#Chem 162-2008 12th week recitation

#Chem 162-2008 12th week recitation - CHEMISTRY 162-2008...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEMISTRY 162-2008 12 th WEEK RECITATION ANNOUNCEMENTS MISCELLANEOUS EXAMS Third hourly exam: Wed., 4/23, 9:40 – 11:00 PM Review session 8 – 10 PM: Wed., April 16 th , Cook Douglass Lecture Hall 102 Thu., April 17 th , Cook Douglass Lecture Hall 109 Coverage (At this time, I don’t know how I’ll split it up.) Chapter 16.6: Complex ions Chapter 22.7-9: Coordination chemistry Chapter 17.1-7: Thermodynamics Chapter 18.1-18.6: Electrochemistry Next week’s recitation: Review only ATTENDANCE Sign in QUIZ Recitation quiz III and old quizzes returned today. MISCELLANEOUS Batteries and corrosion won’t be covered in recitation. Chem 162-2008 12th week recitation 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chem 162-2008 12th week recitation 2
Background image of page 2
PLAN FOR TODAY : CHAPTER 17 - ELECTROCHEMISTRY Oxidation-Reduction Reactions Galvanic cells, Reduction/Oxidation potentials, Free energy The Nernst Equation Electrolysis Chem 162-2008 12th week recitation 3
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 162-2008 12th week recitation 4 REDUCTION OXIDATION
Background image of page 4
∆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 -(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) ∆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 162-2008 12th week recitation 5
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ET: Point out that until now we discussed E under standard conditions.
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 19

#Chem 162-2008 12th week recitation - CHEMISTRY 162-2008...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online