Analytical electrochemistry

Fall - 2010
Chapter 14: Fundamentals of
14-1 Basic concepts
Reducing agent:
Oxidizing agent:
Electric charge, q, is measured in coulombs (C), q = n*F (F is
Faraday constant)
Electric current: the quantity of charge flowing each second through
a circuit, I = electric charge/time
Electric potential, E (volts)
Work = E*q
The free Gibbs energy equals the maximum possible electrical work
that can be done by the reaction on its surroundings, work = -∆G;
∆G = - E*q = -nFE
Ohm’s law I = E/R
Power, work done per unit time (SI unit: J/s, known as watt)
• Exercise A: A mercury cell formerly used to power heart pacemakers
has the following reaction:
If the power required to operate the pacemaker is 0.010 0 W, how
many kilograms of HgO (FM 216.59) will be consumed in 365 days?
The cell voltage will be 1.35 V (because ????)
Use the equation:
P  power   E Volts   I  Ampere 
I = P/E = 0.010 0 W/1.35 V = 7.41 × 10−3 C/s
I = (7.41 × 10−3 C/s)/(9.649 × 104 C/mol)
= 7.68 × 10−8 mol e−/s
for 365 days: I = 2.42 mol e−, which is 1.21 mol HgO
The formula mass HgO = 216.6,
thus we need 1.21 x 216.6 g = 0.262 kg HgO
14.2 Galvanic cells
• Galvanic cell: (also called a voltaic cell) uses a spontaneous
chemical reaction to generate electricity.
• Batteries and fuel cells are galvanic cells that consume their
reactants to generate electricity. A battery has a static compartment
filled with reactants. In a fuel cell, fresh reactants flow past the
electrodes and products are continuously flushed from the cell.
• Half reaction: describes a process occurring at each electrode,
involving only oxidation or only reduction.
• Salt bridge: a conducting ionic medium in contact with two
electrolyte solutions. It maintains electroneutrality throughout the
• Anode: Electrode at
which oxidation occurs.
In electrophoresis, it is
the positively charged
• Cathode: Electrode at
which reduction occurs.
In electrophoresis, it is
the negatively charged
Line notation of an electrochemical cell
Line diagram for the above cell:
Line diagram for the above cell:
14-3 Standard potentials
• Standard reduction potential: E° The voltage that would be
measured when a hypothetical cell containing the desired halfreaction (with all species present at unit activity) is connected to a
standard hydrogen electrode anode.
E° defines the tendency for a reduction process to occur at an
• All ionic species present at a=1
(approximately 1 M).
• All gases are at 1 bar (approximately 1
• No metallic substance is indicated,
the potential is established on an inert
metallic electrode (ex. Pt).
We will write all half-reactions as
reductions. In this way, the voltage
of an electrochemical cell is the
cathode potential – anode potential.
Application of standard potentials
Problem: 14-13. (a) Cyanide ion causes E° for Fe(III) to decrease:
Fe3+ + e- ↔ Fe2+
E°= 0.771 V
Fe(CN)63+ + e- ↔ Fe(CN) 63+ E° = 0.356 V
Which ion, Fe3+, or Fe2+, is stabilized more by complexing with CN?
Solution: Because standard potential describes the tendency of a reduction
to occur.
As a result of complexation, the standard potential of ferric ion
decreases from 0.771 to 0.356 V. It means that the ion becomes more
difficult to be reduced. Therefore, Fe(III) is stabilized more.
If the potential values provided above are not standard, can one still reach
the same conclusion?
14-4 Nernst Equation
Problem 14-16: Write the Nernst equation for the following half-reaction and find
E when pH = 3.00 and PasH3 = 1.0 mbar.
As(s) + 3H+ + 3e- ↔ AsH3(g) Eo = -0.238 V
E = Eo – [RT/(nF)]*ln{(PasH3 /(aH+)3}
Since no temperature is provide, we can assume T = 25 oC, if so we will
Use equation 14-15 to carry out the calculation
E = E0 – (0.05916/3)*log {0.001/ /(aH+)3} = - 0.238 – (0.05916/3)*6 = -0.356 V.
A procedure for writing a net cell
reaction and finding its voltage
• Step 1: Write reduction half-reactions for both half-cells and find E°
for each in Appendix H. Multiply the half-reactions as necessary so
that they both contain the same number of electrons.
