Transcript Lectures on Chapter 4, Part 2 Powerpoint 97 Document

Balancing Redox Equations in Aqueous
Acid and Base Solutions :
ACID : You may add either H+ ( H3O+ ), or water ( H2O ) to either side
of the chemical equation to balance O and/or H atoms.
H+ + OH -
H2O
BASE : You may add either OH -, or water to either side of the
chemical equation to balance O and/or H atoms.
H+ + OH H+ + H 2O
H2 O
H 3O+
REDOX Balancing by Half-Reaction Method-II
Fe+2(aq) + MnO4-(aq)
Fe+3(aq) + Mn+2(aq) [acid solution]
Identify Oxidation and Reduction Half Reactions
Fe+2(aq)
Fe+3(aq) + e-
MnO4-(aq) +5e-
Mn+2(aq)
[oxidation half-reaction]
(Unbalanced redn. half-rxn!)
To balance O: add H2O as product, then H3O+ as reactant, then more H2O!
MnO4-(aq)+ __H3O+(aq) +5e-
Mn+2(aq) + __H2O(l) [redn half-reaction]
Sum the two half-reactions, multiply by integers to get e-s to balance:
{ Fe+2(aq)
MnO4-(aq) + 8H3O+(aq) +5eMnO4-(aq)+ 8H3O+(aq)+__e- +_Fe+2(aq)
Fe+3(aq) +e- } x__
Mn+2(aq) + 12H2O(l)
__Fe+3(aq)+__e- + Mn+2(aq)+12H2O(l)
REDOX Balancing by Half-Reaction Method-II
Fe+2(aq) + MnO4-(aq)
Fe+3(aq) + Mn+2(aq) [acid solution]
Identify Oxidation and Reduction Half Reactions
Fe+2(aq)
MnO4-(aq) +5e-
Fe+3(aq) + eMn+2(aq)
[oxidation half-reaction]
(Unbalanced redn. half-rxn!)
To balance O: add H2O as product, then H3O+ as reactant, then more H2O!
MnO4-(aq) + 8H3O+(aq) +5e-
Mn+2(aq) + 12H2O(l) [redn half-reaction]
Sum the two half-reactions, multiply by integers to get e-s to balance:
{ Fe+2(aq)
MnO4-(aq) + 8H3O+(aq) +5eMnO4-(aq)+ 8H3O+(aq)+5e- +5Fe+2(aq)
Fe+3(aq) +e- } x5
Mn+2(aq) + 12H2O(l)
5Fe+3(aq)+5e- + Mn+2(aq)+ 12H2O(l)
REDOX Balancing Using Ox. No. Method-III
Zinc metal is dissolved in Nitric Acid to give Zn2+ and the ammonium
ion from the reduced Nitric acid. Write the balanced chemical equation!
Zn(s) + H+(aq) + NO3-(aq)
Zn2+(aq) + NH4+(aq)
Oxidation # method
- __ e-
Zn(s) + H+(aq) + NO3-(aq)
Zn2+(aq) + NH4+(aq)
+5
+__ e-3
Multiply Zinc and Zn2+ by __ to balance __ e- from N. Since we have no
oxygen on the product side, add __ water molecules to the product side,
requiring __ + 4 = ___ H+ ions on the reactant side:
__ Zn(s) +__ H+(aq) + NO3-(aq)
__ Zn2+(aq) + NH4+(aq) + __H2O(l)
REDOX Balancing Using Ox. No. Method-III
Zinc metal is dissolved in Nitric Acid to give Zn2+ and the ammonium
ion from the reduced Nitric acid. Write the balanced chemical equation!
Zn(s) + H+(aq) + NO3-(aq)
Zn2+(aq) + NH4+(aq)
Oxidation # method
- 2 e-
Zn(s) + H+(aq) + NO3-(aq)
Zn2+(aq) + NH4+(aq)
+5
+8 e-3
Multiply Zinc and Zn2+ by 4 to balance 8 e- from N. Since we have no
oxygen on the product side, add 3 water molecules to the product side,
requiring 6+4 = 10 H+ on the reactant side:
4 Zn(s) +10 H+(aq) + NO3-(aq)
4 Zn2+(aq) + NH4+(aq) + 3 H2O(l)
REDOX Balancing by Half-Reaction Method-III
Given:
Oxidation:
Zn(s) + H3O+(aq) + NO3-(aq)
Zn(s)
Zn2+(aq) + NH4+(aq)
Zn2+ + 2 e-
H3O+(aq) + NO3-(aq) + 8 e NH4+(aq) + H2O(l)
We will need three waters to pick up the oxygens from the
nitrate ion. For the hydrogens, we will need to have
3x2+4=10 Hydrogen ions. Because the Hydrogen ions came
as hydronium ions, we also need 10 more water molecules.
