Animated Solution for Chemistry - Electrochemistry: The Gibbs energy for the decomposition of ${\text{Al}}_2{\text{O}}_3$ at $500^{\circ }\text{C}$ is as follows $\frac{2}{3}{\text{Al}}_2{\text{O}}_3⟶\frac{4}{3}\text{Al}+{\text{O}}_2$, ${\Delta }_{r}G=+966{\text{ kJ mol}}^{-1}$ The potential difference needed for electrolytic reduction of ${\text{Al}}_2{\text{O}}_3$ at $500^{\circ }\text{C}$ is at least

Extracting Aluminum

Imagine standing inside a massive industrial plant. Before you is a giant vat filled with molten aluminum oxide (${\text{Al}}_2{\text{O}}_3$) at a scorching $500^{\circ }\text{C}$. Aluminum doesn't just willingly separate from oxygen; they are tightly bound. To break this bond and extract pure aluminum, we must force a non-spontaneous reaction to occur.

This is where electrochemistry comes to the rescue. By pumping in electrical energy, we can drive the reaction forward. But the critical question is: exactly how much voltage do we need to apply?

Bridging Thermodynamics and Electrochemistry

To find the required voltage, we need to connect the thermodynamic energy barrier to electrical potential. The bridge between these two worlds is one of the most beautiful equations in physical chemistry:

Here, $\Delta G$ is the Gibbs free energy change, $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant ($96500{\text{ C mol}}^{-1}$), and $E_{\text{cell}}$ is the standard cell potential.

The problem gives us the Gibbs free energy for the specific decomposition reaction:

The energy required is $\Delta G=+966{\text{ kJ mol}}^{-1}$. The positive sign is a clear indicator that this reaction is non-spontaneous and requires an external energy source.

Finding $n$

Before we can plug numbers into our master equation, we need to determine $n$, the number of electrons transferred in this specific balanced equation. This is where many students make a silly mistake!

Let's look at the oxygen atoms. In ${\text{Al}}_2{\text{O}}_3$, oxygen exists as the oxide ion (${\text{O}}^{2-}$). In the products, it becomes elemental oxygen gas (${\text{O}}_2$), where its oxidation state is $0$.

How many oxide ions are we dealing with? In $1$ mole of ${\text{Al}}_2{\text{O}}_3$, there are $3$ moles of oxygen. Therefore, in $\frac{2}{3}$ moles of ${\text{Al}}_2{\text{O}}_3$, there are $\frac{2}{3}\times 3=2$ moles of oxide ions.

The oxidation half-reaction for these two oxide ions is:

Each oxide ion loses $2$ electrons, so $2$ oxide ions lose a total of $4$ electrons. Thus, the number of electrons transferred is $n=4$.

Now we have all the pieces of the puzzle. Let's substitute our values into the master equation. Remember to convert $\Delta G$ from kilojoules to joules to match the units of Faraday's constant!

Solving for $E_{\text{cell}}$:

The negative sign of $E_{\text{cell}}$ confirms that the cell is electrolytic—it consumes electrical energy rather than producing it.

To overcome this natural thermodynamic barrier and force the decomposition to happen, our external battery must supply a potential difference that is at least equal in magnitude to the cell potential.

Therefore, the minimum external potential difference needed is $2.5\text{ V}$.