Chemistry 130
Electrochemistry
Dr. John F. C. Turner
409 Buehler Hall
jturner@ion.chem.utk.edu
Chemistry 130
Redox and electron transfer
A redox reaction is one where electrons are transferred between reactants
and products, changing the formal oxidation states of the species involved.
In order to balance a redox reaction, we use half equations to show the
change in formal oxidation state for the species involved:
2+
Cu aq
_
2e
Fe2aq+ − e_
0
Cu s
Fe3+
aq
reduction
oxidation
Loss of Electrons is Oxidation – LEO – is a useful mnemonic for
remembering the direction of redox reactions.
_
We term reactions such as
Fe2+
−
e
aq
couple.
Chemistry 130
Fe3+
aq
a half-reaction or a
Balancing redox reactions
A stoichiometric redox reaction is composed of two couples – one is an
oxidation and one a reduction. Both the oxidation and reduction couple
must be present in order to conserve charge.
Balancing a redox reaction is lengthy but straightforward. The key to
balancing a reaction is to ensure that the electrons are balanced on both
sides, using acid or base to balance the final equation with respect to acid
numbers.
Chemistry 130
Balancing redox reactions
A stoichiometric redox reaction is composed of two couples – one is an
oxidation and one a reduction. Both the oxidation and reduction couple
must be present in order to conserve charge.
Balancing a redox reaction is lengthy but straightforward. The key to
balancing a reaction is to ensure that the electrons are balanced on both
sides, using acid or base to balance the final equation with respect to acid
numbers.
Example: Write a balanced equation for the reduction of permanganate ion
Mn2aq+
by thiosulphate
in acid solution. The products of the reaction are sulfate ion
and
The unbalanced reaction
is
MnO4 aq
S2 O23 aq
Chemistry 130
Mn2aq+
SO24 - aq
Balancing redox reactions
Example: Write a balanced equation for the reduction of permanganate ion
by thiosulphate in acid solution. The products of the reaction are sulfate ion
2+
and Mn aq
MnO-4 aq
S2 O23 aq
Mn2aq+
SO24 - aq
Step 1: Assign oxidations states to the species
22+
MnO4 aq
S2 O3 aq
Mn aq
VII
Step 2:
II
Mn
S
Write down the half-reactions
-
MnO4 aq
Chemistry 130
Mn
5e
-
Mn aq
S2 O3 aq
2-
2SO4
2-
SO4
II
VI
S
2+
2-
-
aq
8e
aq
Balancing redox reactions
-
2-
MnO4 aq
2+
S2 O3 aq
2-
Mn aq
SO4 aq
Step 1: Assign oxidations states to the species
22+
MnO4 aq
S2 O3 aq
Mn aq
VII
Step 2:
II
Mn
S
Write down the half-reactions
-
MnO4 aq
Mn
5e
-
Mn aq
S2 O3 aq
2-
2SO4
2-
SO4
II
aq
VI
S
2+
2-
-
aq
8e
Step 3: Use acid and water to balance the number of oxygen atoms present
-
MnO4
-
aq
5 H2 O
Chemistry 130
5e
+
Mn aq
S2 O3 aq
2 SO4
8 H aq
2-
2+
4 H2 O
2-
-
aq
8e
+
10 H aq
Balancing redox reactions
-
2-
MnO4 aq
S2 O3 aq
2+
Mn aq
2-
SO4 aq
After steps 1, 2 and 3, we have two balanced half-equations:
-
MnO4 aq
-
5e
+
8H aq
2-
5 H2 O
S2 O3 aq
2+
Mn aq
2-
2SO4 aq
4 H2 O
-
+
8e
10 H aq
Step 4: We now multiply them individually to balance the number of
electrons:
-
8 MnO4
-
aq
25 H 2 O
Chemistry 130
40 e
+
64 H aq
2-
5S2 O3 aq
2+
8 Mn aq
2-
10 SO4 aq
32 H 2 O
-
40 e
+
50 H aq
Balancing redox reactions
-
2-
MnO4 aq
2+
S2 O3 aq
Mn aq
2-
SO4 aq
Step 5: We now add the two half- reactions together:
-
8 MnO4
-
aq
2+
8 Mn aq
40 e
32 H 2 O
+
64 H aq
2-
10SO4
25 H 2 O
-
aq
40 e
2-
5S2 O3 aq
+
50 H aq
Step 6: Species appear on both sides of the reaction, which we cancel. I
particular, 40e- occur on both sides which means that we are correct in
terms of the oxidation and reductions. Also, we64−
have50 H+aq and 32− 25 H 2 O
Chemistry 130
Balancing redox reactions
-
2-
MnO4 aq
-
8 MnO4
S2 O3 aq
-
aq
8 Mn2+
aq
+
2+
Mn aq
2-
SO4 aq
2-
40 e
64 H aq
25 H 2 O
5S2 O3 aq
32 H 2 O
10 SO24 aq
40 e-
50 H +aq
8 Mn2aq+
10SO24 - aq
Step 6: The final reaction is
8 MnO-4
aq
14 H+aq
Chemistry 130
5S2 O23 -aq
7 H2 O
Balancing redox reactions
Example: In basic solution, bromine disproportionates to give bromide ion
and bromate ion. Write a balanced equation for this reaction.
