CHEMISTRY 534
CHAPTER 7
OXIDATION-REDUCTION
Electrochemistry is that branch of chemistry which deals with electric currents associated with
chemical reactions. It is divided into two main branches:
• Electrochemical cells:
The conversion of chemical energy into electrical energy.
Electric current (energy) is obtained from an electrochemical
system.
• Electrolysis: The conversion of electrical energy into chemical energy. Electric current
(energy) is put into an electrolysis system.
Both branches are based on a chemical reaction known as oxidation-reduction or redox for
short. Redox reactions consist of two separate yet dependent and simultaneous reactions. One
reaction is oxidation, the other is reduction. The substance which causes oxidation is called a
"reducing agent" while the substance which causes reduction is called an "oxidizing agent".
Like acid-base reactions, oxidation-reduction reactions also form equilibrium systems. Indeed,
there are a number of similarities between oxidation-reduction (redox) reactions and acid-base
(neutralization) reactions. Just as an acid neutralizes a base, for example, an "oxidizing agent"
neutralizes the "reducing agent". While in an acid-base reaction there is the transfer of protons
(H+ ions), in an oxidation-reduction reaction there is the transfer of electrons (e- charges).
Moreover, much like the acidity constant is an indication of the relative strength of an acid, a
voltage is an indication of the relative strength of an oxidizing or reducing agent.
Originally, oxidation was considered as a reaction in which a substance was "oxidized" thereby
producing an "oxide" The rusting of iron, for example, is an oxidation reaction in which oxygen
in the air (moisture) reacts with iron to produce rust (iron oxide). In fact, although slow, rusting
is a burning process (an exothermic reaction) just like the burning of wood, natural gas, gasoline,
etc. An oxidation reaction, therefore, was defined as a reaction in which a substance "gained"
oxygen.
Today, we know that during an oxidation reaction, an atom loses electrons while in a reduction
reaction an atom gains electrons. As a result, the definition of oxidation has been generalized to
describe any process in which there is a transfer of electrons between atoms regardless of the
presence of oxygen.
In this chapter, we study oxidation-reduction reactions. The study of such reactions provides us
with the opportunity to use and relate the chemical concepts covered in previous chapters.
Page 1
▼ TOPIC-1: AN AFFAIR OF ELECTRONS
In order to discover the characteristics of oxidation-reduction reactions, let's perform an
experiment comparing the reaction of different metals.
➲ EXPERIMENT-33:
Objective:
The capacity to react
To study the reactivity of lead, copper, and zinc. This experiment is divided into
two parts
• PART-A: Visible facts
• PART-B: Hidden facts
PART -A: Visible Facts
Step-1: Place 3 mL of 0.50 M copper(II) nitrate, Cu(NO3)2, into three test tubes.
☞Hint: Identify each test tube as to its contents.
Step-2: Place 3 mL of 0.50 M lead (II) nitrate, Pb(NO3)2, into three test tubes.
Step-3: Place 3 mL of 0.50 M zinc nitrate, Zn(NO3)2, into three test tubes.
Step-4: Into three test tubes each containing Cu(NO3)2, Pb(NO3)2, and Zn(NO3)2
respectively, add a piece of copper metal.
Step-5: Into three test tubes each containing Cu(NO3)2, Pb(NO3)2, and Zn(N03)2
respectively, add a piece of lead metal.
Step-6: Into the remaining three test tubes containing Cu(NO3)2, Pb(NO3)2, and Zn(NO3)2
respectively, add a piece of zinc metal.
1) Observe each test tube for about 2 minutes then record whether or not (yes or no) a reaction
has occurred in each test tube.
METAL
Copper (Cu)
Lead (Pb)
Zinc (Zn)
Page 2
Copper(II) nitrate
Cu(NO3)2
SOLUTIONS
Lead(II) nitrate
Pb(NO3)2
Zinc nitrate
Zn(NO3)2
2) Of the three metals (copper, lead and zinc), which metal reacted:
a) With the greatest number of solutions? ______________
b) With the least number of solutions?
______________
3) Classify the three metals in decreasing order of their reactiveness (tendency to react).
① ________________________________
② ________________________________
③ ________________________________
4) Define a spontaneous reaction:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
5) Give an example of a spontaneous reaction:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
PART-B: Hidden facts
6) For those solutions in which there was a reaction (there should be three), identify the
reactants and the products: (Note: nitrate ions (NO3-) are spectator ions. They remain in the
solution and are unreactive )
METAL
Copper (Cu)
Lead (Pb)
Zinc (Zn)
Copper(II) nitrate
Cu(NO3)2
SOLUTIONS
Lead(II) nitrate
Pb(NO3)2
Zinc nitrate
Zn(NO3)2
Reactants:
Reactants:
Reactants:
Products:
Products:
Products:
Reactants:
Reactants:
Reactants:
Products:
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Reactants:
Reactants:
Reactants:
Products:
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Page 3
7) Consider the transformation of zinc to zinc ions in the presence of copper(II) nitrate.
At the start of the reaction, zinc is a solid. At the end of the reaction, the solid zinc has been
transformed into zinc ions. At the same time, copper(II), Cu2+, ions have been transformed
into solid copper.
a) Write the equation which represents solid zinc being transformed into zinc ions:
__________________________________________________________________________
b) Write the equation which represents copper(II) ions being transformed into solid copper.
__________________________________________________________________________
c) The copper(II) ions have gained two electrons. Where did they come from?
