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Step-by-Step Solution
Step 1: Identify the Relevant Thermodynamic Equation
We know that the standard Gibbs free energy change, $ \Delta G^{\circ} $, is related to the standard electrode potential, $ E^{\circ} $, by the equation
$$
\Delta G^{\circ} = -n F E^{\circ},
$$
where
$ n $ is the total number of moles of electrons transferred per mole of reaction.
$ F $ is the Faraday constant ($ 96500 $ C mol$^{-1}$).
$ E^{\circ} $ is the potential difference (in volts).
Step 2: Write Down the Decomposition Reaction and Gibbs Free Energy
The decomposition reaction of aluminium oxide at $500^\circ \mathrm{C}$ is given by:
$$
\frac{2}{3}\,\mathrm{Al_2O_3} \;\;\to\;\; \frac{4}{3}\,\mathrm{Al} + \mathrm{O_2}
$$
with
$$
\Delta_{r}G = +960 \,\mathrm{kJ\,mol^{-1}}.
$$
To keep units consistent, convert $ 960 \,\mathrm{kJ\,mol^{-1}} $ into joules:
$$
960\,\mathrm{kJ\,mol^{-1}} = 960 \times 10^3\,\mathrm{J\,mol^{-1}}.
$$
Step 3: Determine the Number of Electrons Transferred ($ n $)
From the reaction,
$$
\frac{4}{3}\,\mathrm{Al} \quad\text{is produced}.
$$
Each Al atom is formed from the reduction of $\mathrm{Al^{3+}}$:
$$
\mathrm{Al^{3+}} + 3\,e^- \;\to\; \mathrm{Al}.
$$
Hence, forming $1$ aluminium atom requires $3$ electrons. For $\frac{4}{3}$ aluminium atoms:
$$
n = \frac{4}{3} \times 3 = 4.
$$
Therefore, $4$ moles of electrons are involved in the reaction as written.
Step 4: Calculate the Potential, $ E^{\circ} $
Using the relationship
$$
\Delta G^{\circ} = -\,n\,F\,E^{\circ},
$$
we solve for $ E^{\circ} $:
$$
E^{\circ} = -\,\frac{\Delta G^{\circ}}{nF}.
$$
Substitute the values:
$$
E^{\circ}
= -\,\frac{960 \times 10^3\,\mathrm{J\,mol^{-1}}}{4 \times 96500\,\mathrm{C\,mol^{-1}}}.
$$
Simplify the expression:
$$
E^{\circ} = -\,\frac{960 \times 10^3}{4 \times 96500}
\approx -\,2.5\,\mathrm{V}.
$$
Step 5: Interpret the Result
The negative sign indicates that external energy must be provided to drive this reaction (electrolytic reduction). Thus, to bring about the desired reduction of $\mathrm{Al_2O_3}$ at $500^\circ \mathrm{C}$, at least $2.5\,\mathrm{V}$ is needed.
Final Answer
The minimum potential difference required for the electrolytic reduction of aluminium oxide at $500^\circ \mathrm{C}$ is 2.5 V.
