Question

Aluminium oxide may be electrolyzed at ${{1000}^{\text{o}}}\text{C}$ to furnish aluminium metal (Atomic mass= 27 amu; 1 faraday= 96500 coulomb). The cathode reaction is: $\text{A}{{\text{l}}^{+3}}+3{{\text{e}}^{-}}\to \text{Al}$
To prepare 5.12 kg of aluminium metal by this method we would require.
A. $5.49\times {{10}^{7}}\text{C}$ of electricity
B. $1.83\times {{10}^{7}}\text{C}$ of electricity
C. $5.49\times {{10}^{4}}\text{C}$ of electricity
D. \[~5.49\times {{10}^{10}}\text{C}\] of electricity

Answer 1

Hint: We need to find the equivalent weight of $\text{Al}$ to apply the second law of Faraday. This electrolysis’ law states that when electricity is passed through electrolyte then, the mass of the substance deposited is proportional to its respective equivalent weight. Mathematically, it is represented as; $\text{mass deposited= Zit or Zq}$.

Complete answer:
Let us solve this question using second law equation and the reaction given:
Step (1)- In order to find Z, we need molecular mass and n-factor of aluminium. As,$\text{Z = }\dfrac{\text{molecular mass}}{\text{n-factor}\times 96500}$. n-factor of $\text{Al}$ will be 3 as the reaction is $\text{A}{{\text{l}}^{+3}}+3{{\text{e}}^{-}}\to \text{Al}$ as the number of electrons involved in the reaction will be the n-factor of aluminium.
Step (2)- The molar mass of aluminium is 27 grams. The value of Z will be $\dfrac{27}{3\times 96500}$ or $9.32\times {{10}^{-5}}\text{gram}\text{.}{{\text{C}}^{-1}}$.
Step (3)- The mass of aluminium deposited on the cathode is 5.12 kg, the formula $\text{mass deposited= Zit or Zq}$; the value of q will be $5.12=9.32\times {{10}^{-5}}\times \text{q}$ or $5.49\times {{10}^{4}}\text{C}$ of electricity .

The correct option is option ‘c’.

Additional Information:
Hall–Heroult process is used as the industrial process for the smelting aluminium which takes place by dissolving aluminium oxide in molten cryolite. The Hall–Heroult process produces $99.5-99.8%$ pure aluminium. Recycled aluminum requires no electrolysis.

Note:
In the extraction of aluminium by electrolytic process, $\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}$ is used with fluorspar and cryolite. Cryolite helps in reducing the melting point of the mixture and also this solvent helps in dissolving $\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}$.