Calculate ?G° for the disproportionation reaction of Cu⁺ at 25°C, 2Cu⁺(aq) ? Cu²⁺(aq) + Cu(s) Given the following thermodynamic information.(R = 8.31 J/K?mol,F = 96480 C/mol e?)
$\begin{array}{lccc} \mathrm{Cu}^{+}(a q) +e^{-} \rightarrow \mathrm{Cu}(s) & E^{\circ}=+0.518 \mathrm{~V} \\ \mathrm{Cu}^{2+}(a q) +2 e^{-} \rightarrow \mathrm{Cu}(s) &E^{\circ}=+0.337 \mathrm{~V} \\\end{array}$

Question

Calculate ?G° for the disproportionation reaction of Cu⁺ at 25°C, 2Cu⁺(aq) ? Cu²⁺(aq) + Cu(s) Given the following thermodynamic information.(R = 8.31 J/K?mol,F = 96480 C/mol e?)

\begin{array}{lccc} \mathrm{Cu}^{+}(a q) +e^{-} \rightarrow \mathrm{Cu}(s) & E^{\circ}=+0.518 \mathrm{~V} \\ \mathrm{Cu}^{2+}(a q) +2 e^{-} \rightarrow \mathrm{Cu}(s) &E^{\circ}=+0.337 \mathrm{~V} \\\end{array}

Quizplus · Accepted Answer

The ΔG° for the reaction can be calculated using the formula ΔG° = -nFE°cell, where n is the number of moles of electrons transferred, F is the Faraday constant (96480 C/mol e−), and E°cell is the standard cell potential. The standard cell potential, E°cell, can be found by subtracting the E° value of the oxidation half-reaction from the E° value of the reduction half-reaction. In this disproportionation reaction, Cu+ is both oxidized to Cu2+ and reduced to Cu(s). The E°cell for the reaction is the difference between the reduction potentials: E°cell = 0.518 V - 0.337 V = 0.181 V. Since 2 electrons are transferred in the process (from Cu+ to Cu and Cu2+), n = 2. Therefore, ΔG° = -2 * 96480 C/mol * 0.181 V = -34912 J/mol = -34.9 kJ/mol.

Calculate ?G° for the Disproportionation Reaction of Cu+ at 25°C

Calculate ?G° for the Disproportionation Reaction of Cu⁺ at 25°C