Electrochemistry Terminology
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Oxidizing agent -The substance that is
reduced is the oxidizing agent
Reducing agent - The substance that is
oxidized is the reducing agent
LEO says GER
Electrochemistry Terminology
 Anode -The electrode where oxidation
occurs
 Cathode - The electrode where
reduction occurs
Leo is a
Reduction
at the
Cathode
Step 1:
Step 2:
Step 3:
2ee-
Step 4:
Multiply each half-reaction by a factor
so that the reducing agent supplies as many
electrons as the oxidizing agent requires.
Reducing agent
Cu ---> Cu2+ + 2eOxidizing agent
2 Ag+ + 2 e- ---> 2 Ag
Step 5:
Add half-reactions to give the overall
equation.
Cu + 2 Ag+ ---> Cu2+ + 2Ag
The equation is now balanced for both
charge and mass.
Balance the following in acid solution—
VO2+ + Zn ---> VO2+ + Zn2+
Step 1:
Write the half-reactions
Ox
Zn ---> Zn2+
Red
VO2+ ---> VO2+
Step 2:
Balance each half-reaction for mass.
Ox
Zn ---> Zn2+
Red
2 H+ + VO2+ ---> VO2+ + H2O
Add H2O on O-deficient side and add H+
on other side for H-balance.
Step 3:
Balance half-reactions for charge.
Ox
Zn ---> Zn2+ + 2eRed
e- + 2 H+ + VO2+ --> VO2+ + H2O
Step 4:
Multiply by an appropriate factor.
Ox
Zn ---> Zn2+ + 2eRed 2e- + 4 H+ + 2 VO2+ ---> 2 VO2+ + 2 H2O
Step 5:
Add balanced half-reactions
Zn + 4 H+ + 2 VO2+ ---> Zn2+ + 2 VO2+ + 2 H2O
Draw it!
A great example of a thermodynamically spontaneous reaction is the
thermite reaction. Here, iron oxide (Fe2O3 = rust) and aluminum metal
powder undergo a redox (reduction-oxidation) reaction to form iron
metal and aluminum oxide (Al2O3 = alumina):
Fe2O3(s)
Fe = +3
+
Al(s) ↔ Al2O3(s)
Al = 0
Al = +3
+
Fe(l)
Fe = 0
1)
2)
3)
4)
5)
12
5
2
30
10
1.
2.
3.
4.
5.
Zn → Zn2+ + 2e–
Zn2+ + 2e– → Zn
Zn2+ + Cu → Zn + Cu2+
Zn + Cu2+ → Zn2+ + Cu
two of these
1.
2.
3.
4.
5.
S(s)
NO3–(ag)
Cr2O72–(aq)
I– (aq)
MnO4– (aq)



An apparatus that allows a
redox reaction to occur by
transferring electrons
through an external
connector.
Product favored reaction --> voltaic or galvanic cell ---> electric current
Reactant favored reaction --> electrolytic cell ---> electric
current used to cause
chemical change.
Batteries are voltaic cells
Basic Concepts
of Electrochemical Cells
Anode
Cathode
potential
Eo
+1.10 V
STANDARD CELL POTENTIAL,
Zn(s) ---> Zn2+(aq) + 2eCu2+(aq) + 2e- ---> Cu(s)
-------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Measuring
Standard
Electrode
Potential
Potentials are measured against a hydrogen ion
reduction reaction, which is arbitrarily
assigned a potential of zero volts.
Table of
Reduction
Potentials
Measured
against
the
Standard
Hydrogen
Electrode
oxidizing
ability of ion
Eo (V)
Cu2+ + 2e-
Cu
+0.34
2 H+ + 2e-
H2
0.00
Zn2+ + 2e-
Zn
-0.76
To determine an oxidation from a
reduction table, just take the opposite
sign of the reduction!
reducing ability
of element
+
Anode,
negative,
source of
electrons
Cathode,
positive,
sink for
electrons
Zn(s) ---> Zn2+(aq) + 2eEo = +0.76 V
Cu2+(aq) + 2e- ---> Cu(s)
Eo = +0.34 V
--------------------------------------------------------------Cu2+(aq) + Zn(s) ---> Zn2+(aq) + Cu(s)
Eo = +1.10 V
All ingredients are present. Which way does reaction proceed?
e-
All ingredients are present. Which way does reaction proceed?
Assume I- ion can reduce water.
2 H2O + 2e- ---> H2 + 2 OH2 I- ---> I2 + 2e------------------------------------------------2 I- + 2 H2O --> I2 + 2 OH- + H2
Cathode
Anode
-1.363 V
Negative E˚ means rxn. occurs in opposite
direction (the connection is backwards or you
are recharging the battery)
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
Zn + Cu2+  Zn2+ + Cu
E = +0.76V
E = +0.34V
E0 = + 1.10 V
Line
Notation
An abbreviated
representation of
an electrochemical
cell
Zn(s) | Zn2+(aq)
(1.0M)
Anode
Anode
|
material
solution
|| Cu2+(aq)
||
Cathode
solution
Line notation is cool, just like AC
(1.0M)
|
| Cu(s)
Cathode
material
Zn(s) | Zn2+(aq) (1.0M) || H+(aq) (1.0M) |H2(g)
(1.00 atm)
| Pt(s)
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
Day 3
(dahditdadahditahh…Charge!)
The Nernst Equation
Standard potentials assume a concentration of 1.0 M.
The Nernst equation allows us to calculate potential
when the two cells are not 1.0 M.
RT
EE 
ln( Q)
nF
R = 8.31 J/(molK)
0
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
Try Problem 62
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 x10
37
Let’s Do Problems:
26a, 28a, 32a, 38a
Apply to Problem 70
???
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 n  2
0
(0.10)
Q
(1.0)
0.0591
0.10
E  0.0 
log(
)  0.030 Volts
2
1.0
Nearly same reactions
as in common dry
cell, but under basic
conditions.
Anode (-): Zn + 2 OH- ---> ZnO + H2O + 2eCathode (+): 2 MnO2 + H2O + 2e- ---> Mn2O3 + 2
OH-
Anode:
Cathode:
Anode (-)
Cathode (+)
Anode (-)
Cd + 2 OH- ---> Cd(OH)2 + 2eCathode (+)
NiO(OH) + H2O + e- ---> Ni(OH)2 + OH-
The positive electrode is made of Lithium cobalt oxide, or
LiCoO2
The negative electrode is made of carbon.
The Electrochemical Corrosion of Iro
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
Calculating plating
Step 1 – convert current and time to quantity of charge in
coulombs
a. Amps x time = total charge transferred in coulombs
(Coulomb/sec) x sec = coulombs
Step 2 – convert quantity of charge in coulombs to moles of
electrons
coulombs /(96,485 coulombs/mol e-) = mol eStep 3 – Convert moles of electrons to moles of substance
mol e- x (mole substance/mol e-) = mol substance
Step 4 – Convert moles of substance to grams of substance
mol substance x formula mass of substance = mass of
substance
Suppose that in starting a car on a cold
morning a current of 125 amperes is drawn
for 15.0 seconds from a cell of the type
described above. How many grams of Pb
would be consumed? (The atomic weight of
Pb is 207.19.) 2+
Pb
125 C 15 sec
1 sec
+ 2e  Pb
1 mol e-
1 mol Pb
207.19 g
- 1 mol Pb
2
mol
e
96 485 C
2.01 g Pb
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
Do 79!!!
Do 81!!!