Chapter 4: The Major Classes of

1
C142B Lecture Notes Campbell / Callis
Chapter 4 The Major Classes of
Chemical Reactions
4.1 The Role of Water as a Solvent 4.2
Precipitation Reactions and Acid-Base
Reactions 4.3 Oxidation - Reduction (Redox)
Reactions 4.4 Counting Reactants and Products in
Precipitation, Acid-Base, and Redox
Processes 4.5 Reversible Reactions An
Introduction to Chemical
Equilibrium
2
The Role of Water as a Solvent The solubility
of Ionic Compounds
Electrical conductivity - The flow of electrical
current in a solution is a
measure of the solubility of ionic
compounds or a
measurement of the presence of ions in
solution. Electrolyte - A substance that
conducts a current when dissolved in
water. Soluble ionic compound
dissociate completely and
may conduct a large current, and are called
Strong Electrolytes.
NaCl(s) H2O(l)
Na(aq) Cl -(aq)
When sodium chloride dissolves into water the
ions become solvated, and are surrounded by water
molecules. These ions are called aqueous and
are free to move through out the solution, and
are conducting electricity, or helping electrons
to move through out the solution
3
Fig 4.1 (P135) Electrical Conductivity of
Ionic Solutions
4
Fig. 4.2
5
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem How many moles of each ion are in each
of the following a) 4.0 moles of sodium
carbonate dissolved in water b) 46.5 g of
rubidium fluoride dissolved in water c) 5.14 x
1021 formula units of iron (III) chloride
dissolved in water d) 75.0 mL of 0.56M
scandium bromide dissolved in water e) 7.8
moles of ammonium sulfate dissolved in
water a) Na2CO3(s)
2 Na(aq) CO3-2(aq) moles
of Na 4.0 moles Na2CO3 x
8.0 moles Na and 4.0 moles of CO3-2 are
present
H2O
2 mol Na 1 mol Na2CO3
6
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - II
H2O
b) RbF(s)
Rb (aq) F - (aq)

moles of RbF
H2O
c) FeCl3(s)
Fe3 (aq) 3 Cl - (aq)
moles of FeCl3 9.32 x 1021 formula units x


moles of Cl -
moles Fe3
7
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - II
H2O
b) RbF(s)
Rb (aq) F - (aq)
1 mol RbF 104.47 g RbF
moles of RbF 46.5 g RbF x
0.445 moles RbF
thus, 0.445 mol Rb and 0.445 mol F - are present
H2O
c) FeCl3(s)
Fe3 (aq) 3 Cl - (aq)
moles of FeCl3 9.32 x 1021 formula units
1 mol FeCl3 6.022 x 1023 formula
units FeCl3
x
0.0155 mol FeCl3
3 mol Cl - 1 mol FeCl3
moles of Cl - 0.0155 mol FeCl3 x
0.0465 mol Cl -
and 0.0155 mol Fe3 are also present.
8
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc3(aq) 3 Br -(aq)
Converting from volume to moles
1 L 103 mL
Moles of ScBr3 75.0 mL x x

Moles of Br -
H2O
e) (NH4)2SO4 (s)
2 NH4 (aq) SO4- 2 (aq)
2 mol NH4 1 mol(NH4)2SO4
Moles of NH4 7.8 moles (NH4)2SO4 x
15.6 mol NH4
and 7.8 mol SO4- 2 are also present.
9
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc3(aq) 3 Br -(aq)
Converting from volume to moles
1 L 103 mL
0.56 mol ScBr3 1 L
Moles of ScBr3 75.0 mL x x
0.042 mol ScBr3
3 mol Br - 1 mol ScBr3
Moles of Br - 0.042 mol ScBr3 x
0.126 mol Br -
0.042 mol Sc3 are also present
H2O
e) (NH4)2SO4 (s)
2 NH4 (aq) SO4- 2 (aq)
2 mol NH4 1 mol(NH4)2SO4
Moles of NH4 7.8 moles (NH4)2SO4 x
15.6 mol NH4
and 7.8 mol SO4- 2 are also present.
