Electrochemistry
A. Redox Reactions – Crash course video
electrochemistry is the branch of chemistry that
studies electron transfer in chemical reactions
OXIDATION is a loss of electrons “OIL”
eg)
0
2+
Charge of ion or element (if charge increase,
from its reactant, it’s a oxidation reaction
Mg(s)  Mg2+(aq) + 2e
Electrons are in the product =
losing electrons
-1
0
Charge of ion or element (if charge increase,
from its reactant, it’s a oxidation reaction
2Cl(aq)  Cl2(g) + 2e
Electrons are in the product =
losing electrons
 reduction is a gain of electrons “RIG”
3+
0
eg) Fe3+(aq) + 3e  Fe(s)
Charge of ion or element (if charge decreases,
from its reactant, it’s a reduction reaction
Br2(l) + 2e  2Br(aq)
Electrons are in the reactant =
gaining electrons
 oxidation and reduction reactions occur together,
hence the term
redox
 the reduction and oxidation reactions are called
the half reactions
 “adding” the half reactions together will give you the
net ionic equation
redox reaction
that takes place during the
 the e lost in the oxidation half reaction must equal
the e gained in the reduction half reaction
 you may have to multiply one or both of the
half reactions to balance the e
 spectator ions
(ions not changing) are NOT
included!
 the substance that is reduced
oxidizing agent
by taking e-)
is called the
( OA) (it causes the oxidation
 the substance that is oxidized
is called the
reducing agent (RA) (it causes the reduction
by giving up e-)
Video – REDOX reaction
Example 1
Given the following reaction, write the half reactions
and the net ionic equation.
0
1+ 1–
0
1+ 1–
Na(s) + LiCl(aq)  Li(s) + NaCl(aq)
ox
Ox:
Red:
Net:
red
Cl- is spectator
Na(s)  Na+(aq) + 1e-
Li+(aq) + 1e-  Li(s)
Li+(aq) + Na(s)  Li(s) + Na+(aq)
Step 1: identify the charges for each element or ion.
Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your
spectator ion.
Step 3: write your half reactions (charges must be balanced with same number of electrons.
Step 4: Ensure you have the same number of electrons on both sides (reactants and products)
Step 5: Write the net ionic equation.
Example 2
Given the following reaction, write the half reactions and
the net ionic equation.
0
3+
1–
0
2+
1–
3 Zn(s) + 2 Au(NO3)3(aq)  2 Au(s) + 3 Zn(NO3)2(aq)
ox
red
Ox:
NO3- is spectator
3 [ Zn(s)  Zn2+(aq) + 2e-]
Red:
2 [ Au3+(aq) + 3e-  Au(s)]
Net:
2 Au3+(aq) + 3 Zn(s)  2 Au(s) + 3 Zn2+(aq)
Step 1: identify the charges for each element or ion.
Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your
spectator ion.
Step 3: write your half reactions (charges must be balanced with same number of electrons.
Step 4: Ensure you have the same number of electrons on both sides (reactants and products)
Step 5: Write the net ionic equation.
Practice Questions
Answer: C


Read pages 558 – 564
Hand in introduction to redox lab
Step 1: identify the charges for each element or ion.
Step 2: identify your oxidation (increase in number) and reduction (decrease in number) half reactions. Also identify your
spectator ion.
Step 3: write your half reactions (charges must be balanced with same number of electrons.
Step 4: Ensure you have the same number of electrons on both sides (reactants and products)
Step 5: Write the net ionic equation.
B. Spontaneous Redox Reactions
 chemical reactions which occur on their own, without
the input of additional energy , are called
spontaneous
not all reactions are spontaneous
 in the table of redox half reactions (see pg 7 in Data
Booklet), the strongest oxidizing agent (SOA)
is at the top left and the strongest reducing agent
(SRA) is at the bottom right
 the redox spontaneity
rule states that a
spontaneous reaction occurs if the oxidizing agent
is above the reducing agent in the table of redox
half reactions
Try These:
For each of the following combinations of substances,
state whether the reaction would be spontaneous or
non-spontaneous:
Cr3+(aq) with Ag(s)
non-spontaneous
I2(s) with K(s)
spontaneous
H2O2(l) with Au3+(aq) spontaneous
Sn2+(aq) with Cu(s)
non-spontaneous
Fe2+(aq) with H2O (l)
non-spontaneous
(both ways)
Practice Question
C. Predicting Redox Reactions
Answer: A
 we will be predicting the strongest or most
dominating reaction that occurs when
substances are mixed (other reactions do take place
because of atomic collisions!)
Steps are found at the bottom of examples.
Example 1
Predict the most likely redox reaction when chromium
is placed into aqueous zinc sulphate.
