EXAM3solutions - Version 289 – Exam 3 – flowers –...

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Unformatted text preview: Version 289 – Exam 3 – flowers – (53885) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Exam 3. No books, no notes. You may bring and use as many number 2 pencils as you see fit. PLEASE PLEASE PLEASE make sure to bubble in your UTEID and Exam Version Number properly. 001 10.0 points Calculate the reduction potential for the Zn 2+ | Zn electrode if the Zn 2+ concentra- tion is 4 × 10 − 3 M. The standard potential for Zn 2+ + 2 e − → Zn is- . 7628 V. 1.- . 904 V 2.- 1 . 808 V 3.- . 833 V correct 4.- 1 . 666 V Explanation: [Zn 2+ ] = 4 × 10 − 3 M E = E- parenleftbigg . 0592 n parenrightbigg log Q =- . 7628 V- parenleftbigg . 0592 2 parenrightbigg log parenleftbigg 1 4 × 10 − 3 parenrightbigg =- . 833 V 002 10.0 points Given the oxidation-reduction half-reactions K(s) → K + (aq) Co 2+ (aq) → Co(s) what is the overall balanced reaction equa- tion? 1. K(s) + 2 Co 2+ (aq) → K + (aq) + 2 Co(s) 2. K + (aq) + Co 2+ (aq) → Co(s) + K(s) 3. 2 K(s) + Co 2+ (aq) → 2 K + (aq) + Co(s) correct 4. 2 K + (aq) + Co 2+ (aq) → Co(s) + 2 K(s) 5. K(s) + Co 2+ (aq) → K + (aq) + Co(s) 6. K(aq) + 2 Co 2+ (aq) → 2 Co(s) + K(s) Explanation: 003 10.0 points The potential ( E ) of the cell shown below is +0 . 045 V. Pb(s) | Pb 2+ (aq , ?) || Ni 2+ (aq , . 1 mol / L) | Ni(s) What is the missing concentration of the Pb 2+ ? 1. 1.83546e-11 2. 5.26688e-24 3. 7.33758e-21 4. 1.05119e-09 5. 3.64162e-21 6. 9.41195e-11 7. 1.91456e-24 8. 1.25333e-06 9. 2.1485e-30 10. 2.7905e-22 Correct answer: 1 . 25333 × 10 − 6 mol / L. Explanation: At the cathode, Ni 2+ (aq) + 2 e − → Ni(s) E ◦ =- . 23 V At the anode, Pb(s) → Pb 2+ (aq) + 2 e −- E ◦ = +0 . 13 V Combining both half reactions, we get Pb(s) + Ni 2+ (aq) → Ni(s) + Pb 2+ (aq) E ◦ cell =- . 1 V Using the Nernst equation, E = E ◦- parenleftbigg . 025693 n parenrightbigg ln parenleftbigg [Pb 2+ ] [Ni 2+ ] parenrightbigg . 045 =- . 1- parenleftbigg . 025693 2 parenrightbigg ln parenleftbigg [Pb 2+ ] . 1 parenrightbigg . 145 =- (0 . 0128465) ln [Pb 2+ ] . 1 Version 289 – Exam 3 – flowers – (53885) 2- 11 . 2871 = ln [Pb 2+ ] . 1 e − 11 . 2871 = [Pb 2+ ] . 1 [Pb 2+ ] = 0 . 1 e − 11 . 2871 = 1 . 25333 × 10 − 6 mol / L 004 10.0 points In the cell shown, A is a standard Zn 2+ | Zn electrode connected to a standard hydrogen electrode (SHE)....
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This note was uploaded on 02/04/2009 for the course CH 53885 taught by Professor Flowers during the Fall '08 term at University of Texas at Austin.

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EXAM3solutions - Version 289 – Exam 3 – flowers –...

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