10
Fig. 4.3
11
The Solubility of Ionic Compounds in Water
The solubility of Ionic Compounds in water
depends upon the relative strengths of the
electrostatic forces between ions in the ionic
compound and the attractive forces between the
ions and water molecules in the solvent. There
is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called
insoluble compounds may be several orders of
magnitude less than ones that are
called soluble in water, for example
Solubility of NaCl in water at 20oC 365
g/L Solubility of MgCl2 in water at 20oC 542.5
g/L Solubility of AlCl3 in water at 20oC 699
g/L Solubility of PbCl2 in water at 20oC 9.9
g/L Solubility of AgCl in water at 20oC 0.009
g/L Solubility of CuCl in water at 20oC 0.0062
g/L
12
The Solubility of Covalent Compounds in Water
The covalent compounds that are very soluble in
water are the ones with -OH group in them and
are called Polar and can have strong polar
(electrostatic)interactions with water. Examples
are compound such as table sugar, sucrose
(C12H22O11) beverage alcohol, ethanol
(C2H5-OH) and ethylene glycol (C2H6O2) in
antifreeze.
H
Methanol Methyl Alcohol
C
H
O H
H
Other covalent compounds that do not contain a
polar center, or the -OH group are considered
Non-Polar , and have little or no interactions
with water molecules. Examples are the
hydrocarbons in Gasoline and Oil. This leads to
the obvious problems in Oil spills, where the oil
will not mix with the water and forms a layer on
the surface!
Octane C8H18 and / or Benzene C6H6
13
Acids - A group of Covalent molecules which loose
Hydrogen ions to water molecules in
solution
When gaseous hydrogen Iodide dissolves in water,
the attraction of the oxygen atom of the water
molecule for the hydrogen atom in HI is greater
that the attraction of the of the Iodide ion for
the hydrogen atom, and it is lost to the water
molecule to form an Hydronium ion and an Iodide
ion in solution. We can write the Hydrogen atom
in solution as either H(aq) or as H3O(aq) they
mean the same thing in solution. The presence of
a Hydrogen atom that is easily lost in solution
is an Acid and is called an acidic solution.
The water (H2O) could also be written above the
arrow indicating that the solvent was water in
which the HI was dissolved.
HI(g) H2O(l) H (aq) I
- (aq)
HI(g) H2O(l) H3O(aq) I
-(aq)
H2O(l)
HI(g) H(aq) I
-(aq)
14
Fig. 4.4
15
Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will loose two protons to yield
two Hydronium ions, and one sulfate ion. What
is the molarity of the sulfate and Hydronium ions
in a solution prepared by dissolving 155g of
pure H2SO4 into sufficient water to produce 2.30
Liters of sulfuric acid solution? Plan Determine
the number of moles of sulfuric acid, divide the
moles by the volume to get the molarity of the
acid and the sulfate ion. The hydronium ions
concentration will be twice the acid
molarity. Solution Two moles of H are released
for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
1 mole H2SO4 98.09 g H2SO4
Moles H2SO4
1.58 moles H2SO4
155 g H2SO4
x
Molarity of SO4- 2
Molarity of H3O (or simply H)
16
Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will loose two protons to yield
two Hydronium ions, and one sulfate ion. What
is the molarity of the sulfate and Hydronium ions
in a solution prepared by dissolving 155g of
pure H2SO4 into sufficient water to produce 2.30
Liters of sulfuric acid solution? Plan Determine
the number of moles of sulfuric acid, divide the
moles by the volume to get the molarity of the
acid and the sulfate ion. The hydronium ions
concentration will be twice the acid
molarity. Solution Two moles of H are released
for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
1 mole H2SO4 98.09 g H2SO4
Moles H2SO4
1.58 moles H2SO4
155 g H2SO4
x
1.58 mol SO4-2 2.30 L solution
Molarity of SO4- 2
0.687 Molar in SO4- 2
Molarity of H 2 x 0.687 mol H / 2.30 liters
0.597 Molar in H
17
Fig. 4.5
18
Precipitation Reactions A solid product is formed
When ever two aqueous solutions are mixed, there
is the possibility of forming an insoluble
compound. Let us look at some examples to see
how we can predict the result of adding two
different solutions together.