Cr(s)
S RA
Zn2+(aq) SO42-(aq) H2O(l)
S OA
OA with H2O(l) OA/RA
SOA (Red): Zn2+(aq) + 2e-  Zn(s)
SRA (Ox):
Cr(s)  Cr2+(aq) + 2espont
Net:
Zn2+(aq) + Cr(s)  Zn(s) + Cr2+(aq)
Example 2
Predict the most likely redox reaction when silver is
placed into aqueous cadmium nitrate.
Ag(s)
S RA
H2O(l)
Cd2+(aq)
NO3-(aq)
S OA
OA with H+ (aq) OA/RA
SOA (Red): Cd2+(aq) + 2e-  Cd(s)
SRA (Ox):
Net:
2 [ Ag(s)  Ag+(aq) + e- ]
nonspont
Cd2+(aq) + 2 Ag(s)  Cd(s) + 2 Ag+(aq)
Example 3
Predict the most likely redox reaction when potassium
permanganate is slowly poured into an acidic iron (II)
sulphate solution.
K+(aq) MnO4-(aq) H+(aq) Fe2+(aq) H2O(l) SO42-(aq)
OA S OA with H+ (aq) OA OA/S RA OA/RA OA
with H+ (aq),
H2O(l)
SOA (Red): MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)
SRA (Ox):
Net:
MnO4-(aq)
5 [ Fe2+(aq)  Fe3+(aq) + e- ]
+8H+(aq)
+
spont
 Mn2+(aq) + 4H2O(l) + 5 Fe3+(aq)
5Fe2+(aq)
D. Generating Redox Tables
 you can be given data for certain ions and
elements then be asked to generate a redox table
like the one on pg 7 of you Data Booklet (a smaller
version!)
 you may have to generate a table from real or
fictional elements and ions
 the tables that we use are all written as reduction
half reactions
Video: REDOX tables
Example 1
Generate a redox table given the following data (useful
when all reactions are given:
Cu2+(aq)
Zn2+(aq)
Pb2+(aq)
Ag+(aq)
Cu(s)




Zn(s)




Pb(s)




Ag(s)




 indicates no reaction
 indicates a reaction
SOA
Redox Table
Ag+(aq) + e-  Ag(s)
Cu2+(aq) + 2e-  Cu(s)
Pb2+(aq) + 2e-  Pb(s)
Zn2+(aq) + 2e-  Zn(s) SRA
Put the oxidizing agents in order from strongest to
weakest.
Ag+(aq), Cu2+(aq), Pb2+(aq), Zn2+(aq)
Put the reducing agents in order from strongest to
weakest.
Zn(s), Pb(s), Cu(s), Ag(s)
Example 2:
Generate a redox table given the following data:
Cu(s) + Ag+(aq)  Cu2+(aq) + Ag(s)
Zn2+(aq) + Ag(s)  no reaction
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Hg(l) + Ag+(aq)  no reaction
Label each as OA or Ra
Redox Table
SOA
Hg2+(aq) + 2e-  Hg(l)
Ag+(aq) + e-  Ag(s)
Cu2+(aq) + 2e-  Cu(s)
Zn2+(aq) + 2e-  Zn(s) SRA
Example 2 (continued):
Put the oxidizing agents in order from weakest to
strongest.
Zn2+(aq), Cu2+(aq), Ag+(aq), Hg2+(aq)
Put the reducing agents in order from weakest to
strongest.
Hg(l), Ag(s), Cu(s), Zn(s)
Example 3:
Generate a redox table given the following data:
2X-(aq) + Y2(g)  spontaneous reaction
2Z-(aq) + Y2 (g)  no reaction
2Z-(aq) + W2 (g)  spontaneous reaction
Label each as OA or RA
SOA
Redox Table
W2 (g) + 2e-  2W-(aq)
Z2 (g) + 2e-  2Z-(aq)
Y2 (g) + 2e-  2Y-(aq)
X2 (g) + 2e-  2X-(aq)
SRA
Example 3 (continued):
Put the oxidizing agents in order from strongest to
weakest .
W2(g), Z2(g), Y2(g), X2(g)
Put the reducing agents in order from strongest to
weakest .
X-(aq), Y-(aq), Z-(aq), W-(aq)
Practice Question
Answer: A
E. Oxidation Numbers (States)
 an oxidation number is the charge an atom
appears to have when found in a
neutral molecule or charged polyatomic ion
 can be used when you have a
molecular
compound where there are no ion charges
to determine if oxidation or reduction is occurring
 how do you use a change in the number?
1. if the number decreases
has occurred
then reduction
2. if the number increases then oxidation
occurred
has
Rules for Assigning Oxidation Numbers:
1. In a pure element, the oxidation number is zero.
2. In simple ions, the oxidation number is equal to
the ion charge .
