Electrochemistry
1. Oxidation Numbers
Electrochemistry is the study of oxidation-reduction reactions (redox reactions). These are
reactions in which electrons are transferred or, alternately, where oxidation numbers are changed.
Let us first review the calculation of oxidation numbers.
1.
In most of its compounds oxygen is -2. (The most common exception is hydrogen peroxide,
H2O2, where oxygen is -1.)
2.
In most of its compounds hydrogen is +1. (The most common exception is where hydrogen is
combined with a metal. Here it forms a hydride. for example NaH, and has an oxidation state
of -1.)
3.
The sum of the oxidation states of the atoms in an ion or molecule is equal to its charge.
SO42– = +6
NO–3 = +5
HCl = -1
NH3 = -3
4.
The oxidation state of an atom in an element is zero. This is a consequence of (3).
H2 = 0
O2 = 0
Fe = 0
5.
The oxidation state of a monoatomic ion is equal to its charge. This is a consequence of (3).
Cl– = -1
Ca2+ = +2
S2– = -2
6.
It is often helpful, in ionic solids, to know the charge of one or more of the ions.
If K is known to be +1, then K2VO3 is +4.
If PO4 is known to be -3, then Fe3(PO4)2 is +2.
An oxidation occurs when an atom or molecule loses electrons or increases its oxidation number.
A reduction occurs when an atom or molecules gains electrons or reduces its oxidation number.
You may want to use the mnemonic "LEO says GER." This stands for "losing electrons is
oxidation" and "gaining electrons is reduction." All redox reactions have both an oxidation and a
reduction. But balancing redox equations is a little more difficult than just knowing this fact.
2. Balancing Redox Equations
In a redox reaction one species is oxidized (its oxidation state becomes more positive) and another
species is reduced (its oxidation state becomes less positive). To do these problems, you must
know the common oxidizing and reducing agents mentioned in the worksheet ‘Some Useful
Reactions’. You should realize that oxidizing agents are themselves reduced, while reducing
agents are oxidized. It sounds funny but it makes sense. It helps if you are familiar with the
common oxidizing and reducing agents. For example, hydrogen peroxide is usually an oxidizing
agent (and is therefore reduced):
H2O2 + 2 Fe2+ + 2 H+  2 H2O + 2 Fe3+
However, it can occasionally be a reducing agent (and be oxidized):
3 H2O2 + Cr2O72– + 8 H+  3 O2 + 2 Cr3+ + 7 H2O
Look at the way the oxidation state of oxygen changes in these reactions.
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7b
Permanganate, chlorate, bromate, iodate, chlorine, bromine or oxygen are always oxidizing
agents. Other oxidizing agents are less certain because they can do other things besides oxidize.
You must further remember that permanganate forms Mn2+ in acidic solutions and MnO2 in basic
solutions.
2 MnO–4 + 10 I– + 16 H+  2 Mn2+ + 5 I2 + 8 H2O
2 MnO–4 + 3 SO32– + H2O  2 MnO2 + 3 SO42– + 2 OH–
If the acidity or alkalinity is not specified, assume an acid solution.
Some More Examples
A solution of tin(II) sulfate is mixed with acidified potassium dichromate.
3 Sn2+ + Cr2O72– + 14 H+  3 Sn4+ + 2 Cr3+ + 7 H2O
Hydrogen peroxide solution is added to a solution of iron(II) sulfate
H2O2 + 2 Fe2+ + 2 H+  2 H2O + 2 Fe3+
In grading: “RWO” will mean reduction without oxidation; “OWR” will mean oxidation
without reduction.
Metal Displacement Reactions
These are a particular type of redox reaction in which a metal, in its elemental state is oxidized.
