CJC H2 Chem 2013 Prelim P3 Soln

CHEM Paper 3 F Additiona READ T Write your Write in d NOT ALLO You may u Do not use Answer an You are r e The numb At the end CATH JC2 PR Higher MISTRY Free Respo al Material THESE INS r name and c ark blue or OWED] use a soft pe e staples, pa ny four ques eminded of th ber of marks d of the exam Thi HOLIC JU RELIMINAR r 2 Y onse ls: Data Bo Answer STRUCTIO class on all th black pen o n encil for any d aper clips, hig stions. Write he need for g is given in br mination, fast s document UNIOR C RY EXAMIN ooklet r Paper ONS FIRST he work you n both side s diagrams, gr ghlighters, gl your answer good English rackets [ ] at en all your w t consists o AN EXA COM COLLEGE NATIONS T hand in. s of the pap e raphs or roug lue or correc rs on the ans h and clear p t the end of e work securely of 26 printed NSWE and AMIN MME E er. [PILOT F R gh working. ction fluid. swer paper p presentation i each questio y together. d pages and ERS NERS’ ENTS Thursda RIXION ERAS rovided. in your answ on or part que d 0 blank pa 9 ay 29 Augu SABLE PENS wers. estion. age. [Tu 9647/03 ust 2013 2 hours S ARE urn over 447

2 9647/03/CJC JC2 Preliminary Exam 2013 Answer any four questions. 1. Use of the Data Booklet is relevant to this question. (a) (i) Define the term standard electrode potential, E o . Standard electrode potential , E o , is defined as the potential difference between a standard hydrogen electrode and a metal (the electrode) which is immersed in a solution containing metal ions at 1 mol dm –3 concentration at 25 °C and 1 atmospheric pressure. [values of standard conditions should be given] (ii) For the element M (where M is Co or Ni), suggest an explanation for the difference in the E o values between M 2+ + 2e– M and [M (NH3)6]2+ + 2e– M + 6NH3 [4] E o /V Co2+(aq)+ 2e– Co(s) –0.28 [Co(NH3)6]2+ + 2e– Co + 6NH 3 –0.43 OR Ni2+(aq) + 2e– Ni(s) –0.25 [Ni(NH3)6]2+ + 2e– Ni + 6NH 3 –0.51 E o of [M(NH3)6]2+/M is more negative than that of [M(H2O)6]2+/M. This is because as H2O ligands in the aquated ions are replaced by the stronger NH3 ligands, the +2 oxidation state of M is stabilised and this makes the metal more reducing, or its ion less oxidising. (b) A student set up the following electrochemical cell. (i) What device is used to measure the e.m.f. of the cell? high-resistance voltmeter. salt-bridge + Ag(s) – AgNO3(aq) 0.010 mol dm–3 Ag(s) saturated AgCl(aq) AgCl(s) 448

3 9647/03/CJC JC2 Preliminary Exam 2013 (ii) Name a compound that could be used to prepare a workable salt-bridge for this cell and explain how it helps to maintain electrical neutrality. salt bridge can be prepared by using KNO3(aq) or NaNO3(aq). K+ (or Na +) ions leave the salt bridge into LHS half-cell (where Ag + is reduced to Ag) to 'balance' the loss of positively charged Ag+ ions; NO3 – ions leave the salt bridge into RHS half-cell (where Ag is oxidised to Ag+) to 'balance' the increase in positively charged Ag+ ions. (iii) Amongst other factors, the electrode potential is dependent on the temperature and the concentration of the ions. At 298 K, the reduction potential, E, of Ag+(aq)/Ag(s) can be obtained from the equation E = E o + 0.059 log10 [Ag+(aq)] The student found that the e.m.f. of the cell was +0.17 V at 298 K. Calculate the reduction potential, E, of each half-cell and hence the concentration of Ag +(aq) ions in the saturated AgCl(aq). [Given: E o of Ag+(aq)/Ag(s) = +0.80 V at 298 K] Ag+(aq)/Ag(s) half-cell, Ered = E o + 0.059 log10 [Ag+(aq)] = +0.80 + 0.059 log10 (0.010) = +0.682 V Since e.m.f. of cell = +0.17 V, and Ecell = Ered – Eoxid  Eoxid = Ered – Ecell = +0.682 – 0.17 = +0.512 V AgCl(aq)/Ag(s) half-cell, Eoxid = E o + 0.059 log10 [Ag+(aq)] +0.512 = +0.80 + 0.059 log10 [Ag+(aq)]  [Ag+(aq)] = 1.31 × 10–5 mol dm–3 (iv) State and explain the effect on the cell e.m.f. of adding a small volume of NH3(aq) to the Ag+(aq)/Ag(s) half-cell. [9] Ag+ + e– Ag; E = +0.682 V Ag+ + NH3 + H2O  AgOH(s) + NH4 + or Ag+ + 2NH3  [Ag(NH3)2]+ NH3(aq) reacts with Ag +(aq) to form AgOH ppt (or [Ag(NH 3)2]+ complex ions). Hence, [Ag +] decreases which causes the equilibrium position of above equilibrium reaction to shift to LHS and so, electrode potential, E(Ag+/Ag), becomes less positive; i.e. Ered < +0.682 V. 449

