Chapter 20 Electrochemistry - PowerPoint PPT Presentation

About This Presentation
Title:

Chapter 20 Electrochemistry

Description:

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 20 Electrochemistry John D. Bookstaver – PowerPoint PPT presentation

Number of Views:203
Avg rating:3.0/5.0
Slides: 92
Provided by: JohnB575
Learn more at: https://www.ktufsd.org
Category:

less

Transcript and Presenter's Notes

Title: Chapter 20 Electrochemistry


1
Chapter 20Electrochemistry
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten
John D. Bookstaver St. Charles Community
College St. Peters, MO ? 2006, Prentice Hall, Inc.
2
Electrochemical Reactions
  • In electrochemical reactions, electrons are
    transferred from one species to another.

3
Oxidation Numbers
  • In order to keep track of what loses electrons
    and what gains them, we assign oxidation numbers.

4
Oxidation and Reduction
  • A species is oxidized when it loses electrons.
  • Here, zinc loses two electrons to go from neutral
    zinc metal to the Zn2 ion.

5
Oxidation and Reduction
  • A species is reduced when it gains electrons.
  • Here, each of the H gains an electron and they
    combine to form H2.

6
Oxidation and Reduction
  • What is reduced is the oxidizing agent.
  • H oxidizes Zn by taking electrons from it.
  • What is oxidized is the reducing agent.
  • Zn reduces H by giving it electrons.

7
Assigning Oxidation Numbers
  1. Elements in their elemental form have an
    oxidation number of 0.
  2. The oxidation number of a monatomic ion is the
    same as its charge.

8
Assigning Oxidation Numbers
  • Nonmetals tend to have negative oxidation
    numbers, although some are positive in certain
    compounds or ions.
  • Oxygen has an oxidation number of -2, except in
    the peroxide ion in which it has an oxidation
    number of -1.
  • Hydrogen is -1 when bonded to a metal, 1 when
    bonded to a nonmetal.

9
Assigning Oxidation Numbers
  • Nonmetals tend to have negative oxidation
    numbers, although some are positive in certain
    compounds or ions.
  • Fluorine always has an oxidation number of -1.
  • The other halogens have an oxidation number of -1
    when they are negative they can have positive
    oxidation numbers, however, most notably in
    oxyanions.

10
Assigning Oxidation Numbers
  1. The sum of the oxidation numbers in a neutral
    compound is 0.
  2. The sum of the oxidation numbers in a polyatomic
    ion is the charge on the ion.

11
SAMPLE EXERCISE 20.1 What Chemical Reactions
Occur in a Battery?
Identify the substances that are oxidized and
reduced, and indicate which are oxidizing agents
and which are reducing agents.
Solution   Analyze We are given a redox equation
and asked to identify the substance oxidized and
the substance reduced and to label one as the
oxidizing agent and the other as the reducing
agent. Plan First, we assign oxidation states to
all the atoms in the reaction and determine the
elements that are changing oxidation state.
Second, we apply the definitions of oxidation and
reduction.
Comment A common mnemonic for remembering
oxidation and reduction is LEO the lion says
GER losing electrons is oxidation gaining
electrons is reduction.
12
SAMPLE EXERCISE 20.1 continued
Answer Al(s) is the reducing agent MnO4(aq) is
the oxidizing agent.
13
Balancing Oxidation-Reduction Equations
  • Perhaps the easiest way to balance the equation
    of an oxidation-reduction reaction is via the
    half-reaction method.

14
Balancing Oxidation-Reduction Equations
  • This involves treating (on paper only) the
    oxidation and reduction as two separate
    processes, balancing these half reactions, and
    then combining them to attain the balanced
    equation for the overall reaction.

15
Half-Reaction Method
  1. Assign oxidation numbers to determine what is
    oxidized and what is reduced.
  2. Write the oxidation and reduction half-reactions.

16
Half-Reaction Method
  • Balance each half-reaction.
  • Balance elements other than H and O.
  • Balance O by adding H2O.
  • Balance H by adding H.
  • Balance charge by adding electrons.
  • Multiply the half-reactions by integers so that
    the electrons gained and lost are the same.

17
Half-Reaction Method
  1. Add the half-reactions, subtracting things that
    appear on both sides.
  2. Make sure the equation is balanced according to
    mass.
  3. Make sure the equation is balanced according to
    charge.

