The high overpotential limit The overpotential η is large, but could be positive or negative ! When η is large and positive jc = j0e-afη = j0/eafη becomes very small in comparison to ja Therefore j ≈ ja = j0e(1-a)fη ln(j) = ln(j0e(1-a)fη ) = ln(j0) + (1-a)fη When η is large but negative ja is much smaller than jc then j ≈ - jc = - j0e-afη ln(-j) = ln(j0e-afη ) = ln(j0) – afη Tafel plot: the plot of logarithm of the current density against the over potential.

Applications of a Tafel plot The following data are the anodic current through a platinum electrode of area 2.0 cm2 in contact with an Fe3+, Fe2+ aqueous solution at 298K. Calculate the exchange current density and the transfer coefficient for the process. η/mV 50 100 150 200 250 I/mA 8.8 25 58 131 298 Solution: calculate j0 and α Note that I needs to be converted to J

Self-test 25.7: Repeat the analysis using the following cathodic current data: η/mV -50 -100 -150 -200 -250 I/mA -0.3 -1.5 -6.4 -27.61 -118.6 In general exchange currents are large when the redox process involves no bond breaking or if only weak bonds are broken. Exchange currents are generally small when more than one electron needs to be transferred, or multiple or strong bonds are broken.

The general arrangement for electrochemical rate measurement

25.10 Voltammetry Voltammetry: the current is monitored as the potential of the electrode is changed. Chronopotentiometry: the potential is monitored as the current density is changed. Voltammetry may also be used to identify species and determine their concentration in solution. Non-polarizable electrode: their potential only slightly changes when a current passes through them. Such as calomel and H2/Pt electrodes Polarizable electrodes: those with strongly current-dependent potentials. A criterion for low polarizability is high exchange current density due to j = j0fη

Concentration polarization At low current density, the conversion of the electroactive species is negligible. At high current density the consumption of electroactive species close to the electrode results in a concentration gradient. Concentration polarization: The consumption of electroactive species close to the electrode results in a concentration gradient and diffusion of the species towards the electrode from the bulk may become rate-determining. Therefore, a large overpotential is needed to produce a given current. Polarization overpotential: ηc

(typo in the 8th edition) Consider the reduction half reaction: Mz+ + ze → M The Nernst equation is E = Eө + (RT/zF) ln(a) When using a large excess of support electrolyte, the mean activity coefficients stays approximately constant, E = Eө + (RT/zF)ln(γ) + (RT/zF)ln(c) E = Eo + (RT/zF)ln(c) The ion concentration at OHP decreases to c’ due to the reaction, resulting E’ = Eo + (RT/zF)ln(c’) The concentration overpotential is ηc = E’ – E = (RT/zF)ln(c’/c) (typo in the 8th edition)

The thickness of the Nernst diffusion layer (illustrated in previous slide) is typically 0.1 mm, and depends strongly on the condition of hydrodynamic flow due to such as stirring or convective effects. The Nernst diffusion layer is different from the electric double layer, which is typically less than 1 nm. The concentration gradient through the Nernst diffusion layer is dc/dx = (c’ – c)/δ This concentration gradient gives rise to a flux of ions towards the electrode J = - D(dc/dx) Therefore, the particle flux toward the electrode is J = D (c – c’)/ δ

The cathodic current density towards the electrode is the product of the particle flux and the charge transferred per mole of ions (zF) j = zFJ = zFD(c – c’)/ δ The maximum rate of diffusion across the Nernst layer is when c’ = 0 at which the concentration gradient is the steepest. jlim = zFJ = zFDc/ δ Using the Nerst-Einstein equation (D = RTλ/(zF)2), jlim = cRTλ/(zFδ) where λ is ionic conductivity

Example 25.3: Estimate the limiting current density at 298K for an electrode in a 0.10M Cu2+(aq) unstirred solution in which the thickness of the diffusion layer is about 0.3mm. Solution: one needs to know the following information molar conductivity of Cu2+: λ = 107 S cm2 mol-1 δ = 0.3 mm employing the following equation: jlim = cRTλ/(zFδ) Self-test 25.8 Evaluate the limiting current density for an Ag(s)/Ag+(aq) electrode in 0.010 mol dm-3 Ag+(aq) at 298K. Take δ = 0.03mm.

