• Consider the reduction half reaction: Mz+ + ze → M • The Nernst equation is E = Eө + (RT/zF) ln(a) • When.

• 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
jlim = zFJ = zFDc/ δ
• Using the Nernst-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.
Other applications related to
concentration overpotential
• 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
(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. 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
• The output is the slope of a curve like that obtained by linear-sweep
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 zerocurrent 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
step 3:
= E’ – E:
J = j0fη
• 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.

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