3. In most compounds containing hydrogen, the
oxidation number for hydrogen is +1 . (Exception
is the metal hydrides eg) LiH where the oxidation
number of hydrogen is –1 ).
4. In most compounds containing oxygen, the
oxidation number for oxygen is –2 . (Exception is
the peroxides eg) H2O2, Na2O2 where the oxidation
number of oxygen is –1)
5. The sum of oxidation numbers of all atoms in a
substance must equal the net charge of the
substance. ( Zero for compounds and the charge
of the polyatomic ion)
eg) sum of MgO = 0
sum of PO43- = –3
Common Oxidation Numbers
Atom or Ion
Oxidation Number
Examples
All atoms in elements
0
Na is 0
Cl in Cl2 is 0
Hydrogen in all compounds +1
(except: hydrogen in
-1
hydrides)
H in HCL is +1
H in LiH is -1
Oxygen in all compounds, -2
Except oxygen in peroxides -1
O in H2O is -2
O in H2O2 is -1
All monoatomic ions
Na+ is 1+
S2- is 2-
Charge of Ion
Example
What is the oxidation number (state) for the element
identified in each of the following substances:
a) N in N2O
b) N in
NO3-
+1 –2
N2 O
+2 –2 = 0
+5 –2
N O3–
+5 –6 = –1
individual oxidation
numbers
sum of oxidation
numbers
c) C in C2H5OH
d) C in C6H12O6
–2 +1 –2+1
C2 H5 O H
–4 +5 –2+1 = 0
0 +1 –2
C6 H12 O6
0 +12 –12 = 0
 figuring out oxidation numbers can help to identify
whether a reaction is a redox reaction or not
 for it to be a redox reaction, there has to be both an
increase in oxidation number and a decrease in
oxidation number seen in the reaction
0
+1 -1
0
+1 -1
eg) Ag(s) + NaNO3(aq)  Na(s) + AgNO3(aq)
Ag increases oxidized
redox!!!
Na decreases  reduced…
+2 -2
+1 -1
+2 -1
+1 -2
PbSO4(aq) + 2 KI(aq)  PbI2(s) + K2SO4(aq)
nothing changes  NOT a redox reaction!
Practice Question
Practice question
Answer: a
Answer: C
electron transfer occurs in living systems
eg) photosynthesis, cellular respiration
also occurs in non-living systems
eg) combustion, bleaching, metallurgy
F. Disproportionation
 disproportionation occurs when one element is
both oxidized and reduced in a reaction
eg)
0
-1
-2
0
2 H2O2(aq)  2 H2O(l) + O2(g)
+1
-1
Cl2(g) + 2 OH-(aq)  ClO-(aq) + Cl-(aq) + H2O(l)
Do workbook questions: page 4
G. Balancing Redox Reactions
 sometimes most reactants and products are known
but the complete reaction is not given…called a
skeleton reaction
There are two different ways you can balance
redox reactions: either can be used so find one way
that works best for you…
Half Reaction Method
Oxidation Number Method
Example 1:Of half Reaction Method
Balance the following half reaction :
(+6)
+6 2
4 H+(aq) + 3 e– + CrO42-(aq)
+6 8 = 2
net charge = –1
+3 2
(+3)
 CrO2-(aq) + 2 H2O(l)
+3 4 = 1
net charge = –1
(Cr is already balanced)
Example 2:
Balance the following half reaction:
(+6)
+1 +3 2
0
(0)
6 H+(aq) + 6 e– + 2 HClO2(aq)  Cl2(g) + 4 H2O(l)
+1 +3 4 = 0
net charge = 0
net charge = 0
Practice Question
Answer: A
Example 1:
Balance the following using
oxidation numbers, assuming
acidic conditions:
+6 2
+4 2
+3 2
+6 2
+6 8 = 2
+4 6 = 2
+3 4 = 1
+6 8 = 2
CrO42-(aq) + SO32-(aq)  CrO2-(aq) + SO42-(aq)
(+6)
Red 2 [ 4
(+4)
Ox
(+3)
H+(aq)+ 3 e–+ CrO42-(aq)  CrO2-(aq)+ 2 H2O(l)]
(+6)
3 [ H2O(l)+ SO32-(aq)  SO42-(aq) + 2 e– + 2 H+(aq) ]
8 H+(aq) + 2 CrO42-(aq) + 3 H2O(l) + 3 SO32-(aq)  2 CrO2-(aq) + 4 H2O(l) + 3 SO42-(aq) + 6 H+(aq)
Net
2 H+(aq) + 2 CrO42-(aq) + 3 SO32-(aq)  2 CrO2-(aq) + H2O(l) + 3 SO42-(aq)
Example 1: of Oxidizing Number Method
Balance the following reaction using the oxidation number
method.