__________________________________________________________________________
d) Write the overall equation for this reaction:
__________________________________________________________________________
8) Now consider the transformation of zinc to zinc ions in the presence of lead(II) nitrate. In
this case, as the solid zinc is transformed into ions, the lead ions are transformed into solid
lead.
a) Write the equation which represents the transformation of zinc:
__________________________________________________________________________
b) Write the equation which represents the transformation of lead (Pb2+) ions:
__________________________________________________________________________
c) Write the overall equation for this reaction:
__________________________________________________________________________
9) The last reaction is the transformation of lead in the presence of copper(II) nitrate. Write the
three equations which describe this reaction:
➀ ______________________________________________________________________
➁ ______________________________________________________________________
➂ ______________________________________________________________________
ÄEnd of experiment-33
Page 4
An oxidation-reduction reaction consists of two "half-reactions" each forming a "half-cell". One
half is the oxidation reaction, the other half is the reduction reaction. The two "halves" work
together in that one releases (loses) electrons while the other takes in (gains) the lost electrons.
Don't be confused! Electrons are negative charges. Note that if you lose electrons, you increase
your total numeric value. Conversely, if you gain electrons, you decrease your total numeric
value. Study these examples:
INITIAL
REACTION VALUE
TRANSFER
Reduction
10
+ 3 e= 10 + (-3)
= 10 - (-2)
Oxidation
10
-2e
Reduction
7
+4 e
= 7 + (-4)
= 12 - (-5)
Oxidation
12
-5 e= 6 - (-4)
Oxidation
6
-4 e
Reduction
0
+1e
= 0 + (-l)
= 5 + ( -1)
Reduction
5
+1e
=
=
=
=
=
=
=
10 - 3
10 + 2
7-4
12 + 5
6+4
0-1
5-1
=
=
=
=
=
=
=
FINAL
VALUE
7
12
3
17
10
-1
4
ÄRemember: Oxidation is losing electrons thereby becoming more positive.
Reduction is gaining electrons thereby becoming more negative.
When we place a copper electrode in a solution of silver nitrate (AgNO3), the reaction proceeds
slowly. Soon, we notice the formation of silver- colored "pins" as the solution turns blue. The
blue color is an indication that copper(II) ions are present in the solution and the solid deposit is
silver.
The equations which represent these transformations are:
Loss of 2 electrons by a copper atom:
Cu(s) à Cu2+(aq) + 2e-
Gain of 1 electron by a silver ion:
{ Ag+(aq) + le- à Ag(s) }x2
Overall equation:
Cu(s) + 2 Ag+(aq) à Cu2+(aq) + 2 Ag(s)
Note that in this redox reaction each copper atom loses two electrons while each silver ion gains
one electron. Nonetheless, the electron exchange (gain and loss) occurs in such a way that none
of the electrons appear in the overall equation. As a result, a reaction which liberates electrons is
always accompanied by a reaction which absorbs the liberated electrons.
Page 5
10) As you know, an oxidation reaction is one in which electrons are released. Write the halfreaction which represents the oxidation in the above transformations:
____________________________________________________________________________
11) A reduction reaction is one in which electrons are absorbed. Write the half-reaction which
represents the reduction in the above transformations:
___________________________________________________________________________
By definition, the cathode of an electrochemical cell is the electrode at which
reduction occurs, while the anode is the electrode where oxidation occurs.
AN OX RED CAT
Again, an oxidation-reduction reaction is an exchange of electrons between a donor and a
receiver. A cathodic reaction is the reduction half of a redox reaction. Cathodic reactions take
place at the cathode and consist of receiving electrons. An anodic reaction, on the other hand, is
the oxidation half of a redox reaction. Anodic reactions take place at the anode and consist of the
release of electrons. Furthermore, in all redox reactions, there is no net loss or gain of electrons.
What is lost is what is gained. For example, consider the following redox reaction:
Cu(s) + 2 Ag+(aq) à Cu2+(aq) + 2 Ag(s)
Copper atoms here undergo oxidation (since they lose electrons) while silver ions undergo
reduction (since they gain the lost electrons). Because copper causes the Ag+ ions to gain
electrons, copper is called the reducing agent or reductant. Conversely, because the Ag+ ions
cause the copper to lose electrons, the Ag+ ions constitute the oxidizing agent or oxidant
12) For each oxidation-reduction reaction given below, complete the following:
a)
Page 6
Zn(s) + Cu2+(aq) à Cu(s) + Zn2+(aq)
The oxidation reaction:
__________________________________________
The reduction reaction:
__________________________________________
The oxidizing agent:
__________________________________________
The reducing agent:
__________________________________________
b) Zn(s) + Pb2+(aq) à Pb(s) + Zn2+(aq)
The oxidation reaction:
__________________________________________
The reduction reaction:
__________________________________________
The oxidizing agent:
__________________________________________
The reducing agent:
__________________________________________
c)
Pb(s) + Cu2+(aq) à Cu(s) + Pb2+(aq)
The oxidation reaction:
__________________________________________
The reduction reaction:
__________________________________________
The oxidizing agent:
__________________________________________
The reducing agent:
__________________________________________
13) When a piece of aluminum (Al) is placed in a lead(II) nitrate solution, Pb(NO3)2, a deposit
of lead (Pb) appears on the aluminum. Use this information to complete the following:
2 Al(s) + 3 Pb2+(aq) à 3 Pb(s) + 2 Al3+(aq)
The oxidation half-reaction: __________________________________________
The reduction half-reaction: __________________________________________
Redox reaction:
__________________________________________
The oxidizing agent:
__________________________________________
The reducing agent:
__________________________________________
Page 7
▼ TOPIC-2: HIGH POTENTIAL PAIRS
In the last experiment, we discovered that the capacity to react varies from metal to metal. Zinc,
for example, acted as a good donor of electrons as it gave electrons both to lead ions and to
copper(II) ions. Lead, however, supplied electrons to the Cu2+ ions but not to the Zn2+ ions. It
seems that the ability to donate electrons varies for different metals. In other words, the capacity
to donate electrons is not the same for all metals.
Is the capacity to donate electrons a characteristic property of metals?
Let's find out by an experiment investigating a few different metals.
wExperiment -34 Classifying Metals
Objective: To investigate the capacity of different metals to supply electrons thereby classifying
them according to their ability to donate electrons.
