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Lecture 14
Week 9
Nov. 15, 2002
We are going to need more time to go over the material of Chapter 14 than I thought. So
there is no new reading assignment for next week. We will end the quarter with Chapter
14 and there is no problem assignment for next week.
An Introduction to Electrochemistry
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Now we turn our attention to analytical methods that are based upon oxidation/ reduction
reactions. These methods, which are described in Chapters 14 through 18, include
oxidation/reduction titrations,
potentiometry - where me measure potential or voltage to determine quantities of
substances in solution, the pH meter is a potentiometry device,
coulometry – involves counting the number of electrons needed for a chemical reaction,
electrogravimetric analysis – the analyte is quantitatively deposited as a solid on an
electrode and the mass of analyte determined,
amperometry – current is measured, the oxygen electron uses this method
and voltammetry – involves complex processes where the relationship between current
and voltage is measured to provide information about the quantity of analyte.
The polarograph that you will learn to use next quarter is an example of an instrument that
uses voltammetry.
So this area involves background to analytical methods as well as background to
understand the properties of ions in aqueous solutions.
Iron, for example, can exist as ferrous and ferric ions, Fe(II) and Fe(III) in water. We will
develop methods to speciate ions in water samples to determine the amount of iron in each
redox state.
An oxidation/reduction reaction is one in which electrons are transferred from one
reactant to another.
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An example is the oxidation of iron(II) ions by cerium(IV) ions. The reaction is described
by the equation
Ce4+ + Fe2+ < == > Ce3+ + Fe3+
Notice that the charge balances in the chemical equation – plus 6 on the left and plus six
on the right.
Chemical equations have to conform to both the laws of conservation of mass (atoms) and
Conservation of charge.
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Since the reaction is a transfer of electrons, the Ce(IV) ion must extract an electron from
Fe(II) to form Ce(III) and Fe(III) ions.
A substance, such as Ce(IV), that has a strong affinity for electrons, and thus tends to
remove them from other species, is called an oxidizing agent, or an oxidant.
A reducing agent, or reductant, is a species, such as Fe(II), that readily donates electrons
to another species. With regard to the chemical equation, we say that Fe(II) is oxidized by
Ce(IV)- the oxidizing agent; similarly, Ce(IV) is reduced by Fe(II) – the reducing agent.
Clyde, tell us about LEO/GER. Loss of Electrons is Oxidation. Gain of Electrons is
Reduction.
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Consider the reaction 2 Ca + O2 == > 2 CaO
Which element is reduced and which is oxidized? What are the oxidizing and reducing
agents?
Calcium metal becomes calcium cation, so it must lose electrons to become a positive
cation. We can write what is called a half reaction
Ca == > Ca2+ + 2e`
Because calcium loses electrons, it is oxidized. Since calcium is oxidized, it must be the
reducing agent.
Oxygen becomes the oxide anion, which means oxygen atoms must gain electrons.
O2 + 4e- == > 2 O2Because oxygen gains electrons, it is reduced and must be the oxidizing agent.
Following oxidation and reduction is easy when the reaction involves the actual formation
of ions. However, for other reaction such as the reaction of oxygen and hydrogen to form
water, life is not so simple because there is only a partial transfer of electrons.
O2 + 2 H2 == > 2 H2O
no ions are formed
We know that the H-O bond is polar and that the bonding electrons spend more of their
time around the oxygen atom. Thus we write
(+) H-O (-)
for an O-H bond in water.
How do we keep tabs on these reactions when the electrons are not transferred
completely?
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Chemists have developed a bookkeeping system for all types of reactions that is based on
the concept of oxidation numbers. This is discussed in detail in Appendix D which you
should read.
The concept of oxidation number is based on a model that assumes all bonding is ionic in
nature whether the electrons are transferred completely or not.
Oxidation number is the charge that an atom would have if the electrons in each of its
bonds belonged entirely to the most electronegative element involved in the bond.
Here are rules for assigning oxidation number which are also given in Appendix D.
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1. In free elements each atom has the ON of ? zero
Zn(s), O2(g), I2(s), S8(s), Cl2(g)
2. For each single atom ion, Cl-, Zn2+, Al3+, the ON is the charge on the ion.
3. Fluorine is always
Alkali metals are always
Alkaline earths are always
–1
+1
+2
4. ON for hydrogen is? +1
except in metal hydrides
NaH, MgH2, LiAlH4 it is –1
[Why? Most electronegative element.]