10 H3O+(aq) + NO3-(aq) + 8 e NH4+(aq) + 13 H2O(l)
Reduction:
Finally, if we are to add the two equations, we must multiply the Ox. one
by 4 to be able to cancel out the electrons, so the final balanced
equation is:
10 H3O+(aq) + NO3-(aq) + 4 Zn(s)
4 Zn+2(aq) + NH4+(aq) + 13 H2O(l)
REDOX Balancing using Ox. No. Method - IV
+ 3 e-
+7
MnO4-(aq) + SO32-(aq)
+4
+4
( Acidic Solution )
MnO2 (s) + SO42-(aq)
- 2 e-
+6
Balance the electrons:
_ MnO4-(aq) + _ SO32-(aq) + H3O+(aq)
_ MnO2 (s) + _ SO42-(aq) + H2O(aq)
Account for the oxygen imbalance by adding acid to the reactant side,
and water to the product side.
For the final balance it is necessary to realize that protons needed to bind
up the oxygen atoms must also balance,
and since we have called H+ion - hydronium ions, water must be formed.
2 MnO4-(aq)+ 3 SO32-(aq)+_ H3O+(aq)
2 MnO2 (s) + 3 SO42-(aq) +_ H2O(aq)
REDOX Balancing using Ox. No. Method - IV
+7
+ 3 e-
MnO4-(aq) + SO32-(aq)
+4
( Acidic Solution )
MnO2 (s) + SO42-(aq)
+4
- 2 e+6
To balance the electrons, we must multiply the sulfite by 3, and the
permanganate by 2. We then have to account for the oxygen imbalance
by adding acid to the reactant side, and water to the product side.
2 MnO4-(aq) + 3 SO32-(aq) + H3O+(aq)
2 MnO2 (s) + 3 SO42-(aq) + H2O(aq)
For the final balance it is necessary to realize that protons needed to
bind up the oxygen atoms must be balanced, and since we have called
H+ion - hydronium ions,therefore water will be formed!
2 MnO4-(aq)+ 3 SO32-(aq)+2 H3O+(aq)
2 MnO2 (s) + 3 SO42-(aq) +3 H2O(aq)
REDOX Balancing by Half-Reaction Method-IV
MnO4-(aq) + SO32-(aq)
MnO2 (s) + SO42-(aq) [basic solution]
Oxidation:
SO32SO42-(aq) + 2e Add OH- to the reactant side, and water to the product side to get oxygen
to balance since we have one more oxygen on sulfate than on sulfite.
SO32-(aq) + 2 OH-(aq)
SO42-(aq) + H2O(l) + 2e Reduction:
MnO4-(aq) + 3e MnO2 (s)
Add water to the reactant side and OH- to the product side to take up the
oxygen lost when MnO4- goes to MnO2 and looses two oxygen atoms.
MnO4-(aq) + 2 H2O(l)+ 3e MnO2 (s) + 4 OH-(aq)
Multiply the oxidation equation by 3 to make the electrons 6. Multiply the
reduction equation by 2 to make the electrons 6, and add the two.
3 SO3-2(aq) + 2 MnO4-(aq) + H2O(l)
3 SO4-2(aq) + 2 MnO2 (s) + 2 OH-(aq)
REDOX Balancing by Half-Rxn Method-IV-2
MnO4-(aq) + SO32-(aq)
MnO2 (s) + SO42-(aq) [acidic solution]
Oxidation:
SO32-(aq)
SO42-(aq) + 2 e Add water to the reactant side to supply an oxygen and add two protons
to the product side that will remain plus the two electrons.
SO32-(aq) + H2O(l)
SO42-(aq) + 2 H+(aq) + 2 e Reduction:
MnO4-(aq) + 3 eMnO2 (s)
Add water to the product side to take up the extra oxygen from Mn cpds,
and add Hydrogens to the reactant side .
MnO4-(aq) + 3 e- + 4H+
MnO2 (s) + 2 H2O(l)
Multiply the oxidation equation by 3, and the reduction equation by 2,
and add them canceling out the electrons, protons and water molecules.
