Page 1
Oxidation-Reduction
The transfer of an electron from one compound to another results in the oxidation of the electron donor and the
reduction of the electron acceptor.
Loss of electrons (oxidation)
Zn(s) + Cu+2(aq)
Zn+2(aq) + Cu(s)
Gain of electrons (reduction)
So, reactions involving the transfer of electrons are often called oxidation-reduction or redox reactions.
Electron transfer reactions are numerous including corrosion processes, biological processes and
electrochemical (battery) processes.
Before turning to the study of electrochemical processes, you should review Chapter 3 with regard to balancing
redox reactions and the following definitions.
Oxidation: An atom, ion, or molecule releases electrons and is oxidized. The oxidation number of the atom
oxidized increases.
Reduction: An atom, ion, or molecule accepts electrons and is reduced. The oxidation number of the atom
reduced decreases.
Reducing agent: The atom, ion, or molecule donating electrons.
Oxidizing agent: The atom, ion, or molecule accepting the electron.
Chemical Change Leading to "Current"
The reaction in the figure below represents one that is capable of producing an electric current. A piece of zinc
immersed in a solution containing copper ions causes the copper to eventually "plate out" as the zinc dissolves
as ions. As the copper ions (which appear dark blue when in solution) disappear, the blue color begins to fade.
+
+
+
Page 2
The copper ion has been reduced to copper metal. (note that the Zn metal was the reducing agent,)
Cu+2(aq) + 2 e-1
Cu(s)
The zinc, on the other hand was oxidized to Zn+2. (here, the Cu+2 was the oxidizing agent.)
Zn(s)
Zn+2(aq) + 2 e-1
The net chemical reaction occurring in the beaker, therefore, was the spontaneous reduction of Cu+2 and the
simultaneous spontaneous oxidation of the Zn(s).
Zn(s) + Cu+2(aq)
Zn+2(aq) + Cu(s)
Although an oxidation-reduction reaction did occur here, it could not have been used as a source of current
since the Zn(s) and Cu+2 were in direct contact with one another. However, with a different physical
arrangement, that is, separating the two "half-reactions" and forcing the transferred electrons to travel through
an external circuit is a different story.
In the figure below, electrons flow from the zinc electrode to the solution of Cu+2 ions through an external wire
(and through any device to be powered by our "battery) to another electrode in the Cu+2 solution where the
reduction occurs.
1.10 volts
SO4-2
salt bridge
contains Na2SO4
Zn
Cu
porous plug
Zn+2
SO4-2
Cu+2
SO4-2
This arrangement will only work if we have provided a salt bridge, a device for maintaining a balance of the
ionic charges in the cell compartments.
Page 3
Voltaic Cells and Notation
A voltaic, or galvanic, cell is an electrochemical cell in which a spontaneous reaction produces an electric
current. The example in the last section is a voltaic cell.
Some other useful terminology involves the electrodes. The electrode at which oxidation occurs is called the
anode. Since electrons are being produced at the anode of a battery, this terminal bears a negative charge. The
electrode at which reduction occurs is called the cathode. This terminal on the battery bears a positive charge.
A shorthand method for designating particular voltaic cells involves an abbreviated representation of the two
half-reactions involved in the production of current. For example, our Zn/Cu battery in the first section would
be written:
Zn(s) | Zn+2(aq) || Cu+2(aq) | Cu(s)
In this notation, the anode (where oxidation takes place) is written on the left; the cathode (where reduction
occurs) is written on the right. The two half-cells are connected by a salt bridge, denoted by two vertical lines.
Zn(s) | Zn+2(aq)
anode
||
Cu+2(aq) | Cu(s)
salt bridge
cathode
The cell terminals are at the extreme ends of this notation, and a single vertical bar represents a phase boundary,
such as , between a solid terminal and the electrode solution.
Zn(s)
anode terminal
|
Zn+2(aq)
phase boundary
anode electrolyte
When a half reaction involves a gas, an inert material, such as platinum, serves as the terminal and an electrode
surface upon which the half-reaction takes place. The hydrogen electrode is of this type and its cell notation is
as follows.