4 9647/03/CJC JC2 Preliminary Exam 2013 (c) Putrescine (1,4-diaminobutane) is a foul-smelling compound found in rotting flesh and in "bad breath". It is produced by the breakdown of amino acids in living and dead organisms. Like ammonia, putrescine is a weak base. While ammonia has a p Kb value of 4.74, putrescine has two p Kb values which are found to be 3.20 (p Kb1) and 4.65 (p Kb2) respectively. (i) Explain why putrescine is a stronger base than ammonia. in putrescine - presence of electron-donating alkyl group increases the electron density on the N atom, making the lone pair on the N atom more available to accept protons. (ii) Write equations to represent Kb1 and K b2 of putrescine. Hence explain why pKb1 of putrescine is smaller than pKb2. H2N(CH2)4NH2 + H2O H 2N(CH2)4NH3 + + OH ––––––– equation 1 H2N(CH2)4NH3 + + H2O H 3N(CH2)4NH3 + + OH ––––––– equation 2 Since it is easier to protonate a neutral molecule (equation 1) than to protonate a positively charged ion (equation 2), Kb1 is larger than Kb2. Hence, pKb1 (= log10 Kb1) is smaller than pKb2. Kb2 Kb1 + 450

5 9647/03/CJC JC2 Preliminary Exam 2013 A chemist proposed the following synthetic routes to synthesise putrescine. Route 1: Route 2: (iii) By considering the type of reaction taking place, comment briefly on the yield of putrescine synthesised via route 1. Reaction 1 gives a poor yield since free-radical substitution reaction may occur at any of the C–H bonds and so, many different substitution products are obtained. (iv) Route 2 is found to give only 10 % yield of putrescine. Suggest another synthetic route to obtain a higher yield of putrescine, starting from ethene. State clearly the reagents and conditions required. [7] [Total: 20] CH3CH2CH2CH2Cl ClCH2CH2CH2CH2Cl Cl2, uv light ClCH2CH2CH2CH2Cl H2NCH2CH2CH2CH2NH2 putrescine conc. NH3 (excess) sealed tube CH2=CH2 Cl2 (in CCl4) in the dark ClCH2CH2Cl KCN (ethanol) heat under reflux NCCH2CH2CN H2 / Pt heat H2NCH2CH2CH2CH2NH2 putrescine 451

6 9647/03/CJC JC2 Preliminary Exam 2013 2. (a) Sodium hypochlorite (NaClO) is a chemical commonly found in bleaching agents. It was developed in the 18 th century and was initially used to bleach cotton. It became a popular compound for bleaching clothes when it was found that sodium hypochlorite can remove stains from clothes at room temperature. Sodium hypochlorite is also used as a disinfectant in water treatment and in swimming pools. As compared to chlorine which can cause respiratory problems, sodium hypochlorite is a safer choice of disinfectant. The original method of producing sodium hypochlorite involved passing chlorine gas through a solution of sodium carbonate. The addition of chlorine to water results in the equilibrium shown below. C l 2 + H2O HOCl + H+ + Cl – (i) Some chlorine gas was bubbled through 100 cm 3 of water at 25 °C and allowed to reach equilibrium. The initial concentration of chlorine was 0.10 mol dm –3. At equilibrium, the concentration of chlorine decreased by 30 %. Write an expression for Kc for the above equilibrium and use the data provided to calculate its value. You may assume that the [H 2O] is 55.5 mol dm –3 throughout. Kc= [H+]ሾClିሿ[HOCl] ሾCl2ሿ[H2O] Cl2 H2O HOCl C l – H+ Initial conc/ mol dm–3 0.10 55.5 0 0 0 Change/ mol dm