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 electr