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Chemical Reactions
Chemical Equations and Chemical Reactions
Chemical Reaction – process that causes a chemical change
Reactants – starting materials
Products –substances formed
Reagents –chemicals available in the lab
Symbolizing chemical reactions
Reactants  products
Na + H2O  NaOH + H2
Skeletal equation – not balanced
Equations must be balanced – have same number of each kind of atom in
reactants and products, since atoms are not created or destroyed in chemical
reactions (law of conservation of mass)
2 Na + 2 H2O  2NaOH + H2, balanced
2, 2, 2, (1) are stoichiometric coefficients
- coefficients are relative numbers of atoms or moles in reaction:
“2 atoms/2 mol Na react with 2 molecules/2 mol H2O to produce.”
Include the physical states:
S – solid, precipitate
l- liquid
g- gas
aq – in water solution
high temperature/heat required
Balanced equation: 2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)
Balancing Chemical equations
Many equations can be balanced by simple inspection:
H2 + O2  H2O
- skeletal
2H2 + O2  2H2O
- with coefficients
2H2(g) + O2(g)  2H2O(l)
- with physical states
Never change subscripts to balance equation:
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H2 (g) + O2(g)  H2O2
- other product
Steps in balancing eqautions:
1. First balance the element(s) occurring in fewest formulas.
2. Balance the element in greater number of formulas last
Butane, C4H10(g) burns in oxygen (air) to form carbon dioxide and water vapor
[word equation]. Balance the equation.
C4H10 + O2  CO2 + H2O skeletal
- balance C first, then H, leave O to last
C 4 H 10 + O2
4CO2 + + 5H2O
O in products = 4 x 2 + 5 x 1 = 13 CO or 13/2 O2, so multiply all coefficients by
2.
2 C4H10 (g) + 13O2(g)  8CO2(g) + 10H2O(g)
Example:
Ce(OH)4(aq) + H3PO4(aq)  Ce3(PO4)4(s) + H2O(l)
3Ce(OH)4(aq) + 4H3(PO4)(aq)  Ce3(PO4)4(s) + 12 H2O(l)
Precipitation reactions
Soluble – readily dissolves in solvent (H2O)
Insoluble – slightly/sparingly soluble – do not dissolve significantly, ≤ 0.1 M
Precipitation reaction – insoluble solid product (precipitate) is formed when two
solution are mixed
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Ionic/Net Ionic equations – showing ions present
Complete ionic equation – shows all ions
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)  AgCl(s) + Na+(aq) + NO3(aq)
Hydrated ions Na+(aq) ≡ Na(H2O)+6-8
2
Spectator ions – ions remaining unchanged in reaction: Na+(aq), NO3(aq)
Delete spectator ions to give net ionic equation:
Ag+(aq) + Cl(aq)  AgCl(s) shows actual change.
Writing a net ionic equation
1. Write and balance the complete ionic equation
2. Cancel the spectator ions to obtain the net ionic equation
Example:
Ba(NO3)2(aq) + 2 NH4IO3(aq) mixed to produce Ba(IO3)2(s) and NH4NO3(aq)
Chemical equation:
Ba(NO3)2(aq) + 2NH4IO3(aq)  Ba(IO3)2(s) + 2NH4NO3(aq)
Complete ionic equation:
Ba2+(aq) + 2NO3(aq) 2NH4+(aq) + 2IO3(aq)  Ba(IO3)2(s) + 2NH4+(aq) + 2NO3(aq)
Net ionic equation:
Ba2+ + 2IO3(aq)  Ba(IO3)2(s)
Knowing solubility rules allows prediction of precipitation reactions
Solubility rules for common Inorganic Compounds:
A. Soluble compounds
Ionic salts of group 1 cations: Li+, Na+, K+, Rb+, Cs+
Ammonium salts, NH4+
Chloride (Cl), bromide (Br), iodide (I) salts except Ag+, Hg22+, Pb2+
Nitrate (NO3), acetate (CH3CO2), chlorate (ClO3), perchlorate (ClO4)
salts
5. Sulfates (SO42-) except with Ca2+, Sr2+, Ba2+, Pb2+, Hg22+, Ag+
1.
