CHM 1045: General Chemistry and
Qualitative Analysis
Note that the 1st part of this lesson is
a review of CHM 1045 Redox Rx. New
Unit begins in slide #32
Redox Equations
and Stoichiometry
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Chapter # 11-4 to 11-6
•Module 7
Electrochemistry
Redox Equations & Oxidation Numbers
In redox reactions, electrons are transferred *
from one species to another.
In order to keep
track of what loses
electrons and what
gains them, we
assign oxidation
numbers.
Electrochemistry
Oxidation and Reduction
*
• A species is oxidized when it loses electrons.
 Here, zinc loses two electrons to go from neutral
zinc metal to the Zn2+ ion.
Electrochemistry
Oxidation and Reduction
*
• A species is reduced when it gains electrons.
 Here, each of the H+ gains an electron and they
combine to form H2.
Electrochemistry
Oxidation and Reduction
• What is reduced is the oxidizing agent.
 H+ oxidizes Zn by taking electrons from it.
• What is oxidized is the reducing agent.
 Zn reduces H+ by giving it electrons.
Electrochemistry
Assigning Oxidation Numbers
1. Elements in their elemental form have
an oxidation number of 0.
2. The oxidation number of a monatomic
ion is the same as its charge.
Electrochemistry
Assigning Oxidation Numbers
3. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
 Oxygen has an oxidation number of −2
(except in the peroxide ion in which it has
an oxidation number of −1).
 Hydrogen is −1 when bonded to a metal
(a metal hydride, NaH) and +1 when
bonded to a nonmetal, H2O.
Electrochemistry
Assigning Oxidation Numbers
4. Nonmetals tend to have negative
oxidation numbers, although some are
positive in certain compounds or ions.
 Fluorine always has an oxidation number
of −1.
 The other halogens have an oxidation
number of −1 when they are negative;
they can have positive oxidation
numbers, however, most notably in
oxyanions.
Electrochemistry
Assigning Oxidation Numbers
4. The sum of the oxidation numbers in a
neutral compound is 0.
NaHCO3
(+1)+ (+1)+ (x) + 3(-2) = 0
x = +4
Electrochemistry
Assigning Oxidation Numbers
5. The sum of the oxidation numbers in a
polyatomic ion is the charge on the ion.
PO43(x) + 4(-2) = -3
x = +5
Electrochemistry
Balancing Oxidation-Reduction
Equations
The half-reaction method: This involves treating
(on paper only) the oxidation and reduction as two
separate processes, balancing these half reactions,
and then combining them to attain the balanced
equation for the overall reaction.
Electrochemistry
Half-Reaction Method
1. Assign oxidation numbers to determine what
is oxidized and what is reduced.
2. Write the oxidation and reduction halfreactions.
 Fe 3+ + Cr 3+
Fe 2+ + Cr2O722(x)+7(-2)=-2
x = +6
Oxidation: Fe 2+  Fe 3+
Reduction: Cr2O7
2-
 Cr
3+
Electrochemistry
Half-Reaction Method
Fe 2+ + Cr2O72-
 Fe 3+ + Cr 3+
In acidic medium:
3.
Balance each half-reaction.
a. Balance elements other than H and O.
b. Balance O by adding H2O.
c. Balance H by adding H+.
d. Balance charge by adding electrons.
Oxidation: Fe 2+  Fe 3+ + 1e6+
Reduction: Cr2O72-  2 Cr3+
6 e- + 14 H+  + 7 H2O
Electrochemistry
Cr2O72- + 14 H+ + 6 e-  2Cr3+ + 7 H2O
Half-Reaction Method
In acidic medium:
4. Multiply the half-reactions by integers
so that the electrons gained and lost
are the same.
2+
3+
6(Fe
 Fe + 1e-)
6Fe
 6 Fe + 6eCr2O72- + 14 H+ + 6 e-  2Cr3+ + 7 H2O
2+
3+
Electrochemistry
6Fe 2+ + Cr2O72- + 14H+  6Fe 3+ + 2Cr 3+ + 7H
20
Half-Reaction Method
5. Add the half-reactions, subtracting
things that appear on both sides.
6. Make sure the equation is balanced
according to mass.
7. Make sure the equation is balanced
according to charge.
6Fe 2+ + Cr2O72- + 14H+  6Fe 3+ + 2Cr 3+ + 7H20
Electrochemistry
Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + COElectrochemistry
2(aq)
Half-Reaction Method
First, we assign oxidation numbers.
+7
+3
+2
+4
MnO4− + C2O42-  Mn2+ + CO2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
Electrochemistry
Oxidation Half-Reaction
C2O42−  CO2
To balance the carbon, we add a
coefficient of 2:
C2O42−  2 CO2
Electrochemistry
Oxidation Half-Reaction
C2O42−  2 CO2
The oxygen is now balanced as well.
To balance the charge, we must add 2
electrons to the right side.
C2O42−  2 CO2 + 2 e−
Electrochemistry
Reduction Half-Reaction
MnO4−  Mn2+
The manganese is balanced; to balance
the oxygen, we must add 4 waters to
the right side.
MnO4−  Mn2+ + 4 H2O
Electrochemistry
Reduction Half-Reaction
MnO4−  Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+
to the left side.
8 H+ + MnO4−  Mn2+ + 4 H2O
Electrochemistry
Reduction Half-Reaction
8 H+ + MnO4−  Mn2+ + 4 H2O
To balance the charge, we add 5 e− to
the left side.
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
Electrochemistry
Combining the Half-Reactions
Now we evaluate the two half-reactions
together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O
To attain the same number of electrons
on each side, we will multiply the first
reaction by 5 and the second by 2. Electrochemistry
Combining the Half-Reactions
5 C2O42−  10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
Electrochemistry
Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
Electrochemistry
Balancing in Basic Solution
• If a reaction occurs in basic solution, one can
balance it as if it occurred in acid.
• Once the equation is balanced, add OH− to
each side to “neutralize” the H+ in the
equation and create water in its place.
• If this produces water on both sides, you might
have to subtract water from each side.
Electrochemistry
Balancing in Basic Solution
•
balance it as if it occurred in acid:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
• add OH− to each side to “neutralize” the H+ in the equation and create
water in its place.
+16 OH= 16 H2O
- 8 H2O

