Question 11 Mark
How much electricity in terms of Faraday is required to produce,
40.0g of Al from molten $Al_2O_3$.
40.0g of Al from molten $Al_2O_3$.
Answer
View full question & answer→According to the question,
$\text{Al}^{3+}+3\text{e}^{-1}\ \rightarrow\ \text{Al}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 27\text{g}$
Electricity required to produce 27g of Al = 3 F
Therefore, electricity required to produce 40 g of Al $=\frac{3\times40}{27}\text{F}=4.44\ \text{F}$
$\text{Al}^{3+}+3\text{e}^{-1}\ \rightarrow\ \text{Al}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 27\text{g}$
Electricity required to produce 27g of Al = 3 F
Therefore, electricity required to produce 40 g of Al $=\frac{3\times40}{27}\text{F}=4.44\ \text{F}$