2.
3.
4.
B. Insoluble salts
1. Carbonates (CO32), chromates (CrO42), oxalates (C2O42), phosphates
(PO43); except with group I cations or NH4+
2. sulfides (S2), except with group I and 2 cations or NH4+
3. Hydroxides (OH-) and oxides (O2), which becomes OH when dissolved
in water) except with group 1 and 2 cations
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Slightly soluble (borderline) salts include Ag2SO4, PbCl2, Ca(OH)2, Sr(OH)2, and
Mg(OH)2
Preparing precipitates:
Hg2I2 is insoluble, formed in net ionic equation
Hg22+(aq) + 2I(aq)  Hg2I2(s)
-
so select possible spectator ions that give initially soluble salts: NO3,
K+
Hg2(NO3)2(aq) + 2KI(aq)  Hg2I2(s) + 2KNO3(aq)
Predict results of mixing (NH4)2S(aq) and CuSO4(aq); net ionic equation
CuSO4(aq) + (NH4)2S(aq)  CuS(s) + (NH4)2SO4(aq)
Cu2+(aq) + SO42-(aq) + 2NH4+(aq) + S2-(aq)  CuS(s) + 2NH4+(aq) + SO42-(aq)
Cu2+(aq) + S2-(aq)  CuS(s)
Acid-Base Reactions
-historical recognition
Acid – sour taste, turns litmus dye red
Base – bitter taste, soapy feeling, turns litmus blue
(poor procedures)
Acids/Bases in aqueous solutionsAn acid molecule reacts with water to produce a hydronium ion
HCl(g) + H2O(l)  H3O+(aq) + Cl(aq)
-does HCl (aq) exist?
HCl, HNO3, CH3COOH are typical acids, the acidic H is the atom released as H+
to form H3O+ by adding to H2O molecule
Mineral (inorganic) acids – write acidic H first, H3AsO4
A base is a molecule or ionic compound producing OH in water:
NaOH(s) already contains Na+OH and NH3(g) attracts H+ from H2O:
NaOH(s) + H2O  Na+(aq) + OH-(aq)
NH3(g) + H2O(l)  NH4+ + OH(aq)
(bases)
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Strong and Weak Acids/Bases
Strong acid - effectively totally ionized/dissociated in solution
Weak acid – only partially ionized in solution
Ionization/dissociation means transfer of H+ from acid molecule to H2O forming
hydronium ion, H3O+.
HCl(aq) is strong acid; effectively only H3O+, Cl ion in solution, no HCl(g)
molecules existing water solution:
H3O+(aq) + Cl(aq)
HCl(g) + H2O(l)
CH3COOH(aq) acetic acid, is weak acid, its solution containing mainly CH3COOH
molecules (~ 99%) with small concentrations (~1%) of H3O+ and CH3COO ions
H3O+(aq) + CH3COOaq)
CH3COOH(aq) + H2O(l)
(equilibrium, later discussion)
Most acids are weak; only a few common strong acids: HCl(aq), HBr(aq), (Not
HF(aq), HNO3(aq), H2SO4(aq) – 1st H+ only, HClO4(aq), NOT HClO3(aq)
Strong base is completely ionized in water, while weak base is only partially
dissociated:
NaOH(s) completely ionizes; NH3 partially dissociates
Common strong bases – hydroxides and oxides of grop 1 and 2 metals: MOH, M
= Li+, Na+, Rb+, Cs+; Ca(OH)2, Sr(OH)2, Ba(OH)2 –limited solubility
Oxide ion, O2- reacts with water to produce OH quantitatively
Li2O(s)
water
2Li+(aq) + 2OH-(aq)
{O2- + H2O(l)
2OH(aq)}
Common weak bases – ammonia – NH3, organic amines- RNH2, R2NH, R3N, R =
-CH3 (methyl), -CH2CH3 (ethyl) etc
NH3(aq) + H2O(l)
NH4+(aq) + OH-(aq)
~1%
~99%
No NH4OH
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Acidic or Basic Character of Compounds
Most basic metal oxides from basic solutions in water, while many non-metal
oxides from acidic solutions:
CaO(s) + H2O(l)
Ca(OH)2(aq) limited solubility
CO2(g) + H2O(l)
H2CO3(aq)
SO2(g) + H2O(l)
H2SO3(aq)
SO3(g) + H2O(l)
H2SO4(aq)
-
SO2, SO3 and various oxides cause acidity in rain water
“Acid rain” – dissolve marble (CaCO3)
Most metallic elements form basic oxides and/or hydroxides, non metallic
elements form acidic oxides/hydroxides – acidic oxides react with bases, basic
oxides react with acids:
MgO(s) + 2 HCl(aq)
MgCl2(aq) + H2O(l)
CO2(g) + 2 NaOH(aq)
Na2CO3(aq) + 2H2O(l)
Oxides/hydroxides of some elements (particularly along metalloid diagonal) have
both acidic and basic character (react with bases and acids) – called amphoteric
behavior.
