nernst equation.ppt
TRANSCRIPT
NERNST EQUATION & POURBAIX DIAGRAM
Presented By
Muhammad Faheem Khan (MM-11)Muhammad Ali Siddiqui (MM-03)Muhammad Mehmood Khan (MM-12)
N E R N S T E Q U A T I O N
The Nernst equation was named after the German chemist Walther Nernst who established very useful relations between the energy or potential of a cell to the concentrations of participating ions.
E = E0 - R T lnQ reaction
nF
D E R I V A T I O N The Nernst Equation is derived from the emf and the Gibbs energy
under non-standard conditions.
E0 = E0reduction – E0 oxidation
When E0 is positive, the reaction is spontaneous. E0 is negative, the reaction is not spontaneous. Since the change in Gibbs free energy, , is also related to spontaneity
of a reaction, therefore, and are related. Specifically, ∆G= -nFE0
Where n is number of electrons transferred in the reaction, F is the Faraday constant (96500 C/mol), E0is potential difference. Under standard conditions, this equation is
then ∆G=-nFE0
As we know that Gibbs free energy
∆G= ∆G0 + R T lnQ reaction ……….. (Eq:01)
For a generalized equation aA + bB → mM + nN
[M] m [N]n
Q reaction = ____________
[ A ]a [ B]b
In the case of an electrochemical reaction, substitution of the relationships ∆G = -nFE and ∆G0 = -nFE0 into the expression of a reaction free energy.
After substituting the Eq:01 become
-nFE= -nFE0 + R T lnQ reaction …. (Eq:02)
D E R I V A T I O N…(Con…)
Divide equation (Eq:2) by nF we get
E = E0 - R T lnQ reaction
nF Combining constants at 25oC (298.15 K) gives the simpler form of the Nernst equation for an electrode reaction at this standard temperature
E = E0 - 0.059 log10 Q reaction
n
D E R I V A T I O N…(Con…)
E = E0 - R T lnQ reaction
nF
Relation to equilibrium
At equilibrium, E = 0 and Q = K. …so
E = E0 - R T lnQ reaction
nF
lnk = nFE0
RT
Example of Nernst equationCalculate the voltage produced by the cell Sn(s)|Sn2+||Ag+|
Ag(s) at 25oC given: [Sn2+] = 0.15 M [Ag+] = 1.7 M
Write the Nernst Equation for 25oC: E = Eo -0.0592/n x logQ
SOLUTIONCalculate Eo for the cell anode: Sn(s)→Sn2+ + 2eEo = -0.14V cathode: 2x [e- + Ag+→Ag(s)]Eo = +0.80V ____________________________cell: Sn(s) + 2Ag+→Sn2+ + 2Ag(s) Eo = +0.66V
Write the Nernst Equation for this example: E = Eo -0.0592/n x log([Sn2+]/[Ag+]2)
Substitute the values:
Eo = +0.66V n = 2 (2 moles of electrons transferred during the redox reaction) [Sn2+] = 0.15 M [Ag+] = 1.7 ME = +0.66 -0.0592/2 x log([0.15]/[1.7]2)
Calculate Q:
E = +0.66 -0.0592/2 x log[0.0519] Calculate logQ:
E = +0.66 -0.0592/2 x -1.285 Calculate E:
E = +0.66 -0.0592/2 x -1.285 E = +0.70V
Eh - pH – DIAGRAM or POURBAIX DIAGRAM
INTRODUCTION• Pourbaix diagram (Electrode potential / PH diagram) is a
graphical presentation of the thermodynamic equilibrium states of a metal-electrolyte system.
• Pourbaix diagrams are plotted in the axes Electrode potential of the metal vs. PH of the electrolyte.
• Oxidizing conditions are described by the top part of the diagram (high positive electrode potential)
• Reducing conditions are described by the bottom part of the diagram (high negative electrode potential).
• Acidic solutions are presented in the left side of the diagram (PH lower than 6).
• Alkaline solutions are presented in the right side of the diagram (PH higher than 6).
• Pourbaix diagrams allow determining the corrosion behavior of a metal in water solutions i.e. the direction of electro-chemical processes and the equilibrium state of the metal at a certain electrode potential in a water solution at a certain value of PH
• The lines of the diagrams dividing different zones of the equilibrium states are calculated by the Nernst equation:
How to read Pourbaix Diagram• below the line a-b-j: Solid iron (immunity zone). No corrosion occurs in this zone.• a-b-n-c-d-e: Aqueous solution of ion Fe2+ (corrosion zone). Metallic iron oxidizes in this zone. • e-d-f-g-k: Aqueous solution of ion Fe3+ (corrosion zone). Metallic iron oxidizes (corrodes) in this zone.• h-f-g-m: Aqueous solution of ion FeO42- (corrosion zone). • c-d-f-h-i: Solid ferrous oxide Fe2O3 (passivation zone). Iron oxidizes (corrodes) in this zone however the resulted
oxide film depresses the oxidation process causing passivation (corrosion protection of the metal due to formation of a film of a solid product of the oxidation reaction).
• n-c-i-p: Solid oxide Fe3O4 (Fe2O3*FeO) (passivation zone). The oxide film causes passivation.• b-n-p-j: Solid hydroxide (II) Fe(OH)2 / FeO*nH2O / green rust (passivation zone). • Horizontal lines of the Poubaix diagrams correspond to the redox reactions, which are independent of PH.• Vertical lines of the Poubaix diagrams correspond to the non-redox reactions (electrons are not involved), which are
dependent on PH. Diagonal lines of the Poubaix diagrams correspond to the redox reactions, which are dependent on PH. Here are some of the reactions and the corresponding lines of the Fe-H2O Pourbaix diagram:
• a-b: Fe(s) = Fe2+(aq) + 2e- Redox reaction independent of PH. The equilibrium occurs at the electrode potential value -0.44V, which is equal to the standard electrode potential of iron (see the Electrochemical series).
• e-d: Fe2+(aq) = Fe3+(aq) + e- Redox reaction independent of PH.• d-f: 2Fe3+(aq) + 3O2- = Fe2O3(s) Non-redox reaction dependent on PH.• b-n: Fe2+(aq) + 2OH-(aq) = Fe(OH)2(s) Non-redox reaction dependent on PH.• c-d: 2Fe2+(aq) + 3H2O = Fe2O3(s) + 6H+(aq) +2e- Redox reaction dependent on PH.• b-j: Fe(s) + 2OH-(aq) = Fe(OH)2(s) + 2e- Redox reaction dependent on PH.
Limitations of Pourbaix diagrams• The diagrams provide no information about the kinetic
parameters of the corrosion reactions. • The diagrams consider pure metals and aqueous solutions at
standard conditions (temperature 298K, pressure 1 bar, ion concentration 10-6M).
• Thermodynamic conditions of corrosion for alloys and for non-standard conditions differ from those described by Pourbaix diagrams.
• The diagrams do not take into account non-ideal behavior of aqueous solutions.
• Some thermodynamic data used for building diagrams are not precise enough.
Thanks