➣Reminder: Solid metals which are dipped in solutions are called electrodes.
ÄNote: To be sure that the metals react readily, remove the corrosion (oxidation) on their
surface by sand papering the electrodes.
Step-1:
Pour 100 mL of a 0.50 mol/L sodium chloride (NaCl) solution into a 250 mL beaker.
Step-2:
Place two different metal electrodes into the solution and connect the electrodes to a
voltmeter. Follow the diagram below and make sure that the metals do not touch each
other.
ÄImportant: In using an analog voltmeter, reverse the polarity of the probes if the
needle moves against the zero mark. If you are using a digital
voltmeter, reverse the probes if you obtain a negative voltage.
Step-3:
Page 8
Record the voltage readings for each pair of electrodes in the table below.
(Ä Note: The voltage is independent of the surface area of the electrodes)
14) Record the voltages for each metal pair here:
Electrodes (-) â/ (+) à
Zinc (Zn)
Cu
Mg
Ni
Pb
Sn
Zn
Copper (Cu)
Magnesium (Mg)
Nickel (Ni)
Lead (Pb)
Tin (Sn)
15) How can we classify the metals in order of their ability to donate electrons?
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
16) Classify the metals in decreasing order to their ability to donate electrons:
• ________________________
‚ ________________________
ƒ ________________________
„ ________________________
… ________________________
† ________________________
17) Identify the metal "pair" which produced the highest potential difference (voltage):
_________________________________________________________________
Page 9
18) With reference to the list of STANDARD OXIDATION POTENTIALS, write the
oxidation half-reaction for each metal analyzed. Arrange them in decreasing order of their
ability to release electrons.
• ____________________________________________________________________
‚ ____________________________________________________________________
ƒ ____________________________________________________________________
„ ____________________________________________________________________
… ____________________________________________________________________
† ____________________________________________________________________
➢End of experiment-34
19) Here are more oxidation half-equations:
Al
Cr
Ni
Pb
à
à
à
à
A13+ + 3 eCr3+ + 3 eNi2+ + 2 ePb2+ + 2 e-
With reference to the above list, identify the pair which produces the greatest difference in
potential (voltage). Why?
Al/ Cr
Al/ Ni
Al/Pb
Cr/Ni
Cr/Pb
Ni /Pb
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
Experimentally, we can obtain the difference in potential (voltage) between two metals thus
comparing one metal to another. However, this comparison gives us only the difference between
two metals. This is similar to comparing the difference in heights of a group of people. For
example, we may say, the difference between Jack and Jill is 8 cm, the difference between Jack
and Tom is 3 cm, the difference between Jane and David is 2 cm, etc. Instead, it would be more
useful to simply list the height of each person (from some "zero" reference point such as the
floor).
Page 10
Likewise with metals. It would be more useful to determine and list the difference in potential of
all metals from some zero reference point. To do this, we must select an arbitrary potential of
0.00 volts to be used as a reference.
To establish a potential of 0.00 V as the reference point, chemists have selected the dissociation
of gaseous hydrogen at a pressure of 101.3 kPa and at a temperature of 25 0C. The potential of
this reaction is fixed to be 0.00 V. All potentials relative to this zero voltage point are designated
as E0. Thus, we have:
H2(g) à 2 H+(aq) + 2 eE0 = 0.00 V
Using hydrogen as the standard reference, the potential of a great many metals can be measured. Indeed,
chemists have already done this and call the list the "Standard Oxidation Potentials".
w SOME STANDARD OXIDATION POTENTIALS w
HALF-REACTION
POTENTIAL
Mg
Al
Zn
Cr
Co
Ni
Pb
H2
Sn
Cu
Ag
à
à
à
à
à
à
à
à
à
à
à
Mg2+ + 2 eA13+ + 3 eZn2+ + 2 eCr3+ + 3 eCo2+ + 2 eNi2+ + 2 ePb2+ + 2 e2 H+(aq) + 2 eSn2+ + 2 eCu2+ + 2 eAg+ + 1 e-
E0
E0
E0
E0
E0
E0
E0
E0
Eo
E0
E0
= +2.37 V
= +1.66 V
= +0.76 V
= +0.74 V
= +0.28 V
= +0.26 V
= +0.13 V
= 0.00 V
= -0.13 V
= -0.34 V
= -0.80 V
The oxidation potential of a metal is the tendency of its half-reaction to give up electrons (at
standard conditions) as compared to the tendency of the hydrogen half-reaction. A positive sign
is given to metals whose half-reaction releases electrons more readily than the hydrogen halfreaction, and a negative sign is given to metals whose half-reaction releases electrons less readily
than the hydrogen half-reaction.
20) Explain the arbitrary nature of the oxidation potential table:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
Page 11
21) Find the difference in potential for each of the following metal pairs:
a) Chromium and nickel
_________________________________
b) Chromium and lead
_________________________________
c) Chromium and hydrogen
_________________________________
d) Chromium and copper
_________________________________
e) Chromium and silver
_________________________________
22) Derive a formula for calculating the difference in potential between two metals:
________________________________________________________________________
________________________________________________________________________
23) Identify the metal pair which has the highest difference in potential. Explain:
________________________________________________________________________
________________________________________________________________________
Given any two metal combinations, the oxidation potentials can be used to predict which metal
will undergo oxidation and which will undergo reduction. It's simple, the metal with the higher
oxidation potential undergoes oxidation while the metal with the lower oxidation potential
undergoes reduction.
Also, if the difference in potential between a metal and the metal ion in solution is positive, then
the reaction will take place spontaneously (proceeds by itself). On the other hand, if the
difference in potential between a metal and the metal ion in solution is negative, the reaction will
not occur spontaneously.
Sample Problem-1
Zinc is placed in a solution of nickel sulfate. a) Is the reaction spontaneous?
b) Which metal will undergo oxidation?
