Chapter 20
Electrochemistry and
Oxidation – Reduction
Mg + HCl → MgCl2 + H2
Mg + HCl → MgCl2 + H2
• Redox Reactions:
– Reduction is the decrease of an oxidation
number by the gain of electrons.
– Oxidation is the increase of an oxidation
number by the loss of electrons.
Mg + HCl → MgCl2 + H2
LEO the lion says GER
Lose Electrons Oxidation
Gain Electrons Reduction
Mg + HCl → MgCl2 + H2
• Half-Reactions:
Oxidation ½ reaction:
Mg → Mg2+ + 2eElectron flow
Reduction ½ reaction:
2H+ + 2e- → H2
Half-Cells
• A half-cell represents a half-reaction from
a redox reaction.
• A half-cell is made by placing a piece of
reactant in an electrolyte solution.
• The anode is the half-cell where
oxidation occurs.
• The cathode is the half-cell where
reduction occurs.
Mg + HCl → MgCl2 + H2
• Half-Reactions:
Oxidation ½ reaction:
Mg → Mg2+ + 2e-
•Half-Cells:
A
1
1
1
A
Reduction ½ reaction:
2H+ + 2e- → H2
A
1
1
1
Mg + HCl → MgCl2 + H2
AA
1
1
1
Oxidation (Anode)
Mg → Mg2+ + 2e-
A
A
1
1
AA a
Aaaaa
1
Reduction (Cathode)
2H+ + 2e- → H2
Mg + HCl → MgCl2 + H2
Oxidation (Anode)
Mg → Mg2+ + 2e-
Reduction (Cathode)
2H+ + 2e- → H2
Mg + HCl → MgCl2 + H2
This is a battery.
A battery is a type of
galvanic cell (voltaic cell).
A galvanic cell is a device in
which a chemical reaction
produces electrical energy.
Mg + HCl → MgCl2 + H2
2.37V
Oxidation (Anode)
Mg → Mg2+ + 2e-
Reduction (Cathode)
2H+ + 2e- → H2
Standard Reduction Potentials
Figure 20.1 Page 650
Mg + HCl → MgCl2 + H2
E°cell = E°ox + E°red
• E°cell = electromotive force (emf)
= voltage
= cell potential
Understanding Reduction Potentials
Standard conditions:
•1M for ion concentrations
•1atm for gas pressures
•25°C
The standard hydrogen electrode
Measuring Potential for a Copper Half-Cell
Measuring Potential for a Copper Half-Cell
Zn + CuSO4(aq)  ZnSO4(aq)+ Cu
Zn + CuSO4(aq)  ZnSO4(aq)+ Cu
•
•
•
•
•
Write the half-reactions
Draw the half-cells
Label the anode and the cathode
Label the direction of electron flow
Draw a salt bridge and label the direction of ion
flow within it.
• Describe what you would observe at the anode
as the reaction proceeds.
• Describe what you would observe at the cathode
as the reaction proceeds.
• Calculate the cell potential (E°cell)
Zn + CuSO4(aq)  ZnSO4(aq)+ Cu
A
Zn + CuSO4(aq)  ZnSO4(aq)+ Cu
E°cell and spontaneity
• +E°cell = spontaneous cell reaction
• -E°cell = nonspontaneous cell reaction
What is the potential for the cell
Zn | Zn2+(1.0 M) || Cu2+(1.0 M) | Cu
E°cell = +1.10V
The reaction is spontaneous
Zinc reacts with a solution of copper (II) ions
What is the potential for the cell
Ag | Ag+(1.0 M) || Li+(1.0 M) | Li
• From the table of standard reduction potentials
(Appendix H), you find:
Ag → Ag+ + eE°ox = -0.7991V
Li+ + e- → Li
E°red = -3.09V
E°cell = -3.8891V
E°cell = -3.89V
• The reaction is nonspontaneous
• Silver will not react with a solution of lithium ions
E°red = – 0.440V
Write the cell reaction and calculate
the potential of this galvanic cell.
Co + 2Fe3+ → Co2+ + 2Fe2+
E°cell = +1.048V
Write the cell reaction and calculate the potential of a cell
consisting of a standard bromine electrode as the anode and
a standard chlorine electrode as the cathode.