Pt | H2(g) | H+(aq)
when serving as the anode
and
H+(aq) | H2(g) | Pt
when serving as the cathode
Page 4
Electrochemical Cells and Electromotive Force
Electrons generated at the site of oxidation, the anode, are thought to be "driven" or "pushed" toward the
cathode by an electromotive force or emf. This is analogous to water exerting a pressure as it moves from a
higher point to a lower point. The potential energy of the electron is determined by how much the oxidation
reaction wants to lose electrons coupled to how much the reduction reaction wants to gain electrons.
This change in potential energy of the electron is measured in volts, which is defined as 1 Joule per Coulomb
of charge. That is,
1 volt (V) = 1 joule/coulomb = 1 J/C
The standard voltage potential of a half-cell is defined as its potential with all reactants and products present as
pure solids, or in a solution at a concentration of 1.0 M, or with gases at 1.0 atm. These conditions are
considered standard conditions and the measured voltage is the standard voltage, Eo.
Standard Cell emf's and Standard Electrode Potentials
In looking at the coupling of two half-reactions in a voltaic cell, we can define the driving force at the anode as
"oxidation potential" and the force at the cathode as "reduction potential". This would imply that the overall
potential of the cell is
Ecell = oxidation potential + reduction potential
Since most tables list half-reaction potentials as reductions, we must be careful in tabulating the overall
potential of a cell with regard to 'sign convention'. The oxidation potential for a half-reaction is the negative of
the of the listed reduction potential for that half-reaction written in reverse. That is,
Oxidation potential for a half-reaction = - reduction potential for the reverse half-reaction
To illustrate, let us consider the following cell.
Cd(s) | Cd+2(aq) || Ni+2(aq) | Ni(s)
Referring to a table of reduction potentials, we find that the reduction potential for the nickel half cell is
Ni+2(aq) + 2 e-1
Ni(s)
Eo = -0.23 volts
The cadmium half-cell, however, is also listed as a reduction (even though that is not the way we are using it in
our cell) as:
Cd+2(aq) + 2 e-1
Cd(s)
Eo = -0.40 volts
However, reversing the cadmium reaction in order to represent it as an oxidation, requires that we change the
sign of its "reduction potential".
Cd(s)
Cd+2(aq) + 2 e-1
Eo = +0.40 volts
So now we have our two half-reactions written the way we are using them. All that is left to do is simply sum
the two potentials to obtain the overall cell potential.
Ni+2(aq) + 2 e-1
Cd(s)
overall reaction
Cd(s) + Ni+2(aq)
Ni(s)
Cd+2(aq) + 2 e-1
Cd+2(aq) + Ni(s)
Eo = -0.23 volts
Eo = +0.40 volts
Eocell = +0.17 volts
CHAPTER 19
Page 5
Standard Cell Potentials, ∆Go, and Equilibrium Constants
The standard cell potential can be viewed as a measure of cell spontaneity, in that a positive emf would be
spontaneous but a negative emf would not be.
The standard free energy, ∆ Go, is also the ultimate measure of spontaneity measured in Joules instead of volts.
To convert the Eo of a cell from volts to joules we must use the fact that the charge on a single electron has
been determined to be 1.609 x 10-19 coulombs. This would equate to 96,500 coulombs per mole of electrons.
This value is referred to as Faraday's constant. That is
Faraday's constant (F) = 96,500 coul/mol electrons
So, for a transfer of "n" moles of electrons in an electrochemical process (and remembering that a volt is a joule
per coulomb) we can define the relationship between ∆Go and Eo as
∆ Go = − n
(J)
F
Eo
(mol e-) (C/mol e-) (J/C)
Any cell with a positive emf will result in a negative ∆Go (which is an indicator of a spontaneous reaction)
Thus, all spontaneous electron transfer reactions will have a positive Eo and a negative ∆ Go
and conversely, all nonspontaneous electron transfer reactions will have a negative Eo and a positive ∆ Go.
For example, the cell discussed earlier, Cd(s) | Cd+2(aq) || Ni+2(aq) | Ni(s) , had a potential of +0.17 volts.
We would predict its ∆Go to be negative....but let's calculate the actual value.