+ 16 OH-
• subtract water from each side.

- 8 H2O
8 H2O + 2 MnO4− + 5 C2O42− 
Electrochemistry
2+
2 Mn + 16 OH + 10 CO2
Redox Stoichiometry
Consider the reaction between MnO4− and C2O42− :
MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + COElectrochemistry
2(aq)
purple
clear
clear
clear
Indicators in Redox Titrations
Titrant = Cr2O7-2
6Fe2  Cr2O72  14H   6Fe3  2Cr 3 
(clear)
(clear)
(clear)
(clear) (light green)
7H 2O
(clear)
The indicator used diphenylamine sulfonate indicator (DPAS Ind)
It has a reduced form which is colorless in solution and an
oxidized form is purple in solution.
Excess Cr2O72- + DPAS Ind (red)
(clear)
(clear)
2+
Fe
 DPAS Ind (ox) + Cr+3
(purple)
(clear)
Electrochemistry
Redox Stoichiometry
Balanced Redox Equation:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn2+ + 8 H2O + 10 CO2
At end-point of the titration:
  Subs red
ηox   Subs
ox


 =

ηred
ox
red

 Subs ox  Subs red
(MV)ox (MV)red
x ox = x ox
X = coefficient of oxidizing agent in
balanced redox equation.
X = coefficient of reducing agent in
balanced redox equation.
ηox = moles of oxidizing agent
ηred = moles of reducing agent
Vox = volume of oxidizing agent
Vred = volume of reducing agent
M = molarity
g = grams
Electrochemistry
MW = molecular (formula) weight
L = liters
Redox Stoichiometry
Balanced Redox Equation:
16 H+ + 2 MnO4− + 5 C2O42− 2 Mn2+ + 8 H2O + 10 CO2
How many grams of Na2C2O4 (FW= 134) are required to completely
react with 40.3 mL of a 0.300 M KMnO4 (FW = 158) solution under
acidic conditions.