The unbalanced reaction is
Br 2 l
Br - aq
BrO-3 aq
Step 1: Assign oxidations states to the species
Br 2 l
Br
Step 2:
1
Br 2 l
2
0
Br
Br
aq
−I
-
BrO3 aq
Br
−V
Write down the half-reactions:
Br 2 l
eBr - aq
or
1
Br 2 l
BrO-3 aq
5 e2
Chemistry 130
-
2e
Br 2 l
2 Br
aq
2 BrO-3
aq
10 e-
Balancing redox reactions
Br 2 l
Br -aq
BrO-3 aq
Step 3: Use acid and water to balance the number of oxygen atoms present
Br 2 l
6 H2 O
-
aq
3 aq
2e
2 Br
Br 2 l
2 BrO
-
10 e
+
12 H aq
We now have two individually balanced half reactions.
Step 4: We now multiply them individually to balance the number of
electrons:
-
-
5 Br 2 l
10 e
10 Br aq
6 H2 O
Br 2 l
2BrO3 aq
Chemistry 130
-
-
10 e
+
12 H aq
Balancing redox reactions
Br 2 l
Br - aq
BrO-3 aq
Step 5: We now add the two half- reactions together:
-
-
5 Br 2 l
10 e
10 Br aq
6 H2 O
Br 2 l
2BrO3 aq
6 Br 2 l
or
10 e-
3 Br 2 l
3 H2 O
-
-
10 e
10 Br -aq
6 H2 O
-
5 Br aq
-
BrO3 aq
+
12 H aq
2 BrO-3 aq
10 e-
12 H +aq
+
6 H aq
This is the balanced reaction in acid solution. We need the reaction in basic
solution so we add sufficient hydroxide ion to remove the hydronium ion.
Chemistry 130
Balancing redox reactions
Br 2 l
Br -aq
BrO-3 aq
This is the balanced reaction in acid solution. We need the reaction in basic
solution so we add sufficient hydroxide ion to remove the hydronium ion.
Acid
3 Br 2 l
3 H2 O
Base
3 Br 2 l
6OH -aq
3 Br 2 l
6OH aq
3 Br 2 l
6OH aq
-
-
-
5 Br aq
3 H2 O
3 H2 O
-
BrO3 aq
5 Br -aq
+
6 H aq
BrO-3 aq
-
BrO3 aq
-
BrO3 aq
5 Br aq
5 Br aq
-
6 H2O
-
3 H2 O
Which is the final, balanced equation in basic solution.
Chemistry 130
6 H +aq
6 OH- aq
Redox reactions and electrochemical cells
Redox reactions involve the transfer of electrons from one species to
another. For the two previous examples in acidic and basic solution, there
reactions take place implicitly in the same system.
If we separate the two reaction physically, then no reaction will obviously
take place unless there is a way that the electrons can pass between the
oxidative and reductive parts of the reaction – the half-reactions.
If we connect the two reactions via a conducting wire, the external electrical
contact, then electrons can move through the wire.
This is known as a voltaic cell.
A salt bridge connects the rest of the circuit – the internal electrical contact.
Chemistry 130
Redox reactions and electrochemical cells
We can harness the current that moves through the cell to do work and
when we do so, we are converting chemical energy into work of some
description – a light bulb or a motor are two obvious applications.
In a simple cell based on copper and silver, the half reactions are:
+
Ag aq
_
e
Cu 0s − 2 e_
Ag s
0
reduction
Cu 2aq+
oxidation
The reaction between copper metal and silver ion occurs because the
electrons on copper are less tightly bound – copper is more electropositive
than silver.
Spontaneously, the reaction will proceed at the Cu surface to form silver
metal and copper ion.
Chemistry 130
Electrochemical cells
An electrochemical cell is system that contains an oxidation couple and a
reduction couple, together with two electrodes that connect the two
reactions and allow electrons to pass and a salt bridge that allows the
necessary charge neutrality of the system to be maintained.