(Li, Na, K, Rb, Cs) [ Why? most
(Be, Mg, Ca, Sr, Ba) electropositive
Elements.]
5. ON for oxygen is always –2
except in peroxides
H2O2 where it is –1
and in superoxides
KO2 where it is –1/2
and when it is a cation in the compound OF2 where it is +2
6. The algebraic sum of the oxidation numbers in a neutral compound must be? zero
7. The algebraic sum of the oxidation numbers in a polyatomic ion must be equal to the
charge on the ion.
Consider the dichromate ion, Cr2O72-. Which atom is more electronegative? Oxygen.
So we assign the electrons to the oxygen atoms, which make them negatively charged.
Also, from rule number 7, the sum of the oxidation numbers must be equal to –2.
The oxygens are not in a peroxide or superoxide, so they must be each –2.
7 x ( – 2) = -14
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Thus the chromium atoms must be 2x – 14 = -2

x = +12/2 = +6.
In oxidation-reduction reactions there are changes in oxidation numbers.
An oxidation involves an increase in ON, so the oxidized atom gives up electrons to
become more positively charged.
A reduction involves a decease in ON, so the reduced atom gains electrons and becomes
more negatively charged.
A term that is frequently used interchangeably with oxidation number is oxidation state.
Let us try these ideas on a reaction that takes place when manganese oxide dissolves in
aqueous solution.
KCl(aq) + MnO2(s) + H2SO4(aq) == > K2SO4(aq) + MnSO4(aq) + Cl2(g) + H2O(l)
This reaction may be an oxidation/reduction. How can we tell?
For complex situations, the best way to identify redox changes is to use oxidation
numbers.
We begin by assigning oxidation numbers to each atom on both sides of the equation.
Following the rules, what should we get for each of the atoms of elements in each
compound?
+1-1
+4 –2
+1+6-2
KCl + MnO2 + H2SO4 == >
+1+6-2
+2+6-2
0
K2SO4 + MnSO4 + Cl2 +
+1 -2
H2 O
But the atoms don’t added up! We to have two Cl on the left, two potassium ions and two
sulfates to equal the number on the right. This requires increasing the water on the right to
2.
+1-1 +4 –2
+1+6-2
2 KCl + MnO2 + 2 H2SO4 == >
+1+6-2
+2+6-2
0
+1 -2
K2SO4 + MnSO4 + Cl2 + 2 H2O
Now we have the atoms balanced and know the ONs of each atom in the reactants and
ONs of each atom in the reductants.
Next we look for oxidation/reduction changes, keeping in mind that an increase in
oxidation number is oxidation and a decrease is reduction. Which elements have changed
oxidation number?
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oxidation
reduction
-1
+4
+2
0
2 KCl + MnO2 + 2 H2SO4 == > K2SO4 + MnSO4 + Cl2 + 2H2O
Thus the Cl in KCl is oxidized and the Mn in MnO2 is reduced.
The reducing agent is KCl and the oxidizing agent is MnO2.
Notice the the rest of the materials are spectator ions and that the reaction consumes H+
and liberates chlorine gas.
We can split any oxidation/reduction equation into two half-reactions that show clearly
which species gains electrons and which loses them. Half-reactions are always written as
reductions.
For example, if we take out the spectator ions in the reaction above we get
MnO2(s) + 2e- == > MnSO4(aq)
We can reduce this further to
Cl2(g) + 2 e-
== >
is this the reduction or oxidation half reaction?
Mn4+ + 2 e- === > Mn2+
2 Cl-(aq)
We multiple the half reactions to get the electrons to cancel, turn the oxidation halfreaction around and then add them together.
MnO2(s) + 2e- == > MnSO4(aq)
2 Cl-(aq)
== > Cl2(g) + 2 eMnO2(s) + 2 Cl-(aq) == > MnSO4(aq)
+ Cl2(g)
This is the oxidation/reduction or redox part of the reaction we balanced above.
Remember, an oxidation/reduction reaction is one in which electrons are transferred
from one reactant to another.
Let’s examine another redox reaction
MnO4- + Fe2+ + H+ < == > Mn2+ + Fe3+ + H2O
First let’s balance the atoms and charge without worrying about redox yet.
We need 4 oxygens on the right, which means we need eight hydrogens on the left.
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How do we know how many iron ions are involved? The total charge on both sides of the
chemical equation must be equal. So far we have +9 on the left and +5 on the right.
