Question

The Gibbs’ energy for the decomposition of ${ Al }_{ 2 }{ O }_{ 3 }$ at ${ 500 }^{ \circ }{ C }$ is as follows:
${ 2\div 3Al }_{ 2 }{ O }_{ 3 }{ \rightarrow 4\div 3Al+O }_{ 2 }{ ;\Delta G\circ =+960kJ/mol }$
The potential difference needed for the electrolytic reduction of aluminium oxide ${ Al }_{ 2 }{ O }_{ 3 }$ at ${ 500 }^{ \circ }{ C }$ is at least:
(A) ${ 3.0V }$
(B) ${ 2.5V }$
(C) ${ 5.0V }$
(D) ${ 4.5V }$

Answer 1

Hint:In thermodynamics, the Gibbs free energy is a thermodynamic potential that measures the "useful" or process-initiating work obtainable from an isothermal, isobaric thermodynamic system.

Complete step by step solution:
> The relationship between the Gibbs’ free energy change and the cell potential is given below:
${ \Delta G }^{ \circ }{ =-nFE }^{ \circ }$..........(1)
where, ${ \Delta G }^{ \circ }$ = Gibbs’ free energy change
n = number of electrons transferred
F = Faraday’s constant
${ E }^{ \circ }$ = Electrode potential
It is given that,
${ \Delta G\circ =+960kJ/mol }$ = ${ 960\times 10 }^{ 3 }{ J }mol^{ -1 }$
F = ${ 96500 }$
${ 2\div 3Al }_{ 2 }{ O }_{ 3 }{ \rightarrow 4\div 3Al+O }_{ 2 }$
Here, the value of n = ${ 4 }$
Now, put the values in the equation 1, we get
$ { 960\times 10 }^{ 3 }{ J }mol^{ -1 }= { =\quad -4\times 96500\times E }^{ \circ }$
${ E }^{ \circ }{ =\quad -960000\div 4\times 96500 }$
${ E }^{ \circ }= { =\quad -960000\div 386000 }$
${ E }^{ \circ } = { -2.48V }$
So, difference = ${ 2.5V }$.
Hence, The potential difference needed for the electrolytic reduction of aluminium oxide ${ Al }_{ 2 }{ O }_{ 3 }$ at ${ 500 }^{ \circ }{ C }$ is at least ${ 2.5V }$.

The correct option is B.

> The metals present at the top of the reactivity series such as sodium and potassium can't be reduced using coke, therefore bypassing electric current through their molten salts they are reduced such a process to get metal from its metallic oxide is called Electrolytic Reduction.
> In the spontaneous process, the ${ \Delta G }$ ( free energy change ) must be a negative value.

Note: The possibility to make a mistake is that you have to convert Gibbs’ free energy in J/mol to kJ/mol. Also, the value of electrons transferred in this reaction is ${ 4 }$.