Calculate $\Delta$_rG $\circ$ for the disproportionation reaction of Cu⁺ at 25 $\circ$ C, 2 Cu⁺(aq)$\to$Cu²⁺(aq)+ Cu(s) Given the following thermodynamic information. Cu⁺(aq)+ e^- $\to$ Cu(s)$~~~~~~~~$E $\circ$ = +0.518 V Cu²⁺(aq)+ 2 e^- $\to$Cu(s)$~~~~~~~~$E $\circ$ = +0.337 V A) -165 kJ/mol.rxn B) -135 kJ/mol.rxn C) -34.9 kJ/mol.rxn D) +17.5 kJ/mol.rxn E) +135 kJ/mol.rxn

Question

Quizplus · Accepted Answer

The standard cell potential for the disproportionation reaction can be calculated using the given standard reduction potentials. The reaction is: 2 Cu⁺ → Cu²⁺ + Cu. The standard cell potential (E°) is calculated as E°(Cu²⁺/Cu) - E°(Cu⁺/Cu). Substituting the given values: E° = 0.337 V - 0.518 V = -0.181 V. The Gibbs free energy change (ΔG°) is then calculated using ΔG° = -nFE°, where n = 1 (mole of electrons transferred), F = 96485 C/mol, and E° = -0.181 V. This gives ΔG° = -1 * 96485 * (-0.181) = 34.9 kJ/mol.

Calculate Δ\DeltaΔ rG ∘\circ∘ for the Disproportionation Reaction of Cu+ at 25

Calculate $\Delta$ _rG^$\circ$ for the Disproportionation Reaction of Cu⁺ at 25