In 1786, Italian physician Luigi Galvani observed that a frog’s leg twitched when it was touched
with two different metals. In attempting to explain what happened, Galvani thought that the
animal tissue in the frog’s leg was the source of electricity. Alessandro Volta, an Italian
physicist disputed Galvani’s hypothesis and the resulting controversy resulted in the discovery
that electric currents could be produced by chemical reactions. In 1800, Volta used this
discovery to create the first chemical battery.
Today, we have remote controls for our TVs, DVD players, CD players, VCRs, stereos and the
like, wristwatches with itty bitty teeny tiny batteries, calculators, silverware, and metal plated
jewelry, which all involve the principles of electrochemistry.
In general, all chemistry is electrical in the sense that it involves the behavior of electrons and
other charged particles (remember back to the definition of chemistry! Chemistry is the study
of electrons and how they behave in atoms, molecules, ionic compounds and what happens to
the electrons in a chemical reaction). Electrochemistry is generally reserved, however, for the
processes that convert chemical energy into electrical energy or vice versa.
One branch of electrochemistry deals with galvanic cells such as batteries and fuel cells.
Galvanic cells are also called voltaic cells. The name of this type of cell comes from the work of
Volta and Galvani. These devices use spontaneous oxidation-reduction reactions to produce
electrical currents which can then do electrical work. A galvanic cell is constructed so that the
electrons flow from one terminal to the other when the terminals are connected by an external
circuit. The driving force of the reaction is the spontaneity associated with the reaction in a
particular direction. Electrochemistry studies the relationship between chemical change and
electrical work. There are two different types of electrochemical cells:
One type of cell does work by releasing energy from a spontaneous reaction to produce
electricity. Example: a battery
Another type of cell does work by absorbing energy in order to drive a nonspontaneous
reaction. Example: metal plating
Redox reactions, otherwise known as oxidation-reduction reactions are another type of chemical
reaction to be aware of. These type of reactions include the formation of a compound from its
elements (and vice-versa), all combustion reactions, reactions that generate electricity in
batteries, reactions that produce cellular energy, and many others.
In redox reactions, the key chemical event is the movement of electrons from one species to
another. The driving force for the movement of electrons occurs from the reactant with the
lower attraction for the electron(s) to the reactant with more attraction for the electron(s).
Examine the following reaction:
1
Mg(s) + O2 (g) → 2MgO(s)
Magnesium is a neutral element/atom. O2 is also a neutral molecule. But when these two
neutral species come together in a chemical reaction, they form an ionic compound. On the
product side, we have formed a charged Mg+2 ion which is attracted to the O-2 ion. This means
that Mg metal as the reactant must have given up (lost) its electrons to the oxygen molecule
reactant.
Oxidation is the loss of electrons. When something loses electrons is becomes positive, or more
positive as the case may be.
Reduction is the gain of electrons. When something gains electrons it becomes negative, or less
positive, whatever the case may be.
Here’s the confusing part. Since oxidation can only occur with reduction, and reduction must
occur with oxidation the two are linked together in the chemical reaction. If one species is
oxidized, another must be reduced. Thus, the species that is oxidized is often termed the
reducing agent, and the species that is reduced is called the oxidizing agent.
Mg(s) + O2 (g) → 2MgO(s)
Oxidation: Mg(s) → Mg+2 + 2e-1
Reduction:
O2 + 2e-1 → 2O-1
Since oxygen gained the two electrons that Mg lost, we say that oxygen was reduced.
Therefore, Mg did the reducing, making it the reducing agent. Mg would not have lost its
electrons if oxygen had not wanted them so much, therefore oxygen did the oxidizing making it
the oxidizing agent.
The oxidizing agent becomes reduced when it removes electrons from the other reagent in the
redox reaction, while the reducing agent becomes oxidized as its electrons are lost.
2
Each atom in a molecule is assigned an oxidation number, or the charge it would adopt if the
electrons were actually not shared in the bond but instead were transferred completely. (We
have previously determined this number in CH131 during our Lewis structures unit!!) Below is
a table of general guidelines to follow when determining the oxidation numbers of atoms in a
compound that might make it easier than drawing out the Lewis structure:
Concept Test:
Determine the oxidation number of each atom in the following:
ZnCl2
SO3
HNO3
Sc2O3
The oxidation number for an element in a binary ionic compound has a value based in reality
because it usually is equal to the ionic charge. Oxidation numbers for polyatomic ions or
covalent compounds are not reality based because these atoms are, in fact, sharing their
electrons. But it does provide us with a guide because in a redox reaction, the oxidation
3
numbers of the species change, and it is most important to monitor these species and their
changes.
You should be able to identify the species that was oxidized and the species that was reduced
and label the oxidizing reagent and the reducing reagent.
Several reactions that we have talked about previously are redox reactions.
Combination Reactions:
2H2 + O2  2H2O
neutral hydrogen molecules combine with neutral oxygen molecules to form a
neutral compound where the each hydrogen has an oxidation number of +1
(oxidized) and each oxygen has an oxidation number of -2 (reduced)
Decomposition Reactions:
2KClO3  2KCl + 3O2
oxygen in KClO3 has an oxidation number of -2 while in the neutral molecule
O2 its oxidation number is 0 (oxidized). Cl has an oxidation number of +5 in KClO3
but an oxidation number of -1 in KCl (reduced).