18
Half-Reaction Method
  • Consider the reaction between MnO4- and C2O42-
  • MnO4-(aq) C2O42-(aq) ??? Mn2(aq) CO2(aq)

19
Half-Reaction Method
  • First, we assign oxidation numbers.

Since the manganese goes from 7 to 2, it is
reduced.
Since the carbon goes from 3 to 4, it is
oxidized.
20
Oxidation Half-Reaction
  • C2O42- ??? CO2
  • To balance the carbon, we add a coefficient of
    2
  • C2O42- ??? 2 CO2

21
Oxidation Half-Reaction
  • C2O42- ??? 2 CO2
  • The oxygen is now balanced as well. To balance
    the charge, we must add 2 electrons to the right
    side.
  • C2O42- ??? 2 CO2 2 e-

22
Reduction Half-Reaction
  • MnO4- ??? Mn2
  • The manganese is balanced to balance the
    oxygen, we must add 4 waters to the right side.
  • MnO4- ??? Mn2 4 H2O

23
Reduction Half-Reaction
  • MnO4- ??? Mn2 4 H2O
  • To balance the hydrogen, we add 8 H to the left
    side.
  • 8 H MnO4- ??? Mn2 4 H2O

24
Reduction Half-Reaction
  • 8 H MnO4- ??? Mn2 4 H2O
  • To balance the charge, we add 5 e- to the left
    side.
  • 5 e- 8 H MnO4- ??? Mn2 4 H2O

25
Combining the Half-Reactions
  • Now we evaluate the two half-reactions together
  • C2O42- ??? 2 CO2 2 e-
  • 5 e- 8 H MnO4- ??? Mn2 4 H2O
  • To attain the same number of electrons on each
    side, we will multiply the first reaction by 5
    and the second by 2.

26
Combining the Half-Reactions
  • 5 C2O42- ??? 10 CO2 10 e-
  • 10 e- 16 H 2 MnO4- ??? 2 Mn2 8 H2O
  • When we add these together, we get
  • 10 e- 16 H 2 MnO4- 5 C2O42- ???
  • 2 Mn2 8 H2O 10 CO2 10 e-

27
Combining the Half-Reactions
  • 10 e- 16 H 2 MnO4- 5 C2O42- ???
  • 2 Mn2 8 H2O 10 CO2 10 e-
  • The only thing that appears on both sides are the
    electrons. Subtracting them, we are left with
  • 16 H 2 MnO4- 5 C2O42- ???
  • 2 Mn2 8 H2O 10 CO2

28
SAMPLE EXERCISE 20.2 Balancing Redox Equations in
Acidic Solution
Solution   Analyze We are given an incomplete,
unbalanced equation for a redox reaction
occurring in acidic solution and asked to balance
it. Plan We use the procedure we just learned.
Solve Step 1 We check oxidation state changes.
Each chromium atom is reduced by three electrons,
and each chlorine atom is oxidized by one
electron.
29
SAMPLE EXERCISE 20.2 continued
Recall that we said earlier that each chromium
atom needed to gain three electrons, and now we
can see the six electrons needed for two chromium
atoms.
Recall that we said earlier that each chlorine
atom needed to lose one electron, and now we can
see the two electrons needed for two chlorine
atoms.
Steps 6 and 7 There are equal numbers of atoms
of each kind on the two sides of the equation (14
H, 2 Cr, 7 O, 6 Cl). In addition, the charge is
the same on the two sides (6). Thus, the
equation is balanced.
30
SAMPLE EXERCISE 20.2 continued
31
Balancing in Basic Solution
  • If a reaction occurs in basic solution, one can
    balance it as if it occurred in acid.
  • Once the equation is balanced, add OH- to each
    side to neutralize the H in the equation and
    create water in its place.
  • If this produces water on both sides, you might
    have to subtract water from each side.