+7
+4
+4
+6
2 H+(aq) + 2
3 SO32-(aq)  __
2 MnO2(aq) + __
3 SO42-(aq) + __
1 H2O(l)
__
__MnO4-(aq) + __
 = 3  1 atom = 3  2(Co) =6e = +2  1 atom = +2  3(Co) =+6e-
Example 2:
Balance the following reaction using the oxidation number
method.
+4
3
-1
+3
1 Br(aq)
__ H2O(l) + 3
__N2O4(g) + _
+5
6
6 H+(aq)
 2_ NO2 (aq) + _1 BrO3(aq) + __
 = 1  2 atoms = 2  3(Co) =6e = +6  1 atom = +6 1(Co) =6e-
H. Redox Stoichiometry
1. Calculations
 stoichiometry can be used to predict or analyze a
quantity of a chemical involved in a chemical
reaction
 in the past we have used balanced chemical
equations to do stoich calculations
 we can now apply these same calculations to
balanced redox equations
Example 1
What is the mass of zinc is produced when 100 g of
chromium is placed into aqueous zinc sulphate.
Cr(s)
RA
SRA
H2O(l)
Zn2+(aq)
SO42-(aq)
OA
OA with H2O(l) OA/RA
SOA
SOA (Red): Zn2+(aq) + 2e-  Zn(s)
SRA (Ox):
Net:
Cr(s)  Cr2+(aq) + 2eCr(s) + Zn2+(aq)  Zn(s + Cr2+(aq) )
Cr(s) + Zn2+(aq)  Zn(s) + Cr2+(aq)
m = 100 g
M = 52.00 g/mol
n=m
M
= 100 g
52.00 g/mol
= 1.923… mol
m=?
M = 65.39 g/mol
n = 1.923… mol x 1/1
= 1.923… mol
m = nM
= (1.923…mol)(65.39 g/mol)
= 125.75 g
= 126 g
Example 2
What volume of 1.50 mol/L potassium permanganate is
needed to react with 500 mL of 2.25 mol/L acidic iron (II)
sulphate solution?
K+(aq) MnO4-(aq) H+(aq) H2O(l) Fe2+(aq) SO42-(aq)
OA
OA with H+(aq)OA OA/RA OA/RA OA with H+(aq)
SOA
SRA OA with H2O(l)
SOA (Red): MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) +
4H2O(l)
SRA (Ox):
5 [ Fe2+(aq)  Fe3+(aq) + e- ]
Net: MnO4-(aq) +8H+(aq) + 5Fe2+(aq)  Mn2+(aq)+ 4H2O(l) + 5Fe3+(aq)
MnO4-(aq)+8H+(aq) + 5Fe2+(aq)  Mn2+(aq)+ 4H2O(l) +
5Fe3+(aq)
v=?
c = 1.50 mol/L
n = 1.125 mol x 1/5
= 0.225 mol
v= n
C
v = 0.225 mol
1.50 mol/L
= 0.150 L
v = 0.500 L
c = 2.25 mol/L
n = cv
= (2.25
mol/L)(.500L)
= 1.125 mol
*** Reminder: Use the formulas
n=m/M and C = n/v to find quantities
Practice Question
2. Titrations
Answer:
 a titration is a lab process used to
determine the volumeof a
substance needed to react
completely with another
substance
 this volume can then be used to
calculate an unknown
concentration using
stoichiometry
 one reagent ( titrant- OA ) is
slowly added to another (sample RA ) until an abrupt change
( endpoint ) occurs, usually in
colour
 in redox titrations, two common oxidizing agents are
used because of their colour and strength:
1. permanganate ions (MnO4-(aq)) –
purple
2. dichromate ions (Cr2O72-(aq)) –
orange
 as long as the sample (RA) in the flask is reacting
with the permanganate ions (dichromate ions)
the sample will be colourless (green)
 when the reaction is complete, any unreacted
permanganate ions will turn the sample purple
(pink) (with dichromate, sample goes from orange to
green)
 the volume of titrant (OA) needed to reach the
endpoint is called the equivalence point
 the concentration
accurately known
of the titrant must be
 the concentration of the permanganate solution
must be calculated against a primary standard
(a solution of known concentration) before it can
be used in a titration itself
 this is done just prior to the titration
Example
Find the concentration of (standardize) the KMnO4(aq)
solution by titrating 10.00 mL of 0.500 mol/L acidified
tin (II) chloride with the KMnO4(aq).
