Spring 09 Recitation Quiz 6

Spring 09 Recitation Quiz 6 - CHEM 0120 Golde SPRING 2009...

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CHEM 0120 SPRING 2009 Monday 1:00 PM Golde QUIZ 6 WEEK 8 February 23, 2009 Name_______________________________ E o cell = E o red + E o ox ; G ° = -nFE o ; F = 96500 Coul/mol e - 1. Consider the following condensed cell diagram for a Galvanic cell: Anode (-) Ni(s) | NiSO 4 (aq) || Fe(NO 3 ) 2 (aq) , Fe(NO 3 ) 3 (aq) | Pt(s) (+) Cathode a) Which element is undergoing oxidation and which element is undergoing reduction in this cell? Give your reasoning. Also, identify any spectator ions in the cell. b) Write the balanced equation for the reduction half reaction. c) What is the meaning of the symbols (-) and (+) in the diagram? Explain carefully. 2. For the cell reaction: Cu 2+ (aq) + 2 Tl(s) Cu(s) + 2 Tl + (aq), E o cell = +0.678 V. Calculate G o in kJ for this reaction. Show clearly how you determined the value of n. ANSWERS 1. a) Oxidation occurs at the anode. This half cell includes Ni(s) with oxidation state zero and Ni 2+ . Thus Ni is being oxidized. On the reduction side, iron is present in the Fe 2+ and Fe 3+ states. So, it is Fe which is being reduced. The other interesting elements, S and N, appear only as SO 4 2- and NO 3 - in solution. These must therefore be the spectator ions. b) Fe 3+ (aq) + e - Fe 2+ (aq) c) These represent the polarities of the electrodes. In conventional terms, the cathode (+) has a higher potential than the anode (-). From the perspective of the electron, the (-) electrode has a higher potential than the (+) electrode. 2. n = 2 (Cu gains two electrons and each Tl atom loses one electron) so G o = - 2 mol x 96500 Coul/mol x (0.678) V = -1.31 x 10 5 J = -131 kJ. CHEM 0120 SPRING 2009 Monday 6:00 PM Golde QUIZ 6 WEEK 8 February 23, 2009 For Make-up Only Regular recitation Day: Time: instructor: Quiz Score: Group score: Scores recorded by:
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Name_______________________________ E o cell = E o red + E o ox ; G ° = -nFE o ; F = 96500 Coul/mol e - 1. For the reduction of HClO 3 to Cl - in acid solution, the reduction potential is +1.451V. (a) Write a balanced equation for this half reaction (b) Calculate G o in kJ for your half reaction. 2. It is known that oxidation occurs at the anode in a Galvanic cell. Given that information: a) Explain why electrons flow (through the wire) from the anode to the cathode . b) Explain why the anode has (-) polarity . Be sure to give valid logical reaoning. ANSWERS 1. a) HClO 3 + 5 H + + 6 e - Cl - + 3 H 2 O b) G o = -6mol x 96500 Coul/mol x 1.451V = -8.40 x 10 5 J = -840kJ 2. a) Electrons are produced at the anode, where oxidation takes place, so they must flow from the anode to the cathode.
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