32
SAMPLE EXERCISE 20.3 Balancing Redox Equations in
Basic Solution
Solution   Analyze We are given an incomplete
equation for a basic redox reaction and asked to
balance it. Plan We go through the first steps
of our procedure as if the reaction were
occurring in acidic solution. We then add the
appropriate number of OH ions to each side of
the equation, combining H and OH to form H2O.
We complete the process by simplifying the
equation.
Solve Step 1 We assign oxidation states. This
is a tricky one!
Mn goes from 7 to 4. The sum of the oxidation
states of C and N in CN must be 1, the overall
charge of the ion. In CNO, if oxygen has an
oxidation state of 2 as usual, the sum of the
oxidation states of C and N must be 1. So,
overall, CN is oxidized by two electrons.
33
SAMPLE EXERCISE 20.3 continued
Both half-reactions are now balancedyou can
check the atoms and the overall charge.
34
SAMPLE EXERCISE 20.3 continued
35
Voltaic Cells
  • In spontaneous oxidation-reduction (redox)
    reactions, electrons are transferred and energy
    is released.

36
Voltaic Cells
  • We can use that energy to do work if we make the
    electrons flow through an external device.
  • We call such a setup a voltaic cell.

37
Voltaic Cells
  • A typical cell looks like this.
  • The oxidation occurs at the anode.
  • The reduction occurs at the cathode.

38
Voltaic Cells
  • Once even one electron flows from the anode to
    the cathode, the charges in each beaker would not
    be balanced and the flow of electrons would stop.

39
Voltaic Cells
  • Therefore, we use a salt bridge, usually a
    U-shaped tube that contains a salt solution, to
    keep the charges balanced.
  • Cations move toward the cathode.
  • Anions move toward the anode.

40
Voltaic Cells
  • In the cell, then, electrons leave the anode and
    flow through the wire to the cathode.
  • As the electrons leave the anode, the cations
    formed dissolve into the solution in the anode
    compartment.

41
Voltaic Cells
  • As the electrons reach the cathode, cations in
    the cathode are attracted to the now negative
    cathode.
  • The electrons are taken by the cation, and the
    neutral metal is deposited on the cathode.

42
SAMPLE EXERCISE 20.4 Reactions in a Voltaic Cell
is spontaneous. A solution containing K2Cr2O7 and
H2SO4 is poured into one beaker, and a solution
of KI is poured into another. A salt bridge is
used to join the beakers. A metallic conductor
that will not react with either solution (such as
platinum foil) is suspended in each solution, and
the two conductors are connected with wires
through a voltmeter or some other device to
detect an electric current. The resultant voltaic
cell generates an electric current. Indicate the
reaction occurring at the anode, the reaction at
the cathode, the direction of electron migration,
the direction of ion migration, and the signs of
the electrodes.
Solution   Analyze We are given the equation for
a spontaneous reaction taking place in a voltaic
cell and a description of how the cell is
constructed. We are asked to write the
half-reactions occurring at the anode and at the
cathode, as well as the directions of electron
and ion movements and the signs assigned to the
electrodes. Plan Our first step is to divide the
chemical equation into half-reactions so that we
can identify the oxidation and the reduction
processes. We then use the definitions of anode
and cathode and the other terminology summarized
in Figure 20.6.
43
SAMPLE EXERCISE 20.4 continued
Now we can use the summary in Figure 20.6 to
help us describe the voltaic cell. The first
half-reaction is the reduction process (electrons
shown on the reactant side of the equation), and
by definition, this process occurs at the
cathode. The second half-reaction is the
oxidation (electrons shown on the product side of
the equation), which occurs at the anode. The I
ions are the source of electrons, and the Cr2O72
ions accept the electrons. Hence, the electrons
flow through the external circuit from the
electrode immersed in the KI solution (the anode)
to the electrode immersed in the K2Cr2O7 H2SO4
solution (the cathode). The electrodes themselves
do not react in any way they merely provide a
means of transferring electrons from or to the
solutions. The cations move through the solutions
toward the cathode, and the anions move toward
the anode. The anode (from which the electrons
move) is the negative electrode, and the cathode
(toward which the electrons move) is the positive
electrode.
(a) Indicate which reaction occurs at the anode
and which at the cathode. (b) Which electrode is
consumed in the cell reaction? (c) Which
electrode is positive?
Answer (a) The first reaction occurs at the
anode, the second reaction at the cathode. (b)
The anode (Zn) is consumed in the cell reaction.
(c) The cathode is positive.
44
Electromotive Force (emf)
  • Water only spontaneously flows one way in a
    waterfall.
  • Likewise, electrons only spontaneously flow one
    way in a redox reactionfrom higher to lower
    potential energy.