 1L  5 Na 2 C 2 O 4  134 g

KMnO
4


 = 4.05g


? mL Na 2 C 2O4   0.300

(40 .3 mL)


L


 1000mL  2 KMnO 4  1 Na 2 C 2 O 4 
How many mL of a 1.22 M Na2C2O4 (FW= 134) are required to
completely react with 2.63 mL of a 3.08 M KMnO4 (FW = 158) solution
under acidic conditions.
At end-point of the titration:
(MV)ox = (MV)red
X ox
X red
1.22   V
 3.08   2.63 mL 
L
L

 
5
2
5  3.08  2.63 mL 
L

V
 16.6 mL
Electrochemistry
2 1.22  
L

CHM 1046: General Chemistry and
Qualitative Analysis
Unit 21
Electrochemistry
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Chapter # 21
•Module #11
Electrochemistry
Basic Electricity
• Coulomb (C): a quantity of electricity = 6.24 x 1018
electrons (amount of e- in 1 ampere*sec).
• Faraday (F ) = a MOLE of electrons = 6.02 x 10
= 96,500 x C
23
-
e
Electrochemistry
{e- flow}
Metallic Bonding: Electron-Sea Model
• Metals can be thought of as cations suspended in
“sea” of valence electrons.
• Attractions hold electrons near cations, but not so
tightly as to impede their flow.
This explains
properties of
metals—
 Conductivity
of heat and
electricity
 Deformation
Electrochemistry
Basic Electricity
High flow rate
•
•
•
•
High force
Ampere (I): measure of the flow rate (volume) of electricity ,
measured in coulombs/ second.
I = C/s
Volts (V): measure of electrical force (Joules) per unit of electricity
(coulomb). V = J/C = JC-1
Watt (W) measure of electrical power . W = V x I
{Insulators: substances which resist the flow of electrons.
Resistance (R) is measured in ohms (Ω).}
Electrochemistry
Corrosion of Metals =
Oxidation-Reduction Reactions
• Corrosion, the degradation of
metals as a result of electrochemical
activity, requires an anode and a
cathode in order to occur.
Cathode
O2
H2O
Fe
Anode
• The anode is the substance with a
higher potential to oxidize (lose
electrons) while the cathode is
the substance with a higher
potential for reduction (gaining of
electrons).
Fe0 + ½O2  Fe2+ + O2Anode
Cathode
Fe2+ + ½ O2  Fe3+ + O2Anode
Cathode
• Two things are needed for corrosion: (1) electrical contact (connection)
between the two substances or metals with oxidation potential difference
Electrochemistry
and (2) the presence of an electrolyte (such as water) to conduct
ions
between them.
Table of Standard Reduction Potentials
For Voltaic (Galvanic) Cells: Spontaneous Reactions that produce electricity
Fe0 + O20


