The correct answer is option 1. \(\Delta G^0 = -212.3 \text{ kJ/mol}\).
The standard Gibbs free energy change (\(\Delta G^0\)) for a reaction is related to the standard cell potential (\(E^0_{\text{cell}}\)) by the equation:
\[\Delta G^0 = -nF E^0_{\text{cell}}\]
where \(n\) is the number of moles of electrons transferred in the balanced chemical equation and \(F\) is the Faraday constant (\(96,500 \ \text{C/mol}\)).
The given reaction is:
\[Zn (s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s)\]
The balanced half-reactions are:
\[\begin{align*}\text{Oxidation (anode):} & \ Zn (s) \longrightarrow Zn^{2+}(aq) + 2e^- \ \ \ \ E^0_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \ \text{V} \\\text{Reduction (cathode):} & \ Cu^{2+}(aq) + 2e^- \longrightarrow Cu(s) \ \ \ \ E^0_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \ \text{V}\end{align*}\]
The overall reaction is the sum of these two half-reactions:
\(Zn (s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s)\)
So, \(n = 2\) moles of electrons are transferred. Now, substitute the values into the equation:
\(\Delta G^0 = -nF E^0_{\text{cell}}\)
\(\Delta G^0 = -2 \times (96,500 \ \text{C/mol}) \times (E^0_{\text{Cu}^{2+}/\text{Cu}} - E^0_{\text{Zn}^{2+}/\text{Zn}})\)
\(\Delta G^0 = -2 \times (96,500 \ \text{C/mol}) \times [(+0.34 \ \text{V}) - (-0.76 \ \text{V})]\)
\(\Delta G^0 = -2 \times (96,500 \ \text{C/mol}) \times (1.1 \ \text{V})\)
\(\Delta G^0 = -2 \times (96,500 \ \text{C/mol}) \times (1.1 \ \text{J/C})\)
\(\Delta G^0 = -2 \times (96,500 \ \text{J/mol}) \times (1.1)\)
\(\Delta G^0 \approx -212300 \ \text{J/mol} \ \text{or} \ -212.3 \ \text{kJ/mol}\) |
