Download - Catalyst

Chapter 4: Types of Chemical Reactions
and Solution Stoichiometry
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
Water, the Common Solvent
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
The Composition of Solutions (MOLARITY!)
Types of Chemical Reactions
Precipitation Reactions
Describing Reactions in Solution
Selective Precipitation
Midterm Exam 1
Stoichiometry of Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction Reactions
Balancing Oxidation-Reduction Equations
Simple Oxidation-Reduction Titrations
1
Definitions – Solutes, Solvents and Solutions
Solutes
• The thing being dissolved, mixed, diluted!
Solvents
• The material doing the dissolving, mixing,
dilution!
Solution
• The final combination of the dissolution, mixing,
and dilution!
2
1
Definitions – Solutes, Solvents and Solutions
Solutes
• Compounds extracted
from coffee grounds.
Solvents
• Water
Solution
• Morning coffee
3
WATER
• water is an important solvent – dissolves many substances
• water is a POLAR molecule
• “hydration” breaks ionic compounds into anions and cations
• water dissolves different ionic compounds to different degrees
(more in Ch 8)
• water also dissolves some nonionic substances if they are
polar (ethanol-water)
• nonpolar substances are not dissolved (fats, oils)
4
2
Electronegativity and Polarity
5
Figure 4.1: A space-filling model
of the water molecule.
More on molecular shapes in Ch 13.
6
3
7
Polar water molecules dissolve salts (ionic
compounds) through hydration
Anion
Zumdahl Figure 4.2
Cation
8
4
Ethanol Molecules are Polar
(contain a directional O-H bond)
“LIKE DISSOLVING LIKE”
Zumdahl Figure 4.3
9
The Role of Water as a Solvent:
the Solubility of Ionic Compounds
Electrical conductivity
The flow of electrical current in a solution is an indicator of the
presence of ions in solution and the solubility of ionic compounds.
Electrolyte
A substance that conducts a current when dissolved in water.
Soluble, ionic compounds that dissociate completely conduct a
large current and are called strong electrolytes.
NaCl(s)
H2O
Na+(aq) + Cl -(aq)
When sodium chloride dissolves in water the ions become
solvated/hydrated, and are surrounded by water molecules. These ions
are labeled “aqueous”, are free to move throughout the solution, and
conduct electricity (they help help electrons move through out the
10
solution).
5
Which of the following solutions of
strong electrolytes contains the
greatest number of total ions?
1.
2.
3.
4.
One mole of potassium
chloride dissolved in 1.0 L of
solution
One mole of iron(II) nitrate
dissolved in 1.0 L of solution
One mole of potassium
hydroxide dissolved in 1.0 L
of solution
One mole of sodium
phosphate dissolved in 1.0 L
of solution
11
Electrical Conductivity of Ionic Solutions
12
6
Strong Electrolytes
• Produce ions in aqueous solution and conduct electricity well.
• Strong electrolytes are soluble salts, strong acids, and strong
bases.
• Strong acids produce H+ ions when they dissolve in water.
HCl, HNO3, and H2SO4 are strong acids:
HNO3(aq) → H+(aq) + NO3-(aq)
• Strong bases produce OH- ions when they dissolve in water:
NaOH and KOH are strong bases:
NaOH(s) → Na+(aq) + OH-(aq)
All of the above species are ionized nearly 100%
13
Figure 4.5: HCl (aq) is
completely ionized.
Strong acids fully
dissociate, forming the
anion and a hydrated
proton.
14
7
SIMPLIFIED NOTATION:
+
H
15
Acids - A group of covalent molecules which lose
hydrogen ions to water molecules in solution
When gaseous hydrogen iodide dissolves in water, the attraction
between the oxygen atom of the water molecule and the hydrogen
atom in HI is greater that the attraction of the of the iodide atom for the
hydrogen atom, so…
H+ is lost to the water molecule to form a hydronium ion and an iodide
ion in solution. We can write the hydronium ion in solution as either
H+(aq) or H3O+(aq) - they mean the same thing. The water (H2O) can be
written as a reactant or above the arrow indicating that it is the solvent
in which the HI was dissolved.
HI(g) + H2O(l)
HI(g)
H3O+(aq) + I -(aq)
H2O
H+(aq) + I -(aq)
16
8
Figure 4.6:
An aqueous solution of
sodium hydroxide
(NaOH).
Strong bases fully
dissociate, forming a
cation and the hydroxide
anion.
17
Weak Electrolytes
• Produce relatively few ions in aqueous solutions
• The most common weak electrolytes are weak acids and weak bases:
Acetic acid is a typical weak acid:
HC2H3O2(aq)
⇌
H+(aq) + C2H3O2-(aq)
Ammonia is a common weak base:
NH3(aq) + H2O(l)
⇌
NH4+(aq) + OH-(aq)
Both of these species are ionized only ~1%
18
9
Figure 4.7: Acetic acid
(HC2H3O2) exists in water
mostly as undissociated
molecules.
Weak acids partially
dissociate, forming only a
small number of anions
and hydrated protons.
19
Figure 4.8:
The reaction of NH3 in
water.
Weak bases partially
dissociate (or react with
water to a limited extent),
forming only a small
number of cations and
hydroxide anions.
20
10
Nonelectrolytes
• Dissolve in water but produce no ions in solution.
• Nonelectrolytes do not conduct electricity because when a
sample of the compound dissolves, the substance remains
intact as whole molecules and no ions are produced.
• Common nonelectrolytes include:
ethanol (CH3CH2OH)
table sugar (sucrose, C12H22O11)
21
The Solubility of Covalent Compounds in Water
Polar covalent compounds are very soluble in water. They often have OH groups that can form “hydrogen bonds” with water. Examples are
table sugar (C12H22O11), ethanol (C2H5OH), ethylene glycol (C2H6O2) in
antifreeze, and methanol (CH3OH).
These also are written with “(aq)” (i.e., aqueous) when dissolved in
water.
