Electrochemistry MAE-212 Dr. Marc Madou, UCI, Winter 2015 Class II Thermodynamics of Electromotive Force (II)

Table of Content Standard Redox Potentials Table Thermodynamics Ecell, ΔG, and Keq Derivation of the Nernst Equation Some example problems Structure of the solid/electrolyte interface

Standard Redox Potentials Table E0 is for the reaction as written The more positive E0 the greater the tendency for the substance to be reduced The half-cell reactions are reversible The sign of E0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 (Intrinsic vs extrinsic properties)

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell The free energy function is the key to assessing the way in which a chemical system will spontaneously evolve. In the Gibbs-Duhem formalism of the Gibbs free energy we can write: constant T constant P don’t change shape don’t stretch it If we know the Gibbs function in terms of pressure, temperature and amount of material (T,P, n), then we can, with a variety of partial derivatives, calculate all other thermodynamic properties. If we have the Helmholtz function in terms of (T,V,N) then we have access to all the other parameters also. However, if we only know G in terms of (T,V,n), then we can’t perform the necessary differentials. There are many ways in which the Gibbs function can change: temperature, pressure, amount of material, surface area, elastic stretch. If we keep everything except material constant, then the changes in the Gibbs function is the result of the changes in the composition of the system.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell Every substance has a unique propensity to contribute to a system’s energy. We call this property Chemical Potential: m When the substance is a charged particle (such as an electron or an ion) we must include the response of the particle to an electrical field in addition to its Chemical Potential. We call this the Electrochemical Potential (F is the Faraday constant, z the charge on the particle and f the potential): = m + z F f Z is the charge on the particle, F is Faraday’s constant, phi is the field under consideration.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell The chemical potential or electrochemical potential (if we are dealing with a charged particle) is the measure of how all the thermodynamic properties vary when we change the amount of the material present in the system. Formally we can write: Integration of the expression for the dependence of amount of material on the Gibbs function, leads to the following relationship : G: Gibbs free energy, H: enthalpy, internal energy :U and A: Helmholtz free energy,

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell The internal energy is a natural function of entropy and volume, U(S,V). The Helmholtz free energy is a natural function of temperature and volume, A(T,V). We can also consider the enthalpy, H as a natural function of entropy and pressure and the Gibbs free energy as a natural function of temperature and pressure, G(T,P).

Relation 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: If we start with the First Law of Thermodynamics, we can recall some definitions for heat and work. Under reversible conditions we have the given dq for heat transfer. We also consider for work, either expansion work or electrical work. Substitute these into the First Law equation. Now take the definition for G and, using the definition for enthalpy H, and substitute for internal energy, and cancel terms, we arrive at the important relationship between Gibbs Free Energy and work done in an electrical system. And therefore, the Gibbs function is at the heart of electrochemistry, for it identifies the amount of work we can extract electrically from a system.

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. • 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. Electrical work is just the amount of charge Q and the potential V through which we move it. Recall that a process is spontaneous when its change in its Gibbs function is negative. If we choose to associate a positive potential with a spontaneous process, then we need a negative sign in the equation relating ∆G to E. Must choose a convention for the sign of the cell potential so that it is consistent with these equations and the convention regarding reaction spontaneity and the sign for ∆G. • Note how a measurement of a cell potential directly calculates the Gibbs free energy change for the process.

a  C/C˚ (solution) and a  P/P˚ (gas) Relation between Equilibrium constant, Gibbs free energy and EMF of a cell The propensity for a given material to contribute to a reaction is measured by its activity, a. How “active” is this substance in this reaction compared to how it would behave if it were present in its standard state? • Activity scales with concentration or partial pressure. a  C/C˚ (solution) and a  P/P˚ (gas) BUT… • intermolecular interactions • deviations from a direct correspondence with pressure or concentration Generally speaking, a substance’s activity increases as the amount of that substance available for reaction increases. However, due to intermolecular interactions (which increase as a substance’s concentration or partial pressure increases) there are deviations from a direction correspondence as pressure and concentration increase. These differences can become very signficant.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell Definition of activity is then: Activity coefficients close to 1 for dilute solutions and low partial pressures. • Activity changes with concentration, temperature, other species, etc. Can be very complex. Generally, we ignore activity coefficients for educational simplicity, but careful work always requires its consideration. In the case of pressure, the coefficient used is often called “fugacity” but it is the same thing and behaves in the same way.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell How does chemical potential change with activity? Integration of the expressions for the dependence of amount of material on the Gibbs function, leads to the following relationship : Once we have the chemical potential at the reference state, we can readily calculate changes as the activity changes, usually as a result of changes in amount of material, resulting from a chemical reaction. This will be the foundation by which we analyze any given chemical process to determine its equilibrium position.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell How does Gibbs free energy change with concentration/activity? Same dependence as for the chemical potential: When we apply this to a reaction, the reaction quotient comes into to play, giving us: Say we have the reaction : Once we have the chemical potential at the reference state, we can readily calculate changes as the activity changes, usually as a result of changes in amount of material, resulting from a chemical reaction. This will be the foundation by which we analyze any given chemical process to determine its equilibrium position.

