Electrochemistry MAE-295. Dr. Marc Madou , UCI, Winter 2012 Class II Thermodynamics of Electromotive Force (II) . Table of Content. Standard Redox Potentials Table Thermodynamics E cell , Δ G , and K eq Derivation of the Nernst Equation Some example problems
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Dr. Marc Madou, UCI, Winter 2012
Class II Thermodynamics of Electromotive Force (II)
don’t stretch it
constant PRelation between Equilibrium constant, Gibbs free energy and EMF of a cell
• Start with the First Law of Thermodynamics and some standard thermodynamic relations and we find:
• By convention, we identify work which is negative with work which is being done by the system on the surroundings. And negative free energy change is identified as defining a spontaneous process.Relation between Equilibrium constant, Gibbs free energy and EMF of a cell
• Now we can easily see how this Gibbs function relates to a potential.
• Note how a measurement of a cell potential directly calculates the Gibbs free energy change for the process.
• Activity scales with concentration or partial pressure.
a C/C˚ (solution) and a P/P˚ (gas)
• intermolecular interactions
• deviations from a direct correspondence with pressure or concentration
• Activity changes with concentration, temperature, other species, etc. Can be very complex.
comes into to play, giving us:
• It always has products in the numerator and reactants in the denominator
• It explicitly requires the activity of each reaction participant.
• Each term is raised to the power of its stoichiometric coefficient.
The reaction proceeds, Q changes, until finally DG=0. At that moment the overall reaction stops. This is equilibrium.Relation between Equilibrium constant, Gibbs free energy and EMF of a cell -Chemical Equilibrium
Q* = Keq
Cu | Cu2+ (0.001 M) || Cu2+ (1.00 M) | Cu
Consider the cell
Pt | I– (1.00 M), I2 (1.00 M) || Fe2+ (1.00 M), Fe3+ (1.00 M) | Pt
Standard Cell Potential is (from tables) = 0.771 V - 0.536 V = +0.235 VExample Problems:
This is the free energy change. It leads to the equilibrium constant for the reaction.
V2+ + 2e–® V -1.19
To get a final positive cell potential, the more negative half-reaction (V) must act as the anode.
Fe2+ + V ® Fe + V2+
Ecell = -0.44 - (-1.19) = +0.75 VExample Problems:
Sn2+ + 2e–® Sn -0.14
Ag+ + e–® Ag +0.80
More negative potential reaction is the anode.
Multiply the Ag reaction by 2, but don’t modify the cell potential.
2 Ag+ + Sn ® 2 Ag + Sn2+
Ecell = +0.80 - (-0.14) = +0.94 V
Cathodic Protection of an Iron Storage Tank
Electrolysis is the process in which electrical energy is used to cause a non-spontaneouschemical reaction to occur.
Electrolysis of Water
Electrolysis and Mass Changes
charge (Coulombs) = current (Amperes) x time (sec)
1 mole e- = 96,500 C = 1 Faraday
How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours?
2 mole e- = 1 mole Ca
mol Ca = 0.452
x 1.5 hr x 3600
2Cl- (l) Cl2(g) + 2e-
Ca2+(l) + 2e- Ca (s)
1 mol Ca
1 mol e-
2 mol e-
Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g)
= 0.0126 mol Ca
= 0.50 g Ca
Double layer-(in case of a metal 10-40 µF cm-2)
Inner Helmholtz plane (IHP)
Outer Helmholtz plane (OHP)
Gouy-Chapman layer (GCL)