17-17. The experiment in Figure 17 - 9 required 5.32mA for 864s for complete reaction of a5.00 - mLaliquot of unknown cyclohexene solution.

(a) How many moles of electrons passed through the cell?

(b) How many moles of cyclohexene reacted?

(c) What was the molarity of cyclohexene in the unknown?

1 mole of electrons holds the Avogadro constant, L, electrons - that is, $6.02\times 1023$electrons. You would also be provided that in an exam if you needed to use it. This is known as the Faraday constant.

a) Given data:$=5.32mAt=964sV=5mL$

$n\left(e^{-}\right)=?$The moles of electrons can be calculated using the equation

$$\begin{gathered} n\left(e^{-}\right)=\frac{l.t}{F}=\frac{5.32.{10}^{-3}A.964s}{96485C/mol} \\ n\left(e^{-}\right)=5.32.{10}^{-5}mol \end{gathered}$$

One molecule of cyclohexanol should yield one molecule of cyclohexene. One mole (mol) of cyclohexanol should yield one mole (mol) of cyclohexene.

b) Given data:$=5.32mAt=964sV=5mL$

$n\left(cyclehexene\right)=?$The reaction is:

$$\begin{gathered} B{r}_2+2e^{-}\to 2B{r}^{-} \\ \frac{n\left(B{r}_2\right)}{n\left(e^{-}\right)}=\frac{1}{2} \\ \frac{n\left(cyclohexene\right)}{n\left(B{r}_2\right)}=\frac{1}{2} \end{gathered}$$

So, the n of cyclohexene is:

$$\begin{gathered} n\left(cyclohexene\right)=\frac{n\left(e^{-}\right)}{2}=\frac{5.32.{10}^{-5}mol}{2} \\ n\left(cyclohexene\right)=2.66.{10}^{-5}mol \end{gathered}$$

c) Given data:$=5.32mAt=964sV=5mL$

The molarity of cyclohexene is:

$$\begin{gathered} \left[cyclohexene\right]=\frac{n\left(cyclohexene\right)}{V} \\ \left[cyclohexene\right]=\frac{2.66.{10}^{-5}mol}{5.{10}^{-3}L} \\ \left[cyclohexene\right]=5.32.{10}^{-3}M \end{gathered}$$