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Free Energy and Redox Reactions

Free Energy and Redox Reactions. The emf associated with any redox reaction can be calculated. E o = E o red (reduction) – E o red (oxidation) Spontaneous redox reaction positive E o (standard conditions) positive E (non-standard conditions) negative D G

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Free Energy and Redox Reactions

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  1. Free Energy and Redox Reactions • The emf associated with any redox reaction can be calculated. • Eo = Eored (reduction) – Eored (oxidation) • Spontaneous redox reaction • positive Eo (standard conditions) • positive E (non-standard conditions) • negative DG • Non-spontaneous redox reactions: • negative Eo • negative E • positive DG

  2. Free Energy and Redox Reactions • The change in Gibbs free energy is related to the emf of a redox reaction by the equation: DG = -nFE where DG = change in Gibbs free energy n = number of electrons transferred F = Faraday’s constant = 96,485 J/V.mol E = emf under nonstandard conditions (I will give you this equation and the value of F on your exam.)

  3. Free Energy and Redox Reactions • Under standard conditions, this equation becomes: DGo = -nFEo where DGo = standard Gibbs free energy change n = number of electrons transferred F = Faraday’s constant = 96,485 J/V.mol Eo = emf under standard conditions (I will give you this equation and the value of F on your exam.)

  4. Free Energy and Redox Reactions • You can use the value of Eo to calculate the value of DGo and the equilibrium constant, K, for the reaction. • DGo = -nFEo • DGo = -RTlnK

  5. Free Energy and Redox Reactions Example: Use the standard reduction potentials listed in Appendix E of your text to calculate the equilibrium constant for the following reaction at 298K. 3 Ce4+ (aq) + Bi(s) + H2O (l)  3 Ce3+ (aq) + BiO+ (aq) + 2 H+ (aq) Step 1: Calculate the value for Eo:

  6. Free Energy and Redox Reactions Step 2: Calculate the value of DGo: Step 3: Calculate the value of K: Answer: 2.42 x 1065

  7. Free Energy and Redox Reactions Example: What is the effect on the emf of the cell described by the following equation when the following changes are made: 2 Fe3+ (aq) + H2 (g)  2 Fe2+ (aq) + 2 H+ (aq) The pressure of hydrogen gas in the anode compartment is increased? Iron (III) nitrate is added to the cathode compartment? Sodium hydroxide is added to the anode compartment?

  8. Free Energy and Redox Reactions • The emf of a redox reaction varies with temperature and with the concentrations of reactants and products. • The Nernst equation relates the emf under nonstandard conditions to the standard emf and the reaction quotient. E = Eo- (RT/nF)lnQ

  9. Free Energy and Redox Reactions • Converting from natural log to log base 10 and assuming that T = 298 K, the Nernst Equation becomes: E = Eo – 0.0592 log10 Q n where n = the number of electrons transferred Q = reaction quotient (I will give you this equation. You need to be able to use this equation)

  10. Free Energy and Redox Reactions Example: Calculate the emf generated by the following reaction when [Al3+] = 4.0 x 10-3 M and [I-] = 0.010 M at 298K. 2 Al (s) + 3 I2 (s)  2 Al3+ (aq) + 6 I- (aq) Step 1: Calculate Eo

  11. Free Energy and Redox Reactions • Step 2: Calculate Q • Step 3: Calculate E using the Nernst Eq’n: • Answer: E = 2.36 V

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