Usually this combines with another metal being reduced. (Thus one metal displaces another.) If a
piece of copper is put into a silver nitrate solution, the copper replaces the silver:
Cu + 2 Ag+  Cu2+ + 2 Ag
In real life you would have to know which metal is more active in order to determine whether the
replacement reaction goes. But here, all the reactions go. So you need not worry about the activity
series. Also I’ve included the reaction of metals with acid or water, since the reaction is the same
except now it’s hydrogen which is reduced. So the reaction of aluminum with hydrochloric acid is:
2 Al + 6 H+  2 Al3+ + 3 H2
And the reaction of calcium with water is:
Ca + 2 H2O  Ca(OH)2 + H2
Metal displacement reactions are easy to recognize because they involve a piece of metal.
Dissolving metals in nitric acid or in hot sulfuric acid is more complicated, since these acids also
act as oxidizing agents. In each case the nitrogen or sulfur is reduced to the oxide of a lower
oxidation state. This question usually involves copper, lead, or silver. These metals are less
reactive than H+ and, therefore, too inert to dissolve in ordinary acids (e.g. HCl). They need the
additional oxidizing power of the (acidified) nitrate or the (hot, acidified) sulfate.
4 H+ + NO–3 + 3 Ag  3 Ag+ + 2 H2O + NO (or NO2)
4 H+ + SO42– + Cu  Cu2+ + SO2 + 2 H2O
Some More Examples
A piece of aluminum metal is added to silver acetate solution.
Al + 3 Ag+  Al3+ + 3 Ag
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Combustion
These are easy because the question tells you that something is being burned. You take the
material being burned, add oxygen gas, and convert the carbon to carbon dioxide and the hydrogen
to water.
C2H5OH + 3 O2  2 CO2 + 3 H2O
If they are present, chlorine is converted to HCl and most other elements are converted to their
oxides.
Some More Examples
Carbon disulfide vapor is burned in oxygen.
CS2 + 3 O2  CO2 + 2 SO2
Gaseous silane (SiH4) is burned in oxygen.
SiH4 + 2 O2  SiO2 + 2 H2O
The usual method for balancing redox reactions is to write out two half reactions, one for the
reduction and the other for the oxidation. Each half reaction is balanced and then multiplied by the
number of electrons gained or lost by the other half reaction. These are then divided by a common
denominator, if possible, and then added together to get the final reaction.
We will not use this method but will, instead, use the oxidation state change method. Instead of
balancing electrons, we will balance the changes in oxidation state. It's essentially the same thing
only much faster. After the oxidation number changes have been dealt with, hydrogen and oxygen
must be balanced. First oxygen is balanced by adding water molecules. Then hydrogen is balanced
by adding H+. Actually this is done only in acidic or neutral solutions. In basic solution you cannot
use H+ and must, therefore, balance the hydrogen using OH– and H2O. Look at the steps.
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1.
Determine the oxidation states for the elements which are oxidized or reduced. Remember
that you are dealing with individual atoms, even if the atoms occur in groups of two. Be sure
you have both an oxidation and a reduction.
6
4
3
6
Cr2O72– + SO32–  Cr3+ + SO42–
2.
Calculate the change in oxidation state for each atom which changes.
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Cr2O72–
+ SO32–  Cr3+ + SO42–
+2
3.
Adjust the coefficients so that the total number of increases in oxidation state equals the total
number of decreases in oxidation state.
Cr2O72– + 3 SO32–  2 Cr3+ + 3 SO42–
4.
Balance the oxygen by adding water.
Cr2O72– + 3 SO32–  2 Cr3+ + 3 SO42– + 4 H2O
In acid or neutral solutions
5a. Balance the hydrogen by adding H+.
Cr2O72– + 3 SO32– + 8 H+  2 Cr3+ + 3 SO42– + 4 H2O
6a. Check to see that the charges balance. If they don't, you've made a mistake.
In basic solution
5b. Instead of adding 8 hydrogen atoms to the left side, add 8 water molecules to the left side and
8 hydroxides to the right. Convince yourself that this is really the same thing!