Trial
1
2
3
4
Final Volume
(mL)
18.40
35.30
17.30
34.10
Initial Volume
(mL)
1.00
18.40
0.60
17.30
17.40
16.90
16.70
16.80
pink
light
pink
light
pink
light
pink
Volume of
KMnO4(aq).(mL)
Endpoint
Colour
 endpoint average is calculated by using 3 volumes
within 0.20 mL
Endpoint average = 16.90 mL + 16.70 mL + 16.80 mL
3
= 16.80 mL
Review: Redox reactions
Analysis:
 determine net ionic redox equation
K+(aq) MnO4-(aq) H+(aq) H2O(l) Sn2+(aq) Cl-(aq)
OA
OA with H+(aq)OA OA/RA OA/RA RA
SOA
SRA
SOA (Red):
SRA (Ox):
Net:
2 [ MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)]
5 [ Sn2+(aq)  Sn4+(aq) + 2 e- ]
2MnO4-(aq)+ 16H+(aq)+ 5Sn2+(aq)  2Mn2+(aq)+ 8H2O(l) + 5Sn4+(aq)
 use net redox equation to calculate KMnO4(aq)
concentration
2MnO4-(aq) +16H+(aq) +5Sn2+(aq) 2Mn2+(aq)+ 8H2O(l) +5Sn4+(aq)
v = 0.01680 L
C=?
n = 0.00500 mol x
2/5
= 0.00200 mol
C= n
v
C = 0.00200 mol
0.01680 L
= 0.119 mol/L
v = 0. 01000 L
c = 0.500 mol/L
n = cv
= (0.500 mol/L)(0.01000 L)
= 0.00500 mol
I. Electrochemical Cells
1. Voltaic Cells - animation
 electric cells are devices that convert chemical
energy into electrical energy
 in redox reactions, e- are transferred from the
oxidized substance to the reduced substance
 the transfer of e- can occur through a conducting
wire separating the two substances in
containers called half cells
 a voltaic cell
is an arrangement where
two half cells are joined so that the e- and ions
can move between them
 electrodes are made of good conducting
materials so e- can flow…can be the metal of the
solution or inert such as carbon
 the electrolyte is a solution that contains ions
and will transmit ions (charged particles)
 the electrode where oxidation occurs is called
the anode
 if the anode is a metal, it
loses mass as the cell
operates
 the anode is labelled as negative since it is
the electrode where the electrons originate
 the anions move to the anode since this
electrode loses electrons (leaving a net positive
charge in the electrode)
 the electrode where reduction occurs is
called the cathode
 if the cathode is a metal, it gains mass as the cell
operates
 the cathode is labelled as positive since the
anode is labelled negative
 the cations move to the cathode since this
electrode accepts electrons (leaving a net
negative charge in the electrode)
 electrons flow from the anode (LEOA) to the
cathode (GERC) through a connecting wire
 ions must be able to move
to their attracting
electrode (either through the porous cup or a
salt bridge ) otherwise a buildup of charge will
occur opposing the movement of e-
 the flow of ions through the solution and e-
through the wire maintains overall
electrical neutrality
Cell song
Repeated animation
Summary of Voltaic Cells
Anode (An-ox)
Cathode (Red Cat)
-
+
Is the SRA
Is the SOA
Is being Oxidized
Is being Reduced
“That Cat is a Son Of A Rascal”
Practice Question
2. Standard Reduction Potentials
Answer: 4,6,4,1
 reduction potentials are the ability of a half cell to
attract e these potentials are measured using a voltmeter
 each half reaction listed in the Data Booklet has an
E value measured in volts
assigned to it
 all values in the table are arbitrarily assigned based
on a standard
 the hydrogen cell half reaction has been set as
the standard and has an E value of 0.00 V
3. Predicting Voltage of a Voltaic Cell
 the standard cell potential (Enet) is determined by
adding the E values for the two half reactions
 the sign on the E value for the oxidation
half
reaction must be reversed
 if you multiply an equation to balance e-, you
DO NOT multiply the E value (voltage is
independent of number of e- transferred)
 a positive E net is a spontaneous reaction
 a negative E net is a nonspontaneous reaction
Practice Question
Answer: B
Example
Calculate the E net for the reaction of Zn(s) with CuSO4(aq).
