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
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)