Example: C2H5OH(aq)
Nonpolar covalent compounds can’t form “hydrogen bonds” and have
little or no interactions with water molecules. Examples are the
hydrocarbons in gasoline and oil, which don’t mix with water.
octane = C8H18
and
benzene = C6H6
22
11
Which of the following
statements is true?
1.
2.
3.
4.
If a substance is soluble,
it must be an electrolyte.
If a substance is an
electrolyte, it must be
soluble.
Weak electrolytes must
be less soluble than
strong electrolytes.
Nonelectrolytes are
nonsoluble.
23
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem: How many moles of each ion are formed when the
following compounds are dissolved in water:
a)
b)
c)
d)
4.0 moles of sodium carbonate
46.5 g of rubidium fluoride
9.32 x 1021 formula units of iron(III) chloride
7.8 moles of ammonium sulfate
H2 O
a) Na2CO3 (s)
2 Na+(aq) + CO32-(aq)
moles of Na+ = 4.0 moles Na2CO3
2 mol Na+
1 mol Na2CO3
= 8.0 moles Na+ (and 4.0 moles of CO32-) are present
24
12
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
RbF(s)
H2 O
moles of RbF = 46.5 g RbF
Rb+(aq) + F -(aq)
= 0.445 moles RbF
1 mol RbF
104.47 g RbF
thus, 0.445 mol Rb+ and 0.445 mol F - are present
25
How many moles of iron and chloride ions
are formed by the dissolution of 9.32 x 1021
formula units of iron(III) chloride?
1. Fe+ = 0.0155 mol,
Cl- = 0.0155 mol
2. Fe2+ = 0.0155 mol,
Cl- = 0.0310 mol
3. Fe3+ = 0.0155 mol,
Cl- = 0.0465 mol
4. Fe3+ = 0.155 mol,
Cl- = 0.465 mol
26
13
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - III
c)
FeCl3 (s)
H2 O
Fe3+(aq) + 3 Cl -(aq)
moles of FeCl3 =
9.32 x 1021 formula units
= 0.0155 mol FeCl3
1 mol FeCl3
6.022 x 1023 formula units FeCl3
0.0155 mol Fe3+ are present
moles of Cl - = 0.0155 mol FeCl3 3 mol Cl
= 0.0465 mol Cl 1 mol FeCl3
27
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - IV
d) 7.8 moles of ammonium sulfate dissolved in water
(NH4)2SO4 (s)
H2 O
Moles of NH4+ = 7.8 moles (NH4)2SO4
2 NH4+(aq) + SO42-(aq)
2 mol NH4+ = 15.6 mol NH4+
1 mol(NH4)2SO4
7.8 mol SO42- are also present
28
14
Concentration of Solutions: Molarity = M
M = moles of solute
liters of solution
= moles
L
Units of concentration are mol/L or M
solute = material dissolved in the solvent
In sea water, WATER is the solvent and salts (magnesium chloride,
sodium chloride, etc.) are the solutes
When you add sugar to coffee, water is the solvent and the solutes
are sugar and all the other compounds extracted from the roasted
beans (acids, carbohydrates, proteins, lipids, caffeine, and other
organic compounds).
29
Which solution do you expect to
light the bulb the most brightly?
1.
2.
3.
4.
5.
6.
7.
H2 O
HCl (3M)
HCl (1M)
HC2H3O2 (acetic acid)
(1M)
C2H5OH (ethanol)
NH4OH (1M)
HCl (0.1M)
30
15
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds
Example problem:
How many moles of each ion are in the following:
75.0 mL of 0.56 M scandium bromide in water (ScBr3)
ScBr3(s)
H2 O
Sc3+(aq) + 3 Br -(aq)
Converting from volume to moles:
Moles of ScBr3 = 75.0 mL
1L
0.56 mol ScBr3
1000 mL
L
= 0.042 mol ScBr3
Moles of Br- = 0.042 mol ScBr3 x (3 mol Br -/mol ScBr3)
= 0.13 mol Br Moles of Sc3+ = 0.042 mol ScBr3 x (1 mol Sc3+/mol ScBr3)
= 0.042 mol Sc3+ 31
A sports drink is about 0.14 M NaCl.
Calculate the volume (in mL) of the beverage
that would contain 2.5 mg of NaCl?
MM Na = 22.99 g/mol, MM Cl = 35.45 g/mol
1.
2.
3.
4.
5.
3.1x10-4 mL
0.31 mL
0.31 L
0.14 mL
0.14 mol
32
16
A sports drink is about 0.14 M NaCl. Calculate
the volume (in mL) of the beverage that would
contain 2.5 mg of NaCl?
Convert 2.5 mg NaCl to moles:
2.5 mg NaCl
1 g NaCl
1 mol NaCl
1000 mg NaCl 58.45g NaCl
= 4.28 x 10-5 mol NaCl
What volume of 0.14 M NaCl would contain this amount of NaCl?
V x M = moles
Vx
0.14 mol NaCl
= 4.28 x 10-5 mol NaCl
L solution
Solving for volume gives:
V = 4.28 x 10-5 mol NaCl
L solution
= 3.1 x 10-4 L
0.14 mol NaCl
or 0.31 mL of beverage
33
Preparing a Solution - I
Example problem:
A solution of sodium phosphate is prepared by dissolving
3.95 g of sodium phosphate in water and diluting it to 300.0 mL. What is the
molarity of the salt and each of the ions?
Strategy
(1) Write the chemical equation showing the process of dissolution.
(2) Calculate the moles of each species.
(3) Divide # moles by # L solution to obtain molarity.
34
17
(1) Write the chemical equation showing the process of
dissolution.
Na3PO4 (s)
H2O(solvent)
→
3 Na+ (aq) + PO43- (aq)
(2) Calculate the moles of each species.
Molar mass of Na3PO4 = 163.94 g/mol
mol Na3PO4 = 3.95 g / (163.94 g/mol) = 0.0241 mol
mol Na+ = 3 x mol Na3PO4 = 0.0723 mol
mol PO4-3 = 1 x mol Na3PO4 = 0.0241 mol
(3) Divide # moles by # L solution to obtain molarity.