45
Electromotive Force (emf)
  • The potential difference between the anode and
    cathode in a cell is called the electromotive
    force (emf).
  • It is also called the cell potential, and is
    designated Ecell.

46
Cell Potential
  • Cell potential is measured in volts (V).

47
Standard Reduction Potentials
  • Reduction potentials for many electrodes have
    been measured and tabulated.

48
Standard Hydrogen Electrode
  • Their values are referenced to a standard
    hydrogen electrode (SHE).
  • By definition, the reduction potential for
    hydrogen is 0 V
  • 2 H (aq, 1M) 2 e- ??? H2 (g, 1 atm)

49
Standard Cell Potentials
  • The cell potential at standard conditions can be
    found through this equation

Because cell potential is based on the potential
energy per unit of charge, it is an intensive
property.
50
Cell Potentials
  • For the oxidation in this cell,
  • For the reduction,

51
Cell Potentials
0.34 V - (-0.76 V) 1.10 V
52
(No Transcript)
53
SAMPLE EXERCISE 20.5 continued
Answer 0.40 V
54
Solution   Analyze We are given the equation for
a redox reaction and asked to use data in Table
20.1 to calculate the standard emf (standard
potential) for the associated voltaic cell. Plan
Our first step is to identify the half-reactions
that occur at the cathode and the anode, which we
did in Sample Exercise 20.4. Then we can use data
from Table 20.1 and Equation 20.8 to calculate
the standard emf.
55
SAMPLE EXERCISE 20.6 continued
Check The cell potential, 0.79 V, is a positive
number. As noted earlier, a voltaic cell must
have a positive emf in order to operate.
Answer 2.20 V
56
SAMPLE EXERCISE 20.7 From Half-Reactions to Cell
EMF
By using the data in Appendix E, determine (a)
the half-reactions that occur at the cathode and
the anode, and (b) the standard cell potential.
57
SAMPLE EXERCISE 20.7 continued
PRACTICE EXERCISE A voltaic cell is based on a
Co2/Co half-cell and an AgCl/Ag half-cell. (a)
What reaction occurs at the anode? (b) What is
the standard cell potential?
58
Oxidizing and Reducing Agents
  • The strongest oxidizers have the most positive
    reduction potentials.
  • The strongest reducers have the most negative
    reduction potentials.

59
Oxidizing and Reducing Agents
  • The greater the difference between the two, the
    greater the voltage of the cell.

60
SAMPLE EXERCISE 20.8 Determining the Relative
Strengths of Oxidizing Agents
Using Table 20.1, rank the following ions in
order of increasing strength as oxidizing agents
NO3(aq), Ag(aq), Cr2O72(aq).
Because the standard reduction potential of
Cr2O72 is the most positive, Cr2O72 is the
strongest oxidizing agent of the three. The rank
order is Ag lt NO3 lt Cr2O72.
PRACTICE EXERCISE Using Table 20.1, rank the
following species from the strongest to the
weakest reducing agent I(aq), Fe(s), Al(s).
Answer Al(s) gt Fe(s) gt I(aq)
61
Free Energy
  • ?G for a redox reaction can be found by using
    the equation
  • ?G -nFE
  • where n is the number of moles of electrons
    transferred, and F is a constant, the Faraday.
  • 1 F 96,485 C/mol 96,485 J/V-mol

62
Free Energy
  • Under standard conditions,
  • ?G? -nFE?

63
SAMPLE EXERCISE 20.9 Spontaneous or Not?
Solution   Analyze We are given two equations
and must determine whether or not each is
spontaneous. Plan To determine whether a redox
reaction is spontaneous under standard
conditions, we first need to write its reduction
and oxidation half-reactions. We can then use the
standard reduction potentials and Equation 20.10
to calculate the standard emf, E, for the
reaction. If a reaction is spontaneous, its
standard emf must be a positive number.
64
SAMPLE EXERCISE 20.9 continued
Because the value of E is positive, this
reaction is spontaneous and could be used to
build a voltaic cell.
Answer Reactions (b) and (c) are spontaneous.
65
Nernst Equation
  • Remember that
  • ?G ?G? RT ln Q
  • This means
  • -nFE -nFE? RT ln Q