Combustion Reactions:
C6H6O6 + 6O2  6CO2 + 6H2O
combustion reactions are all redox reactions also as the reaction
incorporates neutral oxygen molecules and produces two compounds
with oxygen having an oxidation number of -2.
Single Replacement Reactions:
3CuCl2 (aq) + 2Al(s)  2AlCl3 (aq) + 3Cu(s)
copper changes its oxidation number from +2 in copper (II) chloride to
0 as a solid metal (reduced). Aluminum changes its oxidation number from 0
as a neutral metal to +3 in aqueous solution (oxidized).
BUT, with single replacement reactions we have another factor to consider. Will the reaction
always occur. We examined this with double replacement reactions: if a reactant was NOT
soluble in water, then the reaction would not occur. With single replacement reactions, the
4
solid metal that is going to lose its electrons (oxidized) must be more reactive than the metal
that it is replacing. If it is not, then the other metal would MUCH rather stay as an ion. Using
the chart, metals that are higher up on the scale can replace the metals that are below it.
In the above example, aluminum was oxidized. That made aluminum the reducing reagent. It
reduced the oxidation number of copper from +2 to 0. Thus, aluminum did the reducing.
The list of metals is arranged with the most active metal (strongest reducing agent) at
the top and the least active metal at the bottom (weakest reducing agent)
In the demo, I performed an oxidation reduction reaction by examining what happened when
Ag(NO3)2 solution surrounded a copper wire . We had Ag+1 ions in solution (remember the lack
of color for the solution?) and then we placed copper wire in the solution. The blue color slowly
appeared and huge chunks of silver “stuck” to the copper wire. Many of you noticed, as I
disturbed the solution by gentle tapping the wire was thinner, or even appeared like part of it
had dissolved. It had. A re-dox reaction had occurred based on the chemical reactivity of the
metals compared to one another.
The solution was in contact with the metal, and as such, the electrons were able to flow directly
from the copper atoms on the surface of the wire to the silver ions that were in contact with the
wire’s surface. This was a spontaneous reaction, we did not have to do anything but put the
wire in and viola – the silver ions were turned into silver metal.
When we examined re-dox reactions in the past we began simply by noticing which species lost
electrons, and which species gained them. Whenever a species loses electrons, there MUST be
another species that is present to gain them. Thus, oxidation and reduction occur together. If a
5
species is reduced, then there is a species that is oxidized. We examined the re-dox system by
adding electrons – always adding electrons. We noticed that if electrons are added to one
species on the reactant side, that they must be added to the product side for the other species.
Concept Test:
Write the species that is oxidized, the species that is reduced.
Identify the oxidizing agent, the reducing agent.
Write the half reactions
CuCl2 (aq) + Zn(s) → ZnCl2 (aq) + Cu(s)
reduction:
oxidation:
What do cars, batteries, bleach, the browning of an apple and photosynthesis all have in
common? They are all dependant upon a type of chemical reaction, oxidation-reduction or
redox for short. In fact, as types of chemical reactions go, it may be the most important. If
suddenly all redox reactions stop occurring, the world would stop and all life would die. So, I
am not understating its importance.
Again, this can only be accomplished by losing and electron, not adding protons.
Here is a little helpful little saying: LEO the lion says GER.
LEO = Loss of Electrons is Oxidation
GER = Gain of Electrons is Reduction
We in fact, balanced many different types of redox reactions without even considering that they
were re-dox reactions! You have balanced single replacement reactions, combustion reactions,
double replacement reactions and the like. However, some re-dox reactions need special
balancing conditions in order to ensure that both the atoms and the electrons are balanced. This
is called charge-balancing the chemical reaction.
A lot of times redox reactions occur under acidic or basic conditions, meaning that there is an
acid or a base present in the “mix”. Redox reactions that occur under these conditions are
balanced in their own unique way. The approaches are similar to one another but not the same.
When a redox reaction occurs in an acidic solution, H2O molecules and H+1 ions are available to
help create a balanced chemical equation. Previously, we have used H3O+1 and H+1
6
interchangeably. And while they are – simply using H+1 instead will make the balancing soooo
much easier! That way you do not have to worry about balancing extra oxygen atoms. It can be
done – using H3O+1 all that will differ between using H3O+1 and H+1 will be the total number of
water molecules that are added.
Rules for balancing redox reactions in acidic solutions:
1.) divide the reaction into half-reaction. Write each half-reaction
2.) balance atoms AND balance charges for each half-reaction. Add in water molecules to
balance O atoms, H+1 ions to balance H atoms, and electrons to balance charge
begin by balancing atoms other than O and H
then balance O atoms using water
balance the H atoms by adding H+1 to the appropriate side
balance the charges using electrons – remember to always add electrons!
3.) multiply each half-reaction by a number so that the number of electrons lost in the
oxidation equals the number of electrons gained in the reduction.
4.) add the reactions together canceling substances that appear on both sides of the reaction
5.) check that the number of atoms AND the charges on both sides of equation are equal.
7
Concept Test:
Balance the following redox reaction in acidic solution
Cr2O7-2(aq) + I-1(aq) → Cr+3(aq) + I2 (s)
8
Using H3O+1 would just mean that we would have more neutral water molecules present in the
final equation. Since the reaction is in water, the water really has no meaning other than to
ensure that the reaction components are balanced. Many of you have previously wondered
where water was in chemical reactions. It is always there! In redox reaction it must be taken
into account in order to help us balance the equation.