∆ Go = − n F Eo = − (2 mol e-) (96500 C/mol e-) (+0.17 J/C) = − 32810 J (or − 32.81 kJ)
The measurement of cell emf's gives us yet another method for determining equilibrium constants (K).
From our chapter on thermodynamics, we saw that at equilibrium ∆Go = −2.303 RT log K. So, combining this
with the relationship between Eocell and ∆Go we get
nFEocell = 2.303RTlogK
or
o
E cell
= 2.303RT
log K
nF
Substituting values for R and F at 25oC gives the equation
E ocell = 0.0592
n log K
Page 6
For example, suppose we allow our cadminm/nickel cell to attain equilibrium and it is delivering its predicted
+0.17 volts. Let's calculate the equilibrium constant, K, for the following system.
Cd(s) + Ni+2(aq)
Cd+2(aq) + Ni(s)
Eocell = +0.17 volts
Since this reaction involved a 2 electron transfer, n = 2. So,
+0.17V =
0.0592
2
log K
or
log K =
so
(2)(+0.17V)
0.0592
= 5.74
K = 10 5.74 = 5.54x10 5
Page 7
Dependence of emf on Concentration
The standard emf of a cell, you recall, is measured with all reactants and products in standard conditions and
concentrations (1.0 M). However, the actual emf of the cell will change if these concentrations are changed.
One can relate actual cell emf's at concentrations other than standard to standard emf's using the Nernst
Equation.
From our chapter on thermodynamics, we saw that
∆G = ∆Go + 2.303RTlogQ
where Q is the reaction quotient representing concentrations (and gas pressures) that exist in a reaction mixture
at a given instant.
If we substitute ∆G
∆ = -nFEcell and
∆ Go = -nFEocell, we obtain
-n F Ecell = -n F Eocell + 2.303 R T logQ
This rearranges to give us the Nernst Equation, an equation relating the actual cell emf to its standard emf and
the reaction quotient.
E cell = E ocell − 2.303RT
log Q
nF
If we substitute values for R and F at 25oC, you get
E cell = E ocell − 0.0592
n log Q
A close examination of the Nernst Equation shows that the emf of the cell decreases as the reaction proceeds,
which is logical that potential would decrease as the concentration of reactants decreased.
For example,
What is the emf of the following cell at 25oC?
The standard emf of the cell is +0.17 V.
Cd(s) | Cd+2(aq) (1.0 x 10-4M)|| Ni+2(aq) (0.10M)| Ni(s)
The cell reaction is Cd(s) + Ni+2(aq)
Cd+2(aq) + Ni(s) The number of electrons transferred is
two, therefore n=2., and the reaction quotient is
[Cd +2 ]
1.0x10 −4
Q=
+2
[Ni ]
=
0.10
= 1.0x10 −3
E cell = +0.17 − 0.0592
log(1.0x10 −3 ) = +0.17 − (−0.09) = +0.26V
2
Page 8
Individual Electrode Potentials for Nonstandard Conditions
Not only can we determine the emf of an entire cell at nonstandard conditions using the Nernst equation, but
individual electrode potentials can also be obtained.
Consider the half-cell Ag+1(aq) (0.010 M) | Ag(s). Since the standard potential of a hydrogen electrode is zero
(see your table of standard reduction potentials), the potential of the following cell
Pt | H2 (1 atm) | H+1(aq)(1 M) || Ag+1(0.010 M) | Ag(s)
would simply be the potential
of the Ag (0.010 M) | Ag(s) half-cell. (Note the standard reduction potential for Ag+|Ag is +0.80 v)
+
So, the Nernst equation for this cell at 25oC becomes
o
0.0592
E Ag +(0.010M) Ag = E Ag
+ Ag −
n log Q
Since n=1 for this reaction
and
then
1/2 H2(g) + Ag+1(aq)
Q=
[H + ]
1
[Ag + ](p H 2 ) 2
=
H+1(aq) + Ag(s)
(1)
1
(0.010)(1) 2
= 1.0x10 2
E Ag+ (0.010M) Ag = +0.80 − 0.0592
log(1x10 2 ) = +0.80 − (+0.12) = +0.68v
1
Page 9
Determination of pH
One can also determine the pH of an unknown solution by measuring its emf using the standard Hydrogen/Pt
electrode in the unknown solution. Since the potential of the H+ | H2 |Pt half-cell is dependent on the
concentration of the H+ ions, a direct correlation can be made.
