Electrochemistry for Engineers 0581.5271 Electrochemistry for Engineers LECTURE 2 Lecturer: Dr. Brian Rosen Office: 128 Wolfson Office Hours: Sun 16:00

The Electrochemical Cell Part II

Lower Stability of Water Starting with H+(aq) + e-  ½H2(g) we write the Nernst equation We set pH2 = 1 atm. Also, Gr° = 0, so E0 = 0. Thus, we have Again, the lower limit of water solubility in Eh-pH space is governed by the same reaction as in pe-pH space. We repeat the same steps as before, except that for this reaction, E0 = 0, because all the reactants and products have free energies of formation that are zero by convention.

Upper Stability of Water ½O2(g) + 2e- + 2H+  H2O but now we employ the Nernst eq. In the next three slides, we repeat the calculation of the stability limits for water, but this time in Eh-pH coordinates. There are two reasons for repeating the calculation in this way. The first is to show how pe-pH diagrams and Eh-pH diagrams relate to each other, and to show the slight differences in how they are calculated. The second reason is so that we can use a published Eh-pH diagram to show where different geological environments sit with respect to pH. The reaction governing the upper stability limit for water on an Eh-pH diagram is exactly the same reaction that we used for the pe-pH diagram. The difference is that, instead of using the mass-action (equilibrium constant) expression, we use the Nernst equation. To employ the Nernst equation given the convention Kehew (2001) has adopted, we must right the reaction as a reduction reaction. Once the appropriate Nernst equation is written, we can proceed to the next step. Note that, in employing the Nernst equation, the convention employed is that ae- = 1. This is a major difference compared to the pe-pH diagram, where ae- is not necessarily equal to 1. The pe = -log ae-, so if we set ae- = 1 we could not plot a pe-pH diagram, because all pe values would be zero. However, such a situation would make many students happy!

Upper Stability of Water Recall pH = -log (aH+) We assume that pO2 = 1 atm. This results in This yields a line with slope of -0.0592. We need to calculate E0 from the value of Gr° as shown above, and then rearrange the Nernst equation so that we have an equation of a straight line with Eh on the Y-axis and pH on the X-axis. We find that we still have to fix the value of the partial pressure of oxygen to plot the boundary, and so we choose 1 atm as for the pe-pH diagram.

The Electrochemical Series

Pourbaix Diagram

Pourbaix Diagram of Water Stability vs. Ag/AgCl?

Recall: A Good Reference is Non-Pol. +0.00 vs. Ag/AgCl +0.197V vs. SHE 0.00V vs. SHE

Pourbaix Diagram vs Ag/AgCl E = 1.03 V vs. Ag/AgCl E = -.197 V vs. Ag/AgCl

Reconsider : Galvanic vs. Electrolytic

Equilibrium (Non-polarized) Cu+2 + 2e- Cu + i - E + E When the electrode is held at its equilibrium potential, the forward and reverse current cancel and the NET current is zero - i E° Cu/Cu+ =0.340 V vs. SHE

Cathodic Polarization Cu+2 + 2e- Cu + i ΔE - E + E When the electrode is negative of its equilibrium potential, reduction is favored and there is a net flow of electrons consumed by the electrode for the reduction reaction. - i E° Cu/Cu+ =0.340 V vs. SHE

Anodic Polarization Cu+2 + 2e- Cu E° Cu/Cu+ =0.340 V vs. SHE When the electrode is positive of its equilibrium potential, oxidation is favored and there is a net flow of electrons produced for the oxidation reaction. - E + E ΔE - i E° Cu/Cu+ =0.340 V vs. SHE

Real Polarization of Electrodes In reality, I-V curves for electrodes take on a variety of shapes and forms dependent on the kinetic and mass-transport parameters

OPEN CIRCUIT POTENTIAL (OCP) E Cell (OCP) = 0.74 V i + E - E i E° Cd/Cd+2 -0.40 vs. SHE + E - E E° Cu/Cu+2 +0.34 vs. SHE 1M Cd+2 1M Cu+2 Cu Cd Cd+2 +2e- Cd Cu+2 +2e- Cu

GALVANIC CELL (SPONTANEOUS) E° Cd/Cd+2 -0.40 vs. SHE + E - E E° Cu/Cu+2 +0.34 vs. SHE E Cell = 0.64 V -0.35V +0.29V i i e- 1M Cd+2 1M Cu+2 Cu Cd e- - + ANODE CATHODE Cd+2 +2e- Cd Cu+2 +2e- Cu

+ - ELECTROLYTIC CELL E Applied = 1.5 V -∆E +∆E + E - E E° Cd/Cd+2 -0.4 vs. SHE + E - E E° Cu/Cu+2 +0.34 vs. SHE +0.75V E Applied = 1.5 V -0.75V i i +0.29V 1.5V e- 1M Cd+2 1M Cu+2 Cu Cd e- + - CATHODE ANODE Cd+2 +2e- Cd Cu+2 +2e- Cu

