Calculate the standard free energy change for the reduction of O2 by NADH. The relevant standard reduction potentials for the two half-reactions are: 1/2 O2 + 2 H+ + 2 e- -> H2O E°' (V) = 0.815 NAD+ + H+ + 2 e- -> NADH E°' (V) = -0.315 The Faraday constant is 96,485 J*V-1*mol-1. Is the reaction spontaneous? Where in an organism does this reaction occur?
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Step 1
This can be done using the Nernst equation: ΔG°' = -nFΔE°' where n is the number of electrons transferred in the reaction, F is the Faraday constant, and ΔE°' is the difference in standard reduction potentials. In this case, n = 2 (since two electrons are Show more…
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Calculate the standard free energy change for the reduction of O2 by NADH: The relevant standard reduction potentials for the two half-reactions are: ½ O2 + 2 H+ + 2 e- → H2O 0.815 NAD+ + H+ + 2 e- → NADH -0.315 The Faraday constant is 96,485 J*V^-1*mol^-1 Extra Credit: Is the reaction spontaneous? Where in an organism does this reaction occur? (Hint: this reaction does not occur in a single step:)
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The standard reduction potentials for two biologically relevant redox reactions are as follows: NAD+ + H+ + 2e- ↔ NADH (E°' = -0.315 V) FAD + 2H+ + 2e- ↔ FADH2 (E°' = -0.219 V) What is the standard reduction potential change (ΔE°') for an overall reaction in which NADH is oxidized and the electrons released are used to reduce FAD? 1. -0.534 V 2. -0.096 V 3. +0.096 V 4. +0.534 V
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Calculate the standard free-energy change at $25^{\circ} \mathrm{C}$ for the following reaction. $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g)+12 \mathrm{H}^{+}(a q) \longrightarrow 4 \mathrm{Al}^{3+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ Use standard electrode potentials.
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