3SO32-(aq) + 2MnO4-(aq) + 2H+(aq)
3 SO42-(aq) + 2MnO2 (s) + H2O(l)
REDOX Balancing by Half-Rxn Method-IV-3
MnO4-(aq) +SO32-(aq)
MnO2(s) + SO42-(aq) [basic solution]
balance the equation as if it were in acid, and then convert it to base:
2MnO4-(aq) + 3SO32-(aq) + 2H+(aq)
2MnO2(s) + 3SO42-(aq) + H2O(l)
To convert to base, add two OH- to each side of the equation:
2MnO4-(aq)+ 3SO32-(aq)+2 H2O(l)
2MnO2(s)+ 3SO42-(aq)+ H2O(l)+2OH-(aq)
On the reactant side, the H+ and the OH- cancel to give water.
2MnO4-(aq)+ 3SO32-(aq)+2H2O(l)
2MnO2(s)+ 3SO42-(aq)+ H2O(l)+2OH-(aq)
Cancel out the water on each side of the equation, and you are done!
2MnO4-(aq) + 3SO32-(aq) + H2O(l)
2MnO2(s) + 3SO42-(aq) +2OH-(aq)
REDOX Balancing by Half-Reaction Method
-V - a
In acid Potassium dichromate reacts with ethanol(C2H5OH) to yield the
blue-green solution of Cr+3, the reaction used in “breathalyzers”.
H3O+(aq) + Cr2O72-(aq) + C2H5OH(l)
Cr3+(aq) + CO2 (g) + H2O(l)
Oxidation:
C2H5OH(l)
CO2 (g)
We need to ballance oxygen by adding water to the reactant side, and
ballance Hydrogen by adding protons to the product side.
C2H5OH(l) + 3 H2O(l)
2 CO2 (g) + 12 H+(aq)
Since we wish to consider H+ as the Hydronium ion - H3O+ , we must
add 12 water molecules to the reactant side, and make the H+ into H3O+.
C2H5OH(l) + 15 H2O(l)
2 CO2 (g) + 12 H2O+(aq) + 12 e -
REDOX Balancing by Half-Reaction Method
-V-b
Reduction:
Cr2O72-(aq)
Cr+3(aq)
Dichromate has two chromium atoms, therefore the products need to
have two Cr+3, and 3 electrons per atom. The oxygen atoms from the
dichromate need to be taken up as water on the product side by adding
protons to the reactant side.
14H+(aq) + Cr2O72-(aq)
Cr+3(aq) + 7 H2O(l)
Each Chromium atom changes oxidation from a +6 to a +3 there by
accepting 6 electrons, so we add 6 electrons to the reactant side.
6e - + 14 H3O+(aq) + Cr2O72-(aq)
2 Cr+3(aq) + 21 H2O(l)
Adding the two equations will give the final equation:
Ox:
C2H5OH(l) + 15 H2O(l)
2 CO2 (g) + 12 H3O+(aq) + 12 e Rd: [6e - + 14 H3O+(aq) + Cr2O72-(aq)
2 Cr+3(aq) + 21 H2O(l)] x 2
C2H5OH(l) + 16 H3O+(aq) + 2 Cr2O72-(aq)
2 CO2 (g) + 4 Cr+3(aq) + 27 H2O(l)
REDOX Balancing Using Ox. No. Method -VI
0
-1 e -
+1
Ag(s) + CN -(aq) + O2 (g)
0
Balance e-s and easy species:
Ag(CN)2-(aq) + OH -(aq)
+ 2 e-
__ Ag(s) + __ CN -(aq) + O2 (g)
-2
__ Ag(CN)2-(aq) + OH -(aq)
Balance H by adding water as a reactant:
_ Ag(s) + _ CN -(aq) + O2 (g) + _ H2O(l)
_ Ag(CN)2-(aq) + _ OH -(aq)
REDOX Balancing Using Ox. No. Method -VI
0
-1 e -
Ag(s) + CN -(aq) + O2 (g)
+1
Ag(CN)2-(aq) + OH -(aq)
0
+ 2 e-2
To balance electrons we must put a 4 in front of the Ag, since each
oxygen looses two electrons, and they come two per O2! That requires
us to put a 4 in front of the silver complex, yielding 8 cyanide ions.
