Question

Calculate the standard Gibbs energy (\( \Delta G^\circ \)) of the following reaction at 25°C: \[ \text{Au}(s) + \text{Ca}^{2+}(1M) \to \text{Au}^{3+}(1M) + \text{Ca}(s) \] \[ E^\circ_{\text{Au}^{3+}/\text{Au}} = +1.5 \, \text{V}, \quad E^\circ_{\text{Ca}^{2+}/\text{Ca}} = -2.87 \, \text{V} \] Predict whether the reaction will be spontaneous or not at 25°C. [1 F = 96500 C mol\(^{-1}\)]

Hint:

A negative value for \( \Delta G^\circ \) indicates that the reaction is spontaneous. The magnitude of \( \Delta G^\circ \) is related to the cell potential, which can be calculated using the Nernst equation.

Answer 1

The standard Gibbs free energy change (\( \Delta G^\circ \)) is related to the standard electrode potential change (\( \Delta E^\circ \)) by the formula: \[ \Delta G^\circ = -nF \Delta E^\circ \] Where: - \( n \) is the number of moles of electrons transferred in the reaction (which is 3 for this reaction). - \( F \) is the Faraday constant (\( 96500 \, \text{C mol}^{-1} \)). - \( \Delta E^\circ \) is the cell potential, which is calculated as: \[ \Delta E^\circ = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.5 - (-2.87) = 4.37 \, \text{V}. \] Now, calculating \( \Delta G^\circ \): \[ \Delta G^\circ = -3 \times 96500 \times 4.37 = -1.26 \times 10^6 \, \text{J/mol}. \] Since \( \Delta G^\circ \) is negative, the reaction is **spontaneous**.