Question

The standard free energy change (\( \Delta G^\circ \)) for the following reaction (in kJ) at 25°C is \[ 3Ca(s) + 2 Au^{+}(aq, 1M) \rightleftharpoons 3Ca^{2+}(aq, 1M) + 2Au(s) \] (given: \( E^\circ_{{Au}^{3+/2+}} = +1.50 \, V \), \( E^\circ_{{Ca}^{2+/Ca}} = -2.87 \, V \), \( F = 96500 \, {C mol}^{-1} \))

• A. \(-2.53 \times 10^3 \)
• B. \( +2.53 \times 10^3 \)
• C. \(-2.53 \times 10^4 \)
• D. \( +2.53 \times 10^4 \)

Hint:

The standard free energy change is related to the cell potential via the equation \( \Delta G^\circ = -nFE^\circ \), where \( n \) is the number of moles of electrons transferred in the reaction.

Answer 1

The Correct Option is A

To find the standard free energy change (\(\Delta G^\circ\)) for the reaction at 25°C, we'll use the relationship between \(\Delta G^\circ\) and the cell potential \(E^\circ\) from electrochemistry: \[ \Delta G^\circ = -nFE^\circ \] where: - \(n\) is the number of electrons exchanged in the balanced reaction. - \(F\) is the Faraday constant \(= 96500 \, \text{C mol}^{-1}\). - \(E^\circ\) is the standard cell potential calculated as: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction: \[ 3Ca(s) + 2Au^{+}(aq) \rightleftharpoons 3Ca^{2+}(aq) + 2Au(s) \] Here, \(Ca\) is oxidized and \(Au^+\) is reduced. The half-reactions are: - Oxidation at the anode: \[ Ca \rightarrow Ca^{2+} + 2e^- \;\; E^\circ_{\text{anode}} = -2.87 \, V \] - Reduction at the cathode: \[ Au^+ + e^- \rightarrow Au \;\; E^\circ_{\text{cathode}} = +1.50 \, V \] Calculating \(E^\circ_{\text{cell}}\): \[ E^\circ_{\text{cell}} = 1.50\, V - (-2.87\, V) = 4.37\, V \] The balanced overall reaction involves 6 electrons (\(n = 6\)) since 3 \(Ca\) lose 6 electrons to become 3 \(Ca^{2+}\), and 2 \(Au^+\) gain 2 electrons to become 2 \(Au\). Substitute into \(\Delta G^\circ\) formula: \[ \Delta G^\circ = -6 \times 96500 \times 4.37 \] \[ \Delta G^\circ = -2.53 \times 10^3 \, \text{kJ/mol} \] Therefore, the standard free energy change for the reaction is \( -2.53 \times 10^3 \, \text{kJ} \).

Answer 2

The standard free energy change (\( \Delta G^\circ \)) for a reaction can be calculated using the following relation: \[ \Delta G^\circ = -n F E^\circ \] where: - \( n \) is the number of moles of electrons transferred in the reaction,
- \( F \) is the Faraday constant (\( 96500 \, {C mol}^{-1} \)),
- \( E^\circ \) is the cell potential.
Step 1: 
First, calculate the cell potential \( E^\circ \). The overall cell reaction is: \[ {3Ca(s) + 2 Au}^{+}(aq, 1M) \rightleftharpoons 3Ca^{2+}(aq, 1M) + 2Au(s) \] The standard cell potential \( E^\circ_{{cell}} \) is given by: \[ E^\circ_{{cell}} = E^\circ_{{cathode}} - E^\circ_{{anode}} \] Here, the cathode is the reduction half-reaction involving \( {Au}^{+} \), and the anode is the oxidation half-reaction involving \( {Ca} \). Thus: \[ E^\circ_{{cell}} = E^\circ_{{Au}^{3+/2+}} - E^\circ_{{Ca}^{2+/Ca}} = 1.50 \, {V} - (-2.87 \, {V}) = 4.37 \, {V}. \] 
Step 2: Next, calculate the number of moles of electrons transferred in the reaction. From the balanced equation, we see that 6 electrons are transferred (3 moles of Ca give 6 electrons). 
Step 3: Now, we can calculate \( \Delta G^\circ \): \[ \Delta G^\circ = -6 \times 96500 \, {C mol}^{-1} \times 4.37 \, {V} = -2.53 \times 10^3 \, {kJ}. \] Thus, the standard free energy change for the reaction is \( -2.53 \times 10^3 \, {kJ} \).