Fe3+ + O2-
Reduction
potentials for
many electrodes
have been
measured and
tabulated.
Cathode
Calculate
the voltage
produced by
reaction!
Anode
Fe2+ + O2-
The substance with
the more positive
reduction potential will
become reduced; the
substances having
less positive
reduction potential
Electrochemistry
become oxidized.
Corrosion of Iron
(1) Initial oxidation of Fe  Fe 2+ yields:
Reduction Reaction
Eo (V)
O2 + 4H+ + 4e- → 2H2O
+1.23
Fe2+ + 2e- → Fe
-0.44
 = Ered
 (cathode) − Ered
 (anode)
ERx
Oxidation Reaction
Eo (V)
+
2H
E 2O → O2 + 4H + 4e
Rx
-1.23
= (1.23) – (- 0.44) = 1.67
Fe → Fe2+ + 2e-
0.44
(2) Further oxidation of Fe 2+  Fe 3+ yields:
Reduction Reaction
O2 +
4H+
+
4e-
→ 2H2O
Fe3+ + e- → Fe2+
Fe
Eo (V)
+1.23
+0.77
Oxidation Reaction
ERx
Eo (V)
= (1.23) – (+ 0.77) = 0.46
2H2O → O2 + 4H+ + 4e- -1.23
Fe2+ → Fe3+ + e-
2.13
-0.77
Electrochemistry
+
2+
(aq)
+ O2(g) + 4H2O + 2xH2O(l)  2Fe2O3 ·xH2O(s) + 8H
(aq)
Corrosion and…
(Fe does not corrode when pH > 9)
Electrochemistry
Fe 2+(aq) + O2(g) + 4H2O + 2xH2O(l)  2Fe2O3 ·xH2O(s) + 8H+(aq)
…Corrosion Prevention
What happens if you place
a half Zn half Fe metal bar
in the presence of water
and oxygen?
 = Ered
 (cathode) − Ered
 (anode)
ERx
Fe + O2 Rx: ERx = Ered (1.23) − Ered (-0.44) = 1.67
Zn + O2 Rx: ERx = Ered (1.23) − Ered (-0.76) = 1.99
Electrochemistry
…Corrosion Prevention
4H+
2H2O
e
-
eReduction Potential: - 0.44 V
Reduction Potential
- 0.76 V
Electrochemistry
…Corrosion Prevention
Oxidation
Mg  Mg 2+
e4H+ + O2
2H2O
Fe
e-
eeElectrochemistry
…Corrosion Prevention
Electrochemistry
Zn
Redox Reactions
In spontaneous oxidation-reduction
(redox) reactions, electrons are
transferred and energy is released.
Cu2+
Cu
Cu2+
Zn2+
Cathode (reduction)
Anode (oxidation)
Now perform
the same
reaction, but
separate the
Zn(s) from
Cu2+(aq) by
wire !
Electrochemistry
Voltaic (Galvanic) Cells
• Electrical energy is generated
from spontaneous redox
reactions.
• We can use that energy to do work
if we make the electrons flow
through an external device (wire).
Beaker 1: Cu metal
with Cu(NO3)2 solution
Beaker 2: Zn metal
with Zn(NO3)2 solution
cathode
anode
Zn
Cu
Zn2+
Which is the Anode and the Cathode?
Cu2+
E0Rx = (0.34) – (- 0.76) = 1.10
Electrochemistry
The first Voltaic (Galvanic) Cell: 1799
Luigi Galvani
(1737-1798)
Alessandro Giuseppe Antonio
Anastasio Volta
(1745–1827)
Electrochemistry
Alkaline Batteries
4+
3+
Lead Batteries
Electrochemistry
{VoltaicGalvanic
Cells*}
-
e
-
e
e-
-
e
e-
-
e
eundergoes
oxidation
producing
cations in
solution
e-
eZn
-
e
{Lemmon
Cells}
Salt Bridge
Voltaic
Cells
Causes
reduction of
the cations
in solution
into solid
metal.
eee
-
Cu
||
Beaker 1: Zn metal anode
with Zn(NO3)2 solution
Electrochemistry
Beaker 2: Cu metal cathode
with Cu(NO3)2 solution
Once electrons flow in
external wire from the
anode to the cathode,
the charges in each
beaker would not be
balanced and the flow of
electrons would stop.
Voltaic Cells & Salt Bridges
Salt Bridge
Therefore, we use a salt
bridge that contains a salt
solution to keep electrical
(ionic) charges balanced.
Cations in salt bridge move
toward the cathode.
Anions move toward the
anode.
Current Balancing
in Salt Bridge:
Electrochemistry
Voltaic
Cell
Notation
Ered = +1.2
Zn2+(aq) + 2e-  Zn(s)
Ered = - 0.76
Phase
separator
Ered = + 0.77
Salt Bridge
2+
3+
Separator of Different
Species of same phase
2+
Zn | Zn (1M) || Fe (1M) , Fe (0.8M) | Pt
Electrochemistry
Anode Compartment
Cathode Compartment
Standard Reduction Potentials (E°red)
Standard Conditions:
• 25°C
• 1 atm (760 torr)
• 1 Molar solution
EMF is the maximum
potential difference
between two electrodes
of a electrochemical
cell. It is also called cell
potential (Ecell).
This quantity is related
to the tendency for a
substance to acquire
(i.e. gain) or Electrochemistry
release
(loss) electrons.
Standard Hydrogen Electrode
• Their values are referenced to a standard hydrogen
electrode (SHE).
• By definition, the reduction potential for hydrogen is 0 V:
2 H+ (aq, 1M) + 2 e−  H2 (g, 1 atm)
Electrochemistry
{Voltaic Zn/H Cell}
Electrochemistry and Free Energy
The greater the
difference between
the two, the greater
the voltage of the
cell.
G for a redox reaction can be
found by using the equation
Under standard conditions,
G = −F Ecell 
where  is the number of moles of
electrons transferred in balanced
redox equation, and F is a
constant, the Faraday.
1 F = 96,485 coulombs /mole
of
Electrochemistry
electrons
Cell Potential under Non-Standard Conditions:
The Nernst Equation
Q
• Remember that
G = G + RT ln Q
• This means
[ Products ]
[ Reactants ]
G = −nF E
−nF E = −nF E + RT ln Q
Dividing both sides by −nF, we get the Nernst equation:
RT
E=E −
nF ln Q
or, using base-10 logarithms,

