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

The minimum cell potential required for an electrolytic process is obtained from the relation between Gibbs free energy change and electrical work:

$$\Delta_r G = -\,nF E_{\text{cell}}$$

where
$$\Delta_r G$$ = Gibbs energy change for the overall reaction (J mol$$^{-1}$$)
$$n$$ = total number of electrons transferred in the balanced reaction (mol e$$^-$$)
$$F$$ = Faraday constant $$\left(96500\ \text{C mol}^{-1}\right)$$
$$E_{\text{cell}}$$ = reversible cell potential (V)

Step 1: Balance the reaction and find $$n$$.
The question gives a fractional decomposition reaction:

$$\frac{2}{3}\,{\text{Al}_2\text{O}_3} \;\rightarrow\; \frac{4}{3}\,\text{Al} \;+\; \text{O}_2$$

• Each $$\text{Al}^{3+}$$ gains $$3$$ electrons to become $$\text{Al}$$.
• The above reaction contains $$\frac{4}{3}$$ aluminium atoms, i.e. $$\frac{4}{3} \times 3 = 4$$ electrons are needed.
Therefore $$n = 4$$.

Step 2: Insert the data in $$\Delta_r G = -\,nF E_{\text{cell}}$$.
Given $$\Delta_r G = +966\ \text{kJ mol}^{-1} = 966000\ \text{J mol}^{-1}$$.

Re-arranging, the magnitude of the minimum potential is

$$E_{\text{cell}} = \frac{\Delta_r G}{nF}$$

$$E_{\text{cell}} = \frac{966000}{4 \times 96500}$$

$$E_{\text{cell}} = \frac{966000}{386000} \approx 2.50\ \text{V}$$

Step 3: Interpret the result.
The positive $$\Delta_r G$$ shows the decomposition is non-spontaneous, so an external voltage of at least $$\approx 2.5\ \text{V}$$ must be applied to drive the electrolysis.

Hence the required minimum potential difference is

Option C which is: $$2.5\ \text{V}$$