H13_Electrochem 2 - krohn (tek335) – H13: Electrochem 2...

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Unformatted text preview: krohn (tek335) – H13: Electrochem 2 – Mccord – (52445) 1 This print-out should have 24 questions. Multiple-choice 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.

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H13_Electrochem 2 - krohn (tek335) – H13: Electrochem 2...

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