To balance the charge we can add iron ions but for each one we add to the left we have to
add one to the right to keep the atoms balanced.
How many iron ions will it take? Need a total of 5 (several ways to get here). This gives
a total charge of +17 on both sides of the equation.
Now we have
MnO4- + 5 Fe2+ + 8 H+ < == > Mn2+ + 5 Fe3+ + 4 H2O
Which element is oxidized?
And which is reduced?
How can you tell?
Iron
Manganese
What is the oxidation number of manganese in the permanganate ion? +7
What is the reduction half-reaction?
Mn7+ + 5 e- ===> Mn2+
What is the oxidation half-reaction?
Fe3+ + 1 e- ===> Fe2+
Which is the reducing agent? Iron is the substance that causes the reduction of
manganese.
Which is the oxidizing agent? Manganese causes the oxidation of iron.
Balancing the half-reactions we get
Mn7+ + 5 e- ===> Mn2+
5[ Fe2+ ===> Fe3+ + 1 e- ]
Mn7+ + 5Fe2+ ===> 5Fe3+ + Mn2+
Many times we can balance redox equations by balancing atoms, but there are also many
complex redox reactions that require a systematic method to balance them. This method is
described in Appendix D.
Before we go into this method, I need to comment that for many redox reactions in aqueous
solutions, H+ or OH- ions play an important role, as do water molecules.
In the last example, when solutions of KMnO4 and FeSO4 are mixed, we find that the
acidity of the mixture decreases as permanganate ion, MnO4-, oxidizes iron(II).
This is because the reaction uses up H+ as a reactant and produces H2O as a product.
In other reactions, OH- is consumed, while in still other reactions H2O is a reactant.
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In fact in many cases the products (or even the reactants) of a redox reaction will differ
depending on the acidity of the solution. For example, in an acidic solution, MnO4- is
reduced to Mn2+ ion, but in a neutral or slightly basic solution the reduction product is
insoluble MnO2.
MnO4- == > Mn2+
how many electrons?
+5 eMnO4- == > MnO2
+3 eBecause of these factors, redox reactions are generally carried out in solutions containing a
substantial excess of either acid or base or are carried out in a buffer solution. Thus before
we can apply the systematic half-reaction method, we have to know the pH of the solution
or system we are examining.
Let's start our development of a systematic approach to balancing redox reactions by
assuming we have an acidic solution in which dichromate and Fe(II) can react.
The dichromate ion, Cr2O7 2- reacts with Fe2+ in an acidic solution to give Cr3+ and Fe3+ as
products. This information gives the skeleton equation.
Cr2O7 2- + Fe2+ == > Cr3+ + Fe3+
We then proceed through the following steps to find the balanced equation.
STEP 1 Divide the skeleton equation into half-reactions. We create the beginnings of
two half-reactions. Except for hydrogen and oxygen, the same elements must appear on
both sides of a given half-reaction.
Cr2O7 2- == > Cr3+
Fe2+ == > Fe3+
STEP 2 Balance atoms other than H and O. There are two Cr atoms on the left and only
one on the right, so we place a coefficient of 2 in front of Cr3+.
The second half-reaction is already balanced in terms of atoms, so nothing need be done to
it.
Cr2O7 2- == > 2 Cr3+
Fe2+ == >
Fe3+
STEP 3 Balance oxygen by adding H2O to the side that needs O. There are seven
oxygen atoms on the left of the first half-reaction and none on the right. Therefore, we add
7 H2O to the right side of the first half-reaction to balance the oxygens.
Cr2O7 2- == > 2 Cr3+ + 7 H2O
Fe2+ == > Fe3+
8
+
STEP 4 Balance hydrogen by adding H to the side that needs H. After adding the
water, we see that the first half-reaction has 14 hydrogens on the right and none on the
left. To balance these, we add 14H+ to the left side of the equation. When you do this step
(or others) be careful to write the charges on the ions. If they are omitted, you will not
obtain a balanced equation in the end.
14 H+ + Cr2O7 2- == > 2 Cr3+ + 7 H2O
Fe2+ == > Fe3+
Now each half-reaction is balanced in terms of atoms (conservation of mass). Next we will
balance the charge.
