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Free Energy and Redox ReactionsPowerPoint Presentation

Free Energy and Redox Reactions

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Free Energy and Redox Reactions

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- The emf associated with any redox reaction (spontaneous or nonspontaneous) 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

- 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.)

- 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.)

- 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

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:

Step 2: Calculate the value of DGo:

Step 3: Calculate the value of K:

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?

- 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

- Converting from natural log to log base 10 and assuming that T = 298 K, the Nernst Equation becomes:
E = Eo – 0.0592 log 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)

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

- Step 2: Calculate Q
- Step 3: Calculate E using the Nernst Eq’n: