• Consider the reduction half reaction: M z+ + ze → M • The Nernst equation is E = E ө + (RT/zF) ln(a) • When using a large excess of support electrolyte, the mean activity coefficients stays approximately constant, E = E ө + (RT/zF)ln( γ) + (RT/zF)ln(c) E = E o + (RT/zF)ln(c) • The ion concentration at OHP decreases to c’ due to the reaction, resulting E’ = E o + (RT/zF)ln(c’) • The concentration overpotential is

η

c

= E’ – E = (RT/zF)ln(c’/c)

(typo in the 8 th 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 (

z

F) 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.

j lim = zFJ = zFDc/ δ • Using the Nernst-Einstein equation (D = RTλ/(zF) 2 ), j lim = cRT λ/(zFδ) where λ is ionic conductivity

• Example 25.3: Estimate the limiting current density at 298K for an electrode in a 0.10M Cu 2+ (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 Cu 2+ : λ = 107 S cm 2 δ = 0.3 mm mol -1 employing the following equation: j lim = cRT λ/(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.

Other applications related to concentration overpotential

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

j



zcFD

 ( 1 

e zf



c

)

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. j lim = zFDc/ δ, maximum current is proportional to the molar concentration of the species. This is why one can the 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 ClC 6 H 4 CN in acidic solution; after reduction to ClC 6 H 4 CN -1 , the radical anion may form C6H5CN irreversibly. ClC 6 H 4 CN + e ↔ ClC 6 H 4 CN -1 ClC 6 H 4 CN -1 + H + + e → C 6 H 5 CN + Cl C 6 H 5 CN + e ↔ C 6 H 5 CN -

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 (IR s ).

• 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’ = ΔФ R ΔФ L • As ΔФ R = E R + η R ; ΔФ L = E L + η L • E’ = E + η R η L

• • To emphasize that a working cell has a lower potential than a zero current cell, we write E’ = E - |η R |- | η L | One should also subtract the ohmic potential difference IRs E’ = E - |η thermal dissipation.

R |- | η L | - IRs • The omhic term is a contribution to the cell’s irreversibility- it is a • E’ = E – IRs – 4RT ln(I/9Aj))/F j = (j oL j oR ) 1/2 where j oL and j oR 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 (8 th edition) or 29.59 )7 th 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 o C and 20-40 atm.

• The cathode reaction is O 2 (g) + 2H 2 O(l) + 4e → 4OH • The anode reaction is oxidation (aq) Eo = 0.40V

H 2 (g) + 2OH (aq) → 2H 2 O(l) + 2e • The overall reaction 2H 2 (g) + O 2 (g) → 2H 2 O(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 → H 2 (g) E o (b) 4H + (aq) + O 2 = 0 + 4e → 2H 2 O(l) E o = 1.23 V In basic solution: (c) 2H 2 O(l) + O 2 (g) + 4e → 4OH (aq) E o • Consider the other half reaction: = 0.40 V Fe 2+ (aq) + 2e → Fe(s) E o = -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)|H 2 (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

in E’;

η = E’ – E:

on the basis that there is no change step 3: J = j 0

f

η

• 25.23b Suppose that the electrode potential is set at 0.50 V. calculate the current density for the ratio of activities α(Cr

3+

)/ α(Cr

2+

) in the range 0.1 to 10.0 and at 25

o

C.