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Chem 162-2011 Lecture 21 truncated

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

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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
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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
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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
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∆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
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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. ∆G = ∆G o + RT ln(Q) ∆G = -nFE cell ∆G o = -nFE o cell -nFE cell = -nFE o cell + RT ln(Q) nFE cell = nFE o cell - RT ln(Q) E cell = E o cell – (RT/nF) ln(Q) @298K: E cell = E o cell – (0.025693/n) ln(Q) or E cell = E o cell – (0.0592/n) log(Q) Chem 162-2011 Lecture 21 5
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Z&Z 51 ET: Point out that until now we discussed E under standard conditions. Now we’ll begin finding E under non-standard condtions, i.e., Nernst equation. Given new concentrations and E o , find qualitative effect on E o .
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