The two couples are termed half cells and in general, an oxidative and a
reductive cell can be assembled to form the full electrochemical cell.
There is a special nomenclature that is used to describe electrochemical
cells.
Oxidation takes place at the anode
Reduction takes place at the cathode
Chemistry 130
Cell diagrams
Cell diagrams are used to represent the cell and is a method that
represents the full cell reaction. We denote phase boundaries in the cell
with a vertical line: |
and the boundary between the two cells, which contains the external
electrical connection
(the wire) and the internal electrical connection (the
∥
salt bridge) by a double line:
For the silver and copper half-reactions,
we write
the cell as
+
_
0
Ag aq
e
Ag s
Cu 0s − 2 e_
0
2+
Cu 2aq+
+
0
Cu s ∣ Cu aq ∥ Ag aq ∣ Ag s
Anode
Cathode
Oxidation
Reduction
Chemistry 130
Cell diagrams
A standard cell is written with the anode on the left and the cathode on the
right.
+
_
Ag aq
0
e
Ag s
Cu 0s − 2 e_
Cu 2aq+
0
2+
+
0
Cu s ∣ Cu aq ∥ Ag aq ∣ Ag s
Anode
Cathode
Oxidation
Reduction
The standard cell potential is then written as
°
°
E = E1/ 2 right
°
− E1/ 2 left
°
E ° = E1/° 2 cathode − E1/2
anode
Chemistry 130
Potentials and the Gibbs function
°
E1 /2
For each half reaction, we can write a half-cell potential,
such that we
can calculate the potential of the full cell by adding these half-cell potentials
together.
The half-cell potential is related to the standard Gibbs function for the half
reaction via
°
G° = − n F E1/2
F is Faraday's constant and F = 96,485 C mol − 1 which is the charge on 1
mole of electrons. n is the number of electrons passed in the half-cell
reaction.
Manipulating half-cell or full cell potentials is equivalent to
manipulating the standard Gibbs function for the half-cell or full cell
reaction.
Chemistry 130
Standard potentials
The zero point for electrode potentials is the Standard Hydrogen
Electrode (SHE)
Pt s ∣ H2 g ∣ H3 O+aq ∥
E1/° 2 = 0.00 V
Platinum acts as a catalyst for the hydrogen-hydronium oxidation, which is
why there are two phase boundaries in the cell.
All other electrode potentials are measured against this reaction, with the
pressure of hydrogen being 1 atm, all concentrations 1 M and the
temperature being 298 K.
Standard half-cell reactions are written as reductions by convention.
Chemistry 130
Standard potentials
Example: Calculate the cell potential for the zinc-copper cell, given the
following cell potentials and construct a full cell diagram for the
spontaneous reaction.
Cu aq
2+
2e
2+
2e
Zn aq
_
Cu
0
s
0
s
E1/2 = 0.340 V
°
_
Zn
E1/2 = − 0.763 V
°
Step 1: Calculate the change in standard Gibbs function for each reaction
Cu aq
2+
2e
2+
2e
Zn aq
_
Cu
0
s
0
s
E1/2 = 0.340 V
°
G = − n F E1 /2 = − 0.68F
_
Zn
E1/2 = − 0.763 V
°
°
°
G = − n F E1 /2 = 1.526F
°
°
We do not need to include the value of F as we will convert back at the end
of the calculation.
Chemistry 130
Standard potentials
Example: Calculate the cell potential for the zinc-copper cell, given the
following cell potentials and construct a full cell diagram for the
spontaneous reaction.
2+
_
0
°
°
°
Cu aq
2e
Cu s
E1/2 = 0.340 V
G = − n F E1 /2 = − 0.68F
2+
_
0
°
°
°
Zn aq
2e
Zn s
E1/2 = − 0.763 V
G = − n F E1 /2 = 1.526F
Step 2: Arrange the equations to give a balanced stoichiometric reaction,
°
changing G as necessary
Zn
0
s
2+
aq
Cu
0
s
Cu
2+
Cu aq
_
2e
2+
Zn aq
_
°
2e
Zn
G =
0
s
2+
Cu aq
0.68F
°
G = 1.526F
Zn
0
s
°
G = 2.206F
G° is positive, so as written, the reaction is non-spontaneous, so we
reverse the reaction.
Chemistry 130
Standard potentials
Example: Calculate the cell potential for the zinc-copper cell, given the
following cell potentials and construct a full cell diagram for the
spontaneous reaction.
