Question

The Gibbs energy for the decomposition of $A{l_2}{O_3}$ at 500℃ is as follows: $\dfrac{2}{3}A{l_2}{O_3} \to \dfrac{4}{3}Al + {O_2};{\text{ }}{\Delta _r}G = + 966{\text{ }}kJmo{l^{ - 1}}$
The potential difference needed for the electrolytic reduction of $A{l_2}{O_3}$ at 500℃ is at least:
A.5.0V
B.4.5V
C.3.0V
D.2.5V

Answer 1

Hint: To solve this question, it is required to have knowledge about the concept of Gibbs free energy. We know that the Gibbs free energy for decomposition of any reaction is related to the potential difference at constant temperature by the formula (given below). So, by putting the appropriate values in the equation we can find the reduction potential, i.e. $E_{cell}^o$.
Formula Used: $\Delta G = - nFE_{cell}^o$ (Eq. 1)
Where$\Delta G$ is Gibbs free energy (in Joules), $n$ is the number of electrons transferred, $F$ is Faraday’s constant ($96500{\text{ }}Cmo{l^{ - 1}}$) and $E_{cell}^o$ is the potential difference (in Volts) required.

Complete answer:
We know that any redox reaction can be written as reduction and oxidation half-reactions from which we can find the number of electrons transferred. So, we write the given reaction as oxidation and reduction half-reactions respectively:
$\dfrac{2}{3} \times (2A{l^{3 + }}) + 4{e^ - } \to \dfrac{4}{3}Al$ (Eq. 2)
$\dfrac{2}{3} \times (3{O^{2 - }}) \to {O_2} + 4{e^ - }$ (Eq. 3)
From the above Eq. 2 and Eq. 3, we can conclude that the number of electrons transferred in both the oxidation as well as reduction half-reaction is 4. Thus, $n = 4$.
Now, we can rearrange Eq. 1 to form the following:
$E_{cell}^o = - \dfrac{{\Delta G}}{{nF}}$
Putting the rest of the values in above equation, we get:
$E_{cell}^o = - \dfrac{{966 \times {{10}^3}}}{{4 \times 96500}}$
By solving this equation for $E_{cell}^o$ we get,
\[ \Rightarrow E_{cell}^o = - \dfrac{{10}}{4}V\]
$ \Rightarrow E_{cell}^o = - 2.5V$

$\therefore $ The correct answer is 2.5V that is option D.

Additional Information:
$\Delta G$ is used to predict the spontaneity of a reaction. If $\Delta G < 0$, the reaction is said to be spontaneous and can occur on its own without any initiation step. For example, rusting or melting of ice at room temperature. If $\Delta G > 0$, the reaction is said to be non-spontaneous and requires initiation to occur, like rise in temperature, use of reagents, addition of a catalyst etc. For example, reactions between ${N_2}$ and ${O_2}$ cannot occur at room temperature, it requires high temperatures.

Note:
As we know that in the case of an electrochemical cell, if $E_{cell}^o > 0$ the cell is considered to be feasible, i.e. the chemical reaction occurring in the half-cells are spontaneous. Here, as $\Delta G > 0$, the reaction is considered non-spontaneous and $E_{cell}^o < 0$.