• It explicitly requires the activity of each reaction participant. Relation between Equilibrium constant, Gibbs free energy and EMF of a cell The above reaction is a generic reaction and in order to analyze this chemical process mathematically, we formulate the reaction quotient Q: • 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. Watch out for the notation here. I am trying to not confuse you with using “a” for activity and “a” for a stoichiometric coefficient and also not confusing with using gamma for activity coefficient and gamma for stoichiometric coefficient. Hence, the w, x, y, z. If you write the reaction in a different way, such as write it backwards, or change the stoichiometric coefficients but multiplying by some factor, the quotient changes correspondingly. Write the quotient for the reaction as it appears.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell- Nernst Equation Take the expression for the Gibbs dependence on activity and rewrite this in terms of cell potential: The relation between cell potential E and free energy gives: Rearrange and obtain the Nernst Equation: This expression relates the dependence of the cell potential on the reaction quotient. This is an exact expression. We start invoking a number of approximations when we use anything other than activity in the reaction quotient.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell- Nernst Equation The equation is often streamlined by restricting discussion to T = 25 °C and inserting the values for the constants, R and F. It is a good thing to remember the relation that the base10 expression tells us that if there are changes in Q such it changes by a factor of 10, this will mean that the cell potential will change by 59.2 mV or about 60 mV for a one electron process (n=1). This rule of thumb relationship will show up often. “n” is the number of electrons which are transferred in the process. This depends upon the stoichiometry and the overall reaction must be dissected into its two half-reactions to be able to clearly identify the value for n. Any reaction equation can be altered by multiplying all the coefficients by a constant. This will affect “n” and it will also affect all of the exponents of the activities of each species in Q. The result of this is to make it such that this equation produces the same result, regardless of how it is written. So, as long as it is written correctly, you will still get the same result. Note the difference between using natural logarithms and base10 logarithms. Be aware of the significance of “n” – the number of moles of electrons transferred in the process according to the stoichiometry chosen.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell -Chemical Equilibrium The reaction proceeds, Q changes, until finally DG=0. At that moment the overall reaction stops. This is equilibrium. This special Q* (the only one for which we achieve this balance) is renamed Keq, the equilibrium constant. Use the definition of ∆G and we find that when Q=1, we have ∆G = ∆G˚. This is the way we designed the equation. What is more interesting is to see what happens when ∆G goes to 0. Here is when we reach equilibrium, the various concentrations in the reaction stop changing. The reaction seems to stop. In this case, we obtain a relation between ∆G˚ and the equilibrium constant. The reaction doesn’t really stop; we simply have reached the situation where the forward and reverse reaction rates are equal. Chemical activity is still ongoing with just as much vigor as before, but it is happening equally in both directions — products are reacting to turn into reactants just as fast as reactants are reacting to turn into products. This is why we call it a dynamic equilibrium, rather than a static equilibrium.

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell -Chemical Equilibrium When the system is at equilibrium, the proper quotient of equilibrium concentrations is equal to the equilibrium constant: The system is at equilibrium when the concentrations are in their equilibrium values, so [A] = [A]e, [B] = [B]e, etc. and thus: Q* = Keq

Relation between Equilibrium constant, Gibbs free energy and EMF of a cell -Chemical Equilibrium

Example problems: Cu is cathode (it is reduced). Zn is anode (it is oxidized). Note that n=2 for this reaction. =1 When we replace activities of solute species with concentrations, we are making the approximation that the activity coefficient is 1. This is wrong but it is widely used to simplify the math. We will put concentrations in here, but don’t forget that we really mean activity. =1 Activity for solid materials is 1; replace activities with concentrations.