ηc = (RT/zF)ln(c’/c) = (RT/zF)ln(1 - jδ/(zFDc)) From j = zFD(c – c’)/ δ, one gets c’ = c - jδ/zFD ηc = (RT/zF)ln(c’/c) = (RT/zF)ln(1 - jδ/(zFDc)) Or j = (zFDc)(1 – ezfη)/δ

Experimental techniques Linear-sweep voltammetry At low potential value, the cathodic current is due to the migration of ions in the solution. The cathodic current grows as the potential reaches the reduction potential of the reducible species. Based on the eqn. jlim = zFDc/ δ, the maximum current is proportional to the molar concentration of the species. This is why one can determine c from this technique

Differential pulse voltammetry The current is monitored before and after a pulse of potential is applied. The output is the slope of a curve like that obtained by linear-sweep voltammetry

Cyclic voltammetry Determine the redox potential Reflect the underlying kinetics

CV spectrum, the sweeping rate and the underlying kinetics

Self-test 25.9 Suggest an interpretation of the cyclic voltammogram shown in the figure. The electroactive material is ClC6H4CN in acidic solution; after reduction to ClC6H4CN-1, the radical anion may form C6H5CN irreversibly. ClC6H4CN + e ↔ ClC6H4CN-1 ClC6H4CN-1 + H+ + e → C6H5CN + Cl- C6H5CN + e ↔ C6H5CN-

25.11 Electrolysis Cell overpotential: the sum of the overpotentials at the two electrodes and the ohmic drop due to the current through the electrolyte (IRs). Electrolysis: To induce current to flow through an electrochemical cell and force a non-spontaneous cell reaction to occur. It requires that the applied potential difference exceed the zero-current potential by at least the cell overpotential. Estimating the relative rates of electrolysis.

Working galvanic cells In working galvanic cells, the overpotential leads to a smaller potential than under zero-current conditions. The cell potential decreases as current is generated because it is then no longer working reversibly. Consider the cell M|M+(aq)||M’+(aq)|M’ and ignore complications from liquid junctions. The potential of the cell E’ = ΔФR - ΔФL As ΔФR = ER + ηR ; ΔФL = EL + ηL E’ = E + ηR - ηL

To emphasize that a working cell has a lower potential than a zero-current cell, we write E’ = E - |ηR |- |ηL| One should also subtract the ohmic potential difference IRs E’ = E - |ηR |- |ηL| - IRs The omhic term is a contribution to the cell’s irreversibility- it is a thermal dissipation. E’ = E – IRs – 4RT ln(I/9Aj))/F j = (joLjoR)1/2 where joL and joR are the exchange current densities for the two electrodes (for single electron transfer and high overpotential) The concentration overpotential also reduces the cell potential see 25.63 (8th edition) or 29.59 )7th edition) The full expression for the cell potential when a current I is being drawn: see eqn 25.64 a or 29.60

The dependence of the potential of a working cell on the current density being drawn (blue line) and the corresponding power output (IE)

Fuel Cells Reactants are supplied from outside. In hydrogen/oxygen cell, the electrolyte used is concentrated aqueous potassium hydroxide maintained at 200 oC and 20-40 atm. The cathode reaction is O2(g) + 2H2O(l) + 4e- → 4OH-(aq) Eo = 0.40V The anode reaction is oxidation H2(g) + 2OH-(aq) → 2H2O(l) + 2e- The overall reaction 2H2(g) + O2(g) → 2H2O(l) E = 1.23V The advantage of the hydrogen/oxygen system is the large exchange current density of the hydrogen reaction, but the oxygen reaction has a small exchange current density.

25.13 Corrosion

Consider the following half reactions: In acidic environment: (a) 2H+(aq) + 2e → H2(g) Eo = 0 (b) 4H+(aq) + O2 + 4e → 2H2O(l) Eo = 1.23 V In basic solution: (c) 2H2O(l) + O2(g) + 4e → 4OH-(aq) Eo = 0.40 V Consider the other half reaction: Fe2+(aq) + 2e → Fe(s) Eo = -0.44V The potential difference suggests that iron can be oxidized under the above three conditions. The thermodynamic discussion only indicates the tendency. The kinetic process shall also be examined.

Self-test: The exchange current density of a Pt(s)|H2(g)|H+(aq) electrode at 298K is 0.79 mAcm-2. Calculate the current density when the over potential is +5.0mV. What would be the current at pH = 2.0, the other conditions being the same? Solution: Step1: using Nernst equation to calculate E, Step 2: calculate η on the basis that there is no change in E’; η = E’ – E: step 3: J = j0fη

25. 23b Suppose that the electrode potential is set at 0. 50 V 25.23b Suppose that the electrode potential is set at 0.50 V. calculate the current density for the ratio of activities α(Cr3+)/α(Cr2+) in the range 0.1 to 10.0 and at 25 oC.