Effect that nonstandard conc. (Q)
have on EO
Some Values for Q:
1M 
0.1M   10  10 log  1
[ Products] 1M 
 1  10 log  0


1
M
[ Reactants ]
Electrochemistry
0.1M   0.1  10 log  -1
1M 
1

E=E −
2.303 RT
log Q
nF
0
1
Nernst Equation

E=E −
2.303 RT
log
Q
nF
At room temperature (298 K),
2.303 RT
=
0.0592
V
F
Thus the equation becomes
0.0592
log Q
E = E −
n

Electrochemistry
2003B Q6
Electrochemistry
Construct voltaic cell for reaction with greatest voltage.
Al
Cu
Al3+
Cu2+
Electrochemistry
Electrochemistry
2006B Q2
3x(
)
2x(
)
n = 6 mol of e-
Electrochemistry
[???]
- [???]
From previous problem:
Present [1.08]
Change +[0.08 ]
Electrochemistry
Table of Standard Reduction Potentials
For Voltaic (Galvanic) Cells: Spontaneous Reactions that produce electricity
Cathode
(reduction)
Reactions are
more
spontaneous
(more likely
to occur) the
further apart
the
chemicals
are on this
Table
G = −nF E
Anode
(oxidation)
Reduction
potentials for
many electrodes
have been
measured and
tabulated.
The substance with
the more positive
reduction potential will
become reduced;
substances having
less positive
reduction potential
Electrochemistry
become oxidized.
Electrolytic Cells
 Process by which electrical energy is used to force
non-spontaneous electrochemical reactions to occur.
 Electrodes used are usually inert electrodes (they do
not participate in the reaction.
½O2 + H2
Electrolysis of Water :
Cathodic Rx:
2 H2O + 2e-
H2 + 2 OH-
Anodic Rx:
H2O
{Electrolysis Movie}
H2O
2 H+ + ½ O2 + 2 e-
3 H2O
H2 + ½ O2 + 2 H+ + 2 OH-
Net Rx:
Electrochemistry
H2O
H2 + ½ O 2
2 H2O
Voltaic vs. Electrolytic Cells
Spontaneous:
electricity produced
Non-spontaneous:
electricity required
vs.
1. NaCl(l)
2. NaCl(aq)
has water
Voltaic (Galvanic) Cell
Electrolytic cells:
1. One container for the
reaction mixture.
2. Two electrodes immersed in
the same reaction mixture.
3. Requires a power supply of
direct current.
Electrolytic Cell
which may
hydrolize
Types of electrolytic cells:
1. Molten salts
2. Solutions of salts
(includes electroplating)
Electrochemistry
Table of Standard Reduction Potentials
For Electrolytic Cells: Non-Spontaneous Reactions forced by electricity
Anode
(oxidation)
In all
electrolytic
cells the
most easily
oxidized
species is
oxidized
and the
most easily
reduced
species is
reduced.
Cathode
(reduction)
Electrolytic
cells are forced
to run in
reverse from
voltaic
(galvanic) cells:
The substance with
the more positive
reduction potential will
become oxidized;
substances having
less positive
reduction potential
Electrochemistry
become reduced.
How do we get pure Sodium?
The Electrolysis of Molten NaCl
2 NaCl (l)
elect
2 Na(l ) + Cl2(g)
Electrochemistry
The Electrolysis of Molten
Potassium
Chloride
(1) Draw cell
(2) Write half
equations for rx.
occurring at the
anode & cathode
(3) Write a bal.
cell equation
Electrochemistry
• In all electrolytic cells, electrons are forced to flow from the
positive electrode (anode) to the negative electrode (cathode).
Electrolysis of Aqueous Salts
More possibilities for electrolysis :
1. The ions of dissolved salts may electrolyze.
2. Water may electrolyze (into O2 or H2).
In all electrolytic cells the most easily
oxidized species is oxidized and the most
easily reduced species is reduced.
Na+(aq) Cl -(aq)
H2O  H2 or ½O2
Electrochemistry
Electrolysis of Aqueous Salts
Examples: NaCl(aq) , Fe(OH)2(aq), AgF(aq) , K2SO4(aq)
Anode
Anion will
oxidize, instead
of water.
(oxidation)
Water oxidizing
Will oxidize, if anions are below.
Metals will
reduce in
water
solutions.
Water reducing
Cathode
(reduction)
In all electrolytic
cells the most
easily oxidized
species is
oxidized and the
most easily
reduced species
is reduced.
Will reduce, if cations are below.
Electrochemistry
Will reduce when there is
no water, if it is melted.
The Electrolysis of Aqueous
Potassium Chloride
Anode reaction
2 Cl  Cl2(g)  2 e
-
-
Cathode reaction 2 H 2 O  2 e  H 2 g   2 OH
-
-
Cell reaction 2 Cl  2 H 2 O  H 2 g   Cl2 g   2 OH
-
-