Cr2O72– + 3 SO32– + 8 H2O  2 Cr3+ + 3 SO42– + 4 H2O + 8 OH6b. Eliminate the number of H2O molecules which appear on both sides of the equation.
Cr2O72– + 3 SO32– + 8 H2O  2 Cr3+ + 3 SO42– + 4 H2O + 8 OH7b. Check to see that the charges balance. If they don't, you've made a mistake.
Cr2O72– + 3 SO32– + 4 H2O  2 Cr3+ + 3 SO42– + 8 OH3. Voltaic (Galvanic) Cells
Galvanic or voltaic cells (named for Luigi Galvani and Allesandro Volta, respectively) are used to
harness the energy of a spontaneous redox reaction. This is done by physically separating the two
half reactions so that the electrons generated by the oxidation half reaction must flow through an
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electrical conductor where they can be used in meters, motors, light bulbs or other devices before
they reach the reduction half reaction.
A galvanic cell can be made using the reaction:
Zn + Cu2+  Zn2+ + Cu
Here zinc is being oxidized and Cu2+ is being reduced. A
cell using this set of reactions is shown in Figure 1. All of
the reactants in the oxidation half reaction are placed in the
left beaker, and all of the reactants in the reduction half
reaction are in the right beaker. The electrodes are
connected to each other with a metal wire, and a device such
as a meter, motor or light bulb may be inserted in the
Figure 1. A galvanic cell
circuit. To complete the circuit, a salt bridge or
semipermeable membrane must be used. Otherwise the flow
of electrons through the wire would (very quickly) make the beaker in which the oxidation occurs
positive and the beaker containing the reduction negative. Since electrons do not spontaneously go
from positive to negative, the reaction would stop. The salt bridge or semipermeable membrane
allows ions to migrate and keep the reaction going by restoring charge neutrality. In this case Zn2+
ions would migrate to the right and NO3– ions would flow to the left.
A slightly more complicated reaction, which is also capable of acting as the basis for a galvanic
cell, is:
5 Fe2+ + MnO–4 + 8 H+  5 Fe3+ + Mn2+ + 4 H2O
The big difference here is that neither the reactant nor the product in either half reaction is a metal.
So what are the electrodes made of? Certainly not Fe2+ ions; that can't be done. The electrodes
must be made from an inert material; exactly which one depends on the circumstances. But we will
spare no expense and use the best. Our inert electrodes will be made of platinum.
It helps to picture what happens. Instead of the zinc
electrode giving up electrons and forming Zn2+ ions, the
Fe2+ ions bump into the electrode, give up one electron
each, and become Fe3+ ions.
Finally there is the matter of what to call the electrodes —
which electrode is the anode and which is the cathode. Many
of you learned in middle school that the cathode is the
positive electrode. This is wrong or, to be more precise, is
Figure 2. Another galvanic cell
correct only half the time. The cathode is the electrode at
which reduction takes place. For a galvanic cell this is the
positive electrode; for an electrolytic cell, which we will discuss shortly, the cathode is negative.
Don't try to remember this. Remember the red cat — reduction at the cathode.
Always remember that the cathode is where reduction takes place!
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Electrochemical cells are typically described using something called "line notation." The two cells
we have discussed would be shown as:
Zn│Zn2+(aq)║Cu2+(aq)│Cu
Pt│Fe2+(aq), Fe3+(aq)║MnO–4(aq), Mn2+(aq)│Pt
You will notice that the anode reaction is on the left, that a single vertical line indicates a
difference in phase and that a double vertical line separates the anode reaction from the cathode
reaction. Multiple species in the same phase are separated by commas and concentrations are given
for species not in their standard states. Line notation lists both a reactive electrode (e.g Cu) and an
inert electrode (e.g. Pt) in the same way.
Reactions involving a gas are easy to describe but difficult to set up
experimentally. Consider, for example, the reaction:
Cl2 + 2 e–  2 Cl–
How does one get chlorine gas into the reaction? The answer is that one
bubbles it over the surface of an inert electrode. In line notation this is:
║ Cl–(aq) │ Cl2(g)│Pt
4. Standard Reduction Potentials
Figure 3. A Cl2 Cell
How do we know whether a reaction goes or in which direction it goes?
