originally present between the R groups. (b) 6 (note that there is a total of 9, but there are repeat amino acids. The question asks for different amino acids) (c)(i) DISPLAYED formula H O C O H H H N C OH C N H H O c(ii) N H d(i) d(ii) H * * C HO O O Add H2SO4 (aq), heat under reflux for several hours (NB: This reaction requires the breaking of the numerous strong covalent peptide bonds, heating for a prolonged period is necessary. Also, as this question asked about “in the laboratory”, a more detailed answer about the procedure will be wise.) O O C HO C H2N NH2 OH R d(iii) H2N C COOH e) H Proteins is made up of amino acid monomers with general formula There are both amine and carboxylic acid functional groups in the amino acid monomer. But monomers of Kevlar consists of only carboxylic acid or amine groups. HOCH2CO2CH(CH3)CO2H CH3CH(OH)CO2CH2CO2H f(i) f(ii) Hydrolysis of the ester linkages The products are able to form hydrogen bonds with water. . Paper 3 Question 1 a(i) Entropy is the measure of the disorder of the system. [1] a(ii) 1 mol of Cl2 added to 1 mol of N2 at 298 K ∆S is positive/entropy increases because there is mixing of gaseous particles, leading to a higher state of disorder in the system. [1] 1 mol of Cl2 heated to 373 K ∆S is positive/entropy increases as the system possess more kinetic energy. This implies that there are more ways to distribute the energy among the particles giving rise to greater disorder. [1] It is NOT enough to state there is an increase in KE. The increase in entropy comes about because in the broadening of the Boltzmann energy distribution as there are more ways of arranging energy quanta in the hotter gas. 1 mol of Cl2 reacts iodine: Cl2 (g) + 1 2 I2 (s)à à ICl3 (l) 3 3 ∆S is negative/entropy decreases as there is a decrease in number of gaseous particles, leading to a higher state of disorder in the system [1] 1 mol of Cl2(g) is photolysed: Cl2 (g) à 2Cl (g) ∆S is positive/entropy increases as there is an increase in number of gaseous particles, leading to a higher state of disorder in the system [1] Be very careful with you terms. Entropy (S) increases/decreases. ∆S is positive or negative. bi) H2 + Cl2 à 2HCl (fast – due to higher bond energy of HCl) H2 + Br2 à 2HBr (slow) [1] The important comparison here is that of the RATE of their reaction, not on the extent of the reaction. No further merit could be gained by detail analysis of the enthalpy changes in bonds being formed or broken. bii) H2 + F2 à 2HF (very fast) [1] ci) Conditions: UV light/Heat [1] cii) Free Radical Substitution. [1] Cl Initiation: Propagation:Cl u.v. Cl H Cl Cl [1] CH3CH2• CH2CH3 CH3CH2• Cl CH3CH2 Cl HCl Cl Cl [1] Termination: CH3CH2• Cl Cl Cl CH3CH2• ciii) CH3CH2 Cl CH3CH2 CH3CH2• Cl Cl CH2CH3[1] Iodoalkanes cannot be formed because the reaction is endothermic. ∆H (iodoalkane) = (+410 + 151) – (240+299) = + 22 kJ mol-1 ∆H (chloroalkane) = (+410 + 244) – (340+431) = – 117 kJ mol-1 [1] Candidates needed to point out that iodination does not occur because the first propagation step is endothermic, or point out the relevance of the BE (H—I) bond. d) CH3 CH3 Cl CH3 Cl CH3 Cl H3C CH2Cl H3C Br CH2Cl B A Br C D [1] for each structure ei) They are inert and stable/non-flammable/non-toxic eii) UV light causes the C—Cl bonds to break and releases chlorine radicals, which attack ozone and destroys