91Ó°ÊÓ

1 mole of electrons holds the Avogadro constant, L, electrons - that is, $6.02×1023$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:

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

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:

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