A galvanic cell is based on the half-reaction

CHEMISTRY · Multi-part · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry

Problem

A galvanic cell is based on the half-reaction Fe3++e−⟶Fe2+  (E∘=0.77 V)\mathrm{Fe^{3+}}+e^-\longrightarrow \mathrm{Fe^{2+}}\;(E^{\circ}=0.77\,\text{V}). Two versions of the cell are prepared with identical initial concentrations but operated at different temperatures. • Cell 1: [Fe3+]=0.150 M[\mathrm{Fe^{3+}}]=0.150\,\text{M}, [Fe2+]=1.00 M[\mathrm{Fe^{2+}}]=1.00\,\text{M}, T=298 KT=298\,\text{K} • Cell 2: [Fe3+]=0.150 M[\mathrm{Fe^{3+}}]=0.150\,\text{M}, [Fe2+]=1.00 M[\mathrm{Fe^{2+}}]=1.00\,\text{M}, T=350 KT=350\,\text{K} Assume ideal behavior for all species. The universal gas constant is R=8.314 J mol−1 K−1R=8.314\,\text{J\,mol}^{-1}\,\text{K}^{-1} and Faraday’s constant is F=96,485 C mol−1F=96{,}485\,\text{C\,mol}^{-1}. (a) Write the balanced cell reaction and state the number of moles of electrons transferred, nn, per mole of overall reaction. (b) Calculate the reaction quotient QQ for the initial mixture shared by Cell 1 and Cell 2. (c) Determine the numerical value of the cell potential for Cell 1 and for Cell 2. Show all calculation steps and report each potential to three significant figures. (d) For each cell, compute the maximum non-PV electrical work (in kJ) that could be obtained per mole of electrons transferred under the stated conditions. (e) Discuss how the change in temperature influences the cell potential and the maximum obtainable work, citing the calculated values as evidence.

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