In a basic solution, there is an abundance of OH-1 ions instead of H+1 ions. In order to balance
the equation in basic solution, you can pretend that it was balanced in an acidic solution but
then add OH-1 ions to EACH side of the reaction equal to the number of H+1 ions that you
added to balance the H atoms. The H+1 ions will combine with the added OH-1 ions to form
additional water. Then the half-reactions can be added together, canceling out the duplicate
water molecules. One side of the reaction will retain the added OH-1 the other side will have
lost its OH-1 in the formation of water molecules.
9
Concept Test:
Permanganate ion is a strong oxidizing agent. It reacts in a basic solution
with the oxalate ion to form the carbonate ion and solid manganese (IV) oxide
MnO4-1(aq) + C2O4-2 (aq) → MnO2 (s) + CO3-2 (aq)
10
Redox reactions reveal a great deal about the reaction. And they follow the same rules that we
have been following when balancing equations. Equations have always been “charge
balanced”, it is just that we never had to pay attention to the electrons before. Now we do!
We can treat the redox reactions as being a total sum of the oxidation half with the reduction
half reactions. The mass of the atoms is conserved in each half-reaction and thus in the overall
chemical reaction. Electrons that are lost on one side of the reaction are gained on the other.
And although we examine the electron loss and gain as if it were happening separately, it is in
fact happening simultaneously.
The voltaic or galvanic cells (batteries and fuel cells) use a spontaneous reaction to generate
electrical energy. Higher energy reactants are converted into lower energy products, and in
doing so, the released energy is electrical energy. This electrical energy is then used to operate
something – such as a light bulb, or a CD player. In other words, the system does the work on
the surroundings.
In an electrolytic cell, extra energy, usually in the form of voltage from a power supply, is used
to drive a nonspontaneous reaction, or a reaction that would not normally occur by itself. The
reactants are lower in energy than the products, thus the reason that extra energy must be put in
– this reaction is not favored as written. The surroundings therefore do the work on the system.
The two types of cells do have some things in common. Each has electrodes, which are the
objects that conduct the electricity between the cell and the surroundings. Each electrode is
submerged in an electrolyte – or a mixture/solution that contains ions. The electrolyte may be
involved in the reaction or the ions may be used to carry the charge. An electrode is identified
as being the anode or the cathode.
Oxidation occurs at the ANODE: electrons are given up by the species being oxidized
and leave the cell at the anode. Anode = Oxidation (AN OX)
Reduction occurs at the CATHODE: electrons are accepted by the species being reduced
and enter the cell at the cathode. REDuction = CAThode (RED CAT)
When one dips a strip of zinc metal into a solution of zinc sulfate, some of the zinc atoms
(neutral) are oxidized. Thus, the neutral zinc atom loses two electrons and enters the solution as
a Zn+2 ion. At the same time, a Zn+2 ion that is present in the zinc sulfate solution, gains those
two electrons and deposits itself (solidifies) on the zinc strip. The opposing oxidation and
reduction reactions reach equilibrium.
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oxidation
Zn (s)
Zn+2(aq) + 2e-1
reduction
This strip of metal is called the electrode. The equilibrium that is established occurs at the
surface of the electrode when it is submerged in a solution of its ions and is called electrode
equilibrium. Another example of electrode equilibrium is copper metal in an aqueous solution
of Cu+2 ions.
oxidation
Cu (s)
Cu+2(aq) + 2e-1
reduction
Loses its electrons easily
(easily oxidized)
Does not lose its electrons
easily
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Zinc is a good reducing agent – and thus, is fairly easily oxidized. In contrast, copper is a poor
reducing agent and thus us not readily oxidized. Thus, when comparing zinc to copper, zinc
would prefer to be oxidized more so than copper. Zinc wants to lose its electrons more so than
copper. This is related to the activity series that we previously examined when we predicted
single replacement reaction.
A half-cell consists of a metal immersed in a solution of its ions. Two half-cells can be connected
together and then joined by a salt bridge. On the left-hand side there is a zinc electrode
submerged in aqueous zinc sulfate solution. On the right hand side, there is a copper electrode
submerged in aqueous copper (II) sulfate solution. The salt bridge contains aqueous sodium
sulfate and is inverted in the solutions of copper (II) sulfate and zinc sulfate. The ends are
plugged, but still allow for the flow of ions across the membrane (usually cotton plugs). Metal
wires connect each electrode to a voltmeter, which is used to measure the flow of electrons.
The voltmeter registers a reading of 1.103 V. The voltmeter reading measure a difference in
electrical potential between two points in an electric circuit. In the case of our voltaic cell, the
two points are the electrodes. The potential difference is the driving force that propels (keeps
them moving!) the electrons from the anode to the cathode and is also called the cell potential
(Ecell). Because the measurements are taken in volts, it is sometimes called the cell voltage. The
cell voltage will depend on a variety of factors, which you will examine in lab!
The overall oxidation-reduction reaction that occurs in the voltaic cell is called the cell reaction.
13
At the zinc electrode, oxidation occurs. Zinc atoms lose electrons and enter the solution as Zn +2
ions.