4 Ag(s) + 8 CN -(aq) + O2 (g)
4 Ag(CN)2-(aq) + OH -(aq)
We have no hydrogens on the reactant side therefore we must add water
as a reactant, and since we also add oxygen, we must add two water
molecules, that well give us 4 hydroxide anions, giving us a balanced
chemical equation.
4 Ag(s) + 8 CN -(aq) + O2 (g) + 2 H2O(l)
4 Ag(CN)2-(aq) + 4 OH -(aq)
REDOX Balancing by Half-Reaction Method
-VI - a
Silver is reclaimed from ores by extraction using basic Cyanide ion.
OH
Ag(s) + CN (aq) + O2 (g)
Ag(CN)2-(aq)
Oxidation:
CN-(aq) + Ag(s)
Ag(CN)2-(aq)
Since we need two cyanide ions to form the complex, add two to the
reactant side of the equation. Silver is also oxidized, so it looses an
electron, so we add one electron to the product side.
2 CN-(aq) + Ag(s)
Ag(CN)2-(aq) + e Reduction:
O2 (g) + H2O(aq)
OH-(l)
Since oxygen is to form oxide ions, 4 electrons need to be added to
the reactant side, and 2 water molecules are needed to supply the
hydrogens to make hydroxide ions, yielding 4 OH- ions.
4 e - + O2 (g) + 2 H2O(aq)
4 OH-(l)
REDOX Balancing by Half-Reaction Method
- VI - b
Adding the Reduction equation to the Oxidation equation will require
the Oxidation one to be multiplied by 4 to eliminate the electrons.
Ox (x4)
8CN-(aq) + 4 Ag(s)
4 Ag(CN)2-(aq) + 4 e -
Rd
4 e - + O2 (g) + 2 H2O(l)
4 OH -(aq)
8 CN -(aq) + 4 Ag(s) + O2 (g) + 2 H2O(l)
4 Ag(CN)2-(aq) + 4 OH -(aq)
Fig. 4.14
Redox Titration- Calculation outline - I
Volume (L) of KMnO4 Solution
a)
M (mol/L)
Moles of KMnO4
b)
Molar ratio
in redox rxn.
Moles of CaC2O4
c)
Chemical Formulas
Moles of Ca+2
Problem: Calcium Oxalate was
precipitated from 1.00 mL blood by
the addition of Sodium Oxalate so
the Ca+2 conc. in the blood could
be determined. This precipitate was
dissolved in a sulfuric acid solution,
which then required 2.05 mL of
4.88 x 10-4 M KMnO4 to reach the
endpoint via the rxn. of Fig. 4.14.
a) Calculate the moles of Ca+2.
b) Calculate the Ca+2 conc. in blood.
Plan: a) Calculate the moles of
Ca+2 in the H2SO4 solution (and
blood sample).
b) Convert the Ca+2 conc.into units
of mg Ca+2/ 100 mL blood.
Redox Titration - Calculation - I
Equation:
2 KMnO4 (aq) + 5 CaC2O4 (aq) + 8 H2SO4 (aq)
2 MnSO4 (aq) + K2SO4 (aq) + 5 CaSO4 (aq) + 10 CO2 (g) + 8 H2O(L)
a) Moles of KMnO4
Mol = Vol x Molarity
Mol = 0.00205 L x 4.88 x 10- 4mol/L
Mol = 1.00 x 10 - 6mol KMnO4
b) Moles of CaC2O4
5 mol CaC2O4
-6
Mol CaC2O4 = 1.00 x 10 mol KMnO4 x
=
2 mol KMnO4
Mol CaC2O4 = 2.50 x 10 -6 mol CaC2O4
c) Moles of Ca+2
Mol
Ca+2
= 2.50 x
10 -6
mol CaC2O4 x
Mol Ca+2 = 2.50 x 10 -6 mol Ca+2
1 mol Ca+2
=
1 mol CaC2O4
Redox Titration - Calculation Outline - II
Moles of Ca2+/ 1 mL of blood
a) Calc of mol Ca+2 per 100 mL
multiply by 100
Moles of Ca2+/ 100 mL blood
M (g/mol)
b) Calc of mass of Ca+2 per 100 mL
Mass (g) of Ca2+/ 100 mL blood
1g = 1000mg
Mass (mg) of Ca2+ / 100 mL blood
c) convert g to mg!