(When you multiply a reaction, you do not multiply E°, why? ).
• Step 2: Write a Nernst equation for the right half-cell, which is
attached to the positive terminal of the potentiometer. This is E+.
• Step 3 Write a Nernst equation for the left half-cell, which is attached
to the negative terminal of the potentiometer. This is E_.
• Step 4 Find the net cell voltage by subtraction: E = E+ - E_.
• Step 5 To write a balanced net cell reaction, subtract the left halfreaction from the right half-reaction. (Subtraction is equivalent to
reversing the left-half reaction and adding.)
Problem 14-18. A nickel-metal hydride rechargeable battery for laptop
computers is based on the following chemistry:
The anode material, MH, is a transition metal hydride or rare earth alloy hydride.
Explain why the voltage remains nearly constant during the entire discharge cycle.
• Solution:
The only substance we need to consider the effect of
activity is OH-. However, when you write the Nernst
equation for the cell reaction, OH- disappears from the
equation, therefore there is no change in potential during
the discharge process.
14-5 Eo and the equilibrium
• At the equilibrium point, the cell potential is 0! the
equilibrium constant can be calculated from rearranging
Nernst equation.
• Example: using standard potential to find the equilibrium
constant for the reaction Cu(s) + 2Fe3+ ↔ 2Fe2+ + Cu2+
• Solution: step 1: identify cathode and anode
step 2: find the standard potential for half reactions
step 3: calculation Eo for the cell reaction
Finding K for net reactions that are not redox
• The general practice is to arbitrarily design one hald reaction (for
cathode) and then use this one to subtract the cell reaction to obtain
the other half reaction (for the anode).
• Problem 14-28: Calculate Eo for the half reaction
Pd(OH)2(s) + 2e- ↔ Pd(s) + 2OHgiven that Ksp for Pd(OH)2 is 3x10-28 and Eo = 0.915 V for the
reaction Pd2+ + 2e- ↔ Pd(s)
• Solution:
Pd(OH)2(s) ↔ Pd2+ + 2OHKsp
Ksp = [Pd2+]([OH-])2
from Ksp one can get Eo for the above reaction (eq 14-23)
Eo = -0.814 V
Since the above reaction can be derived by using Pd(OH)2(s)
+ 2e- ↔ Pd(s) + 2OH- to subtract Pd2+ + 2e- ↔ Pd(s)
we have
-0.814 V = Eo - 0.915 V
Eo = 0.101 V
14-6 Cells as chemical probes
• It is important to distinguish two types of equilibrium in a galvanic
cell: equilibrium between the two half-cells and equilibrium within
each half cell.
A galvanic cell that can be
used to measure the formation
constant of Hg(EDTA)2Hg2+ + 2e- ↔ Hg(l)
Eo = 0.852 V
14-7 Biochemists use Eo’
• Formal potential: Potential of a half-reaction (relative to
a standard hydrogen electrode) when the formal
concentrations of reactants and products are unity. Any
other conditions (such as pH, ionic strength, and
concentrations of ligands) also must be specified.
Finding the formal potential
• Reduction potential of ascorbic
acid, showing its dependence
on pH.
(a) Graph of the function labeled
formal potential.
(b) Experimental polarographic
half-wave reduction potential
of ascorbic acid in a medium of
ionic strength = 0.2 M.
At high pH (>12), the half-wave
potential does not level off to a
slope of 0, as Equation 14-34
predicts. Instead, a hydrolysis
reaction of ascorbic acid
occurs and the chemistry is
more complex.
Example. Cadmium electrode immersed in a solution that is 0.015M in
Cd2+ .
Cd2+ + 2e  Cd(s) 0 = -0.403
ε = εo -
Cd 2+ 
ε = -0.403 log
  0.457 V
• Example: Calculate the potential for a platinum electrode
immersed in a solution prepared by saturating a 0.015 M
solution of KBr with Br2
Br2(l) + 2e  2BrE0 = 1.065 V
 2
Br 
E= 1.065 log
E = 1.065 -
log  0.0150   1.173V
Important notes
1. When you multiply a reaction by a coefficient, the potential of
such a reaction does not change!
2. When you subtract or add reactions, the Gibbs energy of those reactions
can always be directly added or subtracted, not their potentials!

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