M(Na3PO4) = 0.0241 mol Na3PO4 / 0.3000 L = 0.0803 M
M(Na+) = 0.0723 mol Na+ / 0.3000 L = 0.241 M
M(PO4-3) = 0.0241 mol PO43- / 0.3000 L = 0.0803 M
35
Figure 4.9: Steps involved in the
preparation of a standard solution.
36
18
How many grams of ammonium carbonate are
required to prepare 1.00 L of a 0.375 M solution?
MM N = 14.01 g/mol, MM H = 1.008 g/mol,
MM C = 12.01 g/mol, MM 16.00 g/mol
1.
2.
3.
4.
29.3 g
36.0 g
0.375 mol
0.375 g
37
Like Example 4.4
How would you prepare 1.00 L of a 0.375 M solution of ammonium
carbonate?
First, determine the moles of ammonium carbonate required:
1.00 L 0.375 mol (NH4)2CO3
= 0.375 mol (NH4)2CO3
L solution
This amount can be converted to grams by using the molar mass:
0.375 mol (NH4)2CO3 96.09 g (NH4)2CO3 = 36.0 g (NH4)2CO3
mol (NH4)2CO3
To make 1.00 L of solution, 35.3 g of (NH4)2CO3 are weighed out
and transferred to a 1.00 L volumetric flask. DI water is added to
dissolve the solid and dilute the solution to the mark on the neck of
the flask.
38
19
Dilution of Solutions
If we dilute 25.00 mL of 0.0400 M KMnO4 to a final volume of 500. mL ,
what is the new molarity (M) of the diluted solution?
A strategy for calculating final concentration:
The number of moles of solute is the same before and after dilution! (The
difference is the ratio of moles to volume.)
M1 x V1 = moles solute = M2 x V2
M1V1
= M2
V2
0.0400 M* 25.00 mL = 2.00 x 10-3 M
500 mL
39
GRADUATED
and
VOLUMETRIC
pipets.
Zumdahl Figure 4.10
40
20
Figure 4.11:
Dilution of Solutions
KNOWN volume is PIPETTED
into a larger container of
KNOWN volume.
M1
(a) A measuring pipette
(b) Water is added to the flask.
(c) The resulting molarity is
M2 = M1 x V1
V2
V1
V2=
M2
41
Make a Solution of Potassium Permanganate
Potassium Permanganate (KMnO4) has a molar mass of 158.04 g/mol
Problem: Prepare a solution by dissolving 1.58 grams of KMnO4
into sufficient water to make 250.00 ml of solution.
Calculate the molarity of the ions in solution.
1.58 g KMnO4 1 mole KMnO4
158.04 g KMnO4
Molarity =
= 0.0100 moles KMnO4
0.0100 moles KMnO4
=
0.250 liters
0.0400 M
Molarity = [K+] ion = [MnO4-] ion = 0.0400 M
42
21
25.00 mL of 0.0400 M potassium permanganate
solution is diluted to a final volume of 1.00 L.
What is the new concentration?
M1V1=M2V2
1.
2.
3.
4.
0.00100 M
0.001 M
10.0 M
10 M
43
Dilution of Solutions
 Take 25.00 mL of the 0.0400 M KMnO4
 Dilute to 1.000 L
 What is the resulting molarity of the diluted solution?
L x M = # moles
0.02500 L x 0.0400 M = 0.00100 moles
0.00100 mol / 1.000 L = 0.00100 M
44
22
MM (g/mol)
45
TYPES of CHEMICAL REACTIONS
• Precipitation Reactions –
SOLID is formed
• Acid-Base (Neutralization) Reactions –
Often WATER and a SALT are formed
PROTONS are transferred!!
• Reduction-Oxidation (REDOX) Reactions –
ELECTRONS are transferred
46
23
PRECIPITATION REACTIONS
• Double displacement reactions
• Insoluble compound (precipitate) is formed from two
or more solutes
• Precipitate separates from solution
• Small fraction of compound remains ionized
• Compound formed has a net charge of zero
• “Solubility product” is a constant (Ch 8)
47
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 solvent molecules (often
water).
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, consider the solubility (in g/L) of the following compounds in
water at 20oC :
Solubility of NaCl
Solubility of MgCl2
Solubility of AlCl3
Solubility of PbCl2
Solubility of AgCl
Solubility of CuCl
= 365
= 542.5
= 699
=
9.9
=
0.009
=
0.0062
48
24
Figure 4.12:
When yellow aqueous
potassium chromate is
added to a colorless
barium nitrate solution,
yellow barium chromate
precipitates.
49
Figure 4.13:
Reactant
solutions
K2CrO4(aq)
Ba(NO3)3(aq)
50
25
Fig. 4.14: Reaction of K2CrO4 (aq) and Ba(NO3)2 (aq)
Clues:
Precipitate has ZERO net charge.
Chromate is yellow.
Potassium forms soluble salts.
BaCrO4 precipitates!
51
Precipitation Reactions: A solid product is formed
Whenever two aqueous solutions are mixed, there is the possibility that
an insoluble compound will form. We can predict the result of adding
two different solutions together.
Initially:
Pb(NO3)2 (aq) + NaI(aq)
Pb2+(aq) + 2 NO3-(aq) + Na+(aq) + I-(aq)
The ions can combine in the way they came into the solution OR they can
exchange partners. In this case, the product options are lead(II) nitrate
and sodium iodide OR lead(II) iodide and sodium nitrate.
To determine which will happen, we must look at the solubility rules. The
rules indicate that the lead iodide will be insoluble, so a precipitate will
form!
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
52
26
Table 4.1:
Simple Rules for Solubility of Salts in Water
1. Most nitrate (NO3-) salts are soluble.
2. Most salts of Na+, K+, and NH4+ are soluble.
3. Most chloride salts are soluble. Notable exceptions are AgCl,
PbCl2, and Hg2Cl2.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4,
PbSO4, and CaSO4.
5. Most hydroxide salts are only slightly soluble. The important
soluble hydroxides are NaOH, KOH, and Ca(OH)2 (marginally
soluble).