Cu
Zn
Cu
0
s
2+
aq
0
s
2+
aq
Cu
2+
_
Cu aq
_
2e
Zn
2+
Zn aq
Zn
°
2e
G =
0
s
°
G = 1.526F
2+
Zn
0
Zn
Cu aq
0
s
Cu s
0
s
2+
aq
°
G = 2.206F
°
G = − 2.206F
We now have a spontaneous reaction asG°
°
Step 3: Convert the calculated G
°
°
cell
E
0.68F
is negative.
back to the electrode potential:
G
− 2.206F
2.206F
= −
= −
=
= 1.103 V
nF
2F
2F
Chemistry 130
Standard potentials
Example: Calculate the cell potential for the zinc-copper cell, given the
following cell potentials and construct a full cell diagram for the
spontaneous reaction.
Cu 2aq+
Zn0s
Cu 0s
Zn 2aq+
°
G° = − 2.206F Ecell
= 1.103 V
Step 4: Construct the cell diagram from the balanced, spontaneous
reaction:
2+
0
0
2+
°
Cu aq
Zn s
Cu s
Zn aq
Ecell = 1.103 V
0
s
2+
+
0
∣ Zn aq ∥ Cu aq ∣ Cu s
Anode
Cathode
Oxidation
Reduction
Zn
Chemistry 130
°
Ecell = 1.103 V
Standard potentials
This method always gives the correct reaction and the correct direction for
the spontaneous cell reaction, even when the cell reaction is not obvious. It
is equivalent to the Right – Left rule but is simpler as you don't need to
identify the cathodic and anodic reaction initially.
Chemistry 130
Thermodynamics and electrochemistry
The relationships between the equilibrium constant, the cell potential and
the change in the standard Gibbs functions allows us to write a relationship
between the cell potential and the equilibrium constant
°
°
G = − n F Ecell
°
G = − RT ln K eq
°
− n F Ecell = − RT ln K eq
°
cell
E
Chemistry 130
RT
=
ln K eq
nF
Non-standard cell potentials
The dependency of the Gibbs function with concentration is
G=
G°
RT ln Q
where Q is the reaction quotient. Including this in the equation for the cell
potential gives the Nernst equation:
°
G =
G
RT ln Q
°
− n F Ecell = − n F Ecell
RT ln Q
°
n F Ecell = n F Ecell − RT ln Q
RT
Ecell = E −
ln Q
nF
Nernst equation
°
cell
Chemistry 130
Non-standard cell potentials
The Nernst equation allows us to relate changes in concentration to the cell
potential. For the reaction
Cu 2aq+
Zn0s
Cu 0s
Zn 2aq+
°
G° = − 2.206F Ecell
= 1.103 V
°
Ecell is written for concentrations of 1 mol l-1. In general, the potential of the
cell will vary with concentration as
2+
2+
Cu aq
0
Zn s
°
Ecell = Ecell
−
°
Ecell = Ecell −
Chemistry 130
Cu
0
s
2+
Zn aq
RT
ln Q
nF
[Zn 2+
aq ]
{ }
RT
ln
2+
nF
[Cu aq ]
Q =
[Zn aq ]
2+
[Cu aq ]
Non-standard cell potentials
Because of the logarithmic relationship between concentration and the cell
potential, we can measure very precisely very large or very small quantities.
Values of Keq that span 50 or 100 orders of magnitude result in changes in
the cell potential of a few volts, allowing us to measure very subtle changes
in concentration very accurately.
In order to use the Nernst equation, we first calculate the standard cell
potential for the balanced stoichiometric reaction.
Chemistry 130
Non-standard cell potentials
2+
0
0
Example:
Calculate
the
cell
potential
on2+ / Zn
the
Cu /based
Cu and Zn
2+
−1
2+
−1
couples when [Zn aq ] = 2 mol l and [Cu aq ] = 0.05 mol l
Step 1: Write down the half reactions for the two couples:
Cu
Zn
2+
aq
2+
aq
_
Cu
E1/ 2 = 0.340 V
_
Zn
E1/ 2 = − 0.763 V
2e
2e
0
s
0
s
°
°
Step 2: From the balanced half cell reactions, calculate the change in the
standard Gibbs function in terms of F
Cu
Zn
2+
aq
2+
aq
_
Cu
E1/ 2 = 0.340 V
_
Zn
E1/ 2 = − 0.763 V
2e
2e
Chemistry 130
0
s
0
s
°
G = − n F E1/ 2 = − 0.68F
°
°
°
G = − n F E1/ 2 = 1.526F
°
°
Non-standard cell potentials
2+
0
0
Example:
Calculate
the
cell
potential
on2+ / Zn
the
Cu /based
Cu and Zn
2+
−1
2+
−1
couples when [Zn aq ] = 2 mol l and [Cu aq ] = 0.05 mol l
Step 3: Arrange the equations to give a balanced stoichiometric reaction,
changing G° as necessary
Zn
0
s
2+
aq
Cu
0
s
Cu
Cu
2+
aq
_
2e
Zn
2+
aq
_
Zn
2+
0
Zn s
Chemistry 130
G =
0
s
0.68F
°
G = 1.526F
2+
Cu aq
Step 4: Ensure thatG°
spontaneous
0
2+
2+
Cu s
Zn aq
Cu aq
Cu aq
°
2e
0
Cu s
Zn
0
s
°
G = 2.206F
is negative and therefore the reaction is
0
s
2+
aq
°
Zn
G = 2.206F
Zn
G = − 2.206F
°
Non-standard cell potentials
2+
0
0
Example:
Calculate
the
cell
potential
on2+ / Znthe
Cu /based
Cu and Zn
2+
−1
2+
−1
couples when [Zn aq ] = 2 mol l and [Cu aq ] = 0.05 mol l
Step 5: Calculate the standard cell potential from the change in standard
Gibbs function.