Redox Titration - Calculation - II
a) Mol Ca+2 per 100 mL Blood
Mol Ca+2 =
Mol Ca+2 x 100 mL Blood =
100 mL Blood 1.00 mL Blood
-6 mol Ca+2
Mol Ca+2
2.50
x
10
=
x 100 mL Blood =
100 mL Blood
1.00 mL Blood
Mol Ca+2
= 2.50 x 10 -4 mol Ca+2
100 mL Blood
b) mass (g) of Ca+2
Mass Ca+2 = Mol Ca+2 x Mol Mass Ca/ mol =
Mass Ca+2 = 2.50 x 10 -4mol Ca+2 x 40.08g Ca/mol = 0.0100 g Ca+2
c) mass (mg) of Ca+2
Mass Ca+2 = 0.0100g Ca+2 x 1000mg Ca+2/g Ca+2 = 10.0 mg Ca+2
100 mL Blood
Types of Chemical Reactions - I
I) Combination Reactions that are Redox Reactions
a) A Metal and a Non-Metal form an Ionic compound
b) Two Non-Metals form a Covalent compound
c) Combination of an Element and a Compound
II) Combination Reactions that are not Redox Reactions
a) A Metal oxide and a Non-Metal form an ionic compound with
a polyatomic anion
b) Metal Oxides and water form Bases
c) Non-Metal Oxides and water form Acids
III) Decomposition Reactions
a) Thermal Decomposition
i) Many ionic compounds with oxoanions form a metal oxide
and a gaseous non-metal
ii) Many Metal oxides, Chlorates, and Perchlorates release
Oxygen
b) Electrolytic Decomposition
Types of Chemical Reactions - II
IV) Displacement Reactions
a) Single -Displacement Reactions - Activity Series of the Metals
i) A metal displaces hydrogen from water or an acid
ii) A metal displaces another metal ion from solution
iii) A halogen displaces a halide ion from solution
b) Double -Displacement Reactions i) In Precipitation Reactions- A precipitate forms
ii) In acid-Base Reactions - Acid + Base form a salt & water
V) Combustion Reactions - All are Redox Processes
a) Combustion of an element with oxygen to form oxides
b) Combustion of Hydrocarbons to yield Water & Carbon Dioxide
Reactants
Products
Fig.4.15
Reaction of Metals with Non-metals to form
Ionic Compounds - Redox Rxns
Alkali Metals and Alkaline Earth Metals with the Halogens
2 Na(s) + Cl2 (g)
2 NaCl(s)
Ba(s) + Br2 (l)
BaBr2 (s)
A Metal and A Nonmetal (Oxygen) to form ionic compounds
4 Fe(s) + 3 O2 (g)
2 Fe2O3 (s)
4 K(s) + O2 (g)
2 K2O(s)
Metals with the Nonmetals Sulfur and Nitrogen
16 Fe(s) + 3 S8
8 Fe2S3 (s)
3 Ca(s) + N2 (g)
Ca3N2 (s)
Two Non-Metals combine to form a Binary
Covalent Compound - Redox Rxns
Halogens of non-metals
P4 (s) + 6 F2 (g)
4 PF3 (g)
I2 (s) + 5 F2 (g)
2 IF5 (l)
S8 (s) +4 Br2 (l)
4 S2Br2 (l)
N2 (g) + 3 Cl2 (g)
Nitrides and Sulfides
P4 (s) +10 N2 (g)
S8 (s) +2 N2 (g)
8 P4 (s) + 5 S8 (s)
2 NCl3 (g)
4 P3N5 (s)
2 S4N2
(s)
8 P4S5 (s)
Other Elements combine with Oxygen to
form Oxides
Metals combining with Oxygen
4 Na(s) + O2 (g)
2 Ca(s) + O2 (g)
4 Al(s) +3 O2 (g)
2 Na2O(s)
2 CaO(s)
2 Al2O3 (s)
Ti(s) + O2 (g)
TiO2 (s)
N2 (g) + O2 (g)
2 NO (g)
Non-Metals with oxygen
P4 (s) + 5 O2 (g)
P4O10 (s)
S8 (s) +8 O2 (g)
8 SO2 (g)
2 F2 (g) + O2 (g)
2 OF2 (g)
Combination of a compound and an Element:
Non-metal Oxides & Halides react with additional
Oxygen & Halogens to form “higher” Oxides and Halides
1) non-metal oxides with oxygen:
2 NO(g) + O2 (g)
P4O6 (s) +2 O2 (g)
2 CO(g) + O2 (g)
2) non-metal halides with halogens:
ClF(g) + F2 (g)
2 NO2 (g)
P4O10 (s)
2 CO2 (g)
ClF3 (g)
PF3 (g) +F2 (g)
PF5 (l)
IF3 (g) + F2 (g)
IF5 (l)
IF5 (l) + F2 (g)
IF7 (l)
Combination of Two Compounds - I
1) Metal oxide with a non-metal oxide to form an ionic compound
with a polyatomic anion.