66
Nernst Equation
  • Dividing both sides by -nF, we get the Nernst
    equation

or, using base-10 logarithms,
67
Nernst Equation
  • At room temperature (298 K),

Thus the equation becomes
68
Concentration Cells
  • Notice that the Nernst equation implies that a
    cell could be created that has the same substance
    at both electrodes. The more dilute is the anode
    so its Ni2 can increase. The more
    concentrated is the cathode so its Ni2 can
    decrease.
  • Therefore, as long as the concentrations are
    different, E will not be 0. (Until they become
    equal)

69
SAMPLE EXERCISE 20.10 Determining ?G and K
What are the values of E, ?G, and K when the
reaction is written in this way?
Solution   Analyze We are asked to determine ?G
and K for a redox reaction, using standard
reduction potentials. Plan We use the data in
Table 20.1 and Equation 20.10 to determine E for
the reaction and then use E in Equation 20.12 to
calculate ?G. We will then use Equation 19.22,
?G RT in K, to calculate K.
70
SAMPLE EXERCISE 20.10 continued
The positive value of E leads to a negative
value of ?G.
K is indeed very large! This means that we expect
silver metal to oxidize in acidic environments,
in air, to Ag. Notice that the voltage
calculated for the reaction was 0.43 V, which is
easy to measure. Directly measuring such a large
equilibrium constant by measuring reactant and
product concentrations at equilibrium, on the
other hand, would be very difficult.
71
SAMPLE EXERCISE 20.10 continued
Comment E is an intensive quantity, so
multiplying a chemical equation by a certain
factor will not affect the value of E.
Multiplying an equation will change the value of
n, however, and hence the value of ?G. The
change in free energy, in units of J/mol of
reaction as written, is an extensive quantity.
The equilibrium constant is also an extensive
quantity.
72
SAMPLE EXERCISE 20.10 continued
PRACTICE EXERCISE For the reaction
(a) What is the value of n? (b) Use the data in
Appendix E to calculate ?G. (c) Calculate K at T
298 K.
Answer (a) 6, (b) 87 kJ/mol, (c) K 6 ?1016
73
SAMPLE EXERCISE 20.11 Voltaic Cell EMF Under
Nonstandard Conditions
Solution   Analyze We are given a chemical
equation for a voltaic cell and the
concentrations of reactants and products under
which it operates. We are asked to calculate the
emf of the cell under these nonstandard
conditions. Plan To calculate the emf of a cell
under nonstandard conditions, we use the Nernst
equation in the form of Equation 20.16.
74
SAMPLE EXERCISE 20.11 continued
Check This result is qualitatively what we
expect Because the concentration of Cr2O72 (a
reactant) is greater than 1 M and the
concentration of Cr3 (a product) is less than 1
M, the emf is greater than E. Q is about 1010,
so log Q is about 10. Thus, the correction to E
is about 0.06 ??(10)/6, which is 0.1, in
agreement with the more detailed calculation.
PRACTICE EXERCISE Calculate the emf generated by
the cell described in the practice exercise
accompanying Sample Exercise 20.6 when Al3
4.0 ??103 M and I 0.010 M.
Answer E 2.36 V. 
75
SAMPLE EXERCISE 20.12 Concentrations in a Voltaic
Cell
Solution   Analyze We are given a description of
a voltaic cell, its emf, and the concentrations
of all reactants and products except H, which we
are asked to calculate. Plan First, we write the
equation for the cell reaction and use standard
reduction potentials from Table 20.1 to calculate
E for the reaction. After determining the value
of n from our reaction equation, we solve the
Nernst equation for Q. Finally, we use the
equation for the cell reaction to write an
expression for Q that contains H to determine
H.
76
SAMPLE EXERCISE 20.12 continued
Comment A voltaic cell whose cell reaction
involves H can be used to measure H or pH. A
pH meter is a specially designed voltaic cell
with a voltmeter calibrated to read pH directly.
(Section 16.4)
Answer pH 4.19 
77
SAMPLE EXERCISE 20.13 pH of a Concentration Cell
78
SAMPLE EXERCISE 20.13 continued
Comment The concentration of H at electrode 1
is lower than that in electrode 2, which is why
electrode 1 is the anode of the cell The
