2019 NJC Prelim P2 ANS

[Turn over Suggested Answer for SH2 H2 Chemistry 2019 Prelim Paper 2 1 (a) Brass is a mixture of copper and zinc. When a piece of brass is placed in dilute hydrochloric acid, only one of the metals present dissolves. Explain the observation and write an equation for the reaction that occurs. Zn2+ + 2e− Zn Eo = –0.76V Cu2+ + 2e− Cu Eo = +0.34V 2H+ + 2e− H2 Eo = 0.00 V For reaction between Zn and H+, Eocell = Eored – Eoox = 0.00 – (–0.76) = +0.76V (reaction is feasible) Zn + 2HCl → ZnCl2 + H2 OR Zn + 2H+ → Zn2+ + H2 For reaction between Cu and H+, Eocell = Eored – Eoox = 0.00 – (0.34) = –0.34V (reaction is not feasible) Note: It is essential to calculate Eocell between metal and H + to conclude whether the reaction is feasible. Some answers suggest Zn is more likely to be oxidized than Cu but that does not explain why Cu do not react with H+. (b) Copper and zinc are the two electrodes in a Daniell cell. The cell is set up as shown in Fig 1.1. Fig 1.1 (i) Indicate the direction of electron flow on Fig 1.1. [1] Note: Zn2+ + 2e− Zn Eo = –0.76V Cu2+ + 2e− Cu Eo = +0.34V Cu2+/Cu half cell undergoes [R] and gain e− Zn2+/Zn half cell undergoes [O] and lose e− Electrons flow from the anode (oxidation) to the cathode (reduction). e− e−

NJC/H2 Chem Prelim/02/2019 2 For Examiner’s Use (ii) Write the equation for the reaction taking place and calculate ΔGo for the reaction. Cu2+ + Zn → Zn2+ + Cu Eocell = Eored – Eoox = +0.34 – (–0.76) = +1.10V ΔGo = –nFEocell = –2 × 96500 × 1.10 = – 2.12 × 105 J mol–1 [3] (iii) Discuss the effect of the following changes on the cell potential of the above cell. (I) adding excess amount of aqueous ammonia into the Zn2+/Zn half-cell Zn2+ + 2e− Zn Eo = –0.76V When excess amount of aqueous ammonia is added to Zn 2+/Zn half -cell, [Zn(NH3)4]2+ complex ion will be formed.In order to partially offset the decrease in [Zn2+], oxidation of Zn is favoured to form more Zn 2+. Hence E(Zn 2+/Zn) becomes more negative. Ecell = Eo(Cu2+/Cu) – E(Zn2+/Zn) Ecell will become more positive. (II) using a smaller copper electrode Size of electrode will not affect the Eo of the reduction half-equation and Ecell since there is no shift in the position of the equilibrium. Note: Solid does not affect position of equilibrium. (iv) How many hours will it take for t his galvanic cell to plate 15.0 g of metallic copper from 1 mol dm–3 solutions of Cu2+(aq) and Zn2+(aq) using a current of 5.0 A? [R] Cu2+ + 2e− → Cu Amount of Cu = 15.0 63.5 = 0.2362 mol Amount of electrons = 0.4724 mol Q = It =neF t = 0.4724 × 96500 5.0 = 9117s = 2.53h [2]

NJC/H2 Chem Prelim/02/2019 3 For Examiner’s Use [Turn over (v) The Nernst equation is an equation that relates the cell potential, E cell, of an electrochemical reaction to the standard cell potential, Eocell. It can be simplified and written as Nernst equation : Ecell = Eocell – RT zF lnQ where R is the ideal gas constant T is temperature in Kelvin z is the number of electron transferred F is Faraday's constant Q is [Zn2+] [Cu2+] for the Daniell cell Using the above equation, calculate Q for the overall redox reaction of the Daniell cell when it reaches equilibrium at 25 oC. At equilibrium, Ecell = 0 Eocell = RT zF lnQ 1.10 = 8.31 × 298 2 × 96500 lnQ Q = 1.71 × 1037 [2] (vi) In another set-up conducted at 25 oC, the concentration of Cu2+ ions and Zn2+ ions have been adjusted to 0.125 mol dm–3 and 0.002 mol dm–3 respectively. Using the Nernst equation, calculate the e.m.f. of this cell. Ecell = Eocell – RT zF ln [Zn2+] [Cu2+] = 1.10 – 8.31 × 298 2 × 96500 ln [0.002] [0.125] = 1.15V [2] [Total: 17]