Electrons that are lost by the zinc do not just accumulate – they have to go somewhere! – and
they do, they flow through the wires and the voltmeter and to the copper electrode (remember,
copper did not want to lose its electrons compared to zinc). At the copper electrode, the
incoming flux of electrons inhibits the tendency of copper to become oxidized. Instead, the Cu+2
ions in solution accept those electrons and become reduced. In this process, they are reduced to
copper atoms and stick to the copper electrode. The electrodes as such, will develop a charge.
The charge on the electrode will be determined by the source of the electrons and the direction
the electrons flow. The electrons flow from the anode to the cathode in a voltaic cell. Electrons
are thus continuously generated (by the Zn) at the anode and then consumed (by the Cu+2) at
the cathode. Thus, in any voltaic cell, the anode is negatively charged and the cathode is
positively charged.
As a result of each of the half-reactions there is an overall redox reaction that can be examined:
Zn(s) + Cu+2(aq)  Zn+2 (aq) + Cu(s)
The purpose of the salt bridge is to complete the circuit. The cell cannot operate (and thus no
voltage or current can be measured) unless the cell is complete. The oxidation half-cell contains
a neutral solution of Zn+2 and SO4-2 ions. But, as oxidation takes place, some of the Zn metal
(from the electrode) lose their electrons. This creates an excess of newly formed Zn+2 ions.
Similarly, in the reduction half-cell, Cu+2 ions are gaining electrons and becoming neutral Cu(s).
Thus there is an overabundance of the SO4-2 ion in solution. This charge imbalance would arise
and stop the cell operation if it could not be counteracted. To enable the cell to operate, the salt
bridge- and its ions come into play. The salt bridge acts like a liquid wire, transferring its
cations where the abundance of anions occurs and its anions to where the abundance of cations
occurs. Thus, in our model, SO4-2 ions are sent to the anode side where the abundance of Zn+2
ions is and Na+1 ions are sent to the cathode side to counteract the buildup of the SO 4-2 ions as
the Cu+2 ions become neutral Cu(s). As this reaction proceeds, you could remove the electrodes
after a bit of time and mass them. The zinc electrode has lost some of its solid mass as it formed
additional Zn+2 ions while the copper electrode has gained mass as Cu+2 ions plated out onto the
Cu(s) electrode.
There is shorthand notation used to describe electrochemical cells called the voltaic or galvanic
cell diagram. The diagram for the above cell is as follows:
anode
salt bridge
Zn(s)Zn+2(aq)Cu+2(aq)Cu(s)
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cathode
The components of the anode compartment (oxidation) are written on the left hand side. The
reduction compartment is written on the right hand side.
A vertical like represents the phase boundary (e.g. between Zn(s) and Zn+2 (aq)) If several
components of the same phase are present in a particular side of the cell then they are separated
by commas (for example, if Zn+2 and H+1 were both present the beginning notation would be
Zn(s) Zn+2 (aq) , H+1 (aq) . A double vertical line represents the division of the cells by the salt
bridge.
It is often acceptable to put the concentration of solutions in the half-cell notation as well.
Typically half-cell component appear in the same order as in the half reaction, with the solid
zinc coming first (as the solid zinc is oxidized into Zn+2 then the salt bridge, then Cu+2 which is
reduced into Cu(s).
The ions in the salt bridge are eliminated for clarity as they do not participate in the actual redox
reaction, they just complete the circuit.
As a final note, remember that the left/right (anode/cathode) convention in a cell is arbitrary.
In attempting to make it easier, we are defining sides. But when you set up a half-cell, you may
not know which species is going to behave as the reducing agent and which species will behave
as the oxidizing agent. We determine by experiment which species is oxidized and which
species is reduced. Think about your lab benches, what is on your left, appears on the right
according to the person across the bench from you 
Instead of using a voltmeter, a lightbulb could have been placed between the two cells. As the
electrons were transferred, and the salt bridge ions were able to migrate, the light bulb would
turn on. We can therefore see that this type of electrochemical cell generates electrical energy.
But why do the electrons move and why do they move in the direction that they do (away from
Zn). Previously it was just stated, well the activity series of course. But what does that mean??
Let’s pretend that our cells are separate again for a moment.
Zn (s)
Zn+2(aq) + 2e-1
Cu (s)
Cu+2(aq) + 2e-1
From the reaction, Zn gives up its electrons more easily than does copper. This makes zinc a
stronger reducing agent. Therefore the equilibrium position of the reduction reaction lies farrrr
to the right. Zinc will produce more Zn+2 ions than will copper. These electrons can be thought
of as a pressure, a driving force so to speak. They have a greater potential energy than do the
electrons produced by copper oxidizing. This potential energy serves as the drive to push the
electrons through the circuit. There is a higher concentration of electrons on the zinc side than
the copper side. We know that things flow naturally from regions of higher concentration to
lower concentration. From higher energy situations to lower energy situations. The electrons
are therefore going to flow from the zinc anode to the copper anode to attempt to equalize the
15
different in electrical potential. The spontaneous reaction occurs as a result of the different
abilities of these metals to give up their electrons and the ability of the electrons to flow through
the circuit.
We have described how to construct a voltaic cell by combining two half-cells, and how to
measure the voltage with a voltmeter. However, it is also possible to calculate cell voltages
without making specific references to measurement. This is done by establishing the
characteristic potentials for individual half-cells and then determine the difference in potential
(cell voltages) when half-cells are combined into voltaic cells.
