Chem 222 Week 13 notes
Chem 222 Week 13 notes Chem 222
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This 2 page Class Notes was uploaded by Leslie Pike on Friday April 29, 2016. The Class Notes belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 12 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.
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Date Created: 04/29/16
Chem 222 notes Week 13 Sample problem: Calculate ΔG for the following reaction at standard conditions: 2Ag +(aq) Cu (s)> 2Ag (s)+ Cu 2+(aq) The standard reduction potential for silver is 0.799, and the standard reduction potential for copper is 0.337. First step: figure out what you have, what you need, and what relationship gets you from what you have to what you need. You have the E for your two reactions, and 0 you need ΔG. ΔG is related to E by the following equation: ΔG=-nFE. E for the complete reaction is the sum of the E of the two half-reactions. One half reaction is going forward, and the other is going backward. In the equation at the top, silver is written as a reduction; thus, its reduction potential is the same as given in the table. DO NOT multiply by the balancing coefficient. Copper is being oxidized in the above reaction, not reduced, meaning that its E will have an opposite sign of the number given in the table. Add the two half-reactions to get the E for the full reaction: Ecell +0.799 + (-0.337) = 0.462V. Plug this value into the expression for ΔG (n is 2 because you have 2 electrons being transferred in the above reaction, and F is Faraday’s constant, 96485): ΔG=-nFE = -(2)(96485)(.462)= -89000 J In an electrolytic cell, an electric current drives a nonspontaneous reaction. Electrolytic cells have a negative E, this is how you distinguish an electrolytic cell from a galvanic cell. Positive E means spontaneous, negative E means nonspontaneous. (Notice that this is the opposite of what the sign means when you are talking about free energy.) In an electrolytic cell, electrons are added to the cathode and removed from the anode. Metal plates out on the cathode, because it is reduced from its ionic form to its solid form when it takes the extra electrodes from the cathode. Whatever anion the metal was bounded to donates its electrons to the anode. Sample problem: You are purifying copper by electrolysis. You need 2 kg of pure copper. You have a 10 amp current source you can plug your cathode and anode into, and an unlimited supply of impure copper. How long does it take for 2 kg of pure copper to plate out onto the cathode? This problem looks hairy, but trust me, it’s nothing you haven’t already done in Chem 120. This is a regular stoichiometry problem; you just need to learn two new relationships: 96,485 C =1 mole electrons An amp is a coulomb per second (C/s), making a 10 amp current 10 C/ 1 s. 2000 g copper * (mol copper/63.55 g copper) * (2 mol electrons/mol copper) * (96485 coulombs/mol electrons) * (second/10 coulombs) = 607301 seconds = 168 hours = 7 days. You will be waiting a whole week for that copper. Better find a stronger current source.
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