Zn(s)
S RA
H2O(l)
Cu2+(aq)
SO42-(aq)
S OA
OA with H2O(l) OA/RA
SOA (Red): Cu2+(aq) + 2e-  Cu(s)
SRA (Ox):
Zn(s)  Zn2+(aq) + 2e-
E = +0.34 V
E = +0.76 V
Net: Zn(s) + Cu2+(aq)  Cu(s + Zn2+(aq) Enet = +1.10 V
Practice Question
4. Shorthand Notation
Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)
Answer: 1.36V
OR
Zn(s) / Zn2+(aq) // Cr2O72-(aq) , H+(aq) , Cr3+(aq)/
C(s)
***anode // cathode
 line (/) separates
phases
 double line(//) represents the porous cup or
salt bridge and separates the two half reactions
 comma (,) separates chemical species in the
same phase
5. Drawing Cells
 when drawing a cell from the shorthand notation, you
have to be able to label the cathode, anode, positive
terminal, negative terminal, electrolytes, direction of
e flow, and directions of cation and anion flow
 you also have to show and label the reduction half
reaction, oxidation half reaction and net reaction
including E values, E net and spontaneity
Example
Draw and fully label the following electrochemical cell:
Al(s)/ Al3+(aq) // Ni2+(aq) / Ni(s)
e-
V
Ni(s)
cathode
positive
terminal
Al(s)
anode
negative
terminal
anions
Al3+ (electrolyte)
cations
Ni2+ (electrolyte)
Al(s)
S RA
Al3+(aq)
OA
Ni(s)
RA
Ni2+(aq)
S OA
SOA (Red): 3 [ Ni2+(aq) + 2e-  Ni(s)]
SRA (Ox):
H2O(l)
OA/RA
E = –0.26 V
2 [ Al(s)  Al3+(aq) + 3e- ] E = +1.66 V
Net: 2 Al(s) + 3 Ni2+(aq)  3 Ni(s + 2 Al3+(aq) )Enet = +1.40 V
spontaneous:yes
J. Commercial Cells
 batteries are made by connecting two or more
voltaic cells in series (one after the other)
 the voltage of the battery is the sum of the
individual cells
 there are many types of
batteries:
a) Dry Cell
 common 1.5 V and 9 V batteries of clocks, remote
controls, noisy kids toys etc.
Cathode (Red): 2 MnO2(s) + H2O(l) + 2e-  Mn2O3(aq) + 2 OH-(aq) E= +0.79
V
Anode (Oxid):
Zn(s)  Zn2+(aq) + 2eE = +0.76 V
Net:
2 MnO2(s)+ H2O(l) + Zn(s)  Mn2O3(aq) + 2 OH-(aq) + Zn2+(aq)
V
Enet = +1.55
 the OH- produced causes irreversible side reactions
to occur making recharging impossible
b) Nickel-Cadmium
 one type of
rechargeable battery
Cat (Red): 2 NiO(OH)(s)+ 2 H2O(l) + 2e-  2 Ni(OH)2(s) +2 OH-(aq) E= +0.49
V
An (Oxid):
Cd(s) + 2 OH-(aq)  2 Cd(OH)2(s) + 2e- E = +0.76 V
Net: 2 NiO(OH)(s)+ 2 H2O(l) + Cd(s)  2 Ni(OH)2(s)+ 2 Cd(OH)2(s)
Enet = +1.25 V
c) Lead Storage Battery
 typical car battery where lead
serves as the
anode, and lead coated with lead dioxide
serves as the cathode
 both electrodes dip into an electrolyte solution of
sulfuric acid
 six cells are connected in series
Cat (Red): PbO2(s)+ HSO4-(aq) + 3 H+(aq) + 2e-  PbSO4(s)+ 2 H2O(l) E= +1.68
V
An (Oxid):
Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2e- E = +0.36 V
Net: Pb(s)+ PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) 2 PbSO4(s)+ 2 H2O(l)
Enet = +2.04 V
d) Fuel Cells
 cells where reactants are continuously supplied
 the energy from this reaction can be used to run
machines
 one type is the hydrogen-oxygen fuel cell
 hydrogen gas is pumped in at the anode
while oxygen gas is pumped in at the cathode
(which both have a lot of surface area)
 pressure is used to push the H2 through a
platinum catalyst which splits the H2 into 2H+
and 2e-
 the 2e- move through an external circuit towards the
cathode generating electrical energy
 the O2 is also pushed through the platinum catalyst
forming two oxygen atoms
 the H+ ions and oxygen atoms combine to form water
Cathode (Red): O2(g) + 4 H+(aq) + 4e-  2 H2O(l)
Anode (Oxid):
Net:
2 H2(g)  4 H+(aq) + 4e-
O2(g) + 2 H2(g)  2 H2O(l)
E=+1.23 V
E = 0.00 V
Enet = +1.23 V
Hydrogen-oxygen Fuel
Cell
 need a source of hydrogen…reformers are used to
convert CH4 or CH3OH into H2 and CO2
 unfortunately, CO2 is a
greenhouse gas
 about 24-32% efficient where gas-powered car is
