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Reducing Power Electrons of reduced coenzymes flow toward O 2 This produces a proton flow and a transmembrane potential Oxidative phosphorylation is the process by which the potential is coupled to the reaction: ADP + P i ATP
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Reduced Coenzymes Conserve Energy from Biological Oxidations Amino acids, monosaccharides and lipids are oxidized in the catabolic pathways Oxidizing agent - accepts electrons, is reduced Reducing agent - loses electrons, is oxidized Oxidation of one molecule must be coupled with the reduction of another molecule A red + B ox A ox + B red
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Diagram of an electrochemical cell Electrons flow through external circuit from Zn electrode to the Cu electrode
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Reduction Potentials Cathode (Reduction) Half-Reaction Standard Potential E ° (volts) Li + (aq) + e - -> Li(s)-3.04 K + (aq) + e - -> K(s)-2.92 Ca 2+ (aq) + 2e - -> Ca(s)-2.76 Na + (aq) + e - -> Na(s)-2.71 Zn 2+ (aq) + 2e - -> Zn(s)-0.76 Cu 2+ (aq) + 2e - -> Cu(s)0.34 O 3 (g) + 2H + (aq) + 2e - -> O 2 (g) + H 2 O(l)2.07 F 2 (g) + 2e - -> 2F - (aq)2.87
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Standard reduction potentials and free energy Relationship between standard free-energy change and the standard reduction potential: G o’ = -nF E o’ n = # electrons transferred F = Faraday constant (96.48 kJ V -1 ) E o’ = E o’ electron acceptor - E o’ electron donor
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Example Suppose we had the following voltaic cell at 25 o C: Cu(s)/Cu +2 (1.0 M) // Ag + (1.0 M)/ Ag (s) What would be the cell potential under these conditions?
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Example Suppose we had the following voltaic cell at 25 o C: Cu(s)/Cu +2 (1.0 M) // Ag + (1.0 M)/ Ag (s) What would be the cell potential under these conditions? Ag + + e - ---> Ag 0 E 0 red = + 0.80 v Cu +2 + 2e - ----> Cu 0 E 0 red = + 0.337 v
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Example: Biological Systems Both NAD + and FAD are oxidizing agents
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The question is which would oxidize which? OR Which one of the above is the spontaneous reaction? in which G is negative
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To be able to answer the question We must look into the “electron donation” capabilities of NADH and FADH 2 i.e. reduction potentials of NADH and FADH 2
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E o’ = E o’ electron acceptor - E o’ electron donor Remember, For a spontaneous reaction E o ’ must be positive
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Therefore, rearrange Add the two reactions
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electron acceptor electron donor
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