Download - Catalyst

10/20/2011
Acid/Base Titrations
• Read pages 114-117 about titrations.
• We’ll come back to titrations in much more detail in Ch 8 .
• What do you need to know now?
– titration is the volumetric analysis of one solution using
another
– equivalence point (stoichiometric point)…you’ll encounter
these terms in Lab #2
• Examples of titration problems you should be able to solve now:
– 25.0 mL of HCl is titrated with 50.0 mL 0.20 M OH-. What is
the [HCl]? (generic acid is HA, generic base is B-)
– How many mL of 0.35 M HBr are required to neutralize 45 mL
of 0.15 M Ca(OH)2?
Reaction Classes
• Precipitation: synthesis of an ionic solid
– a solid precipitate forms when aqueous solutions of certain
ions are mixed
• Acid-Base: proton transfer reactions
– acid donates a proton to a base, forming a molecule (water or
another weak acid) and an aqueous salt
– Acid: proton-donor; Base: proton-acceptor
• Oxidation-Reduction: electron transfer reactions
– electron transfer from one species to another, causing a change
in the oxidation state of the two species
– OIL RIG: Oxidation Is Loss (of e-), Reduction Is Gain (of e-)
– includes combustion, the reaction of a substance with oxygen
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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
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Redox Reactions
• “Redox” Chemistry: Reduction and Oxidation
• Oxidation: Loss of electrons
• Reduction: Gain of electrons
(a reduction in oxidation number)
OIL RIG
Oxidation Is Loss of electrons
Reduction Is Gain of electrons
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Redox Reactions
• In a redox reaction, one species loses electrons and another
species accepts those electrons.
• Electrons are neither created nor destroyed during the
reaction…charges are conserved.
Example:
Na (s) and Cl2 (g)
reacting to form
NaCl (s), which
contains Na+
and Cl-.
Oxidation Numbers (or States)
•
How can you tell which species is gaining electrons and
which is losing them?
•
We need a way to look at the effective charge on each
species before and after the reaction.
•
This is accomplished by assigning oxidation numbers
(states) to atoms on both the reactant and product sides
of the reaction, then looking for the species that gains
electrons (reduction) and loses electrons (oxidation).
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Oxidation Numbers (or States)
•
An “accounting” of the electrons in a chemical species.
Remember: we are not making ions within covalent
compounds!!
•
Oxidation number (state) represents the number of electrons
required to produce the “effective charge” on a species.
•
Oxidation numbers (states) are chosen so that:
– charges are conserved
– in ionic compounds, the sum of the oxidation numbers
on the atoms coincides with the charge on the ion
Bid idea: the oxidation numbers of the atoms/ions in a species
must sum to the total charge of the species.
See Table 4.3
in Zumdahl
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Period
1
IA
H
+1 -1
IIA
IIIA
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
2
3
4
5
6
Common Oxidation Numbers
IVA
VA
C
N
+4,+2 all from
-1,-4
+5 to-3
Ge
+4,+2
+3, +2 -4
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
Sb
Te
-1 I
+5,+3 +6,+4 +7,+5
-3
-2
+3,+1
Xe
In
+3,+2
+1
Sn
+4,+2,
-4
Ba
Tl
Pb
Bi
+2
+3,+1
+4,+2
+3
Po
-1 At
+6,+4 +7,+5
+2,-2 +3,+1
+2
+6,+4
+2
Rn
+2
Transition Metals
Common Oxidation Numbers
VIIIB
IVB
VB
VIB VIIB
Ti
V
Cr +2 Mn Fe
Co
+3 +4,+3 +5,+4 +6,+3 +7,+6 +3,+2 +3,+2
+2
+3+2 +2
+4,+3
IIIB
Sc
Y
IIB
Zn
+2 +2,+1 +2
Nb
Mo Tc
Ru
Rh
Pd
Ag
+5,+4 +6,+5 +7,+5 +8,+5
+3 +4,+3
+4,+3 +4,+2 +1
+2
+4,+3 +4
+4,+3
La
Zr
Ni
IB
Cu
Cd
+2
Hf
Ta
W
Re +2 Os Ir
Pt
Au
Hg
+5,+4 +6,+5 +7,+5 +8,+6 +4,+3
+3 +4,+3
+4,+2 +3,+1 +2,+1
+3
+4
+4
+4,+3 +1
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Oxidation Numbers
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
Ex 4: barium iodide
Ex 5: ammonium nitrate
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.
Oxidation Numbers
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. in a compound
must add up to zero, the O.N. of N is +4.