To find out, we look up the oxidation potential of both metals. In this case, we have +0.76
V for zinc metal and +0.26 V for nickel ions. Since the difference in potentials is positive
[(+0.76)- (+0.26) = +0.50], reaction is spontaneous.
Moreover, since the oxidation potential of zinc is higher that of nickel, zinc undergoes
oxidation while nickel ions undergo reduction.
Page 12
Sample Problem-2
Silver is placed in a solution of copper (II)sulfate. a) Is the reaction spontaneous?
b) Which metal will undergo oxidation?
We look up the oxidation potential of both metals. For silver, the potential is -0.80 V and
for copper it is -0.34 V.
In this case, the difference in potentials is negative [(-0.80)- (-0.34) = -0.46 V], we
know that the reaction will not occur spontaneously. As a result, we conclude that
oxidation will not occur to any appreciable extent. Conclusion, neither metal undergoes
oxidation.
Page 13
▼ TOPIC-3: GIVING AND RECEIVING
A QUESTION OF POTENTIAL
We know that an oxidation-reduction reaction involves the movement of electrons from one
substance to another. Recall that an electric current is the movement of electrons in a wire. How
may we utilize an oxidation-reduction reaction in order to produce an electrical current in a wire?
Electrons are negatively charged particles which form an integral part of matter. The movement
of electrons is responsible for the operation of all electrical devices, from the light bulb to the
computer. It would be useful to generate electrons on demand.
While there are a number of ways to generate electrons, two particularly convenient methods are
by using a generator and by using a battery. Whereas a generator produces electrons by physical
interaction, a battery produces electrons by chemical reaction. Note that the generation and
control of electrons is so important that a branch of science is dedicated exclusively towards
that end. Electronics is the science of generating and controlling electrons.
In experiment-33, we saw that an oxidation-reduction reaction is a chemical method of
generating and moving electrons What is the mechanism of this electronic movement?
Zn
2+
Cu
Direct exchange of electrons
Since an electrical current is the flow of electrons in a conductor, to obtain an electric current, we
can use a redox reaction. The oxidation process causes the electrons to be generated while the
reduction process attracts the electrons. By connecting an external wire between the two
processes, we have an electric current (electron flow).
Page 14
Review of electric current
Recall that current is the rate of flow of charge and is designated by the letter "I" (Intensity). The unit for current is
the amp•re. One amp•re equals one coulomb per second or 1.04 x 10-5 mol of electrons flowing per second.
In a wire (solid metal), current consists of the flow of electrons and is called metallic conduction. In a solution ,
current consists of ions (both anions and cations ) and is called ionic conduction. Anions are negative ions (attracted
to the anode) and cations are positive ions (attracted to the cathode).
But what causes charges to flow? The answer is the potential difference or voltage.
Do not confuse electrical potential (voltage) with current. By way of analogy, you can think of potential as height,
and current as water flow. Note that it is the height, or more precisely, the difference in height between two points,
which causes water to flow. Likewise with current, it is the potential difference which causes charge to flow
(electrons in a wire and ions in a solution). Where there is no potential difference (no difference in height), no
current can flow. An electrical "load" is any device which requires electricity to operate such as a lamp, a motor, a
heater, etc. All loads require the flow of electrons for their operation. However, in order for electrons (current) to
flow, a voltage must be applied to the load. In effect, the voltage creates a difference in potential (similar to a
difference in height) so that electrons can flow. The two points where a voltage is applied to a load are called the
terminals.
Recall from chapter-6 that all systems have a natural tendency to lower their energy level as much as possible. When
we raise a book in the air, for example, we place the book at a higher energy level. When we let go of the book, it
spontaneously falls down towards a lower energy level. Likewise with electricity, current always flows from a
higher to a lower potential (energy level).
When a potential difference exists between two points, one point is at a higher potential than the other point. Just as
potential energy is a measure of the energy between two heights, voltage is a measure of the difference in electrical
potential energy between two points. Thus, when a voltage is applied to a wire, one end of the wire is at a higher
potential than the other end. Electrons at the higher end have more potential energy than the electrons at the lower
end (just as a book has more potential energy at a higher level than at a lower level). As a result, electrons flow from
the higher to the lower level.
Work is defined as the energy required to perform a useful task. When work is being done, energy is being used. As
electrons flow from a higher to a lower potential, we can use (convert) part of their energy to do work such as
producing heat or light. In doing so, we are in effect lowering the potential energy of the electrons (just as a falling
book can be made to do something useful while falling and in the process lower its potential energy).
A voltmeter is used to measure the difference in potential. The voltmeter is placed across the two points whose
potential (voltage) is to be measured; that is, in parallel with the terminals of a load.
An ammeter is used to measure the intensity of the current (rate of flow of electrons in a wire). Since an ammeter
measures the flow of charges (electrons), it must be connected along the path of the electron flow; that is, in series
with the terminals of a load.
Page 15
In order to generate a current. a potential difference is required. The potential difference can be
generated with an oxidation-reduction reaction. However, to prevent the electrons from flowing
directly from the reducing agent to the oxidizing agent, we must keep the two reactions
separated. We do this by placing the oxidation reaction in one container and the reduction
reaction in another container.
The set up consists of a large container which contains a zinc electrode in a
solution of zinc nitrate, Zn(NO3)2. The zinc nitrate supplies zinc ions,
Zn2+(aq). In another smaller container within the larger one, we have a
copper electrode in a copper nitrate solution, Cu(NO3)2. Because the two
solutions are in separate containers, the zinc ions, Zn2+(aq), are not in direct
contact with copper ions, Cu2+(aq).
As we saw in experiment-33, the reaction consists of a spontaneous
exchange of electrons from the solid zinc metal to the copper(II) ions.
There was no exchange of electrons between solid copper and the zinc
ions.
Due to the set up, the spontaneous tendency of the electrons to go from the zinc to the copper(II)
ions takes place via the connecting wire.
