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\(^{-1}\)]
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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.
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**.
