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Aluminium oxide may be electrolysed at $1000^{\circ} \mathrm{C}$ to furnish aluminium metal (Atomic mass $=27$ amu; 1 Faraday $=96,500$ Coulombs). The cathode reaction is
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}^{\circ}$
To prepare $5.12 \mathrm{~kg}$ of aluminium metal by this method would require
Options:
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}^{\circ}$
To prepare $5.12 \mathrm{~kg}$ of aluminium metal by this method would require
Solution:
1908 Upvotes
Verified Answer
The correct answer is:
$5.49 \times 10^7 \mathrm{C}$ of electricity
$5.49 \times 10^7 \mathrm{C}$ of electricity
$$
\begin{aligned}
& \mathrm{Q}=\frac{\mathrm{mFZ}}{\mathrm{M}}=\frac{5.12 \times 10^5 \times 96500 \times 3}{27} \\
& =5.49 \times 10^7 \mathrm{C}
\end{aligned}
$$
\begin{aligned}
& \mathrm{Q}=\frac{\mathrm{mFZ}}{\mathrm{M}}=\frac{5.12 \times 10^5 \times 96500 \times 3}{27} \\
& =5.49 \times 10^7 \mathrm{C}
\end{aligned}
$$
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