ÄNote: It is important to realize that the receiver of the electrons is not the copper metal
(electrode) but the copper(II) ions in the solution. Metals have a tendency to give off electrons
and in the process become more stable. In fact, when metals gain electrons they become less
stable.
To review, an electrochemical cell is an oxidation-reduction reaction. The
oxidation process, called a half-reaction, generates electrons and forms
'half' of the electrochemical cell (called a half-cell). The reduction process,
also called a half-reaction, receives the electrons generated by the oxidation
process and forms the other half of the electrochemical cell.
Page 16
w EXPERIMENT-35:
Objective:
Potential and work
To study electrochemical cells. [micro approach see page 42 ]
In this experiment, we'll construct two types of electrochemical cells. In one type, the oxidation
and reduction reactions will be linked by a porous material (cup); in the other, a salt bridge will
be used to link the two half reactions. Both set-ups form an electrochemical cell.
Note that although these electrochemical cells do not resemble familiar batteries, they work on
the same principle. Our laboratory cells are intended to make it easier to understand their
operation.
PART-A: An electrochemical cell using a porous cup
Step-1: Fill a porous cup with 1.0 M copper(II) nitrate, Cu(N03)2.
(To approximately 1 cm from the top)
Step-2:
Step-3:
Place the porous cup into a 250 mL beaker.
Fill the beaker with a 1.0 M zinc nitrate solution, Zn(N03)2.
(To about the same level as the copper(II) nitrate solution)
Step-4: Place a strip of copper metal (called the copper electrode) into the
copper(II) nitrate solution.
Ä Note: Sand paper the electrode so as to remove any oxidation.
Page 17
Step-5:
Place a strip of zinc metal into the zinc nitrate solution.
(Again, sand paper the electrode so as to remove any oxidation)
Step-6:
Connect a voltmeter to the electrodes.
(Ä Note: If alligator clips are used, make sure they make a tight grip)
Step-7: Record the voltage of the circuit as well as your observations.
24) Record your observations of the Cu-Zn porous cup cell (including the cell voltage):
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
Next, we'll make a magnesium-zinc electrochemical cell.
Step-8:
Replace the porous cup containing copper(II) nitrate with one containing
magnesium nitrate.
Step-9:
Place a strip of magnesium ribbon into the magnesium nitrate solution.
Step-10: Connect a voltmeter to the zinc and magnesium electrodes.
Step- 11: Record the voltage of the circuit as well as your observations.
25) Record your observations of the Mg-Zn porous cup cell:
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
__________________________________________________________________________
PART-B: An electrochemical cell using a salt bridge
Page 18
Step-1:
Fill a 100 mL beaker with about 80 mL of 1.0 M copper(II) nitrate
solution.
Step-2:
Fill a 100 mL beaker with about 80 mL of 1.0 M zinc nitrate solution.
Step-3:
Place a salt bridge containing a solution of 1.0 M potassium nitrate
between both beakers. (Be sure to fill the ends of the bridge with cotton)
Step-4:
Place a copper electrode in the copper(II) nitrate solution and a zinc
electrode into the zinc nitrate solution.
Step-5: Connect a voltmeter to the electrodes.
Step-6: Record the voltage of the circuit as well as your observation.
26) Record your experimental observations regarding the salt bridge cell:
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
27) Describe the similarities between the two electrochemical cells:
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
2e-
2e-
!
;
(Oxidation)
(Reduction)
Page 19
28) Describe the differences between the two electrochemical cells:
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
29) Explain what happens if:
a) We remove an electrode from solution:
_______________________________________________________________________
_______________________________________________________________________
b) We remove the porous barrier:
_______________________________________________________________________
_______________________________________________________________________
c) We remove the salt bridge:
_______________________________________________________________________
_______________________________________________________________________
30) Provide the following information concerning electrochemical cells used in this experiment:
a) Oxidation reaction:
________________________________________________________________________
b) Reduction reaction:
________________________________________________________________________
c) Oxidation-reduction reaction:
________________________________________________________________________
d) Oxidizing agent: __________________________
e) Reducing agent: __________________________
f) The electrode which increased in mass: ________________________________________
g) The electrode which decreased in mass: _______________________________________
31) Explain why the mass of the electrodes varies:
____________________________________________________________________
____________________________________________________________________
Page 20
32) With reference to the STANDARD OXIDATION POTENTIALS table, determine the
potential difference of this cell.
__________________________________________________________________________
__________________________________________________________________________
33) How does the calculated value compare to the experimental value?
__________________________________________________________________________
__________________________________________________________________________
Ä End of experiment-34
e-
Pb
Zn
Pb2+
Zn2+
34) An electrochemical cell is made by placing a lead electrode in a lead(II) nitrate solution and a
zinc electrode in a zinc nitrate solution.
Fill in the following information concerning this Zn-Pb cell:
a) Oxidation reaction: ___________________________________________________
b) Reduction reaction: ___________________________________________________
c) Oxidation-reduction reaction: ___________________________________________
d) Oxidizing agent: ________________
e) Reducing agent: ________________
f) The electrode which increases in mass: ___________________________________
g) The electrode which decreases in mass: __________________________________
h) The cell voltage: ____________________________________________________
Page 21
35) An electrochemical cell is formed with a copper and a silver electrode. Each electrode is
placed in its respective cation solution.
a) Complete the sketch below by labeling the electrodes and the cations (positive ions) in
each solution. Then fill-in the pertinent information below:
b) Oxidation reaction: ______________________________________________________
c) Reduction reaction: ______________________________________________________
d) Oxidation-reduction reaction: ______________________________________________
e) Oxidizing agent: _____________________ f) Reducing agent:___________________
g) The electrode which increases in mass: ______________________________________
h) The electrode which decreases in mass: ______________________________________
i) The cell voltage: _______________
Page 22
36) A cell is made by using a solution of sodium nitrate as a bridge between two half cells. One
half cell consists of a copper electrode in a copper(II) nitrate solution, while the other half
cell is made with a magnesium electrode in a magnesium nitrate solution. Here is a diagram
of the electrochemical cell:
Mg
Cu
Cu(NO3)2(aq)
Mg(NO3)2(aq)
a ) Oxidation reaction: _________________________________________________________
b ) Reduction reaction: ________________________________________________________
c ) Oxidation-reduction reaction: ________________________________________________
d ) Oxidizing agent: _______________________ e ) Reducing agent: ___________________
f ) The electrode which increases in mass: _________________________________________
g ) The electrode which decreases in mass: ________________________________________
h ) The cell voltage: _______________
Page 23
Although a sketch or drawing of an electrochemical cell is a convenient way of representing an
electrochemical cell, it is inconvenient to make a drawing each time a cell is to be specified.
