Chem Test 3

Chem Test 3 -...

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A voltaic cell is set up at 25 o C with the following half cells Al(0.0010M) Al 3+ Ni(0.50M) Ni 2+ Write an equation for the reaction that occurs when the cell generates an electric current and determine  the cell potential E o cell = .04V E=E o  – (0.0592V/n) x log Q; Q=K @ equilibrium E=0=E o  – (0.0592V/n) x log(k); n=2 log k = (n x E )/0.0592V= (2 x 0.04V)/(0.0592V)=1.350  K=10^1.35=22.4 (K>0) Product Favored Calculate the equilibrium constant for the reaction at 25 o C. Fe(s)+Cd 2+ (aq)->Fe 2+ (aq)+Cd(s) Cd 2+  + 2e -    Cd  E o = -0.40V Fe 2+  + 2e -    Fe E o = -0.44V In so called redox reactions electrons are transferred In a voltaic cell, oxidation occurs at which electrode: anode The electrode at which reduction occurs is assigned a plus sign In a voltaic cell one typically uses a so called salt bridge to balance the charges. The anions in the salt  bridge move toward which electrode? Anode The driving force for electrons to move in a voltaic cell is called the electromagnetive force or cell  potential How is a Volt defined? 1V=1J/C The reduction potential for which element is b definition zero Volts under standard conditions? 
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This note was uploaded on 04/03/2012 for the course CHEM 102 taught by Professor Williamson during the Spring '08 term at Texas A&M.

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Chem Test 3 -...

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