Pb(NO3) (aq) NaI(aq) Pb2(aq) 2
NO3- (aq) Na (aq) I- (aq)
When we add these two solutions together, the
ions can combine in the way they came into the
solution, or they can exchange partners. In this
case we could have Lead Nitrate and Sodium
Iodide, or Lead Iodide and Sodium Nitrate formed,
to determine which will happen we must look
at the Solubility Rules (P 141) to determine what
could form. Rule 3a in Table 4.1 indicates that
Lead Iodide will be insoluble, so a precipitate
will form! The remainder (sodium sulfate) is
soluble (Rule 4a).
Pb(NO3)2 (aq) 2 NaI(aq)
PbI2 (s) 2 NaNO3 (aq)
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Precipitation Reactions Will a Precipitate form?
If we add a solution containing Potassium
Chloride to a solution containing Ammonium
Nitrate, will we get a precipitate?
KCl(aq) NH4NO3(aq) K (aq)
Cl-(aq) NH4(aq) NO3-(aq)
By exchanging cations and anions we see that we
could have Potassium Chloride and Ammonium
Nitrate, or Potassium Nitrate and Ammonium
Chloride. In looking at the solubility rules,
Table 4.1
If we mix a solution of Sodium Sulfate with a
solution of Barium Nitrate, will we get a
precipitate? Table 4.1 shows that Therefore we
get
Na2SO4(aq) Ba(NO3)2(aq)
BaSO4( ) 2 NaNO3( )
20
Precipitation Reactions Will a Precipitate form?
If we add a solution containing Potassium
Chloride to a solution containing Ammonium
Nitrate, will we get a precipitate?
KCl(aq) NH4NO3(aq) K (aq)
Cl-(aq) NH4(aq) NO3-(aq)
By exchanging cations and anions we see that we
could have Potassium Chloride and Ammonium
Nitrate, or Potassium Nitrate and Ammonium
Chloride. In looking at the solubility rules,
Table 4.1 shows all possible products as
soluble, so there is no net reaction!
KCl(aq) NH4NO3(aq) No. Reaction!
If we mix a solution of Sodium Sulfate with a
solution of Barium Nitrate, will we get a
precipitate? Table 4.1 shows that Barium Sulfate
is insoluble, therefore we will get a
precipitate!
Na2SO4(aq) Ba(NO3)2(aq)
BaSO4(s) 2 NaNO3(aq)
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Fig. 4.6
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Predicting Whether a Precipitation Reaction
Occurs (see Table 4.1) Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2(aq) Na2SO4(aq)
CaSO4(s) NaNO3(aq)
Total Ionic Equation
Ca2(aq) 2 NO3-(aq) 2 Na(aq) SO4-2(aq)
CaSO4(s) 2Na(aq) 2 NO3-(aq)
Net Ionic Equation
Ca2(aq) SO4-2(aq)
CaSO4 (s)
Spectator Ions are Na and NO3-
b) Ammonium Sulfate and Magnesium Chloride are
added together.
In exchanging ions, no precipitates will be
formed, accd. To Table 4.1, so there will be no
reactions occurring! All ions are spectator ions!
24
Acid - Base Reactions Neutralization Rxns.
An Acid is a substance that produces H (H3O)
ions when dissolved in water. A Base is a
substance that produces OH - ions when dissolved
in water. Acids and Bases are electrolytes, and
their strength is categorized in terms of their
degree of dissociation in water to make hydronium
or hydroxide ions, resp.. Strong acids and bases
dissociate completely, and are strong
electrolytes. Weak acids and bases dissociate
only to a tiny extent (ltlt100) and are weak
electrolytes.