Al2O3(s) + 6HCl(aq)
2AlCl3(aq) + 3H2O(l) - base
2Na+(aq) + Al(OH)4(aq) - acid
Al2O3(s) + 2NaOH(aq) + 3H2O
More easily seen as hydroxide:
Al(OH3(s) + 3 HCl(aq)
AlCl3(aq) + 2 H2O(l) - base
Al(OH)3(s) + NaOH(aq)
Na+(aq) + Al(OH)4-(aq) - acid
NeutralizationNeutralization is the reaction between an acid and base to produce a salt and
water: acid + base  salt + water
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HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
Cation of salt comes from base, anion from acid; OH from base reacts with H+
(H3O+) from acid to produce H2O:
H3O+(aq) + OH(aq)
2 H2O(l)
Net ionic equation for neutralization of strong acids with strong base
Net ionic equation for reactions of weak acids and bases:
Ionic equations show major species present:
Na+CH3COO(aq) + H2O(l)
CH3COOH(aq) + NaOH(aq)
weak acid
strong base
strong salt
Na+(aq) + CH3COO(aq) + H2O(l)
CH3COOH(aq) + Na+(aq) + OH-(aq)
CH3COO(aq) + H2O(l)
CH3COOH(aq) + OH-(aq)
NH3(aq) + HCl(aq)
weak base
solvent
NH4Cl(aq) + H2O(l)
strong acid
NH3(aq) + H3O+ + Cl-(aq)
salt
water
NH4+ + Cl-(aq) + H2O(l)
What salt is produced from aqueous barium hydroxide and sulfuric acid?
Ba(OH)2(aq),
H2SO4(aq)
BaSO4(s)
Ba(OH)2(aq) + H2SO4(aq)
insoluble
BaSO4(s) + 2 H2O(l)
Ba2+ + 2 OH-(aq) + H3O+ + HSO4(aq)
Formation of GasesSome reactions of acids produce gases:
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BaSO4(s) + 3 H2O(l)
CaCO3(s) + H2SO4(aq)
CaSO4(s) + CO2(g) + H2O(l)
FeS(s) + 2HCl(aq)
FeCl2(aq) + H2S(g)
A hydrogen ion transfer from strong acid to anion of salt producing a gas. The
initial hydrogen compound (acid) may not be a gas, but rapidly decomposes in to
a gas:
CaCO3(s) + H2SO4(aq)
H2CO3(aq)
CaSO4(s) + H2CO3(aq)
CO2(g) + H2O(l)
(test for limestone/marble)
Water itself can act as acid with a very strong base:
CaC2(s) + H2O(l)
carbide
Ca(OH)2(aq) + C2H2(g)
acetylene
Redox Reactions
Oxidation and Reduction – Redox reactions, oxidation-reduction reactions, are
reactions in which oxidation and reduction takes place simultaneously.
