Question

Ascorbic acid is a monoanion at neutral pH. It is a powerful antioxidant, which scavenges free radicals. It is oxidized in a single-electron oxidation to the monodehydroascorbate radical anion (MDHA-) accodring to the reaction Asc- -> MDHA- + H+ + e- The half-reaction has a standard reduction potential (E0') of +0.330V. Ascorbate also undergoes a two-electron oxidation to dehydroascorbate (DHA) Asc- -> DHA + H+ + 2e- This half-reaction has a standard reduction potential (E0') of +0.08V. Finally, MDHA- undergoes a disproportionation reaction 2MDHA- + H+ -> DHA + Asc- Determine the biochemical standard free-energy changeG0' for this reaction.

Answer 1

-48.24 kJ/mol

Step-by-step explanation:

For the global reaction

2MDHA⁻ + H⁺ → DHA + Asc⁻

The reduction potential of the reaction is the reduction potential of the substance that is reducing less the reduction potential of the substance that is oxidizing.

As we can see, MDHA⁻ is reducing (its gaining electrons), and Asc in oxiding(it's losing electrons), so:

E° = 0.330 - 0.08 = 0.250 V

The standard free energy can be calculated by:

ΔG°' = -nFE°

Where n is the number of electrons changed in the reaction (in this case 2), and F is the Faraday constant (96485 C/mol).

ΔG°' = -2*96485*0.250

ΔG°' = -48242.5 J/mol

ΔG°' = -48.24 kJ/mol