For the cell in which of the following occurs:

\(2Au^{3+} (aq) + 6Br^- (aq) \longrightarrow 2Au(s) + 3Br_2 (l)\) has \(E^o_{cell} = 2.49 V\) at \(298 K\)

The standard Gibbs energy change will be:

\(- 1442 kJ\)

The correct answer is option 4. \(- 1442 kJ\).

To determine the standard Gibbs free energy change (\(\Delta G^\circ\)) for the given cell reaction, we use the relationship between the cell potential and the Gibbs free energy change:

\(\Delta G^\circ = -nFE^\circ_{\text{cell}}\) -----(i)

where:

\( n \) = number of moles of electrons transferred in the reaction

\( F \) = Faraday constant (\( F \approx 96500 \, \text{C/mol} \))

\( E^\circ_{\text{cell}} \) = standard cell potential

From the given reaction:

\(2Au^{3+} (aq) + 6Br^- (aq) \longrightarrow 2Au(s) + 3Br_2 (l)\)

Each \(Au^{3+}\) ion gains 3 electrons to become \(Au\), and there are 2 \(Au^{3+}\) ions involved, so:

\(2 \text{ moles of } Au^{3+} \text{ gain } 2 \times 3 = 6 \text{ moles of electrons}\)

Thus, \( n = 6 \).

Applying the values in equation (i), we get

\(\Delta G^\circ = -nFE^\circ_{\text{cell}}\)

Substituting \( n = 6 \), \( F = 96500 \, \text{C/mol} \), and \( E^\circ_{\text{cell}} = 2.49 \, \text{V} \):

\(\Delta G^\circ = -6 \times 9500 \, \text{C/mol} \times 2.49 \, \text{V}\)

\(⇒ \Delta G^\circ = -1441710\, \text{J/mol}\)

\(⇒ \Delta G^\circ = -1441.710 \, \text{kJ/mol}\)

\(⇒ \Delta G^\circ \approx -1442\, \text{kJ/mol}\)