Textbook Question
Chlorine can be prepared in the laboratory by the reaction of hydrochloric acid and potassium permanganate.(b) Calculate E° and ∆G° for the reaction
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20. Electrochemistry
Cell Potential and Gibbs Free Energy
Multiple Choice
Calculate the maximum obtainable work in kJ for the balanced redox reaction below at 25 °C. Given: 3 MnO4–(aq) + 4 H+(aq) + 5 NO(g) --> 3 Mn2+(aq) + 2 H2O(l) + 5 NO3–(aq), E°cell = 0.540 V. (HINT: 1 V = 1 J/C)
A
108.0 kJ
B
27.0 kJ
C
54.0 kJ
D
78.0 kJ
Verified step by step guidance1
Identify the number of electrons transferred in the balanced redox reaction. In this case, the reaction involves the transfer of electrons from NO to MnO4–.
Use the formula for calculating the maximum work obtainable from a redox reaction: \( \Delta G = -nFE^{\circ}_{\text{cell}} \), where \( n \) is the number of moles of electrons transferred, \( F \) is Faraday's constant (approximately 96485 C/mol), and \( E^{\circ}_{\text{cell}} \) is the standard cell potential.
Determine the value of \( n \) by analyzing the balanced equation. For each MnO4– reduced to Mn2+, 5 electrons are transferred, and since there are 3 MnO4– ions, \( n = 15 \) electrons.
Substitute the values into the formula: \( \Delta G = -nFE^{\circ}_{\text{cell}} \). Use \( n = 15 \), \( F = 96485 \) C/mol, and \( E^{\circ}_{\text{cell}} = 0.540 \) V.
Convert the result from joules to kilojoules by dividing by 1000, since 1 kJ = 1000 J. This will give you the maximum obtainable work in kJ.
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