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Unformatted text preview: krohn (tek335) – H13: Electrochem 2 – Mccord – (52445) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The reaction 2 Ag + (aq) + Fe(s) → Fe 2+ (aq) + 2 Ag(s) taking place in a battery generates a current of 2 amp. How much Fe(s) is consumed in 1 hour? 1. 4.16 g 2. 2.08 g correct 3. 1.04 g 4. 8.32 g 5. 3.46 g Explanation: i = 2 A t = 1 h The half equation of interest is Fe(s) → Fe 2+ + 2 e − and the total charge is q = (2 A) (1 h) 60 min 1 h 60 sec 1 min = 7200 C . We can then convert this charge to num ber of electrons and finally the amount of Fe consumed: (7200 C) 1 mol e − 96485 C × 1 mol Fe 2 mol e − × 55 . 847 g Fe 1 mol Fe = 2 . 08374 g Fe 002 10.0 points What mass of copper would be produced by the reduction of copper(II) ions during pas sage of 1 ampere of current through a solution of copper(II) sulfate for 65 minutes? Correct answer: 1 . 2831 g. Explanation: 003 10.0 points How long will it take to deposit 0.00235 moles of gold by the electrolysis of KAuCl 4 (aq) us ing a current of 0.214 amperes? 1. 53.0 min correct 2. 26.5 min 3. 17.7 min 4. 70.7 min 5. 106 min Explanation: 004 (part 1 of 2) 10.0 points Thomas Edison was faced with the problem of measuring the electricity that each of his customers had used. His first solution was to use a zinc “coulometer”, an electrolytic cell in which the quantity of electricity is de termined by measuring the mass of zinc de posited. Only some of the current used by the customer passed through the coulome ter. What mass of zinc would be deposited in 44 days if 1 mA of current passed through the cell continuously? Correct answer: 1 . 28859 g. Explanation: t = 44 d F = 96485 C I = 1 mA = 0 . 001 A = 0 . 001 C/s Zn 2+ + 2 e − → Zn(s) There are 2 mol e − for every mol Zn. The charge is q = I t = (0 . 001 C / s)(44 d) parenleftbigg 24 hr 1 d parenrightbiggparenleftbigg 3600 s 1 hr parenrightbigg = 3801 . 6 C with n − e = (3801 . 6 C) parenleftbigg 1 mol e − 96485 C parenrightbigg = 0 . 0394009 mol e − krohn (tek335) – H13: Electrochem 2 – Mccord – (52445) 2 consumed. Thus n Zn = (0 . 0394009 mol e − ) parenleftbigg 1 mol Zn 2 mol e − parenrightbigg = 0 . 0197005 mol of Zn was reduced and m = (0 . 0197005 mol Zn) parenleftbigg 65 . 409 g Zn 1 mol Zn parenrightbigg = 1 . 28859 g Zn . 005 (part 2 of 2) 10.0 points An alternative solution to this problem is to collect the hydrogen produced by electrolysis and measure its volume. What volume would be collected at 281 K and 1 . 3 bar under the same conditions? Correct answer: 0 . 354038 L. Explanation: The hydrogen reduction is 2 H + + 2 e − → H 2 (g) corresponding to the Zn reduction Zn 2+ + 2 e − → Zn(s) There is 1 mol H 2 for every mol Zn....
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This note was uploaded on 10/25/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Holcombe
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