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Chem 04-24

Course: CHEM 112, Spring 2008
School: Texas A&M
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102 Chemistry 04/24/08 1. The standard cell potential, E (cell), for the reaction of silver ions with copper metal is + .46 V at 25 C. Calculate G for this reaction. Cu^2+(aq) + 2e- Cu(s) E = +.34 Anode (-) (Oxidation) Ag^+(aq) + e- Ag(s) E = .80 Cathode (+) (Reduction) Ox: Cu(s) Cu^2+(aq) + 2eRed: Ag^+(aq) + e- Ag(s) x2 Cu(s) + 2Ag(aq) Cu^2+(aq) + 2Ag(s) Cu: Anode oxidized Ag: Cathode reduced E (cell) =...

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102 Chemistry 04/24/08 1. The standard cell potential, E (cell), for the reaction of silver ions with copper metal is + .46 V at 25 C. Calculate G for this reaction. Cu^2+(aq) + 2e- Cu(s) E = +.34 Anode (-) (Oxidation) Ag^+(aq) + e- Ag(s) E = .80 Cathode (+) (Reduction) Ox: Cu(s) Cu^2+(aq) + 2eRed: Ag^+(aq) + e- Ag(s) x2 Cu(s) + 2Ag(aq) Cu^2+(aq) + 2Ag(s) Cu: Anode oxidized Ag: Cathode reduced E (cell) = E (cathode) - E (anode) = .80V (.34V) = .46 G = -nFT ; F= 96500 C/mol en=2 because there are 2e- transferred G = -(2 mol e-) (96,500 C/mol e-) (+.46V) -89,200 C*V = -89.2 KJ Spontaneous 2. Which of the following metals will react with H+(aq) to produce H2 in a product-favored reaction: Cu,Al,Ag,Fe,Zn? Cu^2+ + 2e- Cu(s) E = .34V Al^3+ + 3e- Al(s) E = -1.66 V Ag+(aq) + e- Ag(s) E = .80V Fe^2+(aq) + 2e- Fe(s) E = -.44 V Zn^2+ + 2e- Zn(s) E = -.76V 3. A voltaic cell is set up at 25 C with the following half-cells: Al (.0010 M Al^3+) Ni (.50 M Ni^2+) Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential Al^3+(aq) + 3e- Al(s) E = -1.66V Na^2+(aq) (anode) + 2e- Ni(s) E = -.25V (cathode) Ox: Al(s) Al^3+ (aq) + 3e- x 2 Red: Ni^2+(aq) + 2e- Ni(s) x3 2Al(s) + 3Ni^2+(aq) 2Al^3+(aq) + 3Ni(s) E (cell) = E (cathode) - E (anode) -.25V (-1.66V) = 1.41V Nernst: E = E - .0257V/n * ln Q Q = [reactants] / [products] Q = [Al^3+]^2/ [Ni^2+]^3 2.303 RT/F - .0592 V E = E - .0592 V/n log Q E = E -.0257 V/n * ln [products]/[reactants] ; n= E = 1.41V - .0257V/6 ln [.0010]^2/ [.050]^3 = 1.41 V - .00428 V * ln (8 x 10^-6) = 1.46V The Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, E (cell) would be 0, but Q would not Therefore, as long as the concentrations are different, E will not be 0. Applications of oxidation-reduction reactions Batteries Dry Cell Battery Primary batteries- uses redox reactions that cannot be restored by recharge. Alkaline Batteries Nearly same reactions as in common dry cell, but under basic conditions. Lead Storage Battery H2SO4 inside and used as an electrolyte. Secondary Battery- can be recharged. Uses redox reactions that can be reversed. Car battery = 12 V
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