Experimental techniques Linear-sweep voltammetry At low potential value, the cathodic current is due...

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. 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.