the ozone layer. [1] eiii) Alkanes are flammable. [1] [1] Question 2 ai) HA + H2O Ka = H3O+ + A— [H3O+ ][A - ] [H O+ ][A - ] ; pK a = - lg 3 [1] [HA] [HA] pKa measures the strength of a weak acid. The lower the pKa, the stronger is the acid. [1] + + NH3 NH3 aii) pH 1 HOOC COOH pH 3 HOOC COO + NH2 NH3 - pH 7 O OC COO - - - pH 11O OC COO - [1] for each structure bi) The protein is denatured during coagulation. The R group interactions are destroyed, disrupting the tertiary and quaternary structures. [1] The protein disintegrates to form random coils of polypeptide chains. [1] Note: Under conditions described, secondary structures remain intact. This is because hydrogen bonds in secondary structures are not affected. The hydrogen bonds are formed between peptide linkages since peptide linkages do not undergo protonation/deprotonation. bii) Reaction with Ca2+ will bind to carboxylic anions [1] as found in glutamic acid residues, hence disrupting the ionic interactions [1] Before denaturation: After denaturation: ------ H3N+−− −−COO −− Ionic bonds −−COO Ca2+ −− H3N+−− (Ionic bonds broken) NB: Only consider the disruption of S—S when cysteine amino acid is present! Only the heavy transition metal ions Cu+ and Hg+ can denature proteins by disrupting the S—S. biii) H+ protonates the ionic R groups, and disrupt the ionic bonds[1] which stabilises the tertiary or quaternary structures. [1] Low pH: biv) [GDL] = Kc = – COO- + H+ à –COO–NH2 + H+ à –NH3+ 1 = 0.02 g dm-3 = 0.112 mol dm-3 50 / 1000 [gluatamic acid] [1] [GDL][H2O] = = [0.0670] [0.112 − 0.0670][55.5] = 0.0269 mol-1 dm3[1] Many candidates omitted water into their Kc expression. Some candidates did not calculate the equilibrium [GDL]. ci) Br HO O Br Br Br Br O OH Br [1] for at least 1 Br substituted into each of two phenolic rings [1] for 2 Br added across C=C cii) HO O HO + H2 O OH OH OH diadzein HO O F HO O O K2Cr2O7 O OH OH F G OH [2] for structures of F and G [1] for all 3 chiral carbon identified. Alkenes and ketones in Diadzein undergoes reduction to give F F contains 3 –OH groups (2 phenol and 1 alcohol) since it reacts with 3 mole of Na 2o alcohol in F then undergoes oxidation with K2Cr2O7 to give G. G contains a carbonyl group since it gives a positive test with 2,4-DNPH. [1] for identifying type of reaction that occurs. Question 3 ai) O2 + 4H+ + 4e− à 2H2O aii) 2CH3OH + 3 O2 à 4H2O + 2CO2 [1] aiii) EO2/H2O = +1.23 +1.18 = + 1.23 – Eoxi Eoxi = +0.05 V aiv) [1] [1] [1] Easy storage and safety – CH3OH and H2O are in liquids and easier to transport. H2 and O2 are flammable gasses and need to be kept under pressure. bi) C2H5OH + 3 O2 à 3H2O + 2CO2 [1] bii) ∆G biii) C2H5OH + 3 H2O à 2CO2 + 12H+ + 12e− [1] = ∆H – T∆S = (–1367) – (298)( –140/1000) = –1325.28 = –1330 kJ mol–1 [1] −∆Gθ −( −1325280) = Ecell = = +1.14V zF (12)(96500) ci) cii) ∆H [1] = = = [1] ∑ bonds broken - ∑ bonds formed (5 C—H + 1C—C + 1 C—O + 1 O—H + 3 O=O) –(4 C=O+ 6O—H) [1] – 1272 kJ mol-1 Bond energies assumes compounds to be in gaseous state. At exists as a liquid. 