The purpose of the voltaic cell is to convert the energy change associated with the spontaneous
reaction into the KE of electrons moving through the electrical circuit. This electrical energy can
then do work (like light up a light bulb, or change the channel on our TVs). How much the
system can do work is going to be dependent on the potential difference between the two
electrodes. This is known as the Ecell. Electrons are negatively charged, so they flow
spontaneously from the negative to the positive electrode. Thus, when the reaction is
spontaneous, there is a positive potential and Ecell is positive. If Ecell = 0, then the redox reaction
has reached equilibrium and the cell can do no more work.
How are the units of electrical potential and electrical charge related? Electrical potential is
measured in volts. Electrical charge is measure in Coulombs.
1 V = 1 J/C
The measured potential of a cell is dependent on concentration as the reaction proceeds.
Therefore, in order to compare the electrical output of different cells, there needs to be some
standard of comparison. The standard used is called the standard cell potential (E ocell) and it is
measured at a specific temperature (usually 298 K), and all components in their standard states
(1 atm for gases, 1M for solutions, and pure solids for electrodes). When cells are connected in
this manner, a standard voltage appears. For the zinc copper cell mentioned previously, the cell
produces 1.10 V
Just like each half-reaction makes up a part of the overall reaction, each potential associated
with the individual half-reactions makes up a part of the whole Ecell.
By definition: standard half-cell potentials always refer to the reaction written under reducing
conditions. Each reduction reaction has an E value associated with that reaction written as a
reduction reaction. BUT remember, in the real cell, one of the reactions is a reducing reaction,
the other is an oxidizing reaction.
All values determined for reduction half-reactions are related back to the standard hydrogen
electrode. These are the half-cell potentials that you can use mathematically to calculate the
Ecell. All values for each reduction were determined by comparing the reactions with the
standard hydrogen electrode (a platinum electrode immersed in a strong acid (H +1) which gets
reduced to H2).
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Ehalf-reaction = Ereference - Ecell
Ereference is defined as being 0
Ehalf-reaction = 0.00 - Ecell
You do not need to know how to calculate the half-reaction values, they are all given to you in
huge tables in text-books, but you need to know how to USE the given reduction potentials!
The above voltaic cell has the Zn half reaction in one half of the cell and the hydrogen reference
reaction in the other. The Zn half reaction is the anode, and is thus negative. The overall Ecell =
0.76 V. The potential of the standard electrode is defined as being zero, so the Ezn = -Ecell so the
Zn reactions is defined as being -0.76
A general relationship exists between the anode and the cathode:
Eocell = Eocathode – Eoanode
Using the above information we can calculate the Ecopper
Ecell = Ecathode - Eanode
Ecell = Ecopper - Ezinc
Ecell = 1.10 V = Ecopper – (-0.76 V)
1.10 V = Ecopper + 0.76 V
Ecopper = 1.10 V – 0.76 V = 0.34 V
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Table of Reduction Potentials
Cathode (Reduction) Half-Reaction
Standard Potential E°
(volts)
Li+(aq) + e-  Li(s)
-3.04
K+(aq) + e-  K(s)
-2.92
Ca2+(aq) + 2e-  Ca(s)
-2.76
Na+(aq) + e-  Na(s)
-2.71
Mg2+(aq) + 2e-  Mg(s)
-2.38
Al3+(aq)
+
-1.66
2H2O(l) + 2e-  H2(g) + 2OH-(aq)
-0.83
Zn2+(aq) + 2e-  Zn(s)
-0.76
Cr3+(aq) + 3e-  Cr(s)
-0.74
Fe2+(aq) + 2e-  Fe(s)
-0.41
2e-
Cd(s)
-0.40
Ni2+(aq) + 2e-  Ni(s)
-0.23
Sn2+(aq) + 2e-  Sn(s)
-0.14
Pb2+(aq) + 2e-  Pb(s)
-0.13
Fe3+(aq) + 3e-  Fe(s)
-0.04
 H2(g)
0.00
Cd2+(aq)
2H+(aq)
+
+
2e-
Sn4+(aq) + 2e-  Sn2+(aq)
0.15
Cu2+(aq) + e-  Cu+(aq)
0.16
AgCl(s) + e-  Ag(s) + Cl-(aq)
0.22
Cu2+(aq) + 2e-  Cu(s)
0.34
Cu+(aq)
e-
+
 Cu(s)
0.52
I2(s) + 2e-  2I-(aq)
0.54
Fe3+(aq) + e-  Fe2+(aq)
0.77
Hg22+(aq) + 2e- 2Hg(l)
0.80
Ag+(aq) + e-  Ag(s)
0.80
Hg2+(aq)
+
2e-
 Hg(l)
0.85
ClO-(aq) + H2O(l) + 2e-  Cl-(aq) + 2OH-(aq)
0.90
2Hg2+(aq) + 2e-  Hg22+(aq)
0.90
Br2(l) + 2e-  2Br-(aq)
1.07
O2(g) + 4H+(aq) + 4e-  2H2O(l)
1.23
Cl2(g) +
Strongest
oxidizing agent
 Al(s)
3e-
2e-

1.36
2Cl-(aq)
Ce4+(aq) + e-  Ce3+(aq)
1.44
H2O2(aq) + 2H+(aq) + 2e-  2H2O(l)
1.78
Co3+(aq) + e-  Co2+(aq)
1.82
F2(g) + 2e- 2F-(aq)
2.87
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Strongest
reducing agent
Given and using the above table, we could have predicted the Ecell knowing the half-reaction
reduction potentials.
