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Calculate the mass of aluminium deposited at cathode \[193C\] of current is passed through molten electrolyte containing alumina. Given the molar mass of Al =\[27g/mol\] ,
1F= \[96500\text{ }C/mol\]

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Last updated date: 20th Jun 2024
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Answer
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Hint: This is based on Faraday’s first and second law of electrolyte. In the calculation, the number of electrons is taken 3 because aluminium forms +3 ions in solution. The mass of the metal produced is directly proportional to the amount of current passed in the solution.

Complete step by step answer:
Faraday’s first law of electrolysis:
The amount of mass liberated at the electrode is directly proportional to the quantity of the current passed into the solution.
\[W\text{ }=\text{ }Z\text{ x }Q\]
W is the mass, Z = proportionality constant called electrochemical equivalent.
Q = amount of electricity.
Faraday’s second law of Electrolysis.
It states that: when electricity is passed to a solution which contains electrolyte that is connected in series, the masses produced at the electrode is directly proportional to their equivalent weight.
When electrolysis occurs, the charge carried by one mole of an electron can be obtained by multiplying the charge present on one electron with Avogadro's number. This equals to 96500 coloumbs. This quantity is called one faraday.
So, by combining both the laws we can formulate:
\[W=\frac{QM}{Fz}\]
F = 1 faraday
Z = valency of the metal.
In the question:
M = Mass of the element =\[~27g/mol\]
Q = amount of electricity = \[193C\]
F = \[96500\text{ }C/mol\]
z = valency of aluminium = 3
by combining all these, we get
\[W=\frac{193\text{x}27}{96500\text{x}3}=0.018g\]

Hence, the amount of aluminium deposited is\[0.018g\] .

Note: You may get confused that in these types of questions usually time is also mentioned, but when the time is not given \[W=\frac{QM}{Fz}\] it is to be used. The actual value of faraday is 96487 but for calculation 96500 is used.