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**Hint:**Electrode potential is a measure of tendency of an electrode in a half cell to lose or gain electrons. The oxidation potentials give the tendency to lose electrons and reduction potentials give the tendency to gain electrons.

Formula used: $\Delta G^\circ = - nFE{^\circ _{cell}}$

Where $\Delta G^\circ = $Gibbs free energy change

\[n = \] Number of electrons change

\[F = \] Faraday

\[E{^\circ _{cell}} = \] Potential difference

**Complete answer:**

Relationship between Gibbs free energy change and the cell potential is given below

$\Delta G^\circ = - nF\,E{^\circ _{cell}}$………………….(i)

Where \[DG^\circ = \] Gibbs free energy change

\[n = \] number of electron transferred

\[F = \] Faraday

\[E{^\circ _{cell}} = \] potential difference

We know that electrolytic reduction of aluminum oxide

$\dfrac{2}{3}A{l_2}{O_3}\xrightarrow{{}}\dfrac{4}{2}Al + {O_2}\,\,\,\Delta G^\circ = + 960KJ/mol$

Number of electrons gain by metal $ = 4$ electrons

\[F = 96500\,C.\]

Substitute the value in Equation (1), we get

$960 \times 1000 = - 4 \times 96500 \times E{^\circ _{cell}}$, solve the equation and find the potential difference

$E{^\circ _{cell}} = - \dfrac{{960000}}{{4 \times 96500}}$

$ = - 2.48V$

So, potential difference is \[ - 2.5V\]

Hence, the correct option is D.

**Note:**The decrease in free energy of the system in a spontaneous redox reaction is equal to the electrical work - done by the system on the surroundings. In thermodynamics, the Gibbs free energy is thermodynamics potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamics system at a constant temperature and pressure.In a galvanic cell the Gibbs free energy related to potential by: $\Delta G^\circ = - nFE{^\circ _{cell}}$ if $E^\circ $cell greater than zero, then the process is spontaneous (galvanic cell). If $E^\circ $cell less than zero, then the process is nonspontaneous (electrolytic cell).

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