NJC/H2 Chem Prelim/02/2019 4 For Examiner’s Use 2 Saccharin is an artificial sweeting agent used in some soft drinks and produced in various ways. Fig 2.1 shows t he original route by Remsen and Fahlberg , starting with methylbenzene undergoing electrophilic substitution reaction using chlorosulfonic acid, C lSO3H. This is also known as aromatic sulfonation process. Fig 2.1 (a) Draw a dot-and-cross diagram of chlorosulfonic acid, ClSO3H and hence state the bond angle for Cl−S−O. Cl−S−O bond angle : 109.5 Note: All acid compounds are covalent molecule. They only dissociate into ions in aqueous state. Acids containing H&O atoms lose H+ from the O−H bonds when ionize in water. Examples of acid molecule structures: S OHO O OH N O OHO C O OHHO P OHO OH OH [2] (b) (i) Reaction I of the aromatic sulfonation process involves the formation of an electrophile. Identify the electrophile. Electrophile : +SO2Cl [1]

NJC/H2 Chem Prelim/02/2019 5 For Examiner’s Use [Turn over (ii) Hence, describe the mechanism for reaction I of the reaction. Show all charges and relevant lone pairs and show the movement of electron pairs by using curly arrows. Note: arrow from benzene  electron to S atom of the electrophile to form arenium cation arrow from C−H bond to center of arenium cation to regenerate aromaticity and lose H+ [2] (iii) State the types of reaction that occur during reactions II and III. reaction II : nucleophilic substitution / condensation reaction III : oxidation [2] (iv) Suggest reagents and conditions for reactions II and III. reaction II : excess ethanolic NH3, heat in sealed tube (behave like C−Cl) (also accept NH3(g) r.t.p. as it could react similar to acyl chloride) reaction III : KMnO4, dilute H2SO4, heat [2] [Total : 9] 3 Perchlorate (ClO4–) compounds are used as oxidisers in some fireworks to aid the combustion reaction. These perchlorates can contaminate bodies of water near fireworks displays. Elevated concentrations of perchlorate in water can affect wildlife and it may also affect human health if it contaminates drinking water. (a) One common perchlorate present in fireworks is ammonium perchlorate, NH4ClO4. Given that the solubility of ammonium perchlorate is 30.6 g per 100 cm 3, express its solubility in mol dm−3 and hence, calculate the solubility product, Ksp, for NH4ClO4. Solubility, s = 30.6 117.5 × 10 = 2.60 mol dm−3 Ksp of NH4ClO4= [NH4+][ ClO4–] =(s)(s) = (2.60)2 = 6.78 mol2 dm−6 [2] slow

NJC/H2 Chem Prelim/02/2019 6 For Examiner’s Use (b) A group of NJC students conducted a study to determine the perchlorate content of the water in the Singapore River before and after the fireworks display in the National Day Parade Rehearsal. They followed the methodology described below one week before the rehearsal:  Collect a 100.0 cm3 sample of the river water and acidify it with 10 cm3 of 1 mol dm3 sulfuric acid.  The resulting solution is added to an excess of 50 cm 3 of 5 × 105 mol dm3 iron(II) sulfate solution. The mixture is then made up to 250 cm3. This is solution X.  A 25.0 cm3 sample of solution X is pipetted into a conical flask. The 25.0 cm3 solution is titrated against 1 × 106 mol dm3 potassium manganate(VII) to determine the amount of remaining iron(II) ions in the sample. (i) Write the half equation for the conversion of perchlorate ions to chlorine in the acidic medium and hence the overall equation for the reaction between perchlorate and iron(II) ions. Half equation : [R] 2ClO4– + 16H+ + 14e– → Cl2 + 8H2O Overall equation : 2ClO4– + 16H+ + 14Fe2+ → Cl2 + 14Fe3+ + 8H2O [2] (ii) Describe how you would recognise