K is a spectator ion. Note that wate r is electrolyz ed!
Electrochemistry
Electrolysis of Aqueous Salts
More possibilities for electrolysis :
1. The ions of dissolved salts may electrolyze.
2. Water may electrolyze (into O2 or H2).
Reduction at the cathode: In an aqueous salt solution there are two possibilities

Mn+ + n e-  M
(all metal ions get reduced except: IA+, IIA2+, or Al3+)
 2 H2O + 2e-  H2 + 2 OH(If metal is IA+, IIA2+, or Al3+, then water gets reduced)
Solution
containing
salts of Ni,
Fe & Zn.
Which will
become
reduced?
Will reduce, in
water solutions.
Will reduced,
if ions are
below.
Electrochemistry
Will not reduce,
water will reduce.
Electrolysis of Aqueous Salts
Oxidation at the anode: In an aqueous salt solution there are two possibilities

X-  ½ X2 + e(of the common anions only Cl-, Br-, or I- get oxidized)

H2O  2 H+ + ½ O2 + 2e(If anions is not Cl-, Br-, or I- , then water gets oxidized)
Solution
containing
salts of F,
Cl, I. Which
will
become
oxidized?
Will oxidize,
water will not.
Will oxidize, if
ions are below.
• In all electrolytic cells the most easily reduced
Electrochemistry
species is reduced and the most easily oxidized
species is oxidized.
The Electrolysis of Aqueous
Potassium Chloride
•
•
•
In this electrolytic cell, hydrogen gas is produced at one electrode.
 The aqueous solution becomes basic near this electrode.
 What reaction is occurring at this electrode? You do it!
Gaseous chlorine is produced at the other electrode.
 What reaction is occurring at this electrode? You do it!
These experimental facts lead us to the following nonspontaneous
electrode reactions:
Anode reaction
2 Cl-  Cl2(g)  2 e -
Cathode reaction 2 H 2 O  2 e -  H 2g   2 OHCell reaction 2 Cl-  2 H 2 O  H 2g   Cl2g   2 OH
K is a spectator ion. Note that wate r is electrolyz ed!
Electrochemistry
Standard Reduction Potentials
Reduction
potentials for
many
electrodes
have been
measured
and tabulated.
The substance with
the more positive
reduction potential will
become reduced;
substances having
less positive
reduction Electrochemistry
potential
become oxidized.
The Electrolysis of Aqueous
Potassium Chloride
Cell diagram
e- flow
- pole of battery
+ pole of battery
Battery, a source
e- flow
of direct current
- electrode
+ electrode
H2 gas
Cl2 gas
aqueous KCl
2 H2O + 2e-  H2 (g) + 2 OHcathode reaction
2Cl-  Cl2 (g) + 2eanode reaction
Electrochemistry
The Electrolysis of Aqueous
Potassium Sulfate
• In this electrolysis, hydrogen gas is produced
at one electrode.
 The solution becomes basic near this electrode.
 What reaction is occurring at this electrode?
You do it!
• Gaseous oxygen is produced at the other
electrode
 The solution becomes acidic near this electrode.
 What reaction is occurring at this electrode?
You do it!
• These experimental facts lead us to the
following electrode reactions:
Electrochemistry
The Electrolysis of Aqueous
Potassium Sulfate
Anode reaction 2 H 2 O  O 2(g)  4 H   4 e Cathodereaction 2(2 H 2 O  2e-  H 2(g)  2OH- )

