Electrochemistry
Electrochemistry Terminology #1
Oxidation – A process in which an
element attains a more positive
oxidation state
Na(s)  Na+ + eReduction – A process in which an
element attains a more negative
oxidation state
Cl2 + 2e-  2Cl-
Electrochemistry Terminology #2
An old memory device for oxidation
and reduction goes like this…
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction
Electrochemistry Terminology #3
 Oxidizing agent
The substance that is reduced is the
oxidizing agent
 Reducing agent
The substance that is oxidized is the
reducing agent
Electrochemistry Terminology #4
 Anode
The electrode where
oxidation occurs
 Cathode
The electrode where
reduction occurs
Memory device:
Reduction
at the
Cathode
Table of
Reduction
Potentials
Measured
against
the
Standard
Hydrogen
Electrode
Measuring
Standard
Electrode
Potential
Potentials are measured against a hydrogen ion
reduction reaction, which is arbitrarily
assigned a potential of zero volts.
Galvanic (Electrochemical) Cells
Spontaneous redox
processes have:
A positive
cell potential, E0
A negative free
energy change, (-G)
Zn - Cu
Galvanic
Cell
From a table
of reduction
potentials:
Zn2+ + 2e-  Zn
Cu2+ + 2e-  Cu
E = -0.76V
E = +0.34V
Zn - Cu
Galvanic
Cell
The less positive,
or more negative
reduction potential
becomes the
oxidation…
Zn  Zn2+ + 2eCu2+ + 2e-  Cu
E = +0.76V
E = +0.34V
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
Line
Notation
An abbreviated
representation of
an electrochemical
cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Anode
Anode
|
material
solution
||
Cathode
solution
|
Cathode
material
Calculating G0 for a Cell
G0 = -nFE0
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
coulombs
Joules
G   ( 2 mol e )(96 485
)(1.10
)

mol e
Coulomb
0

G   212267 Joules   212 kJ
0
The Nernst Equation
Standard potentials assume a concentration of
1 M. The Nernst equation allows us to calculate
potential when the two cells are not 1.0 M.
RT
EE 
ln(Q)
nF
0
R = 8.31 J/(molK)
T = Temperature in K
n = moles of electrons in balanced redox equation
F = Faraday constant = 96,485 coulombs/mol e-
Nernst Equation Simplified
At 25 C (298 K) the Nernst Equation is
simplified this way:
0.0591
EE 
log(Q)
n
0
Equilibrium Constants and Cell Potential
At equilibrium, forward and reverse
reactions occur at equal rates, therefore:
1. The battery is “dead”
2. The cell potential, E, is zero volts
Modifying the Nernst Equation (at 25 C):
0.0591
0 volts  E 
log( K )
n
0
Calculating an Equilibrium Constant from
a Cell Potential
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
0.0591
0 volts 1.10 
log( K )
2
(1.10)(2)
 log( K )
0.0591
37.2  log( K )
10
37.2
 K 1.58 x 10
37
???
Concentration
Cell
Both sides have
the same
components but
at different
concentrations.
Step 1: Determine which side undergoes
oxidation, and which side undergoes reduction.
???
Anode
Concentration
Cell
Cathode
Both sides have
the same
components but
at different
concentrations.
The 1.0 M Zn2+ must decrease in concentration, and
the 0.10 M Zn2+ must increase in concentration
Zn2+ (1.0M) + 2e-  Zn
(reduction)
Zn  Zn2+ (0.10M) + 2eZn2+ (1.0M)  Zn2+ (0.10M)
(oxidation)
???
Concentration Cell
Anode
Cathode
Concentration
Cell
Both sides have
the same
components but
at different
concentrations.
Step 2: Calculate cell potential using the Nernst
Equation (assuming 25 C).
Zn2+ (1.0M)  Zn2+ (0.10M)
0.0591
EE 
log(Q)
n
0
Nernst Calculations
Zn2+ (1.0M)  Zn2+ (0.10M)
0.0591
EE 
log( Q)
n
0
E  0.0 Volts
0
n2
(0.10)
Q
(1.0)
0.0591
0.10
E  0.0 
log(
)  0.030 Volts
2
1.0
Electrolytic
Processes
Electrolytic
processes are
NOT spontaneous.
They have:
A negative
cell potential, (-E0)
A positive free
energy change, (+G)
Electrolysis of
Water
In acidic solution
Anode rxn:

2 H 2O  O2  4 H  4e

Cathode rxn: 4 H 2O  4e  2 H 2  4OH
2 H 2O  2 H 2  O2


-1.23 V
-0.83 V
-2.06 V
Electroplating of
Silver
Anode reaction:
Ag  Ag+ + eCathode reaction:
Ag+ + e-  Ag
Electroplating requirements:
1. Solution of the plating metal
2. Anode made of the plating metal
3. Cathode with the object to be plated
4. Source of current
Solving an Electroplating Problem
Q: How many seconds will it take to plate out
5.0 grams of silver from a solution of AgNO3
using a 20.0 Ampere current?
Ag+ + e-  Ag
5.0 g
1 mol Ag 1 mol e-
96 485 C 1 s
20.0 C
1
mol
e
107.87 g 1 mol Ag
= 2.2 x 102 s