°
G
− 2.206F
2.206F
°
Ecell = −
= −
=
= 1.103 V
nF
2F
2F
Step 6: Calculate the value of the reaction quotient for the reaction
Cu
2+
aq
Zn
0
s
Chemistry 130
Cu
0
s
Zn
2+
aq
Q =
[Zn
[ Cu
2+
aq
2+
aq
]
]
=
2
= 40
0.05
Non-standard cell potentials
2+
0
0
Example:
Calculate
the
cell
potential
on2+ / Znthe
Cu /based
Cu and Zn
2+
−1
2+
−1
couples when [Zn aq ] = 2 mol l and [Cu aq ] = 0.05 mol l
Step 7: Use the Nernst equation to calculate the cell potential
°
Ecell = Ecell −
[Zn 2+
]
aq
{ }
RT
ln
nF
[Cu 2aq+ ]
°
Ecell = 1.103 V
Ecell
{ }
8.314× 298
2
ln
= 1.103 − 0.013 ln {40 }
2× 96,485
0.05
= 1.103 − 0.013 ln {40 } = 1.055 V
Ecell = 1.103 −
Chemistry 130
Chemistry 130
Nuclear Chemistry
Dr. John F. C. Turner
409 Buehler Hall
jturner@ion.chem.utk.edu
Chemistry 130
Radioactivity
The majority of the chemical elements are stable in that their nuclei do not
show any form of decay.
In general, from hydrogen (Z = 1) to bismuth (Z = 83) all elements have at
least one stable nucleus with the exceptions of technetium (Z = 43) and
promethium (Z = 61).
After Bi, all the elements do not have a stable nucleus.
Even for the elements that have at least one stable isotope, many
radioactive isotopes are also known and are produced artificially.
Examples include 3H
1
tritium
Chemistry 130
14
6
C
32
15
P
The nucleus
The nucleus is a very complicated object. Part of the difficulty of describing
the nucleus is the strength of the forces involved and the level at which they
couple to each other.
The two forces in the nucleus that are important in terms of stability are the
strong nuclear force and the electromagnetic force.
Electrons are bound in the atom by the electromagnetic force and the
equilibrium size of the atom represents the magnitude of the force involved
– atoms have a radius of ~ 1-2 Å or ~ 10-10 m.
The nuclear radius is ~ 1-2 fm or ~ 10-15 m or 5 orders of magnitude
smaller than the atom.
Chemistry 130
The nucleus
The constituents of the nucleus are the neutron and the proton.
The neutron has no electric charge, a spin of ½ and a mass very similar to
that of the proton.
The proton has a single positive charge and a spin of ½ as well. Both the
neutron and the proton are therefore fermions and obey the Pauli principle
in a manner analogous to electrons in the atom:
The wavefunction of a fermion must change sign on interchange of
particles
or
No more than two fermions can occupy the same state
Chemistry 130
The nucleus
Though a description of the nucleus is extremely hard, several features are
immediately obvious.
There is a force that is attractive and extremely powerful – the strong
nuclear force.
The strong force is powerful enough to overcome the electrostatic repulsion
of the protons in the nucleus and so the nucleons in the nucleus are bound
– elements after hydrogen are usually stable up to Z = 83.
The size of the nucleus is dictated by the balance between these two
forces.
The definition of the nuclear surface and the nuclear radius and size is
ambiguous and depends on the particle.
Chemistry 130
This week
Revise kinetics and acids, in addition to the electrochemistry
Chemistry 130