298 Hence the difference would be due to the ∆Hvapourisation of ethanol. K, ethanol [1] di) CxHyOH + Na à CxHyO—Na+ + ½ H2 amount of H2, nH2 = amount of CxHyOH = 2 x nH2 Let initial VO2 be z cm3. = 2 x 4.54 x 10-3 = 9.08 x 10-4 mol CxHyOH (l) ( Initial amount/mol 9.08 x 10-4 After combustion and cooling/mol After reaction with NaOH/mol 10.9 × 10−3 = 4.54 x 10-3 mol 24.0 4x + y − 1 )O2 (g) 4 xCO2 (g) ( 0 0 z 0 0 z – 54.4 109 (formed) z – 109 – 54.4 (leftover O2) Volume of O2 used = z – (z– 109 – 54.4) = 163 cm3 Volume of CO2 formed = 109 cm3 Amount of O2 used Amount of CO2 formed 163 x 10 −3 = = 6.79 x 10-3 mol 24.0 109 x 10 −3 = = 4.54 x 10-3 mol 24.0 Comparing ratio: Mole ratio of CxHyOH : CO2 is 1:x. 1 x 4.54 ×10−3 = ∴x = =5 Hence 9.08 ×10−4 4.54 ×10−3 9.08 ×10−4 Comparing mole ratio of CxHyOH and O2 = 1 : 4(5) + y − 1 1 4 = Since x=5, 9.08 ×10−4 6.79 ×10−3 Hence J C5H11OH [3] 4x + y − 1 4 ∴ y = 10.91 ≈ 11 y +1 ) H2O (l) 2 dii) H H C C H OH O C O CH3 H CH3 H H CH3 C C C H OH CH3 H J H H H CH3 C C C H OH CH3 J H H CH3 C C H 3C CH3 K [1] [1] J is a secondary alcohol, hence can be oxidised by K2Cr2O7 K can be obtained by the dehydration of J. Explanation [1] A high proportion of candidates who struggled with (i) abandoned the question at this point, despite the fact that (ii)could be solve directly without reference to (i). Those who tackled it were usually successful. (So don’t give up! Just move on!!) diii) Reagents and Conditions: concentrated H2SO4, heat at 170 oC [1] div) No. K cannot exhibit geometric isomerism. There are 2 methyl groups attached to the same carbon atom in the C=C bond [1] It is necessary to state it was one of the carbon atoms in the C=C that has 2 identical groups. It is WRONG to say these identical groups are on the same side of the C=C bond, or just that there is a carbon having two identical groups Question 4 a) Nucleon number refers to the total number of protons and neutrons in the nucleus of an atom. [1] b) Electrons would deflect towards the positively charged electrode. The angle of deflection would be greater in electron than in proton. [1] for both points c) (i) x x Bond pairs/Lone Pairs (ii) Bond angles ciii) xx O xx N x x xx O xx x x xx O xx O x x xx O xx [1] [1] 2 BP, 1 lone electron (exerts 2 BP 1 LP less repulsion than a lone PAIR) Any value between x and Any value x: 110 < x < 120[1] 170 [1] Cl is in period 3 and can expand its octet to accommodate a maximum of 18 electrons as it has energetically accessible 3d orbitals. [1] Cl O O F is in period 2, thus it is unable to expand its octet. Moreover, F is too electronegative to form covalent bonds with O. [1] Most candidates pointed out that ClO2 would involve chlorine expanding its octet of electrons, which it is able to do through 3d orbitals which is energetically feasible. Very few candidates went further with the discussion. If F cannot expand its octet, then the other alternative would be for F to provide 2 dative bonds, which it cannot because it is too electronegative for dative bonding. di) 2Ca + O2 à 2CaO Heat Ca strongly in air. [1] A brick-red flame is seen, white solid of CaO is seen after the experiment. [1] For Ca to react with oxygen, it is necessary to apply heat, ideally by holding Ca in tongs inside a gas jar. A noticeable red flame should be seen and a white solid should be seen after the experiment. dii) Solubility of Group II hydroxides increases down the group. [OH-] increases, thus pH increases down the group. [1] pH values of solutions increases from Mg to Ba when shaken with water, because each oxide reacts with water to give its hydroxide. The solubility of Group II hydroxides increases down the group, hence the [OH-] increases. e) fi) fii) Ba(O3)2 + H2O à Ba(OH)2 + 5 O2 [1] 2 15.0 × 0.100 = 0.0015mol 1000 Amount of iodine = ½ x 0.0015 = 0.00075 mol [1] Amount of thiosulfate = amount of ozone = amount of iodine = 0.00075 mol Volume of ozone = 0.00075 x 22.4 = 0.0168 dm3[1] 16.8 Percentage of ozone present = × 100 = 3.36% [1] 500 gi) To form aldehydes: Warm a primary alcohol and immediately distill to prevent [1] further oxidation To form carboxylic acids: Heat under reflux for some time to ensure oxidation has been completed. [1] gii) CH3 CH3 H C CH2 C OH H L [1] for each structure M N CH3 Question 5 a) impure copper anode pure copper CuSO4(aq) [1] Correctly assigned cathode and anode [1] Electrolyte o Since E(Cu2+/Cu) is less positive than E(Zn2+/Zn), Zn will also be oxidized at anode. o They dissolve into the solution as cations and migrate to the cathode. o They will not be reduced at the cathode since E(Cu2+/Cu) is less positive than E(Zn2+/Zn) o Since E(Ag+/Ag) is more positive than E(Zn2+/Zn) for (4), Ag (and any other less reactive metals e.g. Au) will not be oxidised. o Ag drops off the electrode as the copper around dissolves, and fall to the bottom of the electrolytic tank to form anode sludge. b) When dilute aqueous NH3 is gradually added, precipitation of Cu(OH)2 (pale blue) occurs. The hydroxide ions comes from aqueous ammonia. NH3(aq) + H2O NH4+ OH- [Cu(H2O)6]2+ (aq) + 2OH– (aq) blue soln • When excess aqueous NH3 is added, ligand exchange takes place. NH3 is a [1] stronger ligand than H2O and thus displaces water from [Cu(H2O)6]2+ to form deep blue complex[Cu(NH3)4(H2O)2]2+. [Cu(H2O)6]2+ (aq) + 4NH3 (aq) blue soln[1] • Cu(OH)2 (s) + 6H2O (l) …………………..(1) blue ppt [1] [Cu(NH3)4(H2O)2]2+ (aq) + 4H2O (l) ……..(2) deep blue complex [1] This causes [Cu(H2O)6]2+ to drop, hence eqm (1) shifts to the left, and the precipitate Cu(OH)2 (s) dissolves. c) O2- + NH3 à OH- + NH2- [1] Since it contains Cu2+, colour of the solution would be blue/ blue-green. [1] d) [Cu(H2O)6]2+ + 4Clblue soln [CuCl4]2yellow + 6H2O [1] When concentrated HCI is added, ligand exchange takes place. Cl- displaces water ligand to form the yellowish [CuCl4]2- complex. [1] Note: The solution appears yellowish green due to the yellow [CuCl4]2complex and blue [Cu(H2O)6]2+. When water is added, the position of equilibrium shits to the left, hence reforming the blue solution [Cu(H2O)6]2+ [1] ei) 2Cu2+(aq) + 4I− (aq) à 2CuI(s) + I2(aq) white ppt brown solution [1] eii) ECu2+/Cu = + 0.15 V E I2/I −= +0.54 V Ecell = +0.15 – 0.54 = – 0.39 V [1] Since Ecell <0, Reaction is energetically not feasible [1] eiii) 2Cu2+(aq) + 4I− (aq) 2CuI(s) + I2(aq) 2+ − + Cu (aq) ----(1) Cu (aq) + e Because CuI is insoluble in aqueous solution, causes the [Cu+(aq)] to be very low and results in the position of equilibrium (1) to shift right, causing more Cu2+ to be reduced to Cu+, allowing the redox reaction to take place. [1] ER: Only a small number of candidates realize that the fact that CuI is insoluble in aqueous solutions would mean that the precipitate would extract the Cu+(aq) ions from the solution, leaving only a low concentration, and disturbing the equilibrium fi) Brick-red ppt form [1] fii) Positive Test: [1] displayed formulae H C O H H H H C C C C H H H H O C H H H H H C C C H H H C H Negative Test [1] displayed formula H H H O H H C C C C C H H H H H H H