Rather than draw the actual cell, we can write the two half reactions instead. In using this
method, a slash (or a vertical line "") is used to separate an electrode from its solution, and a
double slash is used to separate one half cell from the other. Study the example below.
Pb
Ni
Pb(NO3)2(aq)
Ni(NO3)2(aq)
Pb(s) / Pb(NO3)2(aq) // Ni(NO3)2(aq) / Ni(s)
In using the notation illustrated above note that the electrodes are written at the two extremities.
The slash represents the physical contact between the electrode, Pb(s), and its solution,
Pb(NO3)2(aq). The double slash represents the link between the two half cells (in this case, a salt
bridge). Finally, the last slash represents the physical contact between the electrode, Ni(s), and its
solution, Ni(N03)2(aq).
Each barrier (or physical contact) produces a small difference in potential (voltage). Thus, we
have the following potential differences:
Lead and lead(II) nitrate solution
Lead(II) nitrate solution and salt bridge
Salt bridge and nickel nitrate solution
Nickel nitrate solution and nickel electrode
In effect, the sum total of these tiny potential differences equals the cell voltage.
Certain "electric fish", such as the eel, have naturally occurring chemical reactions which generate
electric voltages. In effect, they possess a battery of electrochemical cells connected in series
producing voltages up to 500 volts. They use this electrical energy as a weapon to shock and stun their
enemies as well as a type of radar for orientation purposes.
Page 24
37) An electrochemical cell is depicted by the following notation:
Al(s) / Al(NO3)3(aq) // Co(NO3)2(aq) / Co(s)
Sketch this cell and include the direction of electron flow.
Fill in the following pertinent information concerning this cell:
a ) Oxidation reaction: ________________________________________________________
b ) Reduction reaction: ________________________________________________________
c ) Oxidation-reduction reaction: ________________________________________________
d ) Oxidizing agent: _______________________ e ) Reducing agent: __________________
f ) The electrode which increases in mass: ________________________________________
g ) The electrode which decreases in mass: _______________________________________
h ) The cell voltage: _______________
Page 25
38) Sketch the following cell then fill-in the pertinent information:
Cr(s) / Cr(NO3)2(aq) // Cu(NO3)2(aq) / Cu(s)
a ) Oxidation reaction: _______________________________________________________
b ) Reduction reaction: _______________________________________________________
c ) Oxidation-reduction reaction: _______________________________________________
d ) Oxidizing agent: _______________________ e ) Reducing agent: _________________
f ) The electrode which increases in mass: _______________________________________
g ) The electrode which decreases in mass: ______________________________________
h ) The cell voltage: _______________
Page 26
▼ Topic-4: A Different Point of View
The list of electron donors provides much information. It tells us, for example, that aluminum
is a better donor than zinc, chromium, cobalt, etc.
HALF-REACTION
Mg
Al
Zn
Cr
Co
Ni
Pb
H2
Sn
Cu
Ag
à
à
à
à
à
à
à
à
à
à
à
POTENTIAL
Mg2+ + 2 eA13+ + 3 eZn2+ + 2 eCr3+ + 3 eCo2+ + 2 eNi2+ + 2 ePb2+ + 2 e2 H+(aq) + 2 eSn2+ + 2 eCu2+ + 2 eAg+ + 1 e-
E0
E0
E0
E0
E0
E0
E0
E0
Eo
E0
E0
= +2.37 V
= +1.66 V
= +0.76 V
= +0.74 V
= +0.28 V
= +0.26 V
= +0.13 V
= 0.00 V
= -0.13 V
= -0.34 V
= -0.80 V
The list also tells us that an electrochemical cell made with, say, aluminum and nickel, will react
spontaneously. The aluminum acts as an electron donor and the nickel acts as an electron
acceptor.
Al ➝ Al3+ + 3e-
E0 = + l.66 V
Ni ➝ Ni2+ + 2e-
E0 = + 0.26 V
In a nickel-lead cell, however, it is the nickel which acts as the electron donor and the lead as an
electron acceptor.
Ni ➝ Ni2+ + 2e-
E0 = + 0.26 V
Pb ➝ Pb2+ + 2e-
E0 = + 0.13 V
Obviously, the behavior of nickel is determined by the metal with which it is paired. Note that
while we may classify metals by their ability to donate electrons, it is also possible to classify
them instead by the ability of the cations (positive ions) in solution to accept electrons.
Page 27
In classifying a metal as a good donor of electrons, we are in effect classifying its cations as a
weak acceptor of electrons. For example, since aluminum metal is a good donor of electrons, the
aluminum ion, Al3+(aq), is a poor acceptor of electrons. Thus, it is not surprising to discover that
the reducing potential of aluminum is negative:
Capacity to donate electrons:
Al ➝ Al3+ + 3e-
E0 = + l.66 V
Capacity to receive electrons:
Al3+ + 3e- ➝ Al
E0 = - l.66 V
Indeed, it is possible to use either an oxidation potential list (that is, the ability to donate
electrons), or conversely, a reduction potential list (that is, the ability to receive electrons). Both
lists are valid.
