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Chapter 14
Electrochemistry
Basic Concepts
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Chemical Reaction that involves the transfer
of electrons. A Redox reaction.
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Loss of electrons – oxidation
Gain of electrons – reduction
Oxidizing agent. A species that takes
electrons.
Reducing agent. A species that gives
electrons.
Basics
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Na(s) + H+ -> Na+ + H2(g)
Sodium is a reducing agent
Hydrogen ion is the oxidizing agent.
Basics
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We are donating and gaining electrons. If we could
use these electrons perhaps we could do some useful
work.
If we can make the electron travel in an electrical
circuit then the amount of current can be measured.
Current is related to reaction rate or amount of
reaction
Potential is related to free energy change of the
reaction.
Electron Charge
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q used to denote. Unit is Coulombs (C)
Charge on a single electron is
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1.602x10-19 C which will allow us to determine the
charge on a mole of electrons.
1.602x10-19 C * 6.022x1023 mol-1) = 96490 C mol-1
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This is called the Faraday Constant
q = nF
n is the number of moles
Current
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Charge flowing through a circuit
One ampere, the charge of one coulomb per
second flowing past a given point.
Electrodes
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The interface
between a
solution and an
electrical circuit.
Can be actively
involved or just
serve as a
source or sink
for electons.
Electrical Potential
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Work required when moving and electric
charge from one point to another.
Electrical potential (E) is measured in Volts
(V).
Work is a measure of energy, measured in
joules (J).
Work = E *
q
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Joules
volts
coulombs
Free Energy
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Maximum amount of work that can be done
on the surroundings is equal to the Gibbs
free energy change.
then DG = -work = -Eq
Or DG = -nFE
Ohm’s Law
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Current is proportional to the potential and
and inversely proportional to the resistance.
I = E/R
Electric Circuit
Power
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Work done per unit time. Unit is the J/s
which is know as the watt (W).
P = work/sec = Eq/sec = E(q/sec) = EI
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P = EI = I2R = E2/R
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Galvanic Cells
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Spontaneous chemical reaction used to
generate electricity.
An example might be
Voltmeter
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A device to measure electrical potential. When
electrons tend to flow into the negative terminal then
a positive voltage is measured.
In this cell
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2 AgCl (s) + 2 e- = 2 Ag + 2 Cl- (aq) Red
Cd (s) + = Cd2+ + 2 eOxidation
Cd (s) + 2 AgCl (s) = Cd2+ + 2 Cl- Net
For this reaction we have a DG of -150 kJ/mole per mole of
Cd oxidized.
Potential of this System
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DG = -150 kJ/mole then we have
E = - DG/nF = -150 x 103 J / (2 mol)(9.649x104 C/mol)
E = + 0.777 J/C = +0.777 V
Cathode/anode
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Cathode
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Anode electrode where oxidation occur
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electrode where reduction occurs
Put both terms in alphabetical order to remember
Salt Bridge
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Any bridge in upstate New York in the winter.
Used to isolate the half cells so the work can
be forced out into an external circuit.
The following cell has a problem.
What is it?
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The silver ions in solution can go directly to
the cadmium electrode surface and be
reduced there.
We need to put in a barrier to rapid ionic
transfer.
What about this cell
Isn’t this cute
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Chemistry paper dolls?
Line Notation - Instead of Having to
Draw the Cells
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| phase boundary
For First Cell
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|| salt bridge
Cd(s) | CdCl2(aq) | AgCl(s) | Ag(s)
For Second Cell
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Cd(s) | Cd(NO3)2(aq) || AgNO3(aq) | Ag(s)
A Word of Connectors (Two common in
USA)
Standard Potential
Eo
The energy to a half cell at standard conditions
(1 M and 25 C)
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Let us look at the reduction of silver ion.
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Ag+ + e- = Ag(s)
We will compare this to a fixed reference.
That is the SHE or NHE Standard or Normal
Hydrogen Electrode.
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H+ (aq, A=1) + e- = ½ H2 (g, A = 1)
SHE - All other redox couples are compared to
this half cell. It is assigned a value of 0.000 V
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In our cell the left side electrode (Pt) is
attached to the negative terminal.
(Reference)
Value of E are collected into Tables
(Appendix H)
Nernst Equation
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For the half reaction
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aA + ne- = bB
b
B
a
A
RT A
EE 
ln
nF A
o
Eo = is the standard Potential
R = gas constant (8.314472 (V*C)/(k*mol)
T = Temp (K)
N = # of electrons in the half reaction
F = Faraday
A = Activity
We will often lump the constants and
assume 25 C
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Nernst equation (25 C and converting to
log10
b
0
.
05916
V
A
E  Eo 
log Ba
n
AA
Complete Reaction
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E = E + - E
Steps
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for full cell
Write both half cells as reductions, make electrons equal
Half cell connected to positive terminal is E+
Other half cell is ENet voltage is from the above equation
Balance equation (reversing the left half reaction and
adding to other half cell)
E > 0 spontaneous as written
E < 0 spontaneous in reverse
Eo and K
0.05916
E 
log K
n
o
K  10
nE o
0.05916
Cells as Chemical Probes
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Equilibria between the half cells
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Equilibria within each half cell
A Probe Cell
Probe Cell
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Right side:
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We have our Ksp equilibrium
The electrochemical reaction under this is
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AgCl(s) + e- = Ag(s) + Cl- (aq, 0.10 M)
Eo = 0.222 v
Left side:
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We have our Ka for the weak acid.
The electrochemical reaction
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2 H+(aq) + e- = H2 (g, 1.00 bar)
E = 0.00, but H+ is not fixed at 1 M so E varies with H+
Eo’
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Formal Potential
Since so many redox couples exist in the
body and many have H+ we modify the
potential that we use to pH 7. (A little more
reasonable than 1 M acid.
Homework
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14- 4
13, 14, 15 and 27