about 20% efficient
K. Electrolytic Cells
1. The Basics
 in an electrolytic cell, electrical
energy is used to
force a nonspontaneous chemical reaction to
occur (opposite of a voltaic cell)
 these reactions have a negative Enet
 commonly used to electroplate metals
(eg. gold, silver, bronze, chromium etc),
recharge batteries, and split compounds into
useful gases (eg. H2, O2, Cl2 etc)
 the electrolytic cell is hooked up to a battery or
power supply (instead of load or external circuit)
so the flow of e- is “pushed” by an outside
force
 the cathode of the electrolytic cell is connected to
the anode of the battery and therefore is
negative
anode of the electrolytic cell is connected to
the cathode of the battery and therefore is
Positive
 the
Electrolysis song
Voltaic Cells
Electrolytic Cells
 chemical to electrical energy
 electrical to chemical energy
 cathode +
 cathode –
anode –
 usually contains porous
anode +
 does not (usually) contain a
cup or salt bridge
porous cup or salt bridge
 Enet is positive (spont)
 Enet is negative (nonspont)
 has a voltmeter or external
 has a power supply
load
 e– flow from anode to cathode
 oxidation at anode
 reduction at cathode
 cations migrate to cathode
 anions migrate to anode
Practice Question
Answer: B
 some processes are used in industry to produce
gases, for example:
1. the Hall-Heroult cell for producing Al
…aluminum oxide is electrolyzed using carbon
electrodes …liquid aluminum is collected
http://images.google.ca/imgres?imgurl=http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter18/Text_Im
ages/FG18_18.JPG&imgrefurl=http://wps.prenhall.com/wps/media/objects/602/616516/Chapter_18.html&h=756&w=1600&sz=183&
tbnid=4cFJrFlK4noQXM:&tbnh=70&tbnw=150&hl=en&start=11&prev=/images%3Fq%3Dhallheroult%2Bcell%26svnum%3D10%26hl%3Den%26lr%3D
2. a chlor-alkali plant for producing
chlorine gas …a saturated sodium chloride
solution is electrolyzed …chlorine gas is formed
and collected at the anodes
http://www.cheresources.com/chloralk.shtml
Example 1
An electric current is passed through a solution of nickel
(II) nitrate using inert electrodes. Predict the anode and
cathode reactions, overall reaction, and minimum voltage
required.
H2O(l)
Ni2+(aq)
NO3-(aq)
OA
OA with H+(aq) OA/RA
SOA
SRA
Cathode SOA(Red): 2 [Ni2+(aq) + 2e-  Ni(s)
]
E = -0.26 V
Anode SRA(Ox): 2 H2O(l)  O2(g) + 4 H+(aq) + 4e- E = -1.23 V
Net:
2 Ni2+(aq) + 2 H2O(l)  2 Ni(s)+ O2(g) + 4 H+(aq)
Enet = -1.49 V
Example 2
An electric current is passed through a solution of
potassium iodide using inert electrodes. Predict the anode
and cathode reactions, overall reaction, and minimum
voltage required.
H2O(l)
K+(aq)
I-(aq)
OA
RA
OA/RA
SRA
SOA
Cathode SOA(Red):2 H2O(l) +2 e-  H2(g) + 2 OH- (aq)E = -0.83 V
Anode SRA(Ox):
Net:
2 I-(aq)  I2(s) + 2e-
E = -0.54 V
2 H2O(l) + 2 I-(aq)  H2(g) + 2 OH-(aq) + I2(s)
Enet = -1.37 V
Example 3
An electric current is passed through a solution of
copper(II) sulphate using a carbon electrode and a metal
electrode. Predict the anode and cathode reactions, overall
reaction, and minimum voltage required.
Cu2+(aq)
OA
SOA
Cathode SOA(Red):
SO42-(aq)
OA/RA
H2O(l)
OA/RA
SRA
2 [Cu2+(aq) + 2 e-  Cu(s) ]
E = +0.34 V
+
- E = -1.23 V
Anode SRA(Ox): 2 H O  O
+
4
H
+
4e
2 (l)
2(g)
(aq)
Net: 2 H2O(l) + 2 Cu2+(aq)  2 Cu(s) + O2(g) + 4 H+(aq)
Enet = -0.89 V
*** Note: chlorine (ions) is an exception to our
rules…
when water and chlorine are competing as
reducing agents, water is the stronger RA but
chloride ions is chosen because the
transfer of e- from H2O to O2 is more difficult
…called overvoltage
Example 4
An electric current is passed through a solution of sodium
chloride. Predict the anode and cathode reactions, overall
reaction, and minimum voltage required.