The generalized reaction between an Acid and a
Base is
HX(aq) MOH(aq)
MX(aq) H2O(l)
Acid Base
Salt(aq) Water
25
Fig. 4.7
26
Table 4.2 (P 143) Selected Acids and Bases
Acids
Bases
Strong (100 to H)
Strong (100 to OH-) Hydrochloric, HCl
Sodium hydroxide, NaOH
Hydrobromic, HBr
Potassium hydroxide, KOH Hydroiodoic, HI
Calcium hydroxide,
Ca(OH)2 Nitric acid, HNO3
Strontium hydroxide, Sr(OH)2 Sulfuric
acid, H2SO4 Barium
hydroxide, Ba(OH)2 Perchloric acid,
HClO4 Weak (tiny to H)
Weak (tiny yield of OH-) Hydrofluoric,
HF Ammonia, NH3
Phosphoric acid, H3PO4 Acidic acid, CH3COOH
(or HC2H3O2)
27
Writing Balanced Equations for
Neutralization Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following Chemical reactions
a) Calcium Hydroxide(aq) and HydroIodic
acid(aq) b) Lithium Hydroxide(aq) and
Nitric acid(aq) c) Barium
Hydroxide(aq) and Sulfuric acid(aq) Plan These
are all strong acids and bases, therefore they
will make water and the corresponding
salts. Solution
a) Ca(OH)2(aq) 2HI(aq)
CaI2(aq) 2H2O(l) Ca2(aq) 2 OH -(aq)
2 H(aq) 2 I -(aq)


Ca2(aq) 2 I -(aq) 2 H2O(l) 2 OH
-(aq) 2 H(aq)
2 H2O(l)
28
Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3(aq)
c) Ba(OH)2(aq) H2SO4(aq)

29
Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3(aq)
LiNO3(aq) H2O(l) Li(aq)
OH -(aq) H(aq) NO3-(aq)

Li (aq) NO3-(aq) H2O(l)
OH -(aq) H(aq)
H2O(l)
c) Ba(OH)2(aq) H2SO4(aq)
BaSO4(s) 2 H2O(l) Ba2(aq) 2 OH
-(aq) 2 H(aq) SO42 -(aq) BaSO4(s)
2 H2O(l)
Ba2 (aq) 2 OH -(aq) 2 H(aq) SO42 -(aq)
BaSO4(s) 2 H2O(l)
30
Fig. 4.8
31
Finding the Concentration of Acid from an
Acid - Base Titration
Volume (L) of base (difference in buret readings)
needed to titrate
M (mol/L) of base
Moles of base needed to titrate
molar ratio
Moles of acid which were titrated
volume (L) of acid
M (mol/L) of original acid soln.
32
Potassium Hydrogenphthalate KHC8H4O4
O
C
K
O
O
C
H
O
KHP a common acid used to titrate bases
(M 204.2 g/mol)
33
Finding the Concentration of Base from an
Acid - Base Titration - I
Problem A titration is performed between Sodium
Hydroxide and KHP (204.2 g/mol) to standardize
the base solution, by placing 50.00 mg of solid
Potassium Hydrogenphthalate in a flask with a
few drops of an indicator. A buret is filled with
the base, and the initial buret reading is 0.55
mL at the end of the titration the buret
reading is 33.87 mL. What is the concentration
of the base? Plan Use the molar mass of KHP to
calculate the number of moles of the acid, from
the balanced chemical equation, the reaction is
equal molar, so we know the moles of base, and
from the difference in the buret readings, we
can calculate the molarity of the base. Solution
The rxn is KHP(aq) OH-(aq) --gt KP-(aq)
H2O or
HKC8H4O4(aq) OH -(aq)
KC8H4O4-(aq) H2O(aq)
34
Finding the Concentration of Base from an
Acid - Base Titration - II
50.00 mg KHP

moles KHP
Volume of base Final buret reading - Initial
buret reading
one mole of H one mole of OH- therefore
____________ moles of KHP will titrate
_____________ moles of NaOH.
moles L
molarity of NaOH

molarity of base
35
Finding the Concentration of Base from an
Acid - Base Titration - II
50.00 mg KHP 204.2 g KHP 1 mol KHP
1.00 g 1000 mg
moles KHP x
0.00024486 mol KHP
Volume of base Final buret reading - Initial
buret reading 33.87
mL - 0.55 mL 33.32 mL of base
one mole of H one mole of OH- therefore
0.00024486 moles of KHP will titrate 0.00024486
moles of NaOH.