In the Oxidation Potential table, the equations indicate the donation of electrons. The best donor,
in this scale, is magnesium. This means that silver, at the bottom of the scale, although the worst
donor of electrons is the best receiver of electrons.
On the other hand, in the Reduction Potential table, the equations indicate the acceptance of
electrons. The best acceptor is silver which is at the top of the list. This means that magnesium,
which here is at the bottom of the list, is the worst acceptor of electrons.
Remember: Both tables give the same information but from a different point of view.
Page 28
39) Using the Oxidation Potential table, identify the best donor of electrons for the following
pairs of metals:
a) Zn and Pb
_____________________________________
b) Co and Cr
_____________________________________
c) H2 and Ni
_____________________________________
40) Using the Oxidation Potential table, identify the best acceptor of electrons for the following
pairs of metals:
a) Zn2+ and Pb2+
_____________________________________
b) Co2+ and Cr3+
_____________________________________
c) H+ and Ni2+
_____________________________________
41) Using the Reduction Potential table, identify the best donor of electrons for the following
pairs of metals:
a) Zn and Pb
_____________________________________
b) Co and Cr
_____________________________________
c) H2 and Ni
_____________________________________
42) Using the Reduction Potential table, identify the best acceptor of electrons for the following
pairs of metals:
a) Zn2+ and Pb2+
_____________________________________
b) Co2+ and Cr3+
_____________________________________
c) H+ and Ni2+
_____________________________________
43) Compare your answers for number 40 with those of number 42, and compare your answers
of number 39 with those of number 41. Explain:
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
Page 29
44) Here are two situations. In one beaker, we place a cobalt electrode in a solution of
copper(II) nitrate. In another beaker, we place a copper electrode in a cobalt nitrate solution.
Co
Cu
Cu2+
Cu(N03)2
Co2+
Co(N03)2
Identify the spontaneous reaction and explain your answer:
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
45) Consider the following situation:
Al
Mg2+
Mg (N03)2
After a certain time, is it possible to find aluminum ions (Al3+) in the solution? Explain your
answer:
__________________________________________________________________________
__________________________________________________________________________
46) A copper electrode is placed in an aqueous solution of hydrochloric acid (HCl). Is a gas
released? Explain:
__________________________________________________________________________
__________________________________________________________________________
Page 30
47) A piece of silver jewelry falls into a solution of cobalt nitrate, Co(NO3)2. Is there reason to
be upset? Explain:
________________________________________________________________________
________________________________________________________________________
48) To identify the presence of an acid, pieces of magnesium metal may be used. Explain why:
_______________________________________________________________________
_______________________________________________________________________
49) How may we predict if a reaction is spontaneous?
________________________________________________________________________
________________________________________________________________________
Page 31
t TOPIC 5: Everything is Relative
In constructing an electrochemical cell, we put together two half cells. When the cells are
connected together, a competition begins between the electron donors. In the example below,
magnesium is a better electron donor than lead. Thus, magnesium acts as the donor of electrons
and lead as the receiver. Electrons travel from the magnesium electrode to the lead electrode. As
a consequence, in the half-cell on the left, the Pb2+(aq) cations are the acceptors of the electrons.
E0 = + 0.13 V
E0 = + 2.37 V
Pb
Mg
Pb(NO3)2
Mg(NO3)2
We can represent this system by the following equations:
Oxidation:
Mg(s) ➝ Mg2+(aq) + 2 e-
Reduction:
Pb2+(aq) + 2 e- ➝ Pb(s)
E0 = +2.37V
+E0= -0.13V
Oxidation-reduction:Mg(s) + Pb2+(aq) ÝMg2+(aq) + Pb(s) E0 = +2.24 V
In determining the potential values, we look them up in either the oxidation or the reduction
table. However, it is important to note that the oxidation and reduction reactions are reversed as
are the signs of their values.
We can analyze a system from either point of view. That is, from the point of view of the donors
of electrons or from the point of view of acceptors of electrons.
Page 32
In the Reduction Potential table, lead cations, Pb2+(aq), are better acceptors of electrons than
magnesium cations, Mg2+(aq). The lead cations, therefore, cause the electrons to move from the
cell on the right towards the cell on the left.
E0 = -2.37 V
Eo = -0.13 V
Pb
Mg
Pb(NO3)2
Mg(NO3)2
The lead cation, Pb2+(aq), acts as an electron acceptor thereby causing the magnesium
metal to act as an electron donor. The equations are the same:
Oxidation:
Mg(s) ➝ Mg2+(aq) + 2 e-
Reduction:
Pb2+(aq) + 2 e- ➝ Pb(s)
Oxidation-reduction:
Mg(s) + Pb2+(aq) ➝ Mg2+(aq) + Pb(s)
E0 = +2.37V
+E0= -0.13V
E0 = +2.24 V
50) Determine the potential of the following cell by writing the oxidation, reduction, and the
overall (oxidation-reduction) equations.
Cr(s) / Cr(NO3)3(aq) // Cu(NO3)2(aq) / Cu(s)
Oxidation reaction: _______________________________________________________
Reduction reaction: _______________________________________________________
Overall reaction:
_______________________________________________________
Page 33
t Activity Series of Metals with Half-Reactions
1.
Fill in the blanks to complete the half-reactions on the table.
Activity Series of Metals, with Half-Reactions for Oxidation Processes
Elements
Oxidation half-reactions
Most active Lithium
Li(s) Õ Li+(aq) + ______
and most
Potassium
K(s) Õ ______ + eeasily
_____________
____ Õ Ba2+(aq) + 2eoxidized
Calcium
Ca(s) Õ Ca2+(aq) + _____
Sodium
Na(s) Õ Na+(aq) + _____
Magnesium
Mg(s) Õ ______ + 2eAluminum
Al(s) Õ Al3+(aq) + _____
Zinc
Zn(s) Õ ______ + 2eIron
Fe(s) Õ ______ + 2eNickel
Ni(s) Õ Ni2+(aq) + _____
____________
____ Õ Sn2+(aq) + 2eLead
Pb(s) Õ ______ + 2e*
Hydrogen
H2(g) Õ 2H+(aq) + _____
Least active Copper
Cu(s) Õ Cu2+(aq) + _____
and least
Mercury
Hg(s) Õ ______ + 2eeasily
Silver
Ag(s) Õ Ag+(aq) + _____
oxidized
Gold
Au(s) Õ ______ + 3e*
2.