oxidation of H2 to H(aq) increases H at
electrode 1.
PRACTICE EXERCISE A concentration cell is
constructed with two Zn(s)-Zn2(aq) half-cells.
The first half-cell has Zn2 1.35 M, and the
second half-cell has Zn2 3.75 ? 104 M. (a)
Which half-cell is the anode of the cell? (b)
What is the emf of the cell?
Answer (a) the second half-cell, (b) 0.105 V
79
Applications of Oxidation-Reduction Reactions
80
SAMPLE EXERCISE 20.14 Aluminum Electrolysis
Calculate the number of grams of aluminum
produced in 1.00 h by the electrolysis of molten
AlCl3 if the electrical current is 10.0 A.
Solution   Analyze We are told that AlCl3 is
electrolyzed to form Al and asked to calculate
the number of grams of Al produced in 1.00 h with
10.0 A. Plan Figure 20.31 provides a road map of
the problem. First, the product of the amperage
and the time in seconds gives the number of
coulombs of electrical charge being used
(Equation 20.18). Second, the coulombs can be
converted with the Faraday constant (F 96,485
C/mole electrons) to tell us the number of moles
of electrons being supplied. Third, reduction of
1 mol of Al3 to Al requires three moles of
electrons. Hence we can use the number of moles
of electrons to calculate the number of moles of
Al metal it produces. Finally, we convert moles
of Al into grams.
81
SAMPLE EXERCISE 20.14 continued
Answer (a) 30.2 g of Mg, (b) 3.97 ? 103 s
82
Batteries
83
Alkaline Batteries
84
Hydrogen Fuel Cells
85
SAMPLE EXERCISE 20.15 Calculating Energy in
Kilowatt-hours
Calculate the number of kilowatt-hours of
electricity required to produce 1.0 ? 103 kg of
aluminum by electrolysis of Al3 if the applied
voltage is 4.50 V.
Solution   Analyze We are given the mass of Al
produced from Al3 and the applied voltage and
asked to calculate the energy, in kilowatt-hours,
required for the reduction. Plan From the mass
of Al, we can calculate first the number of moles
of Al, then the number of coulombs required to
obtain that mass. We can then use Equation 20.20,
w nFEext, where nF is the total charge in
coulombs and Eext is the applied potential, 4.50
V.
86
SAMPLE EXERCISE 20.15 continued
Comment This quantity of energy does not include
the energy used to mine, transport, and process
the aluminum ore, and to keep the electrolysis
bath molten during electrolysis. A typical
electrolytic cell used to reduce aluminum ore to
aluminum metal is only 40 efficient, with 60 of
the electrical energy being dissipated as heat.
It therefore requires on the order of 33 kWh of
electricity to produce 1 kg of aluminum. The
aluminum industry consumes about 2 of the
electrical energy generated in the United States.
Because this is used mainly to reduce aluminum,
recycling this metal saves large quantities of
energy.
PRACTICE EXERCISE Calculate the number of
kilowatt-hours of electricity required to produce
1.00 kg of Mg from electrolysis of molten MgCl2
if the applied emf is 5.00 V. Assume that the
process is 100 efficient.
Answer 11.0 kWh
87
Corrosion and
88
Corrosion Prevention
89
SAMPLE INTEGRATIVE EXERCISE Putting Concepts
Together
The Ksp at 298 K for iron(II) fluoride is 2.4 ?
106. (a) Write a half-reaction that gives the
likely products of the two-electron reduction of
FeF2(s) in water. (b) Use the Ksp value and the
standard reduction potential of Fe2(aq) to
calculate the standard reduction potential for
the half-reaction in part (a). (c) Rationalize
the difference in the reduction potential for the
half-reaction in part (a) with that for Fe2(aq).
90
SAMPLE INTEGRATIVE EXERCISE continued
Reaction 3 is still a half-reaction, so we do see
the free electrons.
(Recall that 1 volt is 1 joule per coulomb.)
91
SAMPLE INTEGRATIVE EXERCISE continued
(c) The standard reduction potential for FeF2
(0.61 V) is more negative than that for Fe2
(0.440 V), telling us that the reduction of FeF2
is the less favorable process. When FeF2 is
reduced, we not only reduce the F2 ions but also
break up the ionic solid. Because this additional
energy must be overcome, the reduction of FeF2 is
less favorable than the reduction of Fe2.
Write a Comment
User Comments (0)
About PowerShow.com