Alternately, and this is really the same question, how do we know the
voltage of a given cell? The answer has to do with standard reduction potentials.
Standard reduction potentials are voltages associated with a reduction reaction done under
standard conditions. The terms "voltage," "potential," "electromotive force," and "EMF" are used
interchangeably. It is from EMF that we get the symbol E or E. Since a reduction cannot take place
without an oxidation, reduction potentials correspond to a reaction which is not complete — to a
half-reaction with electrons on its left side. Tables of reduction potentials are readily available; one
is included in the AP test booklet. Cell potentials are calculated by adding the potentials of the two
half cells. Consider the reduction of Cu2+ which takes place in the cell drawn in Figure 1.
Cu2+ + 2 e–  Cu
This has a half-cell potential of 0.337 V.
There is no potential in the table corresponding to the oxidation of zinc.
Zn  Zn2+ + 2 e–
This is, after all, an oxidation. However, we can find a potential for the corresponding reduction:
Zn2+ + 2 e–  Zn
This has a potential of -0.763 V. Zinc, it seems, does not want to be reduced. To get the potential
for the oxidation reaction, one merely reverses the sign of the reduction potential. So -0.763 V
becomes +0.763 V and the potential for the cell is 0.337 + 0.763 = 1.100 V.
Notice that the number of electrons in the half reaction is not used in combining the half-cell
potentials. Consider the reaction 2 Fe3+ + Sn2+  2 Fe2+ + Sn4+. The cell potential is:
Fe3+ + e–  Fe2+
E = 0.77
2+
4+
–
Sn  Sn + 2 e
E = -0.15
2 Fe3+ + Sn2+  2 Fe2+ + Sn4+
E = 0.62
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5. The Faraday
As long as the reaction proceeds, the quantity of material used or produced by a cell has nothing to
do with the cell potential. It has to do with amperage. An ampere is a unit of electrical current; it
measures the rate at which electricity flows. When amperage is multiplied by time, it gives the
quantity of electricity. An ampere flowing for one second gives a quantity of electricity called a
coulomb. It has been determined that 96,500 coulombs equals a mole of electrons, a Faraday. (This
assumes the reactions proceeds with 100% efficiency — that no other reactions occur.)
Using the Faraday to solve problems requires that you cancel units. Trying to remember or figure
out what goes on the top and what goes on the bottom of the fraction is risky. Let us try, as an
example, calculating how many liters of oxygen (at STP) are formed by using 2.2 amps for 10.0
minutes. Forming a molecule of oxygen requires 4 elections: 2 H2O  O2 + 4 H+ + 4 e–. Thus:
V = 2.2 amps x 10.0 min x 60 sec x 1 coul x 1 Faraday x 1 mol O2 x 22.4 L = 0.077 L
1 min 1 amp sec 96,500 coul 4 Faradays 1 mole
For how many minutes can a zinc/copper cell deliver 2.3 amperes of current if it consumes 0.455 g
of zinc? (Recall that a mole of zinc gives up two moles of electrons when it is oxidized.) The
calculation can be done using one string of factors, as above, but this problem is more complex.
You will find it easier to set up if you write an equation with the number of Faradays based on
grams of zinc on one side and number of Faradays based on coulombs on the other.
0.455 g x 1 mol x 2 F = 2.3 amps x X min x 60 sec x 1 coul x
1F
65.38 g 1 mol
1 min 1 amp sec
96,500 coul
Solving this equation for time gives x = 9.7 min.
6. Free Energy and Equilibrium Constants
Cell potentials, like many other chemical quantities, can be viewed in terms of free energy. The
formula which relates them is in the thermochemistry section of the AP formula sheet:
ΔG = - nFE
This can be rearranged to give:
E = – ΔG°
nF
This states that the cell potential is the free energy change per mole of electrons transferred. This
explains why we do not multiply the cell potential by a stoichiometric factor when determining the
potential of a cell.
