In Figure 17-11, 2.00 nmol of fructose were introduced at the time of the arrow.
ID: 706573 • Letter: I
Question
In Figure 17-11, 2.00 nmol of fructose were introduced at the time of the arrow. How many electrons are lost in the oxidation of one molecule of fructose? Compare the theoretical number of coulombs with the observed number of coulombs for complete oxidation of the sample.
(Answer in book says: 2.00 nmol fructose ---> 4.00 nmol e- = 0.386 mC, which agrees with integrated area) Please help me find this!
10 FIGURE 17-11 Controlled-potential coulometric oxidation of the sugar D-fructose in a three-electrode cell by the enzyme o-fructose dehydrogenase adsorbed on carbon particles on a carbon electrode at +500 mV versus Ag AgCI. No current is observed in the absence of enzyme. A Pt auxiliary electrode in Charge 0.3 area beneath current curve bridge completes the circuit and reduces a component of the buffer solution. [Data from S. Tsujimura, A. Nishina, Y. Kamitaka, and K Kano, "Coulometric o-Fructose Biosensor Based on Direct Electron Transfer Using o-Fructose Anal Chem 2009, 81, 9383.] 2 0.0 800 1 000 Time (s) In Figure 17-11, the sugar D-fructose is oxidized at an electrode by controlled-potential coulometry catalyzed by the enzyme D-fructose dehydrogenase adsorbed on a porous carbon electrode held at +500 mV versus Ag | AgCI. Electrons from D-fructose go through the enzyme and into the electrode. Addition of D-fructose at the time indicated by the arrow gives a spike of current followed by exponential decay. The total charge transferred from D-fructose into the electrode is the area under the current-versus-time curve, given by the integral in Equation 17-12. CH OH CH OH D-fructose dehydrogenase HO-C-H HO-C-H H-C-OHHC-OH+2H+2e ??? D-Fructose 5-Kcto-D-fructoseExplanation / Answer
One mole of fructose contains 6.023 x 1023 molecules of fructose. (Avagadro's number)
1 nano = 10-9
So 2 nano moles (2 x 10-9) of fructose contains = 2 x 10-9 x 6.023 x 1023 molecules of fructose = 12.046 x 1014 molecules of fructose.
12.046x 1014 molecules of fructose ( 2nmol of fructose ) -----> 4 nmol of electrons ( 4 x10-9)
1 molecule of fructose -----> 4 x 10-9 mole of electrons x 1molecule / 12.046 x 1014 molecules
1 molecule of fructose -----> 0.332 x 10-23 electrons are lost in the oxidation of one molecule of fructose.
12.046x 1014 molecules of fructose ( 2nmol of fructose ) -----> 0.386 mC
1 molecule of fructose -----> 0.386 mC x 1 molecule/12.046 x 1014 molecule = 0.032 x 10-14 mC
Theoritical number of coulombs with the oxidation of the sample :
Each mole of electron carry 96,500 coulombs
so 4 x 10-9 (4 nanomoles) moles of electrons ----> 4 x 10-9 x 96500 = 386000 x 10-9 = 386 x10-6 C = 386 mC
Theoritical number of coulombs for the complete oxidation of the sample = 386 mC
But the actual number of coulombs observed for the complete oxidation of the sample = 0.386 mC
There is difference of 1000 times between the theortical number of coulombs and actual/observed number of coulombs for the complete oxidation of the sample.
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