Cell reaction 6 H 2 O  2 H 2(g)  O 2(g)  4H

4
OH

 

4 H 2O
T heoverallreactionis 2 H 2 O  2H2(g)  O 2(g)
Electrochemistry
The Electrolysis of Aqueous
Potassium Sulfate
Cell diagram
e- flow
Battery, a source
of direct current
- electrode
e- flow
+ electrode
O2 gas
H2 gas
aqueous K2SO4
2 H2O + 2e-  H2 (g) + 2 OHcathode reaction
2H2O  O2 (g) + 4H+ +Electrochemistry
4eanode reaction
Electroplating
(6+)
{ElectPlating: H2Cr2O7 → Cr0}
Electrochemistry
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Faraday’s Law - The amount of substance
undergoing chemical reaction at each
electrode during electrolysis is directly
proportional to the amount of electricity that
passes through the electrolytic cell.
• A faraday is the amount of electricity that
reduces one equivalent of a species at the
cathode and oxidizes one equivalent of a
species at the anode.
1 faraday of electricit y  6.022 10 e
23
-
Electrochemistry
Stoichiometry and Electrolysis
Ag+(aq) + e-  Ag(s)
Cu2+(aq) + 2e-  Cu(s)
Ampere = coulombs per second (C/s)
Faraday (F ) = 96,500 x C = 1 mole of e-
• How much of a chemical reaction can happen with a certain
current over a certain amount of time?
• Current is measured in amps (Ampere: A : 1 C / s)
• Determine the number of moles of electrons involved by using
the Faraday (1 F = 96,500 x C = 6.02 x 1023 e-)
• Use coefficients of balanced equation to determine moles and
grams of substances involved.
Electrochemistry
Example
• How long must a current of 5.00 A be applied to a solution
of Ag+ ions to produce 10.5 g of Ag metal?
Ag+(aq) + e-  Ag(s)
•
•
How many moles of e- are required?
How many minutes does it take to produce that many moles of e- ?
1molAg
1mol e 
? moles e  10.5gAg

 9.73  10 2 mol e 
107.868gAg 1molAg
-
Ampere = C/s
Faraday (F ) = 96,500 x C = 1 mole of e-
96,485C
1sec 1 min Electrochemistry
? min .  9.73 10 mol e 


 31.3 min

5.00C 60 sec
1mol e
2

Example
• How many grams of Cu will be plated out when 10.0 A of
current is passed for 30.0 min through a Cu2+ solution?
Ampere = C/s
Faraday (F ) = 96,500 x C = 1 mole of eCu2+(aq) + 2e-  Cu(s)
Strategy:
(1) From amperage and minutes, coulombs
(2) From coulombs get moles of e(3) From moles of e- get moles and then grams of Cu
60 sec 10.0 C
1mol e 
? grams Cu  30.0 min 


x
1sec
96,485 C
1 min
1molCu


2mol e
63.546g Electrochemistry
 5.94gCu
1molCu
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• A coulomb is the amount of charge that
passes a given point when a current of one
ampere (A) flows for one second.
• 1 amp = 1 coulomb/second
1 faraday  6.022 1023 e-  96,487 coulombs
Electrochemistry
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Faraday’s Law states that during electrolysis,
one faraday of electricity (96,487 coulombs)
reduces and oxidizes, respectively, one
equivalent of the oxidizing agent and the
reducing agent.
 This corresponds to the passage of one mole of
electrons through the electrolytic cell.
1 equivalent of oxidizing agent  gain of 6.022 1023 e1 equivalent of reducing agent  loss of 6.022 1023 e-
Electrochemistry
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Example 21-1: Calculate the mass of palladium
produced by the reduction of palladium (II) ions
during the passage of 3.20 amperes of current
through a solution of palladium (II) sulfate for 30.0
minutes.
2+
0
Cathode:
Cathode : Pd
Pd 2+++ 2e
2e - 
 Pd 0
11mol
mol 22mol
mol 11mol
mol
106
106gg 2(96,500)
2(96,500)106
106g g
3.20 amp = 3.20 C s
60 s 3.20 C
106 g Pd
? g = 30.0 min 