STEP 5 Balance the charge by adding electrons. First we compute the net electrical
charge on each side. For the first half-reaction we have
14 H+ + Cr2O7 2- == >
Net charge= (14+)+ (2-) = 12+
2 Cr3+ + 7 H2O
Net charge = 2(3+)
+ 0 = 6+
The algebraic difference between the net charges on the two sides equals the number of
electrons that must be added to the more positive (less negative) side. In this instance, we
must add 6e- to the left side of the equation.
6 e- + 14 H+ + Cr2O7 2- == > 2 Cr3+ + 7 H2O
This half-reaction is now complete; it is balanced in terms of both atoms and charge.
To balance the other half-reaction, we add one electron to the right.
Fe2+ == > Fe3+ + eXXXXX
STEP 6 Make the number of electrons gained equal to the number lost and then add
the two half-reactions. At this point we have the two balanced half-reactions
6 e- + 14 H+ + Cr2O7 2- == > 2 Cr3+ + 7 H2O
Fe2+ == > Fe3+ + eSix electrons are gained in the first reaction but only one is lost in the second.
Therefore, before combining the two equations we multiply the coefficients of
the second one by 6.
6 e- + 14 H+ + Cr2O7 2- == > 2 Cr3+ + 7 H2O
6 x (Fe2+ == > Fe3+ + e-)
6 e- + 14 H+ + Cr2O7 2- + 6 Fe2+ == > 2 Cr3+ + 7 H2O + 6 Fe3+ + 6 e-
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STEP 7 Cancel anything that is the same on both sides. Six electrons cancel from both
sides to give the final balanced equation.
14 H+ + Cr2O7 2- + 6 Fe2+ == > 2 Cr3+ + 7 H2O + 6 Fe3+
The final step is to check your answer. We see that both the charge and the atoms balance.
In some reactions, you may have H2O or H+ on both sides, for example, 6 H2O on the left
and 2 H2O on the right. Cancel as many as you can. Thus,
gives
+ 6 H2 O . . .
== > + 2 H2O . . .
+ 4 H2O . . .. . .
== >
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The following is a summary of the steps we've followed for balancing an equation that
occurs in an acidic solution.
Half-Reaction Method - Acidic Solution
................................................... .
Step 1. Divide the equation into two half-reactions.
Step 2. Balance atoms other than H and O.
Step 3. Balance O by adding H2O.
Step 4. Balance H by adding H +.
Step 5. Balance net charge by adding e-.
Step 6. Make e- gain equal e- loss; then add half-reactions.
Step 7. Cancel anything that's the same on both sides.
The simplest way to obtain a balanced equation for a redox reaction that occurs in a
basic solution is to first pretend that the solution is acidic. We balance the equation
using the seven steps just described, and then we use a simple three-step procedure to
convert the equation to the correct form for a basic solution.
As an example, suppose we wanted to balance the following equation for a basic
solution.
SO32- + MnO4- == > SO42- + MnO2
Following Steps 1 through 7 for acidic solutions gives
2 H+ + 3 SO32- + 2 MnO4- == > 3 SO42- + 2 MnO2 + H2O
Conversion of this equation to one appropriate for a basic solution uses the fact that H+
and OH- react in a 1-to-1 ratio to give H2O. The following three steps are involved.
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STEP 8 Add to both sides of the equation the same number of OH- as there are
H+. The equation for acidic solution has 2H+ on the left, so we add 2 OH- to each side.
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This gives
2 OH- + 2 H+ + 3 SO32- + 2 MnO4- == > 3 SO42- + 2 MnO2 + H2O + 2 OHSTEP 9 Combine H+ and OH- to form H2O. The left side has 2 OH- and 2 H+,
which become 2 H2O, so in place of 2 OH- + 2 H+ we write 2 H2O,
2 H2O + 3 SO32- + 2 MnO4- == > 3 SO42- + 2 MnO2 + H2O + 2 OHSTEP 10 Cancel any H2O that you can. In this equation, one H2O can be canceled
from both sides. The final equation, balanced for basic solution, is
H2O + 3 SO32- + 2 MnO4- == > 3 SO42- + 2 MnO2 + 2 OHNow that you've reviewed how to keep track of electrons when they are transferred and to
balance redox equations, you should be able to do stoichiometry with oxidation-reduction
reactions.
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How many milliliters of 0.136 M KMnO4 are requires to react with 25.0 mL of 0.250 M
Na2SO3?
25.0 mL SO32- x 0.250 mmol SO32- x 2 mmol MnO4- x 1 mL MnO4=
221 mL SO3
3 mmol SO3
0.136 mmol MnO4
30.6 mL