Cu+2 + 2e-1  Cu (s) E= 0.34 V
Zn+2 + 2e-1  Zn (s) E = -0.76 V
HOWEVER!! Both reactions are written as reductions!
One of these reactions was NOT a reduction, it was
An oxidation!
The overall reaction was
Zn (s)  Zn+2 + 2e-1
Cu+2 + 2e-1  Cu (s)
Zn + Cu+2 + 2e-1  Cu (s) + Zn+2 + 2e-1
Zn + Cu+2  Cu (s) + Zn+2
Copper was reduced, zinc was oxidized
We can leave the copper reduction potential alone – it was indeed reduced
Zinc however, was oxidized. Therefore, we have to mathematically
Manipulate the reduction value. We take the opposite of the given value!
Cu+2 + 2e-1  Cu (s)
Zn (s)  Zn+2 + 2e-1
E= 0.34 V
E = 0.76 V
METHOD 1
This method relies on you being able to identify who is being
reduced and who is being oxidized:
Then the overall Ecell = the sum of the reduction and oxidation
Reactions!
Ecell = 0.34 V + 0.76 V = 1.10 V
OR METHOD 2
This method relies on you being able to remember who the anode is
and who is the cathode.
Ecell = Ecathode - Eanode
Ecell = 0.34 V – (-0.76 V)
Ecell = 1.10 V
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One of the things that we can learn and examine is the relative strength of the oxidizing and
reducing agents. You can compare the reduction half-reactions given above and use that
knowledge to understand chemical redox reactions
Cu+2 + 2e-1  Cu(s) Eo = 0.34 V
2H+1 + 2e-1  H2 (g) Eo = 0.00 V
Zn+2 + 2e-1  Zn (s) Eo = -0.76 V
The MORE positive the Eo value the MORE likely the reduction is to occur. The more readily
the reaction occurs AS WRITTEN. This means that copper ions gain electrons more readily than
hydrogen ions which both gain electrons more readily than zinc ions. This means, that zinc
really really really does not want to be reduced!
Ranking species as oxidizing agents:
Cu+2 > H+1 > Zn+2
Ranking species as reducing agents:
Zn > H2 > Cu
All values in the table are relative to the standard hydrogen reference electrode.
By convention, all the reactions are written as reduction reactions. This allows for easier
comparison between species. This means that only the reactant are the oxidizing agents and
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only the products are the reducing agents. Again, this allows for an easier comparison between
the species in the table.
The more positive the E value, the more likely the reaction is to occur AS WRITTEN.
The half-reactions are all equilibrium reactions because depending on the half-cell components,
each reaction has the possibility of behaving under one circumstance as a reduction, and under
a different set of conditions as an oxidation.
Reduction potential values are directly related to the oxidation potential values in that the signs
are reversed. (If the reduction value is a positive 0.80, then the oxidation value is -0.80)
Every redox reaction is the sum of an oxidation and a reduction reaction. Thus, the table can be
used to help examine spontaneous redox reactions. In the zinc copper reaction, Zn and Cu are
reducing agents and Zn+2 and Cu+2 are oxidizing agents. However, Cu+2 is a better oxidizing
agent than Zn+2 and Zn is a better reducing agent than Cu.
Therefore, Zn will act as the reducing agent (be oxidized) and Cu+2 will act as the oxidizing
agent (be reduced). The stronger species will react to form the weaker species. Therefore, we
can predict that the reaction will occur between Zn and Cu+2 (as the stronger species) forming
Cu and Zn+2 (the weaker species)
Zn(s) + Cu+2(aq)  Zn+2(aq) + Cu(s)
This parallels the similar ideas presented in the acid-base chemistry – whereby the stronger acid
and the stronger base would react to form the weaker acid and the weaker base. But again, you
need to know how to interpret the data given to you 
The stronger oxidizing agent (the species on the left of the chemical reaction) will have the
MORE positive E value.
The stronger reducing agent (the species on the right of the chemical reaction) has the smaller or
less positive – or even negative E value.
If we know the electrode potentials, we can write spontaneous redox reactions regardless of if
we have the table or not! Here are two half-reactions and their E values
Ag+1(aq) + e-1  Ag (s) E =0.80
Sn+2(aq) + 2e-1  Sn(s) E =-0.14
Which species is the stronger oxidizing agent? The species with the more positive E value
Therefore, Ag+1 wants to be the oxidizing agent. It wants to be reduced. It is already written in
the form of a reduction reaction! Since oxidation MUST be paired with reduction, and Ag+1 will
be the species that is reduced, then we must write the other reaction in terms of its oxidation.
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Ag+1(aq) + 1e-1  Ag (s) E = 0.80
Reverse the second equation, change the sign of E
Sn (s)  Sn+2 + 2e-1 E = 0.14
Ecell will be the overall sum. And then we did not have to
Worry about who was the anode and who was the cathode!
The last thing – remember, redox reactions deal with the loss
and gain of electrons. The same # of electrons!
2Ag+1(aq) + 2e-1  2Ag (s) E = 0.80
Sn (s)  Sn+2 + 2e-1 E = 0.14
2Ag+1 + Sn (s)  Sn+2 2Ag(s) Eocell = 0.0.94 V
It is very important to note the reversing the reaction changes the sign of E
But multiplying the reaction did not change the magnitude of E!!
