Question

17-19. In the Figure, 17-11, 2.00nmol fructose was 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.

Short answer

The n electron is$\mathrm{n}\left({\mathrm{e}}^{-}\right)=4.{10}^{-9}\mathrm{mol}$ and the charge is$\mathrm{q}=3.86.{10}^{-4}\mathrm{C}$

Step-by-step solution

Fructose oxidation:

The body oxidizes a considerable proportion of dietary fructose to create energy. At rest, fructose may be preferentially or comparably consumed to create energy like glucose, however, during exercise, glucose seems to be more preferentially utilized to produce energy by the body.

Find oxidation of one fructose molecule:

The reaction is:

$\mathrm{D}-\mathrm{fructose}\to 5-\mathrm{keto}-\mathrm{D}-\mathrm{fructose}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

The electron is:

$$\begin{gathered} \frac{n\left(e^{-}\right)}{n(fructose)}=\frac{2}{1} \\ n\left(e^{-}\right)=2.n(fructose) \\ n\left(e^{-}\right)=2.2.{10}^{-9}mol \\ n\left(e^{-}\right)=4.{10}^{-9}mol \end{gathered}$$

The charge is:

$$\begin{gathered} \mathrm{q}=\mathrm{n}\left({\mathrm{e}}^{-}\right).\mathrm{Z}.\mathrm{F} \\ \mathrm{q}=4.{10}^{-9}\mathrm{mol}.1.96485\mathrm{C}/\mathrm{mol} \\ \mathrm{q}=3.86.{10}^{-4}\mathrm{C} \end{gathered}$$