Chap4Metab13thedition

# Toreleaseelectronsand hydrogenions h2 2e2h

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Unformatted text preview: ven these conventions, the reduction potential (E0´) of oxygen reduction is written: o 1/2 O2 + 2 H+ + 2e­ ⇒ 2O is =+0.816 volts and H that of hydrogen is written: o 2 H+ + 2e­ ⇒ 2 is ­0.421 volts H o Rewrite direction for more negative (it will be written as an oxidation): o H2 2H+ + 2e­ o ½ O2 + 2H+ + 2e­ H2O Sum: H2 + ½ O2 H2O Oxidation­Reduction Couples Oxidation­Reduction o o o Most molecules can either be electron donors or electron acceptors, depending on the substances with which they react The same atom on each side of the arrow in the half reactions can be thought of as representing a redox couple o 2H+/H2 or 1/2 O2/H2O Remember that the reduced substance of a redox couple whose reduction potential is more negative donates electrons to the oxidized substance of a redox couple whose potential is more positive Electron Tower Electron Tower o o o The tower represents a range of reduction potentials for redox couples from the most negative at the top to the most positive at the bottom The reduced substance in the redox pair at the top of the tower has the greatest tendency to donate electrons, whereas the oxidized substance in the couple at the bottom has the greatest tendency to accept electrons As electrons from donors at the top of the tower fall, they can be “caught” by acceptors at various levels below. Electron Tower Electron Tower o The difference in potential between two substances is expressed as ∆Ε0 ′ o The farther the electrons drop from a donor before they are caught by an acceptor, the greater the amount of energy released; that is, ∆Ε0 ′ is proportional to ∆ G0 ′ o Look at redox tower: Greatest drop is from CO2/Glucose (­0.43) pair to O2/H2O pair (+0.82) Redox Pairs Redox Pairs 1. 2. When writing a redox couple, the reduced member (on right) that is more negative, donates electrons to the oxidized member (on left) of the redox pair that is more positive The greater the difference in redox potential ∆Ε0 ′ between the redox pair that serves as the electron donor and the pair that serves as the acceptor, the greater amount of energy available in the oxidation­reduction reaction Comparing Two Redox Pairs: Comparing Two Redox Pairs: Two Rules • Pick two pairs: • • 2H+/H2 (­0.42) 2e­ 1/2 O2/H2O (+0.82) 2e­ 1/2 Rule I: • • Pick H2 (reduced member of pair that is more negative) will donate to O2 (oxidized member of pair that is more positive) Conclusion: H2 donates 2e­ to 1/2 O2 • • • H2 ⇒ 2H+ + 2e­ (Oxidation half reaction) 2e­ + 1/2 O2 ⇒ O­2 (Reductive half reaction) H2 + 1/2 O2 ⇒ 2H+ + O­2(H2O) Net oxidation­reduction rx ∆ Go’: Rule # 2 • Rule 2: ∆Ε0 ’ is very large, therefore: • • • • • • ∆Ε0 ’= E0’ (oxidizing agent) – E0’ (reducing agent) ∆Ε0 ’= E0’ (more positive E0’ )–E0’ (more negative ∆Ε0 ’) = 0.82­ (­0.42) = 1.24 volts Recall: oxidizing agent gets reduced in reaction Reducing agent gets oxidized in reaction Calculating free energy change from Faraday’s Equation: • • • ∆ G = ­n F ∆Ε0 ’ = ­(2 equiv)(23 kcal/volt equiv)(1.24 volts) = ­ 57 Kcal Coenzyme NADH + H+/ NAD+: Coenzyme NADH +...
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## This document was uploaded on 09/17/2013.

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