0.00024486 moles 0.03332 L
molarity of NaOH
0.07348679 moles per liter
molarity of base 0.07349 M
36
Fig. 4.9
37
Fig. 4.10
38
Fig. 4.11
39
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40
Highest and Lowest oxidation numbers of
Chemically reactive main-group Elements
Fig. 4.12
1A
2A
3A
4A
5A
6A
7A
Period
4
4
5
6
7
1
2
3
-3
-2
-1
F
O
N
C
B
2
Li
Be
Cl
3
S
P
Si
Al
Na
Mg
Br
4
Se
As
Ge
Ga
K
Ca
non-metals
I
5
Te
Sb
Sn
In
Rb
Sr
metalloids
At
6
Po
Bi
Pb
Tl
Cs
Ba
metals
7
Ra
Fr
41
Fig. 4.13
42
Period
IA
VIIIA
Common Ox s Main Group Elements
He
H
1
IIA
IIIA
IVA
VA
VIA
VIIA
1 -1
Ne
F
O
N
C
B
Be
Li
2
4,2 -1,-4
all from
1
2
3
-1
-1,-2
5 -3
Ar
Na
Mg
Al
Si
P
S
Cl
-1
3
7,5 3,1
6,4 2,-2
5,3 -3
1
2
3
4,-4
Kr
K
Ca
Ga
Ge
As
Se
Br
-1
4
4,2 -4
7,5 3,1
2
6,4 -2
5,3 -3
1
2
3, 2
Xe
Rb
Sr
In
Sn
Sb
Te
I
-1
5
3,2 1
4,2, -4
7,5 3,1
6,4 2
6,4 -2
5,3 -3
1
2
Rn
Cs
Ba
Tl
Pb
Bi
Po
At
-1
6
7,5 3,1
6,4 2,-2
1
4,2
2
2
3,1
3
43
Transition Metals
Possible Oxidation States
VIIIB
IIIB
IVB
VB
VIB
VIIB
IB
IIB
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
2
4,3 2
7,6 4,3
6,3 2
5,4 32
3
2
2,1
2
3,2
3,2
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
5,4 2
8,5 4,3
7,5 4
6,5 4,3
3
4,3
2
1
4,2
4,3
Hg
Au
Pt
Ir
Os
Re
W
Ta
Hf
La
2
5,4 3
6,5 4
7,5 4
8,6 4,3
4,3 1
3
4,3
2,1
3,1
4,2
44
Determining the Oxidation Number of an
Element in a Compound
Problem Determine the oxidation number (Ox. No.)
of each element in the following
compounds. a) Iron III Chloride b)
Nitrogen Dioxide c) Sulfuric acid Plan We
apply the rules in Table 4.3, always making sure
that the Ox. No. values in a compound
add up to zero, and in a polyatomic
ion, to the ions charge. Solution a) FeCl3
This compound is composed of monoatomic ions. The
Ox. No. of Cl- is -1, for a total of -3.
Therefore the Fe must be 3. b) NO2 c)
H2SO4
45
Determining the Oxidation Number of an
Element in a Compound
Problem Determine the oxidation number (Ox. No.)
of each element in the following
compounds. a) Iron III Chloride b)
Nitrogen Dioxide c) Sulfuric acid Plan We
apply the rules in Table 4.3, always making sure
that the Ox. No. values in a compound
add up to zero, and in a polyatomic
ion, to the ions charge. Solution a) FeCl3
This compound is composed of monoatomic ions. The
Ox. No. of Cl- is -1, for a total of -3.
Therefore the Fe must be 3. b) NO2 The Ox.
No. of oxygen is -2 for a total of -4. Since the
Ox. No. in a compound must add up to zero,
the Ox. No. of N is 4. c) H2SO4 The Ox. No.
of H is 1, so the SO42- group must sum to -2.
The Ox. No. of each O is -2 for a total of -8.
So the Sulfur atom is 6.
