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Lecture 13: E° cell and D G

Lecture 13: E° cell and D G. Reading: Zuhdahl: 11.3 Outline E° cell and work E° cell and D G. 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.

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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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