Help with Redox Reactions! a) Using the table provided, write the two half react

Question

Help with Redox Reactions!

a) Using the table provided, write the two half reactions (one reduction, one oxidation) for the production of ethanol from acetylaldehyde.

b) Calculate the standard reduction potential for each half-reaction in part a

c) Write the net reaction, and calculate °’

TABLE 15.1 Standard Reduction Potentials of Some Biological Substances HALF-REACTION Cytochrome a, (Fe3+)ecytochrome a, (Fe2*) Cytochrome a (Fe3+) + e- cytochrome a (Fe2+) Cytochrome c (Fe3+) + e- cytochrome c (Fe2+) Cytochrome c1 (Fe3+) + e- cytochrome G (Fe2+) Cytochrome b (Fe3+1+ e- cytochrome b (Fe2+)(mitochondrial) Ubiquinone 2 H+ + 2 e- ubiquinol Fumarate + 2 H+ + 2 e- succinate FAD + 2 H + + 2 e- FADH2 (in flavoproteins) oxaloacetate + 2 H+ + 2 e- malate- Pyruvate 2 H 2lactate Acetaldehyde + 2 H+ + 2 e- ethanol 0.815 0.48 0.42 0.385 0.29 0.235 0.22 0.077 0.045 0.031 0. 0.166 - 0.185 Lipoic acid + 2 H + 2 e dihydrolipoic acid NAD+ H+ + 2 e- NADH NADP+ + H+ + 2 e- NADPH Acetoacetate + 2 H+ + 2 e- 3-hydroxybutyrate Acetate + 3 H+ + 2 e- acetaldehyde + H2O 0.197 0.23 0.29 0.315 0.320 0.346 0.581

Explanation / Answer

producion of ethanol --> from acetaldhyde

clearly we need

Acetaldehyde + 2H+ + 2e- <--> Ethanol E= -0.197 V

now, choose:

Acetate- + 3H+ + 2e- <--> Acetaldehyde + H2O E = -0.581

invert it:

Acetaldehyde + H2O <-->Acetate- + 3H+ + 2e- E = 0.581

Add both

Acetaldehyde + H2O + Acetaldehyde + 2H+ + 2e- <--> Ethanol + Acetate- + 3H+ + 2e-   E= -0.197 + 0.581

simplify

H2O + 2Acetaldehyde<--> Ethanol + Acetate- + H+ E = 0.384 V

then

dG = -nF*E°cell

dG = -2*96500*0.384

dG = -74112 J/mol

dG = -74.112 kJ/mol

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