Question

The quantity of charge required to obtain one mole of aluminium from \[A{{l}_{2}}{{O}_{3}}\] is:
(a)- 1F
(b)- 6F
(c)- 3F
(d)- 2F

Answer 1

Hint: The quantity of charge required is calculated by the quantitative aspects of electrolysis. The quantity of charge required to obtain one-mole aluminium could be equal to the oxidation state of aluminium or the number of electrons required to produce aluminium atoms from its ion form.

Complete step by step answer:
Let us study the quantitative aspects of electrolysis:
Consider the electrolysis of molten$NaCl$, i.e.
$N{{a}^{+}}C{{l}^{-}}\to Na+\dfrac{1}{2}C{{l}_{2}}$
Thus we have,
$N{{a}^{+}}+{{e}^{-}}\to Na$
This means that one electron produces one sodium atom. Therefore, the passage of one mole of electrons will produce one mole of sodium.
Again from the above reaction,
$2C{{l}^{-}}\to C{{l}_{2}}+2{{e}^{-}}$
2 moles of electrons produce one mole of $C{{l}_{2}}$
Similarly looking at the reactions,$C{{u}^{2+}}+2{{e}^{-}}\to Cu$ and $A{{l}^{3+}}+3{{e}^{-}}\to Al$, we find that 2 moles of electrons produce one mole of Cu whereas 3 moles of the electron will produce one mole of Al.
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 96500coloumbs. This quantity is called one faraday.
If n electrons are involved in the electrode reaction, the passage of n faradays of electricity will liberate one mole of the substance.

So, in the question aluminium produced from\[A{{l}_{2}}{{O}_{3}}\], the reaction will be:
$A{{l}_{2}}{{O}_{3}}\to 2Al$
In the electron from the reaction will be:
$2A{{l}^{3+}}+6{{e}^{-}}\to 2Al$

So from the equation, we can see that 2 moles of aluminium is produced from 6 moles of electron hence the charge required will be 6F.
For 1 mole of aluminium, 3 moles of electron will be required hence the charge will be 3F.
So, the correct answer is “Option C”.

Note: The exact value of one faraday is 96487 coulombs but for approximate calculation, the value used is 96500 C. Don’t get confused between 6F and 3F because in the equation:
$2A{{l}^{3+}}+6{{e}^{-}}\to 2Al$
 6 moles of electrons are used to produce 2 moles of aluminium.