you and your research group are studying a new biologically important molecule.

Question

you and your research group are studying a new biologically important molecule. in vivo, Restudoic acid (H2RA) is oxidized to dehydroRestudoic acid(RA). The standard reduction potential was determined to be +0.324V. a. Write the oxidation half reaction of restudoic acid b. Could the experiment you conducted in lab be repeated to determine the amount of restudoic acid in a sample.? Explain why or why not c. Would dissolved oxygen in solution affect the reaction? If so, how would it alter your results? Include any half reactions to support your answer?

Explanation / Answer

Given TA + 2H+(aq) + 2e- ----- > H2TA , E0(red) = + 0.324 V

(a) : The chemical reaction for the oxidation of testudoic acid (H2TA) to dehydrotestudoic acid (TA) is

H2TA ---- > TA + 2H+(aq) + 2e-, E0(oxi) = - (+ 0.324)V = - 0.324 V

Hence the oxidation half-cell reaction should be

H2TA ---- > TA + 2H+(aq) + 2e-, E0(oxi) = - (+ 0.324)V = - 0.324 V

(b) If Ka is the equilibrium constant for the dissociation of H2TA and initial concentration of H2TA be C M, then

----------------------H2TA ---- > TA + 2H+(aq) + 2e- , Ka1

Initial conc(M): C ------------ 0 -------- 0

Eqm.conc(M): (C-x) ---------- x---------x

Since there are 2 variables, we need to repeat the experiment to find Ka value and also the initial concentration, C.

(c): If dissolved O2 is present in the solution, it will react with H+ ion to form water. The half reactions are

H2TA ---- > TA + 2H+(aq) + 2e-

1/2 O2 + 2H+(aq) + 2e- ---- > H2O(l)

Now as H+ reacts with O2 to form H2O, the concentration of H+ (a product for the oxidation of H2TA) will decrease i.e the concentration of the product will decrease. Hence according lechatelier's principle more and more H2TA will oxidise. Hence it will speed up the reaction

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