2Br- + Cl2 → Br2 + 2ClE°cell = +0.2943V
Write the cell reaction and calculate the potential of a cell
consisting of a standard bromine electrode as the anode
and a standard chlorine electrode as the cathode.
Free Energy Change and Cell Potential
∆G = ‒nFEcell
n = number of moles of electrons
F = faraday = 96,485 J V-1 mol-1 ≈ 96,500 J V-1 mol-1
≈ 96.5 kJ V-1 mol-1
Free Energy Change and Cell Potential
∆G = ‒nFEcell
How do ∆G and Ecell relate?
Calculate the cell potential and the ∆G for the reaction:
Cd + Pb2+ → Cd2+ + Pb
A
Calculate the cell potential and the ∆G for the reaction:
Cd + Pb2+ → Cd2+ + Pb
E°cell = +0.277V
∆G = -53.5 kJ
Calculate the cell potential and the ∆G for the reaction:
Sn4+ + 2Fe2+ → Sn2+ + 2Fe3+
Calculate the cell potential and the ∆G for the reaction:
Sn4+ + 2Fe2+ → Sn2+ + 2Fe3+
E°cell = -0.62V
∆G = +120kJ
The Effect of Concentration on Cell Potential
Ecell vs. E°cell
The Nernst Equation (Page 657)
Concentration and Ecell
• Example. Calculate the cell potential for the
following:
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
Where [Cu2+] = 0.30 M and [Fe2+] = 0.10 M
Concentration and Ecell (cont.)
• First, need to identify the 1/2 cells
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
Cu2+(aq) + 2e-
Cu(s)
E°1/2 = 0.34 V
Fe2+(aq) + 2e-
Fe(s)
E°1/2 = -0.44 V
Fe(s)
Fe(s) + Cu2+(aq)
Fe 2+(aq) + 2e-
E°1/2 = +0.44 V
Fe2+(aq) + Cu(s)
E°cell = +0.78 V
Concentration and Ecell (cont.)
• Now, calculate Ecell
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
E°cell = +0.78 V
Ecell = E°cell - (0.05916/n)log(Q)
Fe  (0.10)
Q

 0.33
Cu  (0.30)
2
2
Ecell = 0.78 V - (0.05916 /2)log(0.33)
Ecell = 0.78 V - (-0.014 V) = 0.794 V
Concentration and Ecell (cont.)
• If [Cu2+] = 0.30 M, what [Fe2+] is needed so that
Ecell = 0.76 V?
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
E°cell = +0.78 V
Ecell = E°cell - (0.05916/n)log(Q)
0.76 V = 0.78 V - (0.05916/2)log(Q)
0.02 V = (0.05916/2)log(Q)
0.676 = log(Q)
4.7 = Q
Concentration and Ecell (cont.)
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
4.7 = Q
Q
2
Fe
 
Cu 
2
 4.7
Fe 
Q
 4.7
0.30
2

[Fe2+] = 1.4 M
Comparing Q and Ecell
From the last two problems:
E°cell = +0.78V
Fe(s) + Cu2+(aq)
Fe2+(aq) + Cu(s)
[Cu2+]
[Fe2+]
Q
Ecell
0.30M
0.10M
0.33
+0.794V
0.30M
1.4M
4.7
+0.76v
Ecell = E°cell - (0.05916/n)log(Q)
Example 20.9
Page 657
E°cell = +0.179V
E°cell vs K
This formula can be derived from the Nernst Equation
Example 20.10
Page 659
E°cell from example 20.7 = +0.277V
K = 2.31 x 109
Calculate the value of K for the reaction.
Br2 + 2Cl– → 2Br – + Cl2
K = 1.12 x 10-10
Comparing K and Ecell from the problems:
Reactants
Products
E°cell
K
+0.277V
2.31 x 109
-0.2943V
1.12 x 10-10
A
Table Page 661
Balancing Redox Reactions (See Notes)
•
•
•
The procedure used to balance redox reactions can be broken down into several
steps. These steps are:
Divide the overall reaction into two half equations.
Balance each half equation separately:
–
–
–
–
•
•
•
•
balance the atoms in each half equation
balance the electrons in each half equation
balance any oxygens by adding H2O to one side of the half equation
balance any hydrogens by adding H+ to one side of the half equation
Multiply each half equation so that the number of electrons lost and gained balance.
