During the electrolysis of cryolite , aluminium and fluorine are formed in ……. molar ratio.
1) 1 : 2
2) 2 : 3
3) 1 : 1
4) 1 : 3
Answer
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Hint: The answer to this question depends on the calculation of ratio of their change in oxidation state based on dissociation of $N{{a}_{3}}Al{{F}_{6}}$ to $3NaF+Al{{F}_{3}}$ and the ratio of change in the oxidation states of fluorine and aluminium gives the correct answer.
Complete step by step answer:
In the lower classes of inorganic chemistry, we have come across various methods to extract the pure metal from its metal ore.
Let us see some facts about cryolite and its electrolysis process.
- Cryolite is the main ore of aluminium which exists as an uncommon mineral on the earth which has a formula$N{{a}_{3}}Al{{F}_{6}}$.
- Electrolysis basically involves the process of decomposition of ionic compounds into their elements by passing direct electric current through the compound which is in the form of fluid.
- Electrolysis of this cryolite decomposes $N{{a}_{3}}Al{{F}_{6}}$ into two constituent parts that is, $3NaF+Al{{F}_{3}}$as,
$N{{a}_{3}}Al{{F}_{6}}3NaF+Al{{F}_{3}}$
Here, in this process of decomposition, aluminium obtained here gets reduced from its +3 oxidation state to zero oxidation state at cathode which is represented as below,
\[A{{l}^{3+}}+3{{e}^{-}}\to A{{l}^{0}}\]
And on the other hand, the fluorine obtained as $Al{{F}_{3}}$ is oxidised from two moles of fluorine with -1 oxidation state to ${{F}_{2}}$ with two electrons as shown below,
\[2{{F}^{-}}\to {{F}_{2}}+2{{e}^{-}}\]
Therefore, the ratio of the total moles of fluorine and aluminium formed can be calculated based on the change in the oxidation state which is given as,
Change in oxidation state of fluorine = two moles of fluorine from -1 to 0 that is = change in two units.
Change in oxidation state of aluminium = +3 to 0 that is change in three units.
Therefore the ratio will be 2 : 3
So, the correct answer is “Option 2”.
Note: Note that cryolite is used as solvent for the aluminium ore bauxite that is $A{{l}_{2}}{{O}_{3}}$ because this reduces energy costs in extracting aluminium and also lowers the melting point in a mixture and cryolite is also called as Kryocide which is a trade name and do not be confused if question is asked with this name.
Complete step by step answer:
In the lower classes of inorganic chemistry, we have come across various methods to extract the pure metal from its metal ore.
Let us see some facts about cryolite and its electrolysis process.
- Cryolite is the main ore of aluminium which exists as an uncommon mineral on the earth which has a formula$N{{a}_{3}}Al{{F}_{6}}$.
- Electrolysis basically involves the process of decomposition of ionic compounds into their elements by passing direct electric current through the compound which is in the form of fluid.
- Electrolysis of this cryolite decomposes $N{{a}_{3}}Al{{F}_{6}}$ into two constituent parts that is, $3NaF+Al{{F}_{3}}$as,
$N{{a}_{3}}Al{{F}_{6}}3NaF+Al{{F}_{3}}$
Here, in this process of decomposition, aluminium obtained here gets reduced from its +3 oxidation state to zero oxidation state at cathode which is represented as below,
\[A{{l}^{3+}}+3{{e}^{-}}\to A{{l}^{0}}\]
And on the other hand, the fluorine obtained as $Al{{F}_{3}}$ is oxidised from two moles of fluorine with -1 oxidation state to ${{F}_{2}}$ with two electrons as shown below,
\[2{{F}^{-}}\to {{F}_{2}}+2{{e}^{-}}\]
Therefore, the ratio of the total moles of fluorine and aluminium formed can be calculated based on the change in the oxidation state which is given as,
Change in oxidation state of fluorine = two moles of fluorine from -1 to 0 that is = change in two units.
Change in oxidation state of aluminium = +3 to 0 that is change in three units.
Therefore the ratio will be 2 : 3
So, the correct answer is “Option 2”.
Note: Note that cryolite is used as solvent for the aluminium ore bauxite that is $A{{l}_{2}}{{O}_{3}}$ because this reduces energy costs in extracting aluminium and also lowers the melting point in a mixture and cryolite is also called as Kryocide which is a trade name and do not be confused if question is asked with this name.
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