6. Most sulfide (S2-), carbonate (CO32-), and phosphate (PO43-)
salts are only slightly soluble.
Know these!!!
In general, 1-4 are “solubles” and 5-6 are “insolubles”.
53
Precipitation Reactions: Will a Precipitate form?
Split parent species up into ions and look at all potential new
combinations that could be formed.
Fe(NO3)3(aq) + 3KOH(aq) → Fe3+(aq) + 3NO3- (aq) + 3K+(aq) + 3OH-(aq)
RULE: If any of the possible new species formed by combining
anions with cations is insoluble, then a precipitate will form.
USE TABLE 4.1
In this case, Fe(OH)3 is insoluble and a precipitate forms.
54
27
Precipitation Reactions: Will a Precipitate form?
Example: If a solution containing potassium chloride is added to a
solution containing ammonium nitrate, will a precipitate form?
KCl(aq) + NH4NO3(aq) → K+(aq) + Cl-(aq) + NH4+(aq) + NO3-(aq)
Possible reaction products are KCl and NH4NO3, or NH4Cl and KNO3.
All are soluble, so there is no precipitate.
KCl(aq) + NH4NO3 (aq) = No Reaction!
Example: If a solution containing sodium sulfate is added to a solution
containing barium nitrate, will a precipitate form?
Na2SO4 (aq) + Ba(NO3)2 (aq) →
2 Na+(aq) + SO42- (aq) + Ba2+(aq) + 2 NO3- (aq)
Barium sulfate is insoluble; therefore a precipitate will form.
Na2SO4 (aq) + Ba(NO3)2 (aq)
BaSO4 (s) + 2 NaNO3 (aq)
55
Quantitative Precipitation Problem:
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 0.125 M lead(II) nitrate solution to precipitate ALL of the
lead as PbI2 (s).
The chemical equation for the reaction is:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
56
28
What mass (in g) of solid sodium iodide must be
added to 2.50 L of a 0.125 M lead(II) nitrate
solution to precipitate ALL of the lead as PbI2 (s)?
MM Na = 22.99 g/mol, MM I = 126.9 g/mol
Pb(NO3)2 (aq) + 2 NaI(aq)
1.
2.
3.
4.
PbI2 (s) + 2 NaNO3 (aq)
0.625 g
93.7 g
6.25x10-4 g
9.37 x10-2 g
57
Quantitative Precipitation Problem:
Calculate the mass of solid sodium iodide that must be added to
2.50 L of a 0.125 M lead(II) nitrate solution to precipitate ALL of the
lead as PbI2 (s).
The chemical equation for the reaction is:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
The moles of I- needed to precipitate PbI2 is 2 x moles of Pb2+.
The number of moles of iodide needed:
2.50 L 0.125 mol Pb2+ 2 mol I1.00 L soln. 1 mol Pb2+
= 0.625 mol I-
The mass of sodium iodide required is:
0.625 mol I- 1 mol NaI 149.9 g NaI = 93.7 g NaI
1 mol I- 1 mol NaI
58
29
Predicting Whether a Precipitation Reaction
Occurs and Writing Descriptive Equations
a) calcium nitrate and sodium sulfate solutions are added together
Molecular Equation
Ca(NO3)2 (aq) + Na2SO4 (aq)
CaSO4 (s) +2 NaNO3 (aq)
Complete Ionic Equation
Ca2+(aq)+2 NO3-(aq) + 2 Na+(aq)+ SO4-2(aq)
CaSO4 (s) + 2 Na+(aq)+ 2 NO3-(aq)
Net Ionic Equation
Ca2+(aq) + SO4-2(aq)
CaSO4 (s)
Spectator ions are Na+ and NO3b) ammonium sulfate and magnesium chloride are added together
In exchanging ions, no precipitates will be formed, so no chemical
59
reaction occurs All ions are spectator ions!
Limiting reagent
problems in solution
Only difference:
using molarity and
volume to convert to
moles via
M =
moles
L solution
60
30
Example problem:
Lead has been used as a glaze for pottery for years and, if not fired
properly, is leachable from the pottery. Vinegar is used in leaching
tests, followed by the precipitation of lead as a lead sulfide.
If 257.8 mL of a 0.0468 M solution of lead(II) acetate is added to 156.00
mL of a 0.0950 M solution of sodium sulfide, what mass of solid lead
sulfide will be formed?
This is a limiting-reactant problem because:
the amounts of two reactants are given.
Strategy:
(1)
(2)
(3)
(4)
Write the balanced chemical reaction.
Determine the limiting reactant.
Calculate the moles of product.
Convert moles of product to mass of the product using molar mass.
61
(1) Write the balanced equation
Pb(C2H3O2)2 (aq) + Na2S (aq)
→
PbS (s) + 2 NaC2H3O2 (aq)
(2) Determine the limiting reactant.
mol Pb(C2H3O2)2 = V x M = 0.2578 L x (0.0468 mol/L) = 0.0121 mol
mol Na2S = V x M = 0.15600 L x (0.0950 mol/L) = 0.0148 mol
0.0121 mol Pb(C2H3O2)2 1 mol Na2S
1 mol Pb(C2H3O2)2
= 0.0121 mol Na2S
needed
Therefore, lead(II) acetate is the limiting reactant.
62
31
(3) Calculate the moles of product.
0.0121 mol Pb(C2H3O2)2 x (1 mol PbS / mol Pb(C2H3 O2)2) =
0.0121 mol PbS
(4) Calculate mass of product
0.0121 mol PbS x (239.3 g PbS/mol PbS) = 2.90 g PbS
63
Like Example 4.8
When aqueous solutions of silver nitrate and sodium chloride are mixed,
silver chloride precipitates. What mass of silver chloride is formed by the
addition of 75.00 mL of 3.17 M NaCl to 128 mL of 2.44 M silver nitrate?
The stoichiometric relationship comes from the chemical equation:
AgNO3 (aq) + NaCl(aq)
AgCl(s) + NaNO3 (aq)
There is a 1:1 relationship, therefore, the number of moles are the same,
but which is in the lowest quantity?