Suppose we look at the cell
Cd(s) | Cd+2(aq) (1 M) || H+(aq) (test solution) | H2 (1 atm) | Pt
The cell reaction would be
Cd(s) + 2 H+1
Cd+2(aq) + H2 (1 atm)
According to the table of standard reduction potentials, the standard emf of this cell should be +0.40 v and
Q=
[Cd +2 ]p H 2
[H + ] 2
=
(1M)(1atm)
[H +1 ]2
=
1
[H +1 ] 2
Substituting into the Nernst equation, we get
(1)
+1
E cell = +0.40 − 0.0592
log
=
+0.40
+
0.0592
log[H
]
+1
2
2
[H ]
Rearranging to obtain the relationship between cell emf and pH, we must substitute the following into the
preceding equation.
pH = -log [H+1]
The result is
Ecell = +0.40 -0.0592 pH
which we can rearrange to give the pH in terms of the cell emf
pH =
+0.40−E cell
0.0592
Note that the +0.40 volts is
particular to the Cd/Cd+2 half-cell
and would be a different value had
we measured our test solution
against any other half-cell
So, measurement of the cell emf will give you the pH of the test solution. This is essentially how a pH
electrode, used in the laboratory, functions. Although we seldom use a hydrogen/platinum electrode, the
principle is the same.
Page 10
Stoichiometry of Electrolysis
Cells with negative emf's are not spontaneous reactions as we saw in our thermodynamic view of emf and ∆Go.
These nonspontaneous cells (called electrolytic cells), however, can be "driven" by applying a voltage greater
than their negative emf. This, essentially, is how electroplating is accomplished.
The amount of material to be "plated" can be easily determined by relating the amount of total charge
transferred (in coulombs) during the electrolysis to the number of electrons transferred using Faraday's constant
(1 mol electrons = 96,500 C). Since the number of electrons transferred relates directly to the number of atoms
reduced to solid metal, the mass of metal plated can be calculated.
First, a couple definitions. Current flow is usually measured in Amperes (A). An ampere is defined as
1 coulomb/sec. Given the number of seconds and the current used in an electrolysis, we can easily determine
the number of coulombs of charge transferred. That is,
Coulombs of charge transferred (C) = Current (A) x time (sec)
To find the number of moles of electrons transferred, we simply divide the coulombs of charge transferred by
Faraday's constant.
moles of electrons transferred =
coulombs of charge transferred
96,500(
C
mol e −1
)
For example, suppose an aqueous solution of silver cyanide (AgCN) is electrolyzed and silver metal is "plated
out".
Ag+1 + 1 e-1
Ag(s)
How many grams of silver could we plate out when a current of 15.0 mA flows through the cell for 5.00
minutes?
First, let's calculate the total charge transferred.
C
Charge (C)= 1.50x10 −2 ( sec
) x 300 seconds= 4.5 coulombs
Note that 1 mole of electrons is equivalent to 1 mole of silver deposited, so
4.5 C
1 mol e −1
x 96,500C
x
1 mol Ag (s)
1 mol e
−1
x
107.9 g
1 mol Ag
= 5.03x10 −3 g Ag
Conversely, one can also calculate the amount of current flow in an electrolytic cell given the amount of
material deposited in a given amount of time.
Page 11
Exam Review Topics
terms
skills/operations
oxidation/reduction
voltaic cell
electrolytic cell
half-cell
salt bridge
anode/cathode
volt
Faraday's constant
standard emf
standard electrode potential
Nernst equation
ampere
Sketch/label voltaic cell
Cell reaction from cell notation
Maximum work
Calculate standard cell emf
Calculate ∆Go from Eo
Calculating Eo from ∆Go
Calculate Ecell at nonstandard conditions
Calculating K from Ecell
Relating amount of charge and product in electrolysis