Na2O(s) + CO2 (g)
Na2CO3 (s)
K2O(s) + SO2 (g)
K2SO3 (s)
CaO(s) + SO3 (g)
CaSO4 (s)
2) Metal oxides react with water to form hydroxides
Na2O(s) + H2O(l)
2 NaOH(aq)
BaO(s) + H2O(l)
Ba(OH)2 (s)
2 Sc2O3 (s) + 6 H2O(l)
FeO(s) + H2O(l)
4 Sc(OH)3 (s)
Fe(OH)2 (s)
Combination of Two Compounds - II
3) Non-metal oxides react with water to form acids.
CO2 (g) + H2O(l)
H2CO3 (aq)
SO2 (g) + H2O(l)
3 NO2 (g) + H2O(l)
H2SO3 (aq)
2 HNO3 (aq) + NO(g)
4) Hydrates result from the reaction of anhydrous (without water)
compounds with water.
CuSO4 (s) + 5 H2O(l)
MgSO4 (s) + 7 H2O(l)
Na2CO3 (s) + 10 H2O(l)
.
.
Na CO .10H O
CuSO4 5H2O (s)
MgSO4 7H2O(s)
2
3
5) Addition reactions with carbon compounds.
C2H4 (g) + Cl2 (g)
C2H4Cl2 (g)
Ethylene + Chlorine
dichloroethane
2
(s)
Fig. 4.16
Thermal Decomposition Reactions - I
Carbonates
Oxides and Carbon Dioxide
Na2CO3 (s)
Na2O(s) + CO2 (g)
MgCO3 (s)
MgO(s) + CO2 (g)
Sulfites
MgSO3 (s)
Oxides and Sulfur Dioxide
MgO(s) + SO2 (g)
K2SO3 (s)
K2O(s) + SO2 (g)
Metal oxides, chlorates, and perchlorates
oxygen
2 Na2O(s)
4 Na(s) + O2 (g)
KClO4 (s)
KCl(s) + 2 O2 (g)
Thermal Decomposition Reactions - II
Hydroxides, Hydrates, and some oxoacids
Ca(OH)2 (s)
Na2CO3 10H2O (s)
H2SO3 (l)
water
CaO(s) + H2O(g)
Na2CO3 (s) + 10 H2O(g)
SO2 (g) + H2O(g)
2 Fe(OH)3 (s)
Fe2O3 (s) + 3 H2O(g)
MgSO4 7H2O(s)
MgSO4 (s) + 7 H2O(g)
H2CO3 (aq)
CO2 (g) + H2O(g)
2 NaOH(s)
Na2O(s) + H2O(g)
Fig. 4.17
Fig. 4.18
Fig. 4.20
Metals Displace Hydrogen from Water
Metals that will Displace Hydrogen from Cold Water:
2 Cs(s) + 2 H2O
H2 (g) + 2 CsOH(aq)
Ba(s) + H2O(l)
Na(s) + H2O(l)
Metals that will Displace Hydrogen from Steam:
Mg(s) + 2 H2O(g)
Cr(s) +
H2O(g))
Zn(s) + H2O(g)
H2 (g) + Mg(OH)2 (s)
Metals Displace Hydrogen from Water
Metals that will Displace Hydrogen from Cold Water:
2 Cs(s) + 2 H2O
H2 (g) + 2 CsOH(aq)
Ba(s) +2 H2O(l)
H2 (g) + Ba(OH)2 (aq)
2 Na(s) + 2 H2O(l)
H2 (g) + 2 NaOH(aq)
Metals that will Displace Hydrogen from Steam:
Mg(s) + 2 H2O(g)
2 Cr(s) + 6 H2O(g)
Zn(s) + 2 H2O(g)
H2 (g) + Mg(OH)2 (s)
3 H2 (g) +2 Cr(OH)3 (s)
H2 (g) + Zn(OH)2 (s)
Metals Displace Hydrogen from Acids
Reactions of Metals above hydrogen in the Activity series
Mg(s) + 2 HCl(aq)
MgCl2 (aq) + H2 (g)
Zn(s) + H2SO4 (aq)
Al(s) + HCl(aq)
Cd(s) + HBr(aq)
Sn(s) +
H2SO4 (aq)
Reactions of Metals below hydrogen in the Activity series
Cu(s) + HCl(aq)
Au(s) + H2SO4 (aq)
No Reaction!