VAgNO3 x MAgNO3=0.128 L x 2.44 M =0.312 mol AgNO3 =0.312 mol Ag+
VNaCl x MNaCl = 0.07500 L x 3.17 M = 0.238 mol NaCl =0.238 mol ClSince the amount of chloride ion is smaller, it is limiting, and we use it to
calculate the mass of AgCl produced. Since we can only obtain 0.238
mol of AgCl:
0.238 mol x 143.35 g AgCl/ mol = 34.1 g AgCl 64
32
Gravimetric Analysis for % Ca in Phosphate Rock
0.2920 g CaC2O4 .H2O
1 mol CaC2O4 .H2O = 1.998 x 10-3 mol
146.12 g CaC2O4 . H2O
CaC2O4 . H2O
1.998 x 10-3 mol Ca2+
40.08 g Ca2+ = 8.009 x 10-2 g Ca2+
1 mol Ca2+
Mass % Ca is:
8.009 x 10-2 g Ca x 100% = 18.34%
0.4367 g sample
65
Figure 4.16:
Selective precipitation
of Ag/Ba/Fe
Separating a mixture of
silver, barium and iron
ions.
66
33
TYPES of CHEMICAL REACTIONS
• Precipitation Reactions –
SOLID is formed
• Acid-Base (Neutralization) Reactions –
Often WATER and a SALT are formed
PROTONS are transferred!!
• Reduction-Oxidation (REDOX) Reactions –
ELECTRONS are transferred
67
Acids = Covalent molecules that donate H+ cations
(or protons) in solution
When gaseous hydrogen iodide dissolves in water, the water molecules
help the HI dissociate into H+ and I- ions.
We can write the hydrogen ion in water as either H+(aq) or H3O+(aq). They
mean the same thing.
A molecule which has a H atom that is easily lost to the water solution as
H+ (or proton) is called an “acid”, (PROTON DONOR) and the resulting
solution is called an “acidic” solution. The acid is said to be
“deprotonated” in solution:
HI(g)
H+(aq) + I -(aq)
HI(g) + H2O(l)
H3O+(aq) + I -(aq)
The water (H2O) could also be written above the arrow indicating that
the solvent was water in which the HI was dissolved:
H2 O
HI(g)
H+(aq) + I -(aq) 68
34
H3 O+ is called the hydronium ion
69
What is the molarity of the nitrate and hydronium
ions in a solution prepared by dissolving 155 g of
concentrated nitric acid in sufficient water to
produce 2.30 L of acid solution?
Step 1: Write the balanced equation.
1.
2.
3.
4.
1.07 M
2.46 M
1.43 M
3.30 M
70
35
Strong Acids and the Molarity of H+ Ions in
Aqueous Solutions of Acids
Problem: In aqueous solutions, each molecule of nitric acid will lose
one proton to yield one H+ ion and one nitrate ion. What is the molarity
of the nitrate and hydronium ions in a solution prepared by dissolving
155 g of concentrated nitric acid in sufficient water to produce 2.30 L
of acid solution?
Plan: Determine the number of moles of nitric acid; divide the moles by
the volume to get the molarity of the acid and each ion.
Solution: One mole of H+ is released for every mole of acid:
HNO3 (l) + H2O(l)
H3O+(aq) + NO3- (aq)
Moles HNO3 = 155 g HNO3 1 mol HNO3
= 2.46 moles HNO3
63.02 g HNO3
Molarity of NO3- = 2.46 mol NO3
= 1.07 M in NO32.30 L solution
71
Molarity of H3O+ = 1 x 1.07 M = 1.07 M in H3O+ (or H+)
Acids and Bases
An acid is a substance that produces H+ (H3O+) ions when dissolved
in water, and is a proton donor.
A base is a substance that produces OH - ions when dissolved in water:
Example: NaOH(aq) → Na+(aq) + OH-(aq)
The OH- ions react with the H+ ions (if an acid is present) to produce
water, H2O, and are, therefore, proton acceptors.
Acids and bases are electrolytes. Their strength is categorized in
terms of their degree of dissociation in water to make hydronium or
hydroxide ions. Strong acids and bases dissociate completely and are
strong electrolytes. Weak acids and bases dissociate partially (some
small % of the molecules dissociate) and are weak electrolytes.
72
36
Selected Acids and Bases
Acids
Bases
Strong: H+(aq) + A-(aq)
hydrochloric, HCl
hydrobromic, HBr
hydroiodoic, HI
nitric acid, HNO3
sulfuric acid, H2SO4
perchloric acid, HClO4
Strong: M+(aq) + OH-(aq)
lithium hydroxide, LiOH
sodium hydroxide, NaOH
potassium hydroxide, KOH
calcium hydroxide, Ca(OH)2
strontium hydroxide, Sr(OH)2
barium hydroxide, Ba(OH)2
*(M is Group I or II metal)
Weak
hydrofluoric, HF
phosphoric acid, H3PO4
acetic acid, CH3COOH
(or HC2H3O2)
Weak
ammonia, NH3
accepts proton from water to make
NH4+(aq) and OH-(aq)
73
Acid - Base Reactions: Neutralization Rxns.
A generalized reaction between an acid and a base is:
HA(aq) +
Acid
+
MOH(aq)
Base
MA(aq) +
→
Salt
+
H2 O(l)
Water
The salt product can either be
dissolved as ions or form a
precipitate.
Not ALL A/B reactions
produce water.
Look for the transfer of H+.
74
37
Writing Balanced Equations for
Neutralization Reactions - I
Problem: Write balanced descriptive equations (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, the products will
be 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)
75
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)
76
38
What are the coefficients for the neutralization
reaction between barium hydroxide(aq) and
sulfuric acid(aq)?
MOH(aq) +
1.
2.
3.
4.
5.
6.
HA(aq)
MA(aq) +
H2 O(l)
1,1,1,1
1,1,1,2
1,2,1,2
2,1,1,2
2,2,1,1
2,2,2,4
77
Writing Balanced Equations for
Neutralization Reactions - III
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)
2 OH -(aq) + 2 H+(aq)
2 H2O(l)
78
39
Like Example 4.10
What volume of 0.125 M HCl is needed to neutralize 200.0 mL
of a 0.00955 M Ca(OH)2 solution?