Oxidation – loss of electron, increase in oxidation number
Reduction – gain of electrons, decrease in oxidation number
Oxidation and reduction can be related to oxidation numbers of elements, the
real or arbitrary charge assigned to an atom
For monoatomic ions, ionic charge = oxidation number
For the free element, oxidation number is zero: He, H2, P4, S8
Mg2+, oxidation number = +2
Cl, oxidation number = -1
Rules for Oxidation Numbers of Elements in Compounds
1. Oxidation number for element in the uncombined (free state) is zero
2. Sum of Oxidation numbers of all the atoms in a species is equal to total
charge on the species.
H2PH4, 2 x ox. no. H + ox. No P + 4 x ox no O = -1
Na2SO4, 2 x ox. no. Na + ox no S + 4 x ox no O = 0
3. oxidation number for alkali metal, group 1 in compounds, is always +1:
Li+, Na+, K+, Rb+, Cs+ (in compounds)
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4. Oxidation number of alkaline earth, group 2 metals in compounds is
always +2: Be2+, Mg2+, Ca2+, Sr2+, Ba2+
5. Oxidation number of F in compounds is always -1: F
6. Oxidation number of H is usually +1, except in binary compounds with
active metals when it is -1 (hydride ion); NaH, CaH2.
7. Oxidation number for O is usually -2 in compounds, except as peroxide,
O22 (-1), superoxide, O2 (-1/2), or in binary compounds with F: OF2,
O2F2
8. Halogens, Cl, Br, I; tend to have -1 oxidation number except when
combined with O or another more active halogen, F > Cl > Br > I eg.
IF5, BrCl3
9. Some metal cations are reliably one charge: Al3+, Zn2+, Cd2+, Sc3+, Ni2+
(usually)
Example:
S in NaHSO4:
+1 +1
-2
Na H S O4
+1 +1
-8
x + 1 + - 8 = 0, x = +6
Oxidation number of P in H2PO4-:
+2 x
-8
H2 P
O4
+1
-2
X+ 2-8 = -1, x = +5
Oxidation number of Cr in K2Cr2O7
+2 2x
-14
K2 Cr2 O7
+1 x
-2
2x + 2 – 14 = 0
2x = 12
X=6
Oxidizing and Reducing Agents
An oxidizing agent in a redox reactioni. received electrons from species being oxidized, thereby “causing”
oxidation;
ii. Is the species being reduced;
iii. Contains the element undergoing a decrease in oxidation number
A reducing agent in a redox reactioni. donates electrons to species being reduced, thereby “causing
reduction”;
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ii. Is the species being oxidized:
iii. Contains the element undergoing an increase in oxidation number
Example:
Zn(s) + CuSO4(aq)
ZnSO4(aq) + Cu(s)
Net;
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
Zn is oxidized, therefore it is the reducing agent
Cu2+ is reduced, therefore it is the oxidizing agent, or CuSO4
Example:
Acidic solution of potassium dichromate reacts with iron(II) chloride solution to
form iron(III) and chromium(III) ions. Identify the oxidizing and reducing agents.
Cr3+(aq) + Fe3+(aq)
K2Cr2O7(aq) + FeCl2(aq)
(not balanced)
Cr atom has been reduced: Cr in +6 oxidation state in K2Cr2O7 (+6 to +3)
So K2Cr2O7 is the oxidizing agent.
Fe atom has been oxidized: Fe in +2 oxidation state in FeCl2 (+2 to +3)
So FeCl2 is the reducing agent.
Balancing Simple Redox Equations
In a redox reaction, all electrons lost in an oxidation step must be gained in the
reduction step; total change of reactants must equal that of products
Consider:
Cu(s) + Ag+(aq)
Cu2+(aq) + Ag(s)
Atoms balance, but charges don’t:
Cu(s) + 2 Ag+(aq)
Cu2+(aq) + 2 Ag(s)
Also,
Sn(s) + Fe3+(aq)
Sn(s) + 2 Fe3+(aq)
Sn2+(aq) + Fe2+(s)
Sn2+(aq) + 2 Fe2+(s)
Half-Reaction Method for Balancing Redox Reactions
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


The oxidation and reduction processes are written as two, separated halfreactions
Each half-reaction is balanced separately and then combined, so that
electrons lost (oxidation) = electrons gained (reduction)
Any redundancies, inconsistencies are corrected in final step
Procedure
1.