Add the half equations together and simplify.
If asked to balance the reaction in “basic solution” do so by adding OH- ions. Add
enough OH- to neutralize any H+ ions. Make sure to add an equal number of OHions to each side of the equation.
Check your final answer to make sure that
– there are the same number of atoms of each element on both sides
– the net charge is the same on both sides
– the coefficients are in the simplest whole number ratio possible
Cr3+ + Cl-  Cr + Cl2
MnO4- + Fe2+  Mn2+ + Fe3+
MnO4- + 8H+ + 5Fe2+  Mn2+ + 4H2O + 5Fe3+
MnO4- + Fe2+  Mn2+ + Fe3+ (basic solution)
MnO4- + 8H+ + 5Fe2+  Mn2+ + 4H2O + 5Fe3+
MnO4- + 4H2O + 5Fe2+  Mn2+ + 8OH- + 5Fe3+
Sn2+ + Fe3+ → Sn4+ + Fe2+
2+
Sn
+
3+
2Fe
→
4+
Sn
+
2+
2Fe
Zn + NO3- → Zn2+ + NH4+
4Zn + 10H+ + NO3- → 4Zn2+ + NH4+ + 3H2O
Zn + NO3- → Zn2+ + NH4+ (basic solution)
4Zn + 10H+ + NO3- → 4Zn2+ + NH4+ + 3H2O
4Zn + 7H2O + NO3- → 4Zn2+ + NH4+ + 10OH-
Galvanic and Electrolytic Cells
• See your notebook for supplemental notes on
electrolysis.
• Oxidation-reduction or redox reactions take
place in electrochemical cells.
• There are two types of electrochemical cells.
– Spontaneous reactions (+Ecell) occur in galvanic
(voltaic) cells.
– nonspontaneous reactions (-Ecell) occur in
electrolytic cells.
Electrolytic Cells
• Electrolysis is the process by which an
electrical current (voltage) causes a
chemical reaction to occur.
Electrodes & Charge
• The anode of an electrolytic cell is positive
(cathode is negative), since the anode attracts
anions from the solution.
• However, the anode of a galvanic cell is
negatively charged, since the spontaneous
oxidation at the anode is the source of the cell's
electrons or negative charge. The cathode of a
galvanic cell is its positive terminal.
• In both galvanic and electrolytic cells, oxidation
takes place at the anode and electrons flow from
the anode to the cathode.
Zn + Cu2+ → Cu + Zn2+
Ecell = +1.10V
e-
A
Galvanic
Cell
Figure 20.21
2Na+ + 2Cl- → 2Na + Cl2
Ecell = - 4.07 V
Page 667
Electrolytic Cell
Electrolysis of NaCl
• Cathode (-): Na+ + e- → Na
-2.71 V
• Anode (+): 2Cl‾ → Cl2 + 2e-1.36 V
2Na+ + 2Cl‾→ 2Na + Cl2
-4.07V
• The battery used to drive this reaction must therefore
have a potential of at least 4.07 volts.
Electrolysis
• For any compound undergoing electrolysis
it is the negative ion which is oxidized and
the positive ion which is reduced.
• Cathode (-): Na+ + e- → Na
-2.71 V
• Anode (+): 2Cl- → Cl2 + 2e-1.36 V
-4.07V
Rules for the electrolysis of aqueous solutions.
• Rule 1: At the anode the negative ion will
be oxidized unless it is easier to oxidize
water. Anions that do not oxidize as easy
as water are F-, NO3- and SO42-.
Rules for the electrolysis of aqueous solutions.
• Rule 2: At the cathode the positive ion will
undergo reduction unless it is easier to
reduce water. Cations that do not reduce
as easy as water are the metal cations
from groups I and II and Al3+.
Rules for the electrolysis of aqueous solutions.
• The following half reactions are important for
you to know in this process.
– Oxidation of water: 2H2O → O2 + 4H+ + 4e– Reduction of water: 2H2O + 2e- → H2 + 2OH‾
– Oxidation of hydroxide: 4OH- → O2 + 2H2O + 4e-
Oxidation of water:
2H2O → O2 + 4H+ + 4e-
Reduction of water:
2H2O + 2e- → H2 + 2OH‾
Oxidation of hydroxide:
4OH- → O2 + 2H2O + 4e-
Electrolysis of Aqueous NaCl
• Reduction occurs at the cathode. But, now there are two
substances that can be reduced at the cathode: Na+ ions and
water molecules.
