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Lecture 13: E° cell and D G

This lecture explores the relationship between cell potential (E°cell) and Gibbs free energy (ΔG°) in galvanic electrochemical cells. We review the calculation of work done by these cells and their capacity to generate electrical energy. The course material covers definitions, relevant equations, and examples illustrating how to determine E°cell from half-cell potentials and calculate ΔG°. Additionally, the links between spontaneity, electrochemical equilibrium (K), and thermodynamic principles are discussed to highlight their significance in electrochemistry.

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Lecture 13: E° cell and D G

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  1. Lecture 13: E°cell and DG • Reading: Zuhdahl: 11.3 • Outline • E°cell and work • E°cell and DG

  2. E°cell and work • We saw in Lecture 12 how to construct a galvanic electrochemical cell that was capable of generating a flow of electrons. • This flow of electrons (current) can be used to perform work on the surroundings.

  3. E°cell and work (cont.) • From the definition of electromotive force (emf): Volt = work (J)/charge (C) • In English: 1 J of work is done when 1 C of charge is transferred between a potential difference of 1 V.

  4. E°cell and work (cont.) • Recall, we have a system-based perspective such that work done by the system is negative: E°cell = -w/q q = charge (not heat!) w = work • Rearranging: qE°cell = -w

  5. E°cell and work (cont.) • An example: What is the amount of work that is done by a galvanic cell in which 1.5 moles of electrons are passed between a potential of 2 V? qE°cell = -w (1.5 mol e-)(2 V) = -w What do we do with this?

  6. E°cell and work (cont.) • Define the “Faraday” (F): 1 F = the amount of charge on 1 mol of e- 1 F = 96,485 C/mol e- • Then: q = (1.5 mol e-) x (1 F = 96,485 C/mol e-) = 144,728 C • Finally: -w = qE°cell = (144,728 C)(2 J/C) = 298.5 kJ

  7. E°cell and DG • Since there is a relationship between DG and work, there is also a relationship between DG and E°cell. DG° = -nFE°cell • The above relationship states that there is a direct relationship between free energy and cell potential. • For a galvanic cell: E°cell > 0 Therefore, DG° < 0 (spontaneous)

  8. E°cell and DG (cont.) • For the following reaction, determine the overall standard cell potential and determine DG° Cd+2(aq) + Cu(s) Cd(s) + Cu+2(aq) E°1/2 = -0.40 V Cd+2 + 2e- Cd E°1/2 = +0.34 V Cu+2 + 2e- Cu E°1/2 = -0.34 V Cu Cu+2 + 2e-

  9. E°cell and DG (cont.) E°1/2 = -0.40 V Cd+2 + 2e- Cd E°1/2 = -0.34 V Cu Cu+2 + 2e- Cd+2(aq) + Cu(s) Cd(s) + Cu+2(aq) E°cell = -0.74 V 2 DG° = -nFE°cell = (-2 mol e-)(96485 C/mol e-)(-0.74 J/C) = 142.8 kJ (NOT spontaneous, NOT galvanic)

  10. E°cell and DG (cont.) • We now have two relationships for DG°: DG° = -nFE°cell = -RTln(K) -nFE°cell = -RTln(K) E°cell = (RT/nF) ln(K) E°cell = (0.0257 V) ln(K) n

  11. E°cell and DG (cont.) E°cell = (0.0257 V) ln(K) = (0.0591) log(K) n n • The above relationship states that by measuring E°cell, we can determine K. • The above relationship illustrates that electrochemical cells are a venue in which thermodynamics is readily evident

  12. E°cell and DG (cont.) • Developing the “big picture” DG° = -RTln(K) E°cell = (0.0591 V) log(K) DG° = -nFE°cell n • It is important to see how all of these ideas interrelate.

  13. An Example • Balance, determine E°cell and K for the following: S4O62- (aq) + Cr2+(aq) Cr3+(aq) + S2O32-(aq) 2e- + S4O62- S2O32- 2 Cr2+ Cr3+ + e- x 2 S4O62- + 2Cr2+ 2Cr3+ + 2S2O32-

  14. An Example (cont.) • Determining E°cell 2e- + S4O62- S2O32- 2 E°1/2 = 0.17 V E°1/2 = 0.50 V 2Cr2+ 2Cr3+ + 2e- S4O62- + 2Cr2+ 2Cr3+ + 2S2O32- E°cell = 0.67 V

  15. An Example (cont.) • Determining K S4O62- + 2Cr2+ 2Cr3+ + 2S2O32- E°cell = 0.67 V E°cell = (0.0257 V) ln(K) = (0.059 V) log K n n n(E°cell) 2 (0.67 V) = 22.7 = log K = (0.059 V) (0.059 V) K = 1022.7 = 5 x 1022

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