Summary η = (E - E°) = “overpotential” Galvanic (spontaneous cells) operate below their maximum potential (OCP) due to the electrode polarization required to produce current (anode E is less negative than E°, and cathode E is less positive than E°) Electrolytic cells require an external voltage more than the minimum required OCP predicted by thermodynamics (anode E is more positive than E°, cathode E is more negative than E°) η = (E - E°) = “overpotential”

Electrode Processes MASS-TRANSFER diffusion convection migration KINETICS

Electrochemical Kinetics

Electrode Processes MASS-TRANSFER diffusion convection migration KINETICS (FAST) (SLOW) In our modeling of kinetics, we will assume that mass transfer to the electrode is fast compared to the reaction rate

Our Goal i = f(E) How can we model the current at an electrode as a function of its potential assuming the reaction is the rate limiting step? (i.e. mass transfer to the electrode is FAST)

Surface vs. Bulk Concentration Surface concentration of A CA f(x,t) ..at surface x=0, therefore surface concentration is CA (0,t) or (CA for short here) CA* is the symbol we will use for “bulk” concentration

Basic Rate Laws (1st order rxn) kf Ox + ne- Red kb NOTE Cox is really Cox(0,t), the surface concentration of “Ox”! Likewise Nox is the number of mols Of “Ox” on the surface Therefore v is in units of mols/(time*area)

Measured Current #mols Ox on surface i [=] Coulombs / s v [=] mol/(sec*cm2) F is the “molecular weight equivalent of electrons in Coulombs / mol

Mercury Drop Electrode Na+ + 2e-  Na(Hg) E° = - 2.71 V vs. SHE Dangerous! Solid Na is NOT stable, hence the VERY negative E° Reacts exothermically with moisture Na+ + 2e-  Na E° = - 2.71 V vs. SHE Na+ Hg

The Energy Barrier Gibbs Free Energy AB + C Reasons for Energy Barriers Reconfiguration of atomic position Repulsion forces Desolvation of ions in solution Even though the reaction is favorable according to THERMODYNAMICS, it must overcome an energy barrier. LARGER BARRIER = SLOWER REACTION

- + Energy increases as we bring the sodium ion closer to the Na(Hg) surface due to repulsion forces. Vice-versa also true for energy of amalgamated sodium atom + Changing the potential changes the energy of the participating electron

- + Na+ +e- Na+ +e- Na (Hg) Na (Hg) The Hg drop electrode is ANODICALLY polarized by ΔE = E - E° The energy of the electrode changes by F ΔE - Charge x Voltage = Energy Na+ +e- Na (Hg) +

- + How do the anodic and cathodic activation barriers Na+ +e- Na (Hg) How do the anodic and cathodic activation barriers change with overpotential (η = E-E°)?

Symmetry Factor, α The symmetry factor, α, describes what fraction of the overpotential goes towards changing the CATHODIC barrier. (1-α) , same for ANODIC barrier For multistep/multi-electron reactions, the symmetry factor is called the transfer coefficient

- Na+ +e- Na (Hg) +

R - (1-α)F Δ E αFΔE +

Cathodic barrier goes up by the same amount that the anodic barrier goes down Cathodic barrier goes up by less than the anodic barrier goes down Cathodic barrier goes up by more than the anodic barrier goes down

Defining the Rate Constant Substituting our expression for how ΔG changes with η we get We assume Af = Ab Where f = F/RT defining a standard rate constant

Butler-Volmer Equation Inserting our expression for the rate constants into our expression for total current We get: We now have our relationship for how the current at an electrode varies with overpotential!

At Equilibrium Ox + ne- Red Recall that at equilibrium, the kinetic kf Ox + ne- Red kb Recall that at equilibrium, the kinetic model and the thermodynamic model must match!!!

At Equilibrium Ox + ne- Red kf Ox + ne- Red kb but….. νf and νr are NOT ZERO, rather, they cancel!

Equilibrium At equilibrium, the surface concentration C(0,t) equals the bulk concentration in solution C* Which is just an exponential form of the Nernst equation Kinetic model breaks down into thermodynamic model at equilibrium

Exchange Current Density Although the NET current is zero, the anodic and cathodic currents still have equal but opposite values known as the exchange current density Raise both sides to -α Substitute into to get Note i0 is dependent only on kinetic parameters!

Current-Overpotential Equation Divide the Butler-Volmer Equation by i0 to get: Simplifies to : ..where f = F/RT

i0 Visualized

Effect of i0 on Polarization

Effect of Symmetry Factor on Polarization

Tafel Plot ..where f = F/RT at large η either the left or the right side becomes negligible.

What happens when we account for mass transport?