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