 316
. g Pd
min
s
296,500 C
Electrochemistry
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
• Example 21-2: Calculate the volume of
oxygen (measured at STP) produced by the
oxidation of water in example 21-1.
++
-Anode
:
2
H
O

O
+
4H
+
4e
Anode
22O  O 22gg  + 4H + 4e
2 mol
? LSTP
mol 44mol
mol 44mol
mol
11mol
22.4L
496,500
22.4
LSTP
96,500CC
STP
60 s 3.20 C 22.4 LSTP O 2
O 2  30.0 min 


min
s
496,500 C 
 0.334 LSTP O 2 or 334 mL STP O 2
Electrochemistry
Commercial Applications of
Electrolytic Cells
Electrolytic Refining and Electroplating of
Metals
• Impure metallic copper can be purified
electrolytically to  100% pure Cu.
 The impurities commonly include some active
metals plus less active metals such as: Ag, Au,
and Pt.
• The cathode is a thin sheet of copper metal
connected to the negative terminal of a direct
Electrochemistry
current source.
• The anode is large impure bars of copper.
2007A Q3
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Hydrogen Fuel Cells
Electrochemistry
2007B Q3
Electrochemistry
Electrochemistry
Applications of
Oxidation-Reduction
Reactions
Electrochemistry
Batteries
Electrochemistry
Alkaline Batteries
4+
3+
Electrochemistry
pH Meters
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Voltaic Cells
•
•
In the cell, then, electrons leave the anode and flow through the wire to the
cathode.
As the electrons leave the anode, the cations formed dissolve into the solution in
the anode compartment.
Electrochemistry
Voltaic Cells
• As the electrons reach the cathode, cations in the cathode are attracted to
the now negative cathode.
• The electrons are taken by the cation, and the neutral metal is deposited
on the cathode.
Electrochemistry
Electromotive Force (emf)
• Water only spontaneously
flows one way in a waterfall.
EMF is the maximum
potential difference
between two electrodes
of a electrochemical cell.
This quantity is related
to the tendency for a
substance to acquire
(i.e. gain) or release
(loss) electrons.
• Likewise, electrons only
spontaneously flow one
way in
Electrochemistry
a redox reaction—from higher
to lower potential energy.
Electromotive Force (emf)
• It is the electrical potential difference (voltage)
between the anode and cathode in an
electrochemical cell.
• It is also called the cell potential, and is designated
Ecell. Cell potential is measured in Joules of energy
per Coulomb of charge = volt (V).
J
1V=1
C
Electrochemistry
Oxidizing and Reducing Agents
• The strongest
oxidizers have the
most positive
reduction potentials.
• The strongest
reducers have the
most negative
reduction potentials.
Electrochemistry
Corrosion and…
Electrochemistry
2004A Q6
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
2005 B electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
2001
Electrochemistry
Electrochemistry
2002
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
Electrochemistry
2002 B
Electrochemistry
Electrochemistry
Electrochemistry
2003
A
Electrochemistry
Electrochemistry
Electrochemistry
200
3B
Electrochemistry
Electrochemistry
2004 A
Electrochemistry
Electrochemistry
200
4B
Electrochemistry
Electrochemistry
200
5A
Electrochemistry
Anode: oxidation, where Anions (A-)
go to loose electrons
Cathode: reduction, where Cations
(Ca+) go to gain electrons
Electrochemistry
Concentration Cells
• The Nernst equation implies that a cell could be created
that has the same substance at both electrodes.
• For such a cell,

Ecell
would be 0, but Q would not.
• Therefore, as long as the concentrations are
different, E will not be 0.
Electrochemistry
2.303
RT

E=E −
nF
log Q