Question

The number of moles of electrons involved in the cell reaction of the cell Al/Al3+(0.01M)∥Fe2+( 0.02M)/Fe are ____. *​

Answer 1

$\huge{\mathfrak{\underline{\red{Answer!}}}}$

Answer 2

Answer:

The number of moles of electrons involved in the cell reaction of the cell Al/Al3+(0.01M)∥Fe2+( 0.02M)/Fe is Six

Explanation:

Given: $\mathrm{E}_{\mathrm{Al}}^{\circ}{ }^{+3} / \mathrm{A} 1=-1.66 \mathrm{~V}$

$\mathrm{E}^{0} \mathrm{Fe}^{+2} / \mathrm{Fe}=-0-44 \mathrm{~V}$

we know the cell reaction

$E_{\text {cell }}^{0}=E_{\text {Cathode }}^{0}-E_{\text {anode }}^{0}$

Here Fe is cathode and $\mathrm{Al}$ is anode

$\begin{aligned}&\therefore E_{\text {cell }}^{0}=-0.44-(-1.66) \\&=-0.44+1.66\end{aligned}$

Cell reaction $=+1.22 \mathrm{~V}$

$2 \mathrm{Al}-6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al}^{3+}$

At anode

$3 \mathrm{Fe}^{2+}+6 \mathrm{e}^{-} \rightarrow 3 \mathrm{Fe}$

At cathode

$2 \mathrm{Al}+3 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Al}^{3+}+3 \mathrm{Fe} \quad$

Cell reaction

Here

$n=6$ (Number of electron in cell reaction)

$\begin{aligned}&E_{\text {cell }}^{0}=E_{\text {cell }}^{0}+\frac{0.059}{n} \log \frac{\left[\mathrm{Fe}^{2+}\right]^{3}}{\left[\mathrm{Al}^{3+}\right]^{2}} \\&=1.22+\frac{0.059}{6} \log \frac{[0.02]^{3}}{[0.01]^{2}} \\&=1.22+9.8 \times 10^{-3} \log \left[\frac{8 \times 10^{-6}}{1 \times 10^{-4}}\right] \\&=1.22+9.8 \times 10^{-3} \times 1.25 \\&=1.22+0.0122 \\&=1.232 \mathrm{~V}\end{aligned}$