Hydrogen is included for reference purposes.
Use the table to determine if the following statements are true or false.
________ a.
Tin is more easily oxidized than magnesium.
________ b. Aluminum is more easily oxidized than mercury.
________ c.
Zinc becomes copper-plated when dipped in a copper sulfate solution.
________ d. Silver metal is oxidized by lead ions.
________ e. Iron metal is oxidized by nickel ions.
________ f. An acid will release H2 gas when copper is placed in the acid.
Page 34
▼ Voltaic Cell
1. Determine the location where each process occurs in this electrochemical cell. Write the
letter for that location in the blank provided.
Locations
a. in the wire and light bulb
c. in the salt bridge
b. at the cathode
d. at the anode
Processes
______ Electrons pass through the external circuit to the copper strip.
______Positive and negative ions move through the aqueous solution to maintain electrical
neutrality.
______Electrons are passed to copper ions, and reduction takes place.
______Electrons are produced by oxidation.
2. Write the half-reactions and overall cell reaction for this voltaic cell.
Oxidation half-reaction: _______________________________________________________
Reduction half-reaction: _______________________________________________________
Overall cell reaction: _________________________________________________________
Page 35
t Electrochemical Cells
Zn(s) / Zn2+(aq) // Cr2O72-(aq), H+(aq), Cr3+(aq) / C(s)
Cell 1
2. reduction half cell:
3. oxidation half reaction:
4. net reaction:
5. spontaneous: yes/no
Fe(s) / Fe2+(aq) // Cr2O72-(aq), H+(aq), Cr3+(aq) / C(s)
Cell 2
2. reduction half cell:
3. oxidation half reaction:
4. net reaction:
5. spontaneous: yes/no
Page 36
Cu(s) / Cu2+(aq) // Cr2O72-(aq), H+(aq), Cr3+(aq) / C(s)
Cell 3
2. reduction half cell:
3. oxidation half reaction:
4. net reaction:
5. spontaneous: yes/no
Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)
Cell 4
2. reduction half cell:
3. oxidation half reaction:
4. net reaction:
5. spontaneous: yes/no
Page 37
Zn(s) / Zn2+(aq) // Fe2+(aq) / Fe(s)
Cell 5
2. reduction half cell:
3. oxidation half reaction:
4. net reaction:
5. spontaneous: yes/no
Fe(s) / Fe2+(aq) // Cu2+(aq) / Cu(s)
Cell 6
2. reduction half cell:
3. oxidation half reaction:
4. net reaction:
5. spontaneous: yes/no
Page 38
▼ Battery Technology
Examples
(1) Dry Cell (LeclanchŽ)
Battery characteristics:
-not rechargeable
-low cost
-low energy density
-low efficiency under
high current drain
(2) Alkaline
Battery characteristics:
-3x cost of LeclanchŽ cell
-3x capacity of LeclanchŽ cell
-longer shelf life
-steadier voltage under
high current
Page 39
(3) Mercury
Battery characteristics:
-not rechargeable
-constant voltage
-long shelf life
used in hearing aids,
watches,
calculators,
pacemakers, etc.
(4) Lead-acid Storage Battery
Battery characteristics:
-rechargeable by
inputting energy
-low cost vehicle
usage
-high rate performance
-good charge retention
-long lifespan (6 yrs.)
(5) Nicad (Nickel-cadmium)
Battery characteristics:
-only water is consumed
on discharge resulting in
little change in electrolyte
concentration
-stable voltage
-long shelf life
used in calculators, flashlights,
tools, etc.
Page 40
(6) The Molicel
Battery characteristics:
-high-energy
-rechargeable
-3x energy density
and 5x the power
capability of the
lead-acid cell in
a car battery.
invented in British Columbia
(7) Aluminum-air fuel cell
Battery characteristics:
-reactants supplied externally (usually gaseous) redox reductions are used to
convert chemical energy directly to electrical energy
-requires electrolyte (ion conductor)
-compact, lightweight, mobile, safe
-high energy, high power density (8-10x energy density of lead-acid battery)
-refuelable (aluminum anode oxidized by circulating alkaline electrolyte, air
breathing cathode results in oxygen gas being reduced)
- ideal supply of reserve and/or emergency power long shelf life (10 yrs.)
Page 41
w Mirco Approach to Redox Experiment on page
Objective: To study electrochemical cells.
In this experiment, we'll construct electrochemical cells. In the oxidation and reduction
reactions a salt bridge (consisting of a strip of filter paper dipped in a salt solution)will be used
to link the two half reactions. This will form an electrochemical cell.
Each of the following cells will be
built in adjoining reaction wells with
the salt bridge suspended in both
wells. The two separate electrodes,
each connected to the appropriate
terminal of the voltmeter, will then be
inserted into their respective well.
Salt bridge
Possible half cells to be used:
1.0 M Zn(NO3)2 (aq) with Zn electrode
1.0 M Cu(NO3)2 (aq) with Cu electrode
1.0 M Mg(NO3)2 (aq) with Mg electrode
1.0 M Al(NO3)3 (aq) with Al electrode
1.0 M Pb(NO3)2 (aq) with Pb electrode
Anode Half-cell/ Cathode Half-cell
Zn/Zn2+
Cu/Cu2+
Mg/Mg2+
Al/Al3+
Page 42
Zn/Zn2+
Cu/Cu2+
Mg/Mg2+
Al/Al3+