Chem 162-2011 Lecture 21 truncated

Chem 162-2011 Lecture 21 truncated - CHEMISTRY 162-2011...

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Unformatted text preview: CHEMISTRY 162-2011 LECTURE 21 ANNOUNCEMENTS MISCELLANEOUS ATTENDANCE • Sign in E-MAIL EXAMS • Exam III – How did we do? :>) QUIZ MISCELLANEOUS Chem 162-2011 Lecture 21 1 PLAN FOR TODAY : CHAPTER 18 - ELECTROCHEMISTRY 18.6 Effect of concentrations on cell voltage Nernst equation Concentration cells pH measurement 18.7 Batteries: Using chemical reactions to make electricity The Dry Cell The Lead-Acid Storage Battery Other Secondary Cells Fuel Cells Chem 162-2011 Lecture 21 2 ELECTRODE REDUCTION POTENTIALS ET: OXLEA; discuss oxidation in terms of reacting with electronegative oxygen; discuss oxidation and reduction in terms of oxidation number; discuss oxidation and reduction vs oxidizing and reducing agents. Some Selected Standard Electrode Reduction Potentials E o , volt (red’n) F 2 + 2 e-----> 2 F- +2.87 H 2 O 2 + 2 H + + 2 e-----> 2 H 2 O +1.78* MnO 4- + 8H + + 5e- → Mn 2+ + 4H 2 O +1.51 Au 3+ + 3e- ----> Au +1.50 PbO 2 + 4 H + + 2 e-----> Pb 2+ + 2 H 2 O +1.46 Cl 2 + 2 e-----> 2 Cl- +1.36 Cr 2 O 7 2- + 14 H + + 6 e-----> 2 Cr 3+ + 7 H 2 O +1.33 O 2 + 4 H + + 4 e-----> 2 H 2 O +1.23* Br 2 + 2e-----> 2Br- +1.09 Ag + + e-----> Ag +0.80 Fe 3+ + e-----> Fe 2+ +0.77 MnO 4- + 2 H 2 O + 3 e-----> MnO 2 + 4 OH- +0.60 I 2 + 2 e-----> 2 I- +0.54 Cu 2+ + 2 e-----> Cu +0.34 Cu 2+ + e-----> Cu + +0.16 2H + + 2 e-----> H 2 0.00 Fe 3+ + 3e-----> Fe-0.036 Pb 2+ + 2 e-----> Pb-0.13 Ni 2+ + 2 e-----> Ni-0.23 Cd 2+ + 2e- → Cd-0.40 Cr 3+ + e-----> Cr 2+-0.50 Zn 2+ + 2 e-----> Zn-0.76 2 H 2 O + 2 e-----> H 2 + 2 OH--0.83* Al 3+ + 3e- → Al-1.66 Mg 2+ + 2 e-----> Mg-2.37 Na + + e-----> Na-2.71 K + + e- ← → K-2.92 Li + + e-----> Li-3.05 * H 2 O half-cell reactions Chem 162-2011 Lecture 21 3 REDUCTION OXIDATION ∆G o , E o , K, Interconversions E o cell = E o redn + E o oxidn E cell = E o cell-(RT/nF)ln(Q) = Nernst equation (E cell = E o cell – ((8.314 x 298.15)/(n x 96485)) x (2.303 x log(Q)) E cell = E o cell-(0.0592/n)log(Q) = Nernst equation At equilibrium: 0 = E o cell-(0.0592/n)log(K) E o cell = (0.0592/n)log(K) E cell = (0.0592) x pH ∆G = ∆G o + RT ln(Q) At equilibrium: 0 = ∆G o + RTln(K) ∆G o = -RTln(K) ∆G o = -nFE o cell ∆G = -nFE cell w max = ΔG Definitions: 1 ampere = 1 coulomb of charge/second 1 mole of electrons caries a charge of 1 Faraday = 96485 coulombs Calculations : (1) Amperes x ((Coulombs/sec)/Ampere) x sec x (1 mol e- /96485 coulomb) x (mol subst./mol e- ) = mol substance or (2) (amperes x seconds)/(96485 x electrons) = mol substance Chem 162-2011 Lecture 21 4 NERNST EQUATION DERIVATION E o cell = (0.0592/n)log(K) Used only for standard conditions of 1 atm of 1M. But most cell measurements under nonstandard conditions....
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This document was uploaded on 11/02/2011 for the course GEN CHEM 162 at Rutgers.

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Chem 162-2011 Lecture 21 truncated - CHEMISTRY 162-2011...

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