Example problems: What is the potential in the cell if [Cu2+] = 0.01 M and [Zn2+] = 1.00 M? Note that the cell potential decreased by about 60mV. This was a change in concentration of TWO orders of magnitude, but since it was also a TWO electron process, we saw the same 60 mV change in potential.

Cu | Cu2+ (0.001 M) || Cu2+ (1.00 M) | Cu Example Problems: Nernst equation demonstrates that potential depends upon concentration. A cell made of the same materials, but with different concentrations, will also produce a potential difference. Cu | Cu2+ (0.001 M) || Cu2+ (1.00 M) | Cu What is standard cell potential E˚ for this cell? What is the cell potential E? What is “n”, the number of electrons transferred? Which electrode, anode or cathode, will be in numerator? The standard cell potential in a concentration cell clearly must be 0 for the two electrodes, at standard conditions are identical. The number of electrons transferred is not difficult in this problem because the two reactions are identical and therefore the number of electrons passed is the same. What does require some thought is knowing which concentration to place in the numerator and which in the denominator of the Nernst Equation. Remember that we are calculating the cell potential for the cell as written in the definition. The right hand side is the cathode and so refers to a reduction process. The other is for oxidation. If you write out those two reactions, we see that the aqueous solution from the cathode appears as a reactant. Hence the cathode aqueous solution is in the denominator. Conversely for the numerator coming from the anode side.

Example Problems: The equations we have derived allow us to relate measured cell potentials to Standard Gibbs Free Energies of reaction. These in turn are related to a reaction’s equilibrium constant. 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 V This is the free energy change. It leads to the equilibrium constant for the reaction.

Example Problems: Sn2+ + 2e– ® Sn -0.14 Fe2+ + 2e– ® Fe -0.44 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 Fe2+ + 2e– ® Fe -0.44 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 V Take two reactions. Let the more negative (least positive) reaction be that for the anode. That means that its direction will be reversed. Balance the number of electrons but scaling the reactions appropriately, but DO NOT modify their cell potentials. The cell potential is a “per charge” quantity, so it does not change with the amount of material making the change. This is in contrast to all other thermodynamic properties such as free energy, enthalpy, or entropy, which are scaled by the same amount as the reaction.

Example Problems: Anode: 2H2 (g) + 4OH- (aq) 4H2O (l) + 4e- Cathode: A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: 2H2 (g) + 4OH- (aq) 4H2O (l) + 4e- Cathode: O2 (g) + 2H2O (l) + 4e- 4OH- (aq) 2H2 (g) + O2 (g) 2H2O (l)

Example Problems: Corrosion

Cathodic Protection of an Iron Storage Tank Example Problems: Cathodic Protection of an Iron Storage Tank

Example Problems: Electrolysis is the process in which electrical energy is used to cause a non-spontaneous chemical reaction to occur.

Example Problems: Electrolysis of Water

Electrolysis and Mass Changes Example Problems: Electrolysis and Mass Changes charge (Coulombs) = current (Amperes) x time (sec) 1 mole e- = 96,500 C = 1 Faraday

Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) 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? Anode: 2Cl- (l) Cl2 (g) + 2e- Cathode: 2 mole e- = 1 mole Ca Ca2+ (l) + 2e- Ca (s) Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) mol Ca = 0.452 C s x 1.5 hr x 3600 s hr 96,500 C 1 mol e- x 2 mol e- 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca

Structure of the solid/electrolyte interface Every solid/liquid has a interface that one would like to control --the better one can control that interface the better one can say something about the environment Interfaces are very complex often involving fractals (beach, trees, snow flakes, etc.) rather than smooth transitions, this implies , for example, that perfect selectivity will be hard to achieve (too many different binding sites)

Structure of the solid/electrolyte interface Charge carriers in electrode materials: Metals (e.g. Pt) : electrons Semiconductors (e.g. n-Si) : electrons and holes Solid electrolytes (e.g. LaF3 ) : ions Insulators (e.g. SiO2):no charge carriers Mixed conductors (e.g. IrOx) : ions and electrons Solution (e.g. 1 M NaCl in H2O): solvated ions Double layer-(in case of a metal 10-40 µF cm-2) Inner Helmholtz plane (IHP) Outer Helmholtz plane (OHP) Gouy-Chapman layer (GCL)