Question

The Gibb's energy for the decomposition of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at \(500^{\circ} \mathrm{C}\) is as follows:
$$
2 / 3 \mathrm{Al}_{2} \mathrm{O}_{3} \longrightarrow 4 / 3 \mathrm{Al}+\mathrm{O}_{2}, \Delta_{r} G=+966 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
The potential difference needed for electrolytic reduction of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) at \(500^{\circ} \mathrm{C}\) is at least
(a) \(5.0 \mathrm{~V}\)
(b) \(4.5 \mathrm{~V}\)
(c) \(3.0 \mathrm{~V}\)
(d) \(2.5 \mathrm{~V}\)

Answer 1

(d) : The ionic reactions are :
$$
\begin{aligned}
&2 / 3 \mathrm{Al}_{2}^{3+}+4 e^{-} \longrightarrow 4 / 3 \mathrm{Al} \\
&2 / 3 \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{O}_{2}+4 e^{-}
\end{aligned}
$$
Thus, no. of electrons transferred \(=4=n\).
$$
\Delta G=-n F E=-4 \times 96500 \times E
$$
$$
\begin{aligned}
&\text { or } 966 \times 10^{3}=-4 \times 96500 \times E \\
&\Rightarrow E=-\frac{966 \times 10^{3}}{4 \times 96500}=-2.5 \mathrm{~V} .
\end{aligned}
$$

Answer 2

tadalafil canada <a href="https://ordergnonline.com/">tadalafil ca</a> ed pills that work quickly

Answer 3

price cialis <a href="https://ordergnonline.com/">cialis price</a> cheapest ed pills online