Balancing the half-reactions by changing the co-efficient did not effect E. E is an intensive
property that does NOT depend on amount of substance present. You could have 35 moles of
Ag needed and the E value would still be the same.
General Rules:
When pairing reactions. The reaction with the more positive (higher) E value will stay written
as the reduction reaction.
The reaction that you pair it with must therefore be reversed and the sign changed
Make sure that the # of electrons is equal
Add the reactions together
Add the E values together to get the Eocell
Spontaneous Redox Reactions
If Eocell is positive, then the reaction is spontaneous in the forward direction – AS WRITTEN
If Eocell is negative, then the reaction is nonspontaneous in the forward reaction – AS WRITTEN
and is instead spontaneous in the reverse direction
If Eocell = 0 then the reaction is at equilibrium
When reactions are reversed, the sign of Eocell is reversed as well
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Thus far, we have been examining reactions that occur under standard conditions. But many
times when creating electrochemical cells, you are not dealing with chemical species always at 1
M. Therefore, one must be able to examine Ecell under nonstandard state conditions. (Eocell
means standard states!!) Ecell means any condition NOT at standard states (not at 1 M, not at 298
K, not at 1 atm)
The Nernst Equation:
Ecell = Eocell – RT ln Q
nF
R = 8.3145 J/moleK
T = temp in K
F = 96485 C/mole
n = # electrons in either half-reaction AFTER the half-reactions have
been adjusted to balance the equation
The Nernst equation states that the cell potential at some other condition depends on the
potential under standard state conditions and also takes into consideration the reaction quotient
When Q < 1 [reactant] > [product]
When Q = 1 [reactant] = [product]
When Q > 1 [reactant] < [product]
ln Q < 0 so Ecell > Eocell
ln Q = 0 so Ecell = Eocell
ln Q > 0 so Ecell < Eocell
The equivalent expression using the common logarithm scale is
Ecell = Eocell – 0.0592 log Q
n
Remember that the Q expression does not contain liquids or solids! Even though solids may be
acting as electrodes.
Concept Test:
Write the Q expression for the following:
Cd(s) + 2Ag+1  Cd+2 + Ag(s)
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Concept Test:
A chemist creates a new voltaic cell consisting of a Zn/Zn+2 half-cell and an H2/H+1
half-cell under the following conditions:
[Zn+2] = 1.1 M
H+1
[H+1] = 2.5 M
[H2] = 0.013 M
Calculate Ecell at 25oC
has a more positive value therefore it will be written as the reduction 
To use the Nernst equations, we need Eocell and Q.
Eo
Using the ln expression would give the same answer 
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Remember that when Ecell = 0 the reaction was defined as being at equilibrium. Also, at
equilibrium, Q = K. Therefore the equation changes ever so slightly at 298 K:
Ecell = 0 = Ecello - 0.0592 V log K
n
0 = Ecello - 0.0592 V log K
n
solving for Eocell
Eocell = 0.0592 V log K
n
At equilibrium, no more energy is released, and we say that the system is doing no more work –
or in plain English – the battery went dead. Given the above equation you should be able –
since you can calculate Eocell values, to solve for K. The cell will continue to do work until the
products overwhelm the reactants – the ratio of [products]/[reactants] gets very large.
The main point of this is that as the cell operates, its potential to do work slowly decreases. And
that should make sense to all of you. Batteries do not last forever. As we use them, they begin
to put out less and less electrical energy until pretty soon, they do not work anymore. And
what do we do then?? We bang on the TV controller, we roll the batteries, we swear, but, in
fact, we must get up and change the channel manually and put a new set of batteries in 
You know that if you place two solutions on either side of a permeable membrane that the
solution of higher concentration will diffuse to the region of lower concentration until the two
regions are of equal concentrations. This idea – of species’ desire to move from regions of
higher concentrations to lower concentrations is the basis of a concentration cell. In the
concentration cell, this spontaneous tendency is taken advantage of in order to generate
electrical energy.
The two solutions are kept in separate half-cells and therefore do not mix, but their
concentrations can become equal through the redox reaction and the transfer and gain of
electrons. Since we are speaking of the same species, examining the E ocell means that the
reduction and oxidation reactions have equal but opposite values which means that E ocell = 0.
But, since the concentrations are different, a voltage CAN be measured as the magnitude of the
nonstandard Ecell value will depend on the ratio of the concentrations. The cell will operate
until the concentrations of ions are equal.
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Ecell = Eocell – 0.0592 V log [X]dilute
2
[X]concentrated
Eocell = 0
Ecell = 0 - 0.0592 V log [X]dilute
2
[X]concentrated
This means that the concentrated solution is written as the reduction
reaction while the dilute solution is written as the oxidation reaction
For example, examining Cu+2 (1.0 M) in half-cell 1
And Cu+2 (0.10 M) in half-cell 2
In order to make the concentrations equal, the 1.0 M solution must make
MORE Cu+2 ions. Therefore, it must be reduced so that more Cu(s)
in the dilute solution can be oxidized to make more Cu+2!!
