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HW 10 Solutions

# HW 10 Solutions - Wright Jacqueline Homework 10 Due 11:00...

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Wright, Jacqueline – Homework 10 – Due: Nov 21 2007, 11:00 pm – Inst: James Holcombe 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Calculate the potential for the cell indicated: Fe | Fe 2+ (10 - 3 M) || Pb 2+ (10 - 5 M) | Pb Pb 2+ + 2 e - Pb E 0 = - 0 . 126 V Fe 2+ + 2 e - Fe E 0 = - 0 . 440 V 1. 0.255 V correct 2. 0.284 V 3. 0.196 V 4. 0.373 V 5. 0.432 V Explanation: The overall reaction is Fe + Pb 2+ Fe 2+ + Pb Please notice that since the concentrations are not 1 M, the Nernst equation must be used. In this cell notation, the anode is located on the left of the salt bridge || and the cathode on the right. So first calculate E 0 cell = E cathode - E 0 anode = - 0 . 126 V - ( - 0 . 440) V = 0 . 314 V Using the Nernst Equation E cell = E 0 cell - 0 . 05916 n log Q = 0 . 314 V - 0 . 05916 2 log [Fe 2+ ] [Pb 2+ ] = 0 . 314 V - 0 . 05916 2 log 10 - 3 10 - 5 = 0 . 25484 V 002 (part 1 of 1) 10 points An evil scientist has created a lightening weapon to destroy the entire world unless they give him a herring (a fish) and a very nice shrubbery. As a power source, he has created a fuel cell that he thinks will run on protons (H + ) and fluorine molecules (F 2 ). (He cre- ates them and holds them in special magnetic bottles made of superconductive material.) Given the standard reduction potentials 2 H + + 2 e - H 2 E = 0.00 2 F 2 + 2 e - F - E = 2.87 will his power source work? 1. Ni! Ni! Ni! 2. Choose me! Choose me! 3. Nope. He’s a defective evil genius. There is no H 2 to be oxidized. correct 4. M R ducks! M R not! M R 2! 5. Yes. The E cell is positive so he will have plenty of power to run his weapon. Explanation: There is no H 2 for the F 2 to oxidize. 003 (part 1 of 1) 10 points You turn on a flashlight containing brand new NiCad batteries and keep it lit for a minute or two. Which of the following can be considered TRUE regarding the chemical state of these batteries? I. Δ G for the battery reaction is negative. II. E cell > 0. III. The batteries are at equilibrium. IV. E cell is substantially decreasing during this time. 1. All 2. II and IV only 3. III only 4. All but IV 5. I and II only correct 6. Maybe IV and I

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Wright, Jacqueline – Homework 10 – Due: Nov 21 2007, 11:00 pm – Inst: James Holcombe 2 7. All but III Explanation: (I and II: TRUE) Δ G must be negative because the light IS on and therefore a spon- taneous change is occuring. Same logic goes for the E cell except opposite in sign (positive).
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HW 10 Solutions - Wright Jacqueline Homework 10 Due 11:00...

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