2.
3.
4.
5.
6.
7.
8.
Write the unbalanced half-reactions for the oxidation and reduction step
Balance all atoms, except H and O
Balance O by adding H2O to the opposite side of the equation
Balance H by adding H+ (instead of cumbersome H3O+) to the appropriate
side of the equation
For acidic solutions, can have H+(aq), H2O(l), not OH(aq), never e(aq) or
O2(aq); - for basic solutions can have OH(aq), H2O(l), not H+(aq) never e(aq)
or O2(aq); so must correct for H+ from step (4). – For reactions in basic
media add appropriate number of OH to each side of equation so that all
H+ ions are converted to H2O molecules; ie H+ + OH = H2O. Cancel
H2O molecules that are on both sides of equation.
Balance ionic charges by adding adequate numbers of electrons to
appropriate side of equation (right side for oxidation, left side for reduction)
[Check: electrons lost = increase in oxidation number, electron gained =
decrease in oxidation number]
Equalize number of electrons lost and gained in the two half-reactions with
appropriate factors (ie multiply half-reactions with factors making e lost =
egained)
Add the two half-reactions and cancel equal amounts of species
appearing on both sides of equation; check the coefficients are smallest
whole-numbers
Example 1
Use half-reaction method to balance in acid medium:
H2S(aq) + NO3(aq)
Oxidation:
H2S(aq)
Reduction:
NO3(aq)
[1]
H2S(aq)
H2S(aq)
S(s) + NO(g)
S(s)
NO(g)
S(s) + 2H+(aq)
[1]
[2]
atoms
S(s) + 2H+(aq) + 2e charges
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1
[2]
NO3(aq)
NO(g)
NO3(aq)
NO(g) + 2H2O(l)
4H+(aq) + NO3(aq)
NO(g) + 2H2O(l)
3e + 4H+(aq) + NO3(aq)
atoms
2
NO(g) + 2H2O(l)
multiply eqn 1 with 3 and eqn 2 with 2 would equal the number of electrons
3x 1
3H2S(aq)
2x
2
3 S(s) + 6H+(aq) + 6e
6e + 8H+(aq) + 2 NO3(aq)
2 NO(g) + 4H2O(l)
3H2S(aq) + 8H+(aq) + 2NO3(aq)
2 NO(g) + 4H2O(l) + 3 S(s) + 6H+(aq)
3H2S(aq) + 2H+(aq) + 2NO3(aq)
2 NO(g) + 4H2O(l) + 3 S(s)
Example 2
Use half-reaction method to balance in basic solution:
Bi2O3(s) + ClO(aq)
Oxidation
Bi2O3(s)
BiO3(aq) + Cl(aq)
BiO3(aq)
Bi2O3(s)
2BiO3(aq)
3H2O(l) +Bi2O3(s)
2BiO3(aq)
3H2O(l) +Bi2O3(s)
2BiO3(aq) + 6H+
6OH(aq) + 3H2O(l) +Bi2O3(s)
2BiO3(aq) + 6H2O(l)
6OH(aq) + Bi2O3(s)
2BiO3(aq) + 3H2O(l)
6OH(aq) + Bi2O3(s)
2BiO3(aq) + 3H2O(l) +4e
Reduction
ClO(aq)
ClO(aq)
Cl(aq)
Cl(aq) + H2O(l)
2H+(aq) + ClO(aq)
Cl(aq) + H2O(l)
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1
2OH(aq) + 2H+(aq) + ClO(aq)
H2O(l) + ClO(aq)
Cl(aq) + H2O(l) + 2OH(aq)
Cl(aq) + 2OH(aq)
2e + H2O(l) + ClO(aq)
Cl(aq) + 2OH(aq)
2
1 x (4e) = 2 x (2e)
6OH(aq) + Bi2O3(s) +2 H2O(l) + 2ClO(aq)
2OH(aq) + Bi2O3(s) + 2ClO(aq)
2Cl(aq) + 4OH(aq) + 2BiO3(aq) + 3H2O(l)
2Cl(aq) + 2BiO3(aq) + H2O(l)
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