4. Ascorbic acid (Vitamin C) is a mono-anion at neutral pH. It is a powerful ant

Question

4. Ascorbic acid (Vitamin C) is a mono-anion at neutral pH. It is a powerful antioxidant, which scavenges free radicals. It is oxidized in a single-electron oxidation to the mono-hydroascorbate radical anion (MDHA-) according to the reaction: The half-reaction has a standard reduction potential of +0.30 V. Ascorbate also undergoes a two- electron oxidation to dehydroascorbate (DHA): Asc DHA+ H++ 2e. This half-reaction has a standard reduction potential of +0.10 V. Finally, MDHA undergoes a disproportionation reaction: H+ DHA + Asc. Determine the biochemical standard free-energy change (AG°) for this reaction. a) +28.2 kJ mol-1 b) -28.2 kJ mol-1 c) +38.6 kJ mol-1 d) -38.6 kJ mol-1 e) +48.9 kJ mol-1

Explanation / Answer

G = -nFE°cell

Here the value of F is 96458 C

Now to get the desired chemjcal reaction,

We subtract the double of equation 1 from equation 2.

Thus we get

2MHDA + 2H+ +2e- +Asc- => DHA+ 2Asc- +2e- +H+

And finally removing 2e- and equalising the H+ and Asc-

We get the desired equation.

Thus Ecell= 0.10 - 2x 0.30 = -0.50 V

Here because the number of electrons being changed is 2 thus the value of n=2.

We get the answer -2*96458*-0.5/2 ( I have divided it by 2 because there are 2 moles of electrons)

The answer according to me is 48.9 KJ ( after converting joules into kilo joules by dividing g by 1000)

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