Calculate the number of moles of base:
Vbase x Mbase = 0.2000 L x 0.00955 M = 0.00191 mol Ca(OH)2
From the balanced equation find the moles of acid needed:
Ca(OH)2(aq) + 2 HCl (aq)
2 H2O(l) + CaCl2 (aq)
Since there are two hydroxide ions per molecule of base and one proton
per molecule of acid, we need twice as much acid as we have base:
0.00382 mol HCl
Volume of acid:
moles acid
0.00382 mol
Vacid =
=
= 0.0306 L HCl
Macid
0.125 mol
79
L
TITRATIONS
•
Volumetric analysis - technique by which one solution is used to analyze
another solution
•
Titrant, also called a standard solution (M known exactly), is delivered from
a buret.
•
The sample and an indicator are in an Erlenmeyer flask.
•
The indicator changes color at the equivalence point, the point at which
titrant and sample are stoichiometrically equal.
•
The concentration of standard must be known exactly (in order to determine
moles, millimoles, or equivalents added).
•
The concentration of an unknown sample is calculated from the volume
and concentration of the standard used to reach the equivalence point,
along with the volume of the sample.
80
40
81
Determining the Concentration of Acid by
an Acid-base Titration with Known Base
Volume (L) of base used to titrate
(difference in buret readings)
M (mol/L) of base
Moles of base used to titrate
molar ratio
Moles of acid that were titrated
volume (L) of acid
M (mol/L) of original acid solution
82
41
Finding the Concentration of a Base from an
Acid - Base Titration - I
Problem: A titration is performed between sodium hydroxide and
potassium hydrogen phthalate (KHP) to standardize the base solution.
50.00 mg of solid KHP are placed in a flask with deionized water and 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: 1) Use the molar mass of KHP (204.2 g/mol) to calculate the number
of moles of the acid. 2) From the balanced chemical equation, the
reaction is equal molar (1:1), so we know the moles of base needed. 3)
From the difference in the buret readings, we can calculate the volume
of base added and, therefore, the molarity of the base.
KHC8H4O4 (aq) + OH -(aq)
K+(aq) + C8H4O42-(aq) + H2O(l)
83
Potassium Hydrogen Phthalate KHC8H4O4
O
O
C
O
K+
C
K+
O
O
C
O
H
O
C
H+
O
ONE molecule of KHP releases ONE proton
84
42
Finding the Concentration of a Base from an
Acid - Base Titration - II
Solution:
moles KHP = 50.00 mg KHP
1.00 g
1000 mg
1 mol KHP
204.2 g KHP
= 0.0002449 mol KHP
Volume of base = Final buret reading - Initial buret reading
= 33.87 mL - 0.55 mL = 33.32 mL of base delivered
one mole of acid = one mole of base
therefore 0.0002449 moles of acid will NEUTRALIZE 0.0002449
moles of base in a volume of 33.32 mL
molarity of base =
0.0002449 moles
0.03332 L
= 0.07349 M
85
Example: Aluminum hydroxide reacts with hydrochloric acid
according to the balanced equation
Al(OH)3 (s) + 3 HCl (aq) → 3 H2O(l) + AlCl3 (aq)
What volume of 1.50 M HCl(aq) is required to neutralize
10.0 g Al(OH)3(s)?
Strategy:
(1) Calculate moles of Al(OH)3(s).
(2) Calculate moles of HCl needed using the balanced equation.
(3) Calculate volume HCl from moles of HCl and known molarity.
86
43
What volume of 1.50 M HCl(aq) is required to
neutralize 10.0 g Al(OH)3(s)?
Al(OH)3 (s) + 3 HCl (aq) → 3 H2O(l) + AlCl3 (aq)
1.
2.
3.
4.
0.128 L
0.385 L
0.256 L
0.0853 L
87
(1) Calculate moles of Al(OH)3(s).
10.0 g Al(OH)3
= 0.128 mol Al(OH)3
78.00 g/mol
(2) Calculate moles of HCl needed using the balanced equation.
Al(OH)3 (s) + 3 HCl (aq)
0.128 mol Al(OH)3
→ 3 H O(l) + AlCl
3 mol HCl
mol Al(OH)3
2
3
(aq)
= 0.385 mol HCl
(3) Calculate volume HCl from moles of HCl and known
molarity.
V=
mol HCl
molarity
= 0.385 mol HCl
1.50 mol/L
= 0.256 L
88
44
TYPES of CHEMICAL REACTIONS
• Precipitation Reactions –
SOLID is formed
• Acid-Base (Neutralization) Reactions –
Often WATER and a SALT are formed
Protons are transferred
• Reduction-Oxidation (REDOX) Reactions –
ELECTRONS are transferred
89
Formation of NaCl is a REDOX Reaction
• 2 Na(s) + Cl2(g) → 2 NaCl(s)
• NaCl contains the ions Na+ and Cl• Electrons are transferred
• Energy is released
90
45
Oxidation-Reduction Reactions
Many important chemical reactions involve oxidation and
reduction. In fact, most reactions used for energy
production are redox reactions:
In humans the oxidation of sugars, fats, and proteins provides
the energy necessary for life.
Glucose + O2 → Energy
Combustion reactions, which provide most of the energy to
power our civilization, also involve oxidation and reduction.
CxHy + O2 → CO2 + H2O + Energy
91
Metallic Sodium Reacts with Chlorine Gas!!!
Zumdahl Fig 4.19
92
46
Oxidation - A substance gives up electrons to another
substance. 2 Mg + O2 → 2 Mg2+ + 2 O2- Mg is oxidized
Reduction - A substance accepts electrons from another
substance. 2 Mg + O2 → 2 Mg2+ + 2 O2- O is reduced
In a chemical reaction, the total number of electrons and the
number of charges are conserved. It is convenient to assign
fictitious charges to the atoms in a molecule and call them
“oxidation states” (O.S.) or “oxidation numbers”
(O.N.).