Metals Displace Hydrogen from Acids
Reactions of Metals above hydrogen in the Activity series
Mg(s) + 2 HCl(aq)
MgCl2 (aq) + H2 (g)
Zn(s) + H2SO4 (aq)
ZnSO4 (aq) + H2 (g)
2 Al(s) + 6 HCl(aq)
2 AlCl3 (aq) + 3 H2 (g)
Cd(s) + 2 HBr(aq)
Sn(s) + 2 H2SO4 (aq)
CdBr2 (aq) + H2 (g)
Sn(SO4)2 (aq) + 2 H2 (g)
Reactions of Metals below hydrogen in the Activity series
Cu(s) + HCl(aq)
No Reaction!
Au(s) + H2SO4 (aq)
No Reaction!
Fig. 4.19
Fig. on p. 163
Single Replacement Reactions Metals Replace Metal Ions from Solution
Ba(s) + Co(NO3)2 (aq)
Cd(s) + AgNO3 (aq)
Mg(s) + Pb(NO3)2 (aq)
Al(s) + Ba(NO3)2 (aq)
Cr(s) + PtCl4 (aq)
Ca(s) + Hg(NO3)2 (aq)
Li(s) + Na2SO4 (aq) + H2O(l)
Co(s) + Ba(NO3)2 (aq)
Single Replacement Reactions Metals Replace Metal Ions from Solution
Ba(s) + Co(NO3)2 (aq)
Cd(s) + AgNO3 (aq)
Co(s) + Ba(NO3)2 (aq)
2 Ag(s) + Cd(NO3)2 (aq)
Mg(s) + Pb(NO3)2 (aq)
Pb(s) + Mg(NO3)2 (aq)
Al(s) + Ba(NO3)2 (aq)
No Reaction
4 Cr(s) + 3 PtCl4 (aq)
Ca(s) + Hg(NO3)2 (aq)
2 Li(s) + Na2SO4 (aq) + H2O(l)
3 Pt(s) + 4 CrCl3 (aq)
Hg(l) + Ca(NO3)2 (aq)
2 NaOH(aq) + H2 (g) + Li2SO4 (aq)
Double Replacement Reactions
1) Precipitation Reactions - An insoluble product is formed:
Pb(NO3)2 (aq) + 2 NaI(aq)
PbI2 (s) + 2 NaNO3 (aq)
Ba(NO3)2 (aq) + Na2SO4 (aq)
2) Acid - Base Neutralization Reactions - water is formed:
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
H2SO4 (aq) + Ca(OH)2 (s)
3) A Carbonate or Sulfite reacts with acid to form a gas:
Na2CO3 (aq) + 2 HBr(aq)
2 NaBr(aq) + H2O(l) + CO2 (g)
K2SO3 (aq) + 2 HI(aq)
2 KI(aq) + H2O(l) + SO2 (g)
Double Replacement Reactions
1) Precipitation Reactions - An insoluble product is formed:
Pb(NO3)2 (aq) + 2 NaI(aq)
Ba(NO3)2 (aq) + Na2SO4 (aq)
PbI2 (s) + 2 NaNO3 (aq)
BaSO4 (s) + 2 NaNO3 (aq)
2) Acid - Base Neutralization Reactions - water is formed:
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
H2SO4 (aq) + Ca(OH)2 (s)
CaSO4 (s) + 2 H2O(l)
3) A Carbonate or Sulfite reacts with acid to form a gas:
Na2CO3 (aq) + 2 HBr(aq)
2 NaBr(aq) + H2O(l) + CO2 (g)
K2SO3 (aq) + 2 HI(aq)
2 KI(aq) + H2O(l) + SO2 (g)
Combustion Reactions: Redox Reactions
Elements combining with Oxygen to form Oxides:
2 Al(s) + 3 O2 (g)
2 Al2O3 (s)
S8 (s) + 8 O2 (g)
8 SO2 (g)
C(s) + O2 (g)
Compounds combining with Oxygen to form Oxides
2 Fe2S3 (s) + 9 O2 (g)
2 Ca3N2 (s) + 7 O2 (g)
2 Fe2O3 (s) + 6 SO2 (g)
6 CaO(s) + 4 NO2 (g)
Hydrocarbons combining with Oxygen to form CO2 and H2O
CH4 (g) + 2 O2 (g)
2 C4H10 + 18 O2 (g)
Identifying the Type of Chemical Reaction - I
Problem: Classify the following as combination, decomposition, or
displacement reactions, identify the underlying chemical process as
precipitation, acid-base, or redox, and write a balanced molecular
equation for each. for Redox, identify the oxidizing and reducing agents.