Na+(aq)
OA
Cl-(aq)
RA
SRA
H2O(l)
OA/RA
SOA
Cathode SOA(Red):2 H2O(l) +2 e-  H2(g) + 2 OH- (aq)E = -0.83 V
Anode SRA(Ox):
2 Cl-(aq)  Cl2(g) + 2e-
E = -1.36 V
Net: 2 H2O(l) + 2 Cl-(aq)  H2(g) + Cl2(g) + 2 OH- (aq)
Enet = -2.19 V
2. Quantitative Study of Electrolysis
 quantitative analysis (stoich) provides information
on necessary quantities, current and/or time for
electrolytic reactions
 the unit for charge (q) is the Coulomb (C)
 one e- carries 1.60 x 10-19 C
 this means that one mole
of charge
of e- carry 9.65 x 104 C
of charge
 9.65 x 104 C/mol
is called the
Faraday constant (F) (see Data Booklet pg
3)
ne- = q
F
q = It
where: ne- = number of moles of electrons (mol)
q = charge in Coulombs (C)
F = Faraday constant
= 9.65 x 104 C/mol
I = current in C/s or Amperes (A)
t = time in seconds (s)
 the above equations can be combined into one
equation:
ne- = It
F
 we can use these equations in stoichiometric
calculations for current, time, mass, moles of a
substance and moles e-
Example 1
An electrochemical cell caused a 0.0720 mol of e- to flow
through a wire. Calculate the charge.
ne- = 0.0720 mol
F = 9.65 x 104 C/mol
ne- = q
F
0.0720 mol =
q
9.65 x 104 C/mol
q = 6948 C
= 6.95  103 C
Example 2
Determine the number of moles of electrons supplied by
a dry cell supplying a current of 0.100 A to a radio for
50.0 minutes.
ne- = It
I = 0.100 A (C/s)
F
t = 50.0 min  60
= (0.100 C/s)(3000 s)
s/min
4 C/mol
9.65
x
10
= 3000 s
= 0.00311 mol
F = 9.65 x 104 C/mol
Example 3
If a 20.0 A current flows through an electrolytic cell
containing molten aluminum oxide for 1.00 hours, what
mass of Al(l) will be deposited at the cathode?
Al3+(l) + 3 e-

ne- = It
F
= (20.0 A)(1.00 h  3600
s/h)
9.65 x 104 C/mol
= 0.746…mol
Al(l)
n = 0.746…mol  1/3
= 0.248…mol
M = 26.98 g/mol
m = nM
=
(0.248…mol)(26.98g/mol)
= 6.71 g
Practice Question
L. Rust and Corrosion
Answer: D
 corrosion can be viewed as the process of
returning metals to their natural state (ore)
 the metal is oxidized causing the loss of
structural integrity
 most metals develop a thin oxide coating which
then protects their internal atoms against
further oxidation
 commonly, the oxide coating will scale off leaving
new metal exposed an susceptible to corrosion
 salt will speed up the oxidation by acting as a
salt bridge
O2(g)
H2O
droplet
Fe(s)
Fe(OH)2(s)
rust
cathode
anode
Cathode SOA(Red): O2(g) + 2H2O(l) + 4e-  4 OH-(aq)
Anode SRA(Ox):
2 [ Fe(s)  Fe2+(aq) + 2e-]
O2(g) + 2H2O(l) + 2Fe(s)  4 OH-(aq) + 2 Fe2+(aq)
Net:
O2(g) + 2H2O(l) + 2Fe(s)  2 Fe(OH)2(s)
Practice Question
Answer: D
M. Prevention of Corrosion
 applying a coating of paint to protect metal from
oxidation
 other metals (eg. Zn, Cr, Sn) can be plated onto
metals that you don’t want to corrode (eg. steel
(Fe))
 this coating is of a metal that is a stronger
reducing agent than the metal that is to be
protected…the coating metal will react instead and is
called the sacrificial anode
Fe
Zn coating
 this method is also called cathodic protection
 has been in use before the science of
electrochemistry was developed
 Sir Humphrey Davy first used cathodic protection
on British naval ships in 1824!
 can be used to protect any metal but steel (iron) is
most commonly protected
 we use steel (iron) for many structures such as
buried fuel tanks, septic tanks, pipelines, hulls of
ships, bridge supports etc
 to protect these structures by cathodic protection, an
active metal (eg. Mg, Zn, Al) is connected by a
wire to the structure
 because the attached metal is stronger reducing
agent a than the iron in the steel, the more
active metal supplies the e- for reduction
and therefore the steel (iron) becomes the cathode
and is protected
Fe (tank)
more active
metal eg)
Mg, Zn, Al
Video – sacrificial anode
 another protection method is alloying pure
metals, which changes their reduction potential
 stainless steel contains chromium and nickel,
changing steels reduction potential to one
characteristic of noble metals like gold
(basically unreactive)
 electroplating is the process of depositing the
neutral metal on the cathode by
metal ions in solution
reducing
 an object can be plated by making it the cathode
in an electrolytic cell
plating metal
Video electroplating
containing ions of the
Electrolytic Cell
http://player.d
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d=D5A677D5947C-46E9BEFF875236C12D46
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