Question

(a) Calculate the standard Gibbs energy (\(\Delta G^\circ\)) of the following reaction at 25°C: 
\(\text{Au(s) + Ca\(^{2+}\)(1M) $\rightarrow$ Au\(^{3+}\)(1M) + Ca(s)} \)
\(\text{E\(^\circ_{\text{Au}^{3+}/\text{Au}} = +1.5 V, E\)\(^\circ_{\text{Ca}^{2+}/\text{Ca}} = -2.87 V\)}\)
\(\text{1 F} = 96500 C mol^{-1}\)

Hint:

A positive cell potential (\(E^\circ_{\text{cell}}>0\)) indicates a spontaneous reaction, as \(\Delta G^\circ\) will be negative.

Answer 1

The standard Gibbs energy change (\(\Delta G^\circ\)) is related to the standard cell potential (\(E^\circ_{\text{cell}}\)) by the equation: \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] First, calculate the cell potential: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For this reaction: - The cathode (reduction half-reaction) is the gold electrode (\(\text{Au}^{3+}/\text{Au}\)) with \(E^\circ = +1.5 \, V\). - The anode (oxidation half-reaction) is the calcium electrode (\(\text{Ca}^{2+}/\text{Ca}\)) with \(E^\circ = -2.87 \, V\). Thus, \[ E^\circ_{\text{cell}} = 1.5 - (-2.87) = 1.5 + 2.87 = 4.37 \, V \] Next, calculate the Gibbs energy: \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] For this reaction, \(n = 6\) (since 6 electrons are transferred). Therefore, \[ \Delta G^\circ = -(6)(96500)(4.37) = -2.53 \times 10^6 \, \text{J/mol} = -2530 \, \text{kJ/mol} \] Since \(\Delta G^\circ\) is negative, the reaction will be spontaneous at 25°C.