a) Barium chloride(aq) and Ammonium sulfate(aq)
b) Manganese metal and Tin IV chloride (aq)
c) Strontium hydroxide and Bromic acid
d) Cobalt metal and Nitrogen gas to yield Cobalt II Nitride
e) Sodium peroxide to give sodium oxide and oxygen gas
Plan: Identify the reaction type, and identify the products and equation.
Solution:
a) Displacement (metathesis): two substances form two others, one of
which is a precipitate (as can be told from looking at Table 4.1):
BaCl2 (aq) + (NH4)2SO4 (aq)
BaSO4 (s) + 2 NH4Cl(aq)
Identifying the Type of Chemical Reaction -II
b) Manganese metal and Tin IV chloride (aq).
Hint: Ck. The activity series of the two metals for a displacement.
2 Mn(s) + SnCl4 (aq)
2 MnCl2 (aq) + Sn(s)
c) Strontium hydroxide and Bromic acid.
Hint: This is an acid - base reaction, the base Sr(OH)2 reacts with twice
as much Bromic acid to give water and the salt, Strontium Bromate
dissolved in water.
Sr(OH)2 (aq) + 2 HBrO3 (aq)
Sr(BrO3)2 (aq) + 2 H2O(L)
Identifying the Type of Chemical Reaction -II
b) Manganese metal and Tin IV chloride (aq).
Displacement Reaction(single): This “redox” reaction occurs when
the more active manganese displaces the less reactive tin.
2 Mn(s) + SnCl4 (aq)
2 MnCl2 (aq) + Sn(s)
Mn is the reducing agent, and SnCl4 is the oxidizing agent.
c) Strontium hydroxide and Bromic acid.
Displacement Reaction(metathesis): Two substances form two others.
This is an acid - base reaction, the base Sr(OH)2 reacts with twice as
much Bromic acid to give water and the salt, Strontium Bromate
dissolved in water.
Sr(OH)2 (aq) + 2 HBrO3 (aq)
Sr(BrO3)2 (aq) + 2 H2O(L)
Identifying the type of Chemical Reaction -III
d) Cobalt metal and Nitrogen gas to yield Cobalt II Nitride.
Co(s) + N2 (g)
e) Sodium peroxide to give sodium oxide and oxygen gas.
Na2O2 (s)
+ O2 (g)
Identifying the type of Chemical Reaction -III
d) Cobalt metal and Nitrogen gas to yield Cobalt II Nitride.
Combination Reaction: This “Redox” reaction occurs when Cobalt
metal is heated in Nitrogen gas and forms the solid compound
Cobalt II Nitride.
3 Co(s) + N2 (g)
Co3N2 (s)
Nitrogen is the oxidizing agent, and is reduced; Cobalt is the
reducing agent, and is oxidized.
e) Sodium peroxide to give sodium oxide and oxygen gas.
Decomposition Reaction: One substance forms two substances. This
is a :Redox” reaction that forms Oxygen gas.
2 Na2O2 (s)
2 Na2O(s) + O2 (g)
Sodium peroxide is the reducing and oxidizing agent, as oxygen
is both oxidized, and reduced.
Fig. 4.21
Many Chemical Reactions are in a state of
Dynamic Equilibrium
Solid - Gas Equilibrium processes
CaCO3 (s)
CaO(s) + CO2 (g)
Solution Equilibrium processes involving weak acids and bases
CH3COOH(aq) + H2O(l)
NH3 (aq) + H2O(l)
CH3COO -(aq) + H3O+(aq)
NH4+(aq) + OH -(aq)