As can be seen from these equations, a large, positive cell potential corresponds to a large negative
free energy change. For example, calculate the standard free energy change in the cell:
Pt│Mn2+, MnO4–║IO4–, IO3–│Pt
First we calculate the standard cell potential:
2e– + 2 H+ + IO4–  IO3– + H2O
E = 1.60
Mn2+ + 2 H2O  MnO4– + 8 H+ + 5 e– E = -1.51
This gives a cell potential of 0.09 V. (Notice that regardless of what else we do in this problem, the
rest of our calculations will not be good to better than one significant digit.)
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The free energy change is:
ΔG = - nFE = -10  96,500  0.09 = -87,000 joules/mol = -9  104 joules/mol
How do we know that the units here are joules rather than kilojoules? Because no matter how
convenient we may find kilojoules to be, the metric unit is joules.
Since we already know that ΔG is related to K, it is no surprise that the equilibrium constant for a
redox reaction is also related to its E. Note that you must use E, which is invariant for a given
reaction, and not E, which depends on the concentrations, in order to calculate K. The equation
which relates K with E is in the electrochemistry section of the AP formula sheet:
nEo
log K = 0.0592
The equilibrium constant for the reaction can be calculated as follows:
log K = 10 x 0.09 and therefore K = 2 x 1015
0.0592
The significance of the equilibrium constant is that it determines where the equilibrium will lie
after the reducing and oxidizing agents have been mixed together. It also tells the point at which
the cell will have zero voltage. (A dead battery, one with a potential of 0 volts, is at equilibrium.)
7. The Nernst Equation
It should be clear that STANDARD reduction potentials are measured under standard conditions.
That means at 25 C and with reactants and products having concentrations of 1 M or pressures of
1 atm. What happens at non-standard conditions? For this you need the Nernst Equation.
As we said before, the free energy of an electrochemical process is given by:
ΔG = - nFE or ΔG = - nFE
Now recall from thermodynamics the equation which relates ΔG and ΔG:
ΔG = ΔG + RT ln Q
Combining these equations gives:
- nFE = - nFE + RT ln Q
This can be rearranged to give:
E = E° – RT ln Q
nF
This is the Nernst Equation, as given on your AP formula sheet.
E is the cell potential while E is the standard cell potential. R is the gas constant which, although
this is not obvious, must be in metric units — 8.31 J/moldeg. T is the temperature in Kelvin; n is
the number of electrons transferred; and F is the Faraday — 96,500 coulombs. Q is the reaction
quotient which, as you recall, has the form of a mass action expression but is used for a system
which is not at equilibrium. For the reaction A + 2B  3C, the equilibrium quotient would be:
[C]3
Q=
[A][B]2
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Most Nernst Equation problems involve correcting for changes in concentration and pressure, not
changes in temperature. So the usual form of the Nernst Equation is obtained by plugging in values
of R, T and F and by converting the "ln" to a "log." This gives another formula which appears on
the AP formula sheet:
E = Eo -
0.0592
n log Q
Let us calculate the potential for the cell shown on the right.
First we calculate the standard cell potential for:
Al│Al3+ (0.01 M)║Cu2+ (1.4 M)│Cu
Note, again, that we do not multiply the potential by the number
of electrons.
Al  Al3+ + 3 e–
E = 1.66
Cu2+ + 2 e–  Cu
E = 0.34
E = 2.00
Figure 4. A Non-Standard Cell
This leads to the balanced equation:
2 Al + 3 Cu2+  2 Al3+ + 3 Cu
Notice that the first half reaction is written as an oxidation, as it actually occurs in the cell reaction.
The oxidation potential is the same as the reduction potential, but with a different sign..