Oxidation numbers are chosen so that:
(a) charges are conserved
(b) in ionic compounds the sum of oxidation numbers on the
atoms coincides with the charge on the ion
93
See Table 4.3
in Zumdahl
94
47
Period
IA
H
1
+1 -1
2
3
4
5
6
Possible OXIDATION NUMBERS
IIA
IIIA
IVA
Li
Be
B
+1
+2
+3
Na
Mg
Al
Si
+1
+2
+3
+4,-4
K
Ca
Ga
+1
+2
Rb
Sr
+1
+2
Cs
+1
VA
C
N
+4,+2 all from
-1,-4
+5 -3
Ge
+4,+2
+3, +2 -4
In
+3,+2
+1
Sn
+4,+2,
-4
Ba
Tl
+2
+3,+1
Pb
+4,+2
VIA
VIIA
O
F
-1,-2
-1
VIIIA
He
Ne
P
+5,+3
-3
S
-1 Cl
+6,+4 +7,+5
+2,-2 +3,+1
Ar
As
+5,+3
-3
Se
-1 Br
+6,+4 +7,+5
-2
+3,+1
Kr
+2
Sb
Te
-1 I
+5,+3 +6,+4 +7,+5
-3
-2
+3,+1
Bi
+3
Po
-1 At
+6,+4 +7,+5
+2,-2 +3,+1
Xe
+6,+4
+2
Rn
+2
95
Transition Metals
Possible Oxidation States
IIIB
Sc
+3
Y
+3
La
+3
VIIIB
IVB
VB VIB VIIB
Ti
V
Cr +2 Mn Fe
Co
+4,+3 +5,+4 +6,+3 +7,+6
+3,+2 +3,+2
+2
+3+2 +2
+4,+3
Zr
Ni
+2
IB
Cu
IIB
Zn
+2,+1 +2
Nb
Mo
Tc
Ru
Rh
Pd
Ag
+5,+4 +6,+5 +7,+5 +8,+5
+4,+3
+4,+3 +4,+2 +1
+2
+4,+3 +4
+4,+3
Cd
+2
Hf
Ta
W
Re +2 Os Ir
Pt
Au
Hg
+5,+4 +6,+5 +7,+5 +8,+6 +4,+3
+4,+3
+4,+2 +3,+1 +2,+1
+3
+4
+4
+4,+3 +1
96
48
Determine the oxidation number (O.N.) of each element
in the following compounds.
Ex 1: iron(III) chloride
Ex 2: nitrogen dioxide
Ex 3: sulfuric acid
Strategy: We apply the rules in Table 4.3, always making
sure that the O.N. values add up to zero in a compound
or, for a polyatomic ion, to the ion’s charge.
Solutions:
Ex 1: FeCl3 This compound is composed of monoatomic
ions. The O.N. of Cl- is -1, for a total of -3. Therefore, the
O.N. of Fe is +3.
Ex 2: NO2
The O.N. of oxygen is -2 for a total of -4. Since the O.N.
97
in a compound must add up to zero, the O.N. of N is +4.
What are the oxidation numbers for H, S,
and O in H2SO4 (sulfuric acid)?
1.
2.
3.
4.
+1, -2, -2
+1, -2, +2
+1, +6, -2
+1, -6, +2
98
49
Ex 3: H2SO4
The O.N. of H is +1, so the SO42- group must sum to -2.
The O.N. of each O is -2 for a total of -8. Therefore S has
the O.N. +6.
Ex 4: BaI2
I = -1
Ba = +2
Ex 5: NH4NO3
H = +1
N = -3
O = -2
N = +5
99
Mnemonic devices
“LEO says GER”
“OIL RIG”
100
50
Recognizing Oxidizing and Reducing Agents - I
Problem: Identify the oxidizing and reducing agent in each of the rxns
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. The reactant is the reducing agent if it
contains an atom that is oxidized (O.N. increased in the reaction). The
reactant is the oxidizing agent if it contains an atom that is reduced
( O.N. decreased).
Solution:
a) Assigning oxidation numbers:
-1
+1
0
Zn(s) + 2 HCl(aq)
-1
0
+2
ZnCl2 (aq) + H2 (g)
HCl is the oxidizing agent, and Zn is the reducing agent!
101
Recognizing Oxidizing and Reducing Agents - II
b) Assigning oxidation numbers:
0
+6
0
S8 (s) + 12 O2 (g)
-2
8 SO3 (g)
S [0]
S[+6]
S is Oxidized
O[0]
O[-2]
O is Reduced
S8 is the reducing agent and O2 is the oxidizing agent
c) Assigning oxidation numbers:
-2
-2
+2
+2
NiO(s) + CO(g)
0
+4
Ni[+2]
Ni[0]
Ni is Reduced
C[+2]
C[+4]
C is Oxidized
-2
Ni(s) + CO2 (g)
CO is the reducing agent and NiO is the oxidizing agent
102
51
Demo: Activity Series
Mg(s) + 2 HCl(aq) → Mg2+(aq) + 2 Cl– (aq) + H2(g).
Mg is a very active metal (reduces H in both water and acid).
Zn(s) + 2 HCl(aq) → Zn2+(aq) + 2 Cl–(aq) + H2(g).
Zn is an active metal (reduces H in acid but not in water).
Cu(s) + HCl(aq) → no reaction
Cu is an inactive metal (does not reduce H in acid nor water).
103
Activity
Series
Lab 3:
You use Mg instead of Zn
104
52
Balancing REDOX Equations:
The oxidation states (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) Choose coefficients for these species to make the electrons lost
equal the electrons gained; in other words…
total increase in O.N. = total decrease in O.N.
Step 5) Complete the balancing by inspection.
105
REDOX Balancing using Ox. No. Method - I
0
+2 e- per O = 4 e-
___
2 H2 (g) +___ O2 (g)
0
-2
___
2 H2O(g)
- 1 e- per H
+1
4 x (- 1 e- per H) = 4 eelectrons lost must = electrons gained
Therefore, we need 4 H = 2 * H2.