– It is much easier to reduce water than Na+ ions, therefore
water will reduce at the cathode.
Cathode (-): 2H2O + 2e- → H2 + 2OH‾
• Oxidation occurs at the anode: There are two substances that
can be oxidized at the anode: Cl‾ ions and water molecules.
– Chloride oxidizes easier than water. Therefore Cl‾ oxidizes
at the anode.
Anode (+): 2Cl‾ → Cl2 + 2eThe overall reaction for this process is 2H2O + 2 Cl‾ → H2 + Cl2 + 2OH‾
Electrolysis of Aqueous NaCl
Figure 20.22 Page 668
Electrolysis of aqueous sodium chloride doesn't give the
same products as electrolysis of molten sodium chloride.
Electrolysis of aqueous sodium chloride doesn't give the
same products as electrolysis of molten sodium chloride.
Write the half reaction that occurs at each electrode and the
overall balanced equation for the electrolysis of NiBr2(aq).
• Anode: 2Br – → Br2 + 2e–
• Cathode: Ni2+ + 2e– → Ni
• Overall: 2Br – + Ni2+ → Br2 + Ni
Write the half reaction that occurs at each electrode and the
overall balanced equation for the electrolysis of KOH(aq).
• Anode: 4OH– → O2 + 2H2O + 4e–
• Cathode: 2H2O + 2e– → H2 +2OH–
• Overall: 2H2O → 2H2 + O2
Write the half reaction that occurs at each electrode and the
overall balanced equation for the electrolysis of Na2SO4 (aq).
• Anode: 2H2O → O2 + 4H+ + 4e–
• Cathode: 2H2O + 2e– → H2 + 2OH–
• Overall: 2H2O → 2H2 + O2
Notice that a salt bridge is not necessary in an
electrolytic cell because you are supplying the
electricity rather than needing to have the reaction
continue in order to produce the voltage.
Uses of Electrolysis
• Production of many pure elements.
Uses of Electrolysis
•
These inexpensive decorative
carabiners have an anodized
aluminum surface that has been
dyed and are made in many colors.
• Anodizing is an
electrolytic process used
to increase the thickness
of the natural oxide layer
on the surface of metal
parts. Anodizing
increases corrosion
resistance and wear
resistance, and provides
better adhesion for paint
primers and glues than
bare metal.
Uses of Electrolysis
• Electroplating is a
plating process that
uses electrical current
to reduce cations of a
desired material from
a solution and coat a
object with a thin
layer of the material,
such as a metal.
Uses of Electrolysis
• Electrolysis is also used
in the cleaning and
preservation of old
artifacts. Because the
process separates the
non-metallic particles
from the metallic ones, it
is very useful for cleaning
old coins and even larger
objects.
Coulombs
• A Coulomb (C) is the amount of electricity
carried in 1 second by a current of 1 amp.
C=A•s
Amperes (Amps)
• The amp is a measure of the amount of electric
charge passing a point per unit time.
A=?
• A current of one amp is one coulomb of charge
going past a given point per second.
Faradays
• The faraday (F) is the charge on 1 mole of
electrons.
• F = mol e− = 96,485C ≈ 96,500C.
Amounts of various metals deposited at the cathode
by 1 faraday of electricity (Figure 20.27 Page 671)
Faradays Law of Electrolysis
• The mass of a substance produced or
consumed at an electrode during
electrolysis is directly proportional to the
amount of electricity that passes through
the cell.
What mass of Cu is produced at the
cathode when 1.600A flows for 1.000 hour
through a solution of CuSO4?
How much time is required to produce
1.000 x 106g of Mg when a current of
1.500 x 105A flows through a solution of MgCl2?
Calculate the volume of H2 gas at 25°C and 1.00 atm that will
collect at the cathode when an aqueous solution of Na2SO4 is
electrolyzed for 2.00 hours with a 10.0-amp current.
4H2O + 4e– → 2H2 + 4OH–
or
2H2O + 2e– → H2 + 2OH–