Now let's plug this into the Nernst Equation:
E = Eo -
0.0592
[Al+3]2
0.0592
[0.01]2
log
=
2.00
log
= 2.04
+2
3
n
6
[Cu ]
[1.4]3
The solids Cu and Al (in the electrodes) are omitted because they each have a value of one. More
of a problem for students learning the Nernst Equation is how to determine the value of "n." It
could be 6, consistent with the equation:
2 Al + 3 Cu2+  2 Al3+ + 3 Cu
Or it could be 3, consistent with the equation:
Al + 3/2 Cu2+  Al3+ + 3/2 Cu
Which is it? The answer is that it doesn't matter. Either value of "n" has a different Q, and the
differences cancel.
An additional type of cell, one which can be understood only in
terms of the Nernst Equation, is a cell in which both the anode
and cathode reactions are the same. The only difference
between the two half cells is that the concentrations are
different. This cell is called a "concentration cell" and its
potential can be calculated as below. Notice that the larger
copper concentration is the reactant and is therefore on the
bottom. This is a simple application of Le Chatelier's Principle.
Figure 5. A Concentration Cell
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E=-
0.0592 [Cu+2]
n log [Cu+2]
E=-
0.0592
0.001
2 log 1.4 = 0.09V
You can see that in all these cases the Nernstian term, which we go to a lot of trouble to calculate,
is very small. Yet small though the term is, it is important. Most electrochemistry problems on the
AP test contain a Nernst Equation calculation. Typically it's the hardest part of the question.
8. Electrolytic Cells
Most of what you have learned has involved galvanic cells — cells
with positive potentials and in which, therefore, the reaction "wants"
to go. An electrolytic cell is one in which the reaction has a negative
potential and which, therefore, is non-spontaneous. Note that the
cathode is now negative.
Most of what you need to know about electrolytic cells (i.e.
electrolysis) is covered in the "reactions" section of this course. The
only electrolysis calculations you are likely to see involve the
amount of material which can be deposited by using a certain
number of coulombs of electricity (section 5 of this chapter.) You
Figure 6. An Electrolytic Cell
should know that electroplating, where a thin layer of a metal (e.g.
gold) is electrolytically deposited on the surface of another metal, is a type of electrolysis. But you
don't need to know much more than that.
One particular application of electrochemistry which has
appeared on a previous AP tests is called electrolytic refining.
This is an electrolysis where a piece of impure metal serves as
the anode and pure metal plates out on the cathode. Let us
consider the electrolytic purification of copper. When impure
copper is oxidized, those impurities which are less active than
copper (e.g. Ag) don't dissolve but fall to the bottom of the
tank. However, the impurities which are more active than
copper (e.g. Zn) dissolve but don't plate out. Thus the cathode
is made of relatively pure copper. Look at the price of copper
in the commodity section of the newspaper and you will probably see "copper cathodes" listed.
This is why.
9. Corrosion
The most common meaning of "corrosion" is the oxidation of a metal by air, often in the presence
of water. Because of the huge cost of corrosion to our economy, it is industrially one of the most
important areas of electrochemistry.
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Since corrosion is an electrochemical process, its tendency to proceed is closely related to the
standard reduction potentials with which you are familiar. Metals with large negative reduction
potentials (e.g. alkali and alkaline earth metals) are easily oxidized and hence corrode readily.
Metals with large positive reduction potentials (e.g. gold, platinum) are not easily oxidized and
hence do not corrode. Most metals lie between these extremes. The chemistry here involves
finding ways to prevent these metals from corroding.
The most common method of preventing corrosion is to create a barrier between the metal and the
air. Typically in cars and bridges this is done using some sort of paint. However, paint will chip,
creating an unprotected site where corrosion will occur.
A more effective barrier will be in some way active. For example aluminum, despite an E of
-1.66 V, corrodes hardly at all. This is because aluminum forms an oxide barrier which adheres
tightly to the metal surface. If the oxide is scratched or chipped, the coating immediately reforms.