106
53
Demo: Thermite Reaction
Reaction Equation:
__ Fe2O3(s) + __ Al(s) → __ Fe(s) + __ Al2O3(s)
ΔHo = –849 kJ/mol
ΔSo = –37.48 J/mol-K
ΔGo = –838 kJ/mol
Uses:
welding, purifying an ore
(U in the Manhattan Project)
In this reaction, unfavorable entropy (ΔS) is offset by a very
large negative ΔH (heat is released). The heat is sufficient to
raise the temperature of the products past the melting point of
Fe (1530 oC).
Can go horribly wrong…the heat has to go somewhere and a
sand pit is the best.
“Mythbusters”: thermite and ice…ice chunks flew 150 feet, or
107
half a football field.
What are the coefficients of the balanced
reaction of thermite?
__ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(s)
Step 1: Assign Oxidation States
Step 2: Identify oxidized and reduced elements
Step 3: Balance - total O.N. increase = total O.N. decrease
1.
2.
3.
4.
1,1,1,1
1,2,1,2
2,1,2,1
2,2,2,2
108
54
Balance the thermite reaction:
__ Fe2O3(s) + __ Al(s) → __ Al2O3(s) + __ Fe(s)
Assign O.N. to each element:
Reactants
O: -2
Fe:+3
Al:0
Products
O: -2
Al:+3
Fe:0
Identify the elements that are oxidized and reduced:
Fe3+ →Fe0 : iron is reduced (O.N. decreases by 3)
Al0 →Al3+ : aluminum is oxidized (O.N. increases by 3)
Balance atoms and ensure total O.N. increase = total O.N.
decrease:
Fe2O3 + 2Al → Al2O3 + 2Fe
109
Balancing the Thermite Reaction
-2
Fe2O3(s) + 2 Al(s) →
+3
-2
Al2O3(s) + 2 Fe(s)
0
→ +3
2 x (- 3e- per Al)
0
2 x (+3e- per Fe)
BALANCED!
110
55
REDOX Balancing Using Ox. No. Method - II
-1e- per Fe
+2
+3
+1
-2
Fe2+(aq) + MnO4-(aq) + H+(aq)
+7
+1 -2
Fe3+(aq) + Mn2+(aq) + H2O(l)
+5 e- per Mn
+2
Balance the number of each redox element and then # electrons.
Multiply Fe+2 & Fe+3 by five to balance the electrons gained by Mn:
5 Fe2+(aq) + MnO4-(aq) + H+(aq)
5 Fe3+(aq) + Mn2+(aq) + H2O(l)
Balance O: Need 4 H2O on right to balance 4 O from the MnO4-.
Balance H: Need 8 H+ on the left to balance 8 H in the 4 H2O.
5 Fe2+(aq) + MnO4-(aq) +8 H+(aq)
5 Fe3+(aq) + Mn2+(aq) +4 H2O(l)
BALANCED!
111
Balancing REDOX Equations: The half-reaction method
Step 1) Write the half-reactions for the chemical equation.
Step 2) For each reaction, balance the atoms other than O and H.
Step 3) Add H2O to balance O, then H+ to balance H.
Step 4) Balance the charge by adding electrons. The net charge of
the reactants should equal the net charge of the products.
Step 5) Add the two half-reactions together, making sure e- lost
equal e- gained, and canceling any species that appear on
both sides
of the reaction. The reaction is now balanced in an acidic solution.
Step 6) If you need to balance in a basic solution, first balance in acidic
solution (!), then add OH- to both sides to neutralize any H+ present.
Cancel any species that appear on both sides of the reaction.
112
56
Balancing redox eqn using half-cell method in acidic solutions
113
Cu(s) + HNO3(aq) → Cu2+(aq) + NO(g)
Identify half-reactions → one is ox., other is red.
Cu(s) → Cu2+(aq)
HNO3(aq) → NO(g)
Balance all atoms that are neither H nor O
OK as is
Balance O by adding H2O to side deficient in O
Cu(s) → Cu2+(aq)
HNO3(aq) → NO(g) + 2H2O(l)
114
57
Balance H by adding H+ to side deficient in H
Cu(s) → Cu2+(aq)
3H+ +HNO3(aq) → NO(g) + 2H2O(l)
Balance charge by adding e- to side that has + charge
Cu(s) → Cu2+(aq) + 2e+
3e- + 3H + HNO3(aq) → NO(g) + 2H2O(l)
Multiply each equation by factors so electrons cancel out
3[ Cu(s) → Cu2+(aq) + 2e- ]
2[ 3e- + 3H+ + HNO3(aq) → NO(g) + 2H2O(l) ]
Add equations and cancel spectators (none here)
3Cu(s)+ 6H+ (aq) + 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)
115
Balancing redox
equations in
basic solutions:
First balance in
acidic solution
116
58
Balancing in Basic Solution
3Cu(s)+ 6H+(aq)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
+6 OH-(aq)
+6 OH-(aq)
3Cu(s)+ 6H2O(l)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
+6 OH-(aq)
3Cu(s)+ 2H2O(l)+ 2HNO3(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O
+6 OH-(aq)
117
REDOX Balancing by Half-Reaction Method-I
Fe2+(aq) + MnO4-(aq)
Fe3+(aq) + Mn2+(aq) [acid solution]
Identify Oxidation and Reduction Half Reactions
Fe2+(aq)
Fe3+(aq) + e-
MnO4-(aq)
Mn2+(aq)
what shall we do with oxygen?
iron is oxidized
manganese is reduced
Add H2O to the products to balance O, add H+ to balance the H added,
add electrons to balance the charge.
5e- + MnO4-(aq) + 8H+(aq)
Mn2+(aq) + 4H2O(l)
Sum the two half-reactions
{ Fe2+(aq)
MnO4-(aq) + 8H+(aq) +5eMnO4-(aq) + 8H+(aq) +5e- +5Fe2+(aq)
Fe3+(aq) +e- } x5
Mn2+(aq) + 4H2O(l)
5Fe3+(aq) +5e- + Mn2+(aq)+ 4H2O(l)
118
59