Chem 162-2011 Lecture 20

Chem 162-2011 Lecture 20 - CHEMISTRY 162-2011 LECTURE 20...

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Unformatted text preview: CHEMISTRY 162-2011 LECTURE 20 ANNOUNCEMENTS EXAMS Exam III Wednesday, April 13 th , 9:40 11:00 PM Coverage: Chapters 16.4-16.7, 20.7-20.9, 17.1-17.7 ATTENDANCE Sign in QUIZ Chem 162-2011 Lecture 20 1 PLAN FOR TODAY : CHAPTER 18 - ELECTROCHEMISTRY 18.3. A qualitative description of voltaic cells 18.4. Standard electrode potentials 18.5. Electrode potentials, spontaneous change and equilibrium Chem 162-2011 Lecture 20 2 Chem 162-2011 Lecture 20 3 PERIODIC TABLE AND PERIODIC PROPERTIES OXIDATION-REDUCTION REACTIONS Fe Fe 3+ + 3e- OXLEA: Oxidation is loss of electrons at the anode. OXION: Oxidation is increase in oxidation number. Fe is oxidized by O; O 2 is the oxidizing agent. Note that the atom is what is oxidized or reduced, but the agent is the molecule. O + 2e- O 2- (REGEC): Reduction is gain in electrons at the cathode. (RERON): Reduction is reduction in oxidation number. O is reduced by Fe; Fe is the reducing agent. Note that oxidation and reduction always occur together. If an element gets oxidized (gives off electrons) there must be another element getting reduced (gaining electrons. Thus, oxidation-reduction reactions are electron transfer reactions. You cannot have oxidation without reduction or reduction without oxidation. They are complimentary processes. or OILRIG: Oxidation is loss of electrons; reduction is gain of electrons. Fe + O 2 Fe 2 O 3 Fe oxidized O reduced Fe reducing agent O 2 oxidizing agent o Periodic table: Most easily reduced (highest potential to be reduced; best oxidizing agents) are electronegative substances (upper right- hand corner of periodic table. They pull electrons from other substances to form an octet.) Most easily oxidized (highest potential to be oxidized; best reducing agents) are electropositive substances (lower left- hand corner of periodic table. They donate electrons to other substances to form an octet.) Chem 162-2011 Lecture 20 4 ELECTROCHEMISTRY FORMULAS 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) 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 = G 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 20 5 ET : Draw a voltaic cell Hypothetical Voltaic Cell of Iron Oxidation Fe + O 2 Fe 2 O 3 Fe Fe 3+ + 3e- OXLEA: Oxidation is loss of electrons at the anode....
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Chem 162-2011 Lecture 20 - CHEMISTRY 162-2011 LECTURE 20...

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