Thus the metal surface is always protected. Contrast this to iron oxide, which does not adhere to
the metal surface and, therefore, provides no protection.
A similar process occurs in stainless steel. Nickel and chromium, which are components of the
alloy, both form adherent oxide layers on the surface. Cheaper grades of stainless contain less
nickel and chromium and are, therefore, less resistant to corrosion.
Galvanized steel, which you often see in chain link fences, electrical conduit, and corrugated
sheets, has been coated with zinc. Originally the zinc was applied electrolytically (galvanically)
but today the steel is dipped in molten zinc. Zinc, with a cell potential of -0.76, is more reactive
than iron. So even when the zinc no longer covers the iron entirely, the zinc will continue to
corrode. The corrosion of the zinc creates a negative potential on the metal and, therefore, prevents
the iron from corroding.
A related process is called cathodic protection. Here a very active metal (e.g. magnesium) is
welded or bolted to a less active metal (e.g. iron). The magnesium oxidizes giving the iron a
negative potential (i.e. making the iron cathodic). This prevents the iron from corroding. Cathodic
protection is often used on fences and boat hulls.
10. Redox Titrations
The titrations we have done in this course have involved the reaction of an acid with a base. We
can also do a titration in which an oxidizing agent reacts with a reducing agent. This is called a
redox titration. In many cases one of the reactants or products is sufficiently strongly colored that it
functions as an indicator. For example, when MnO4– is used as the oxidant, its color disappears at
first because it is being reduced. At the equivalence point, however, there is no longer any
reductant present and the solution turns purple because the permanganate is no longer being
reduced. In other cases, where neither the reactants nor the products are colored, a redox indicator
is used. For example a drop of Fe3+/SCN– mixture gives a strong color in an oxidizing solution.
However when the environment becomes reducing and Fe3+ is reduced to Fe2+, the colored
complex is no longer formed.
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Consider an experiment in which we wish to determine the concentration of a potassium
permanganate solution. We pipet 10.00 mL of the permanganate solution into a flask, dilute it, and
titrate it with 0.01134 M SO32– solution. If the titration requires 33.6 mL of sulfite solution, what is
the molarity of the permanganate?
First we need to determine the changes in oxidation number which occur in both reactants so that
we can determine how many moles of electrons are taken up or given off by each mole of
compound. We can get the same number from the number of electrons in the balanced half
reaction, but using the change in oxidation states is easier.
In the reduction reaction MnO4– (the oxidizing agent) goes to Mn2+ and the change in oxidation
number is -5. In the oxidation reaction SO32– (the reducing agent) goes to SO42– and the change in
oxidation number is +2. Now we can calculate:
MO  VO  nO = MR  VR  nR
Here nO and nR are the number of moles of electrons gained or lost by a mole of oxidizing or
reducing agent.
MO  10.00  5 = 0.01134  33.6  2
MO = 0.0152 M
The significant digits are limited by the accuracy of the titrant volume. The values of n are, of
course, exact.
11. Identifying Which Species Will be Oxidized and Which Will be Reduced
Something like this is often part of a multiple choice question.
Given the following three equations:
1) F2 + 2 H+ 2 e–  2 HF
E = 3.03
2) 2 H+ + 2 e–  H2
E = 0.00
–
–
3) F2 + 2 e  2 F
E = 2.87
Which of the following is the best reducing agent (under standard conditions)?
F2
F–
H+
H2
HF
A reducing agent is oxidized; so that immediately rules out F2 and H+, neither of which is capable
of being oxidized.
The oxidation reaction for HF (the reverse of equation 1) has a potential of -3.03 V. Clearly HF
does not want to be oxidized.
The oxidation reaction for F– (the reverse of equation 3) has a potential of -2.87. Like HF, F– does
not want to be oxidized.
So this leaves H2, with an oxidation potential of zero, as the most easily oxidized species (i.e. the
best reducing agent) on the list.
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