46
Recognizing Oxidizing and Reducing Agents - I
Problem Identify the oxidizing and reducing
agent in each of the Rx a) Zn(s) 2 HCl(aq)
ZnCl2 (aq) H2 (g) b) S8 (s)
12 O2 (g) 8 SO3 (g) c) NiO(s)
CO(g) Ni(s) CO2 (g) Plan First
we assign an oxidation number (O.N.) to each atom
(or ion) based on the rules in Table 4.3. A
reactant is the reducing agent if it contains an
atom that is oxidized (O.N. increased in the
reaction). A reactant is the oxidizing agent if
it contains an atom that is reduced ( O.N.
decreased). Solution a) Assigning oxidation
numbers
-1
-1
1
0
0
2
Zn(s) 2 HCl(aq) ZnCl2(aq)
H2 (g)
HCl is the oxidizing agent, and Zn is the
reducing agent!
47
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers
S8 (s) 12 O2 (g) 8
SO3 (g)
____ is the reducing agent and ____ is the
oxidizing agent.
c) Assigning oxidation numbers
NiO(s) CO(g)
Ni(s) CO2 (g)
___ is the reducing agent and ____ is the
oxidizing agent.
48
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers
S 0 S6
-2
S is Oxidized
6
0
0
O0 O-2
S8 (s) 12 O2 (g) 8
SO3 (g)
O is Reduced
S8 is the reducing agent and O2 is the oxidizing
agent
c) Assigning oxidation numbers
C2 C4 C is oxidized
Ni2 Ni0 Ni is Reduced
-2
-2
-2
0
2
4
2
NiO(s) CO(g)
Ni(s) CO2 (g)
CO is the reducing agent and NiO is the oxidizing
agent
49
Balancing REDOX Equations
The oxidation number method
Step 1) Assign oxidation numbers to all
elements in the equation. Step 2) From the
changes in oxidation numbers, identify the
oxidized and reduced species.
Step 3) Compute the number of
electrons lost in the oxidation and
gained in the reduction from the oxidation
number changes. Draw tie-lines
between these atoms to show electron
changes. Step 4) Multiply one or both of these
numbers by appropriate factors to
make the electrons lost equal the electrons
gained, and use the factors as
balancing coefficients. Step 5) Complete the
balancing by inspection, adding states of matter.
50
REDOX Balancing using Ox. No. Method - I
2 e-
0
-2
___ H2 (g) ___ O2 (g)
___ H2O(g)
2
2
1
- 1 e-
0
electrons lost must electrons gained therefore
multiply Hydrogen reaction by 2! and we are
balanced!
51
REDOX Balancing Using Ox. No. Method - II
-1e-
2
3
Fe2(aq) MnO4-(aq) H3O(aq)
Fe3(aq) Mn2(aq) H2O(aq)
2
5 e-
7
Multiply Fe2 Fe3 by ___ to correct for the
electrons gained by the Manganese.
__Fe2(aq) MnO4-(aq) H3O(aq)
__ Fe3(aq) Mn2(aq) H2O(aq)
Next balance the O and H atoms
5 Fe2(aq) MnO4-(aq) __ H3O(aq) 5
Fe3(aq) Mn2(aq) __ H2O(aq)
52
REDOX Balancing Using Ox. No. Method - II
-1e-
2
3
Fe2(aq) MnO4-(aq) H3O(aq)
Fe3(aq) Mn2(aq) H2O(aq)
2
5 e-
7
Multiply Fe2 Fe3 by five to correct for the
electrons gained by the Manganese.
5 Fe2(aq) MnO4-(aq) H3O(aq)
5 Fe3(aq) Mn2(aq) H2O(aq)
Next balance the O and H atoms Make 4 water
molecules from the oxygens of the MnO4-, which
requires 8 protons from the acid (H3O ions).
These 8 H3O ions will also make 8 more water
molecules, for a total of 48 12 water
molecules.
5 Fe2(aq) MnO4-(aq) 8 H3O(aq) 5
Fe3(aq) Mn2(aq) 12 H2O(aq)