Question

Given,$A{l_2}{O_3}$ is reduced by the electrolysis at low potential and high currents. If $4.0 \times {10^{4\;}}$ amperes of current is passed through the molten $A{l_2}{O_3}$​ for $6$ hours, what mass of aluminium is produced? (Assume $100\%$ current efficiency, At. mass of $Al = 27gmo{l^{ - 1}}$ )
$(A)1.3 \times {10^4}g$
$(B)9.0 \times {10^3}g$
$(C)8.1 \times {10^4}g$
$(D)2.4 \times {10^5}g$

Answer 1

Electrolysis is a process in which electricity is passed through a solution of a substance in order to proceed a chemical reaction which is actually non-spontaneous in nature. In order to solve this question we need to understand Faraday's law of electrolysis and remember its formula. This will enable us to solve this numerical problem.

Complete answer:
We know that in electrolysis, electricity is passed through the solution of a substance in order to proceed with a chemical reaction which is non-spontaneous in nature.
So, Faraday's law states that the amount of a substance released, formed or produced during electrolysis on an electrode is directly proportional to the amount of charge passed through it.
Hence, the formula we will use here is given as:
$w = Z \times I \times t........(1)$
Where, w is the amount or mass deposited, $I$ is the amount of current and t is the time which is taken in seconds.
Z is the electrochemical equivalent of the substance which is given as:
$z = \dfrac{E}{{96500}}$
Where, E is the equivalent weight of the substance.
Now, z for aluminium is:
$z = \dfrac{{27}}{{3 \times 96500}}$
$t = 6 \times 60 \times 60\sec$
Put these values in first equation:
$w = \dfrac{{27}}{{3 \times 96500}} \times 4.0 \times {10^{4\;}} \times 6 \times 60 \times 60$
On solving this, we get:
$w = 8.1 \times {10^4}g$
Hence, the correct option is $(C)8.1 \times {10^4}g$ .

Note:
Equivalent weight or mass of an element or compound is different for different reactions because in different reactions it behaves differently. In other words, suppose in one reaction a specific element exchanges three electrons and in another reaction, it donates two electrons. So, in both these cases, the equivalent weight will be different.