Cu+2 (1.0 M)  Cu+2 (0.10 M)
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Because they are so compact – and useful! – batteries influence our lives on a daily basis. This
device, the battery, stores chemical energy for the release of electricity some time later. In most
portable and electronic devices we use the dry-cell batteries. The cell reactions that occur in the
dry cell batteries are irreversible. Reactants are converted into products until all the reactants
are used up. Then the battery is “dead”. Alkaline cells undergo the same types of halfreactions that dry cell batteries undergo but in an alkaline medium. These batteries are able to
last longer than the typical dry cell battery, but again, they will go “dead”.
Rechargeable batteries use electrical energy (usually you plug the service center in to the wall)
to run the reaction in reverse – thus regenerating the reactants. It TAKES energy to create more
reactants – thus it is very similar to a dry cell battery in terms that when in use, reactants are
being formed into products. This battery too goes dead, but by inputting energy the reaction
can be reversed. The lead-acid battery in your car is a rechargeable battery. The cell reaction
can be reversed and the battery restored to its “original” condition. Which is why- if you leaver
your lights on – you can get jumped and “recharge” your battery! The battery is very durable
and can be used through repeated cycles of discharging and recharging (aka leaving your lights
on and jumping!) By connecting the cells to an outside electrical source, you can force the
battery to run in reverse – sending electrons in the opposite (nonspontaneous) direction. The
half-cell reactions and the cell reactions are reversed and the battery is recharged.
In an automobile, the battery is discharged when the engine is started. While running, the
engine powers the alternator, which produces electrical energy to recharge the battery. Thus, if
your alternator goes – so does your battery - and then your car does not run  In principle, a
lead-acid battery should last forever. But they do not. During discharge, PbSO 4 (s) deposits on
the electrodes. When the PbSO4 is in this form, that battery can no longer be recharged. Buy a
new one! This condition, called sulfating, is the major cause of battery failure.
Fuel cells are a voltaic cell in which the cell reaction is equivalent to a combustion reaction.
Combustion reactions, if you remember are indeed oxidation reduction reactions anyway.
Harnessing the energy of this reaction creates an efficient fuel cell. The hydrogen-oxygen fuel
cell is the cell that is showing the most promise for powering automobiles as their byproduct is
simply humid air. Government agencies classify these fuel cells in vehicles as zero emission.
More “Helpful” Information:
Looking at this galvanic cell you should say to yourself, I have seen many a battery but none
that look like that. Today’s batteries are much smaller, but the
same components are found in all batteries. Instead of having
beakers filled with solution, a battery replaces the beakers
with a can inside of another can. The solutions are replaced
with a paste or gel. Since these batteries do not have a
solution, they are named “dry” cell batteries.
The first dry cell developed by the German chemist Carl
Gassner in 1888, consisted of a zinc can that served both as a
container and as the anode. The cathode, a carbon (graphite)
rod, was immersed in a manganese dioxide/carbon black
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mixture. It was separated from the zinc container by a folded paper sack, soaked in a solution
of ammonium chloride, which acted as the electrolyte. During use the zinc casing of the battery
was gradually consumed by the chemical reaction. A sealant prevented evaporation of water
from the electrolyte and the admission of oxygen.
I cannot say I have ever seen a dry cell battery consisting of copper and silver. Why do you
think they do not exist? Cost, weight and voltage. The cost and weight would be too much,
while the voltage output would be too small.
Most car batteries are of the lead acid type. These batteries have the ability to produce large
amounts of power. But you wouldn’t want to drag around a 40-pound cell phone. The most
recent battery development in mobile phones is the lithium ion battery. Why do you think
lithium makes a good battery component? It is the lightest metal, it is the most active metal and
it is not too expensive.
Looking back to the activity series, can you see why gold is an important and expensive metal?
Gold plated cables or computer parts are not as susceptible to oxidation even when running
current through them at high temperatures.
The Statue of Liberty gets its green colored skin from the copper that its exoskeleton is made of.
But the statue gets most of its structural support from steel beam, steel is mostly iron.
Oxidation or rusting occurs when oxygen gas is allowed to contact a metal when the metal is
wet. The copper skin is not “copper” colored, its green color comes from the copper (II) oxide
that forms on its surface. Hence, the skin of the statue is covered in CuO (s). Once the entire
surface has been oxidized the interior copper is protected.
Iron on the other hand is a metal that continues to undergo a reaction after the surface of the
iron has oxidized. This oxidation or rust can permeate through the beams destroying the
structural integrity of the statue.
Which of the metals would you consider to be most active? Which metal weakens fastest, the Fe
or the Cu? Fe will react before Cu. But the New York port authority does not want the Cu or Fe
to rust. How can they stop this from occurring? Employ a metal that is more active than either
Cu or Fe. To prevent rusting, paint with zinc dust was used. Why?
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Iron beams are coated with this zinc paint so that oxygen reacts with the zinc and not the iron.
Zn has higher activity series thus protects the iron beams. The oxygen will not rust the iron
until all of the zinc has been oxidized. The zinc is sacrificing itself for the iron. And since the
zinc is being oxidized, the zinc is referred to as a sacrificial anode.
Often a person will hammer a rod of zinc or magnesium into the ground close to metal structure
and attach a wire from the structure to the rod, so that any oxides of Fe will be exchanged by
having the O2 prefer to react with Zn to form ZnO or Mg to form MgO. Zinc and magnesium
have a greater affinity to O2 and thus draws O2 away from the metal structure and towards
itself. This is another example of a metal sacrificing itself to spare another metal from corrosion.
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