VJC H2 CHEM P3 Answers Prelim

1  VJC 2014 9647/03/PRELIM/14 2014 Preliminary Examination H2 Chemistry Paper 3 Answer any 4 questions. 1 (a) Use of the Data Booklet is relevant in this question. The following cell has been used to light up a light bulb. (i) Suggest, with a reason, how the cell potential chan ges when the current has passed through for a while. Anode: Mg(s)  Mg2+(aq) + 2e– Cathode: Cu2+(aq) + 2e–  Cu(s) OR Cu2+(aq) + Mg(s)  Cu(s) + Mg2+(aq) As [Mg2+] increases, E(Mg2+/Mg) will be less negative. As [Cu2+] decreases, E(Cu2+/Cu) will be less positive. The cell potential will become less positive. The electrode potential of the cell is related to the equilibrium constant by the following equation, Eo= c RT lnzF K where T is the temperature measu red in Kelvin, z is the number of moles of electrons transferred during the redox reaction and F is the Faraday constant. (ii) By using relevant data from the Data Booklet, determine the equilibrium constant Kc for the reaction between Cu2+ and Mg. Ecell = Ered – Eox = (+0.34) – (–2.38) = +2.72 V ln Kc = 298 x 8.31 2.72 x 96500 x 2 (ecf) = 212 Kc = 1.16 x 1092 (ecf) salt bridge Cu electrode Cu2+(aq), 1 mol dm–3 light bulb X Mg electrode Mg2+(aq), 1 mol dm–3

2  VJC 2014 9647/03/PRELIM/14 (iii) Hence, determine the ratio [Mg 2+]:[Cu2+], when there is no current flowing in this cell. When there is no current flowing in the electric cell, it has reached equilibrium. [Mg2+]:[Cu2+] = Kc = 1.18 x 1092 (iv) Suggest the significance of the magnitude of your answer in (a)(iii). Almost all the Cu2+ has been consumed. OR Reaction goes to completion when the equilibrium has been reached. [5] (b) Copper minerals often contain suifides of magnesium and silver as impurity. After minerals are reduced with carbon, the solid impure copper is purified by electrolysis. (i) Describe the electrode reactions that take place during the electrolysis and explain in detail how each of the two impurity metals is removed from copper. Two electrodes + Electrolyte OR diagram above At the anode, oxidation takes place and Cu dissolves. Cu(s)  Cu2+(aq) + 2e– Mg with Eo which is less positive than that of Cu dissolves as ions. Ag with Eo which is more positive than that of Cu remain s undissolved and drop to the bottom of the vessel as ‘anode sludge’. At the cathode, reduction of copper ions occurs due to more positive Eo (Cu2+/Cu) compared to Eo (H2O/H2) and Eo (Mg2+/Mg) Cu2+(aq) + 2e–  Cu(s) Hence, only pure Cu is formed at the cathode. An impure copper rod is purified by electrolysis using a constant current. After 1 hour, mass of one electrode decreased by 19.0 g while mass of the other electrode increased by 17.8 g. The electrolyte was further analysed and found that the amount of Cu2+ ions is reduced by 0.020 mol. Power source electron flow Anode (+) Impure Cu 2Cl-(l)  Cl2(g) + 2e– Cathode (–) Pure Cu Na+(l) + e–  Na(s) + – CuSO4(aq)

3  VJC 2014 9647/03/PRELIM/14 (ii) Calculate the current used for this process. Amount of Cu formed = 5.63 8.17 = 0.280 mol Amount of electron transferred = 0.280 x 2 = 0.560 mol (ecf) Charge transferred = 0.560 x 96500 = 54100 C (ecf) Current = 3600 54100 = 15.0 A (ecf) (iii) Determine the percentage composition by mass of the three elements. Amount of Mg oxidised = amount of Cu2+ changed = 0.020 mol Mass of Mg = 0.020 x 24.3 = 0.486 g Percentage composition by mass of Mg = %100 x 0.19 486.0 = 2.56% Amount of Cu oxidised = 0.280 – 0.020 = 0.260 mol Mass of Cu = 16.51 g Percentage composition by mass of Cu = %100 x 0.19 51.16 = 86.9 % Percentage composition by mass of Ag = 10.54 % (iv) Suggest how magnesium can be recovered as magnesium oxide near the end of the electrolytic process. Write balanced equations for any reactions involving magnesium that might occur. Filter the electrolyte to remove the metallic silver. Add aqueous ammonia until excess. Filter the mixture to obtain the residue which is magnesium hydroxide. Heat the solid until no more change in mass. Mg2+ + 2OH–  Mg(OH)2 Mg(OH)2  MgO + H2O [Accept reaction of Mg(OH) 2 with HNO3 and followed by thermal decomposi tion of Mg(NO3)2]. [11]

4  VJC 2014 9647/03/PRELIM/14 (c) The methods of synthesis shown below are faulty. In each case  Explain why this is so.  Suggest how the synthesis from the initial reactant to the final product could be satisfactorily achieved. (i) (ii) [4] (c) (i) When HCl is used: Dehydration will not occur with dilute HCl. Ester will be hydrolysed to give alcohol and carboxylic acid. Correct reaction reagent and condition: Al2O3 catalyst with heating. (Conc. H2SO4 is not a good choice as it may lead to hydrolysis of ester.) (ii) NaOH causes hydrolysis of CN instead of reduction. Amine is a stronger nucleophile as nitrogen is less electronegative , thus it will react with CH3COCl first to give amide instead. OR Both alcohol and amine will react with CH 3COCl, when excess amount of CH3COCl is used. React with CH3COCl at room temperature and followed by reduction with H2 with Pt catalyst at room temperature. [4] [Total: 20] 2 (a) (i) When CHI2CHO was warmed with alkaline aqueous iodine, two organic products are formed. Construct an equation for this reaction. CHI2CHO + I2 + 2OH–  CHI3 + HCO2 – + I– + H2O (ii) There are suggestions on using CF 3CH2F to replace CFCs. One problem however is that CF 3CH2F can be converted to CF 3CO2H, which is toxic when ingested. When CF3CH2F eventually reaches the atmosphere, it is thought that CF 3CH2F is initially attacked by •OH radicals. A student suggested the following equation to represent the above reaction: CF3CH2F + •OH  CF3CH2 + HOF By considering the •

5  VJC 2014 9647/03/PRELIM/14  bonds broken and  bonds formed in the above reaction, explain why the above reaction will not occur. Bond broken: C-F bond is very strong / high bond energy due to the extensive overlap between the C and F atomic orbitals. Bond formed: O-F bond is very weak / low bond energy O and F atoms are small / strongly electronegative, and the electron pairs will be close to each other and repel strongly. (iii) In light of your answers to (a)(ii), suggest a more appropriate equation for the reaction between CF3CH2F and •OH. CF3CH2F + •OH  CF3CHF + H2O (iv) Ethanedioic acid, HO2C-CO2H is a dibasic acid, with pKa1 = 1.3 and pKa2 = 4.3. Methanoic acid is a monobasic acid with pKa = 3.7. Account for the lower p Ka1 value of ethanedioic acid as compared to the p Ka of methanoic acid. Lower pKa1 value implies stronger acid. HO2C-CO2 – is more stable than HCO2 –. CO2H group exerts an electron withdrawing inductive effect leading to better charge dispersal in HO2C-CO2 – compared to HCO2 –. Hence, greater tendency for HO2C-CO2H to dissociate OR equilibrium position of HO 2C-CO2H HO2C-CO2 – +H+ lies more to the right. HO2C-CO2 – can also be stabilised by intramolecular hydrogen bonding. OR equiv