Consider the reduction half reaction: M z+ + ze → M The Nernst equation is E = E ө + (RT/zF) ln(a

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# Consider the reduction half reaction: M z+ + ze → M The Nernst equation is E = E ө + (RT/zF) ln(a - PowerPoint PPT Presentation

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)

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## Consider the reduction half reaction: M z+ + ze → M The Nernst equation is E = E ө + (RT/zF) ln(a

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Presentation Transcript
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 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
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
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.
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 ηon 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. 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.