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
Ba = +2
I = -1
Ex 5: NH4NO3
H = +1
O = -2
N = -3
N = +5
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Mnemonic devices
“LEO says GER”
“OIL RIG”
Redox Examples
Fe2O3(s) +
+3 -2
iron is REDUCED
Mg(s) +
2Al(s) 
2Fe(s) +
0
0
aluminum is OXIDIZED
2HCl(aq) 
Al2O3(s)
+3 -2
nothing happens to oxygen
H2(g) +
MgCl2(aq)
+1 -1
+2 -1
0
0
hydrogen is REDUCED magnesium is OXIDIZED nothing happens to chlorine
Cu(s) + 2AgNO3(aq)
0
silver is REDUCED
+1

2Ag(s) + Cu(NO3)2(aq)
-1
copper is OXIDIZED
0
+2
-1
nothing happens to nitrate
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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)
First we assign an oxidation number (O.N.) to each atom (or ion). 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).
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!
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
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10/20/2011
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).
Activity
Series
Lab 3:
You use Mg instead of Zn
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A “Classic” Redox Example
Solid copper is immersed in
aqueous silver nitrate.
Blue color is formation of Cu2+
Ag is formed from solution of Ag+
net charge = +2
Balanced Net
Ionic eqn.:
Ox. #:
net charge = +2
Cu(s) + 2Ag+(aq)
0
+1
Cu2+(aq) + 2Ag(s)
+2
oxidation
0
reduction
At the particulate level…
Before rxn
occurs...
During rxn
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Writing Half-Reactions
We can separate an electron-transfer reaction between Cu(s)
and Ag+(aq) into two “half-reactions,” where the electron
loss or gain is explicitly stated.
The half-reaction idea provides us with a reliable method of
balancing redox reactions in aqueous solution.
Electrons just
2+
change
Cu(s)
Cu (aq) + 2e
places…
Ag+(aq) + eAg(s) x 2
charge is
+
conserved!!!
Cu(s) + 2Ag+(aq)
net charge = +2
Cu2+(aq) + 2Ag(s)
net charge =+2
Balancing by Half-Reactions
• Recall, a properly balanced redox reaction means
both mass and charge are conserved.
• Oftentimes the solvent (in this course, we will only
consider water) will be explicitly involved in the
redox reaction.
• For example, consider the following reaction:
CuS(s) + NO3-(aq)
Cu2+(aq) + SO42-(aq) + NO(g)
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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.
Balancing by Half-Reactions (cont.)
• Step 1: Identify and write down the unbalanced
half-reactions.
CuS(s) + NO3-(aq)
-2
Cu2+(aq) + SO42-(aq) + NO(g)
+5
+6
oxidation
+2
reduction
CuS(s)
Cu2+(aq) + SO42-(aq) oxidation
NO3-(aq)
NO(g) reduction
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Balancing by Half-Reactions (cont.)
• Step 2: Balance atoms (other than H and O).
• Step 3: Use H2O to balance O, and H+ to balance H
(assume acidic media).
• Step 4: Use e- to balance charge (won’t always be 0).
4H2O + CuS(s)
Cu2+(aq) + SO42-(aq) + 8H+ + 8e-
3e- + 4H+ + NO3-(aq)
NO(g) + 2H2O
Balancing by Half-Reactions (cont.)
• Step 5: Add the reactions together - multiply each
half-reaction by an integer such that the
number of electrons cancels.
4H2O + CuS(s)
Cu2+(aq) + SO42-(aq) + 8 H+ +8e- x 3
3e- + 4 H++ NO3-(aq)
8
3CuS +
8H++
8NO3-
NO(g) + 2H2O x 8
3Cu2+
4
+ 3SO42- + 8NO + 4H2O
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Balancing by Half-Reactions (cont.)
• Step 6: For reactions that occur in basic solution, first
balance in acid. At the end, add OH- to both
sides for every H+ present, neutralizing acid
present and leaving OH- on the other side.
3CuS + 8H++ 8 NO3+ 8OH-
3Cu2+ + 3SO42- + 8NO + 4H2O
+ 8OH-
3CuS + 8H2O+ 8 NO34
3Cu2+ + 3SO42- + 8NO + 4H2O
+ 8OH-
Example
Balance the following redox reaction occurring
in basic media:
BH4- + ClO3-
H2BO3- + Cl-
BH4-
H2BO3-
ClO3-
Cl-
-5
+5
+3
-1
oxidation
reduction
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Example (cont)
3H2O + BH4-
H2BO3- + 8H+ + 8e- x 3
6e- + 6H+ + ClO3-
Cl- + 3H2O x 4
3
3 BH4- +4 ClO3-
3H2BO3- +4 Cl- + 3H2O
Done!
Demo: Thermite Reaction
__ 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
half a football field; in another episode, they split a car in half
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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 transfers.
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. If you need to add O
and H, use the same process as in the half-reaction method:
add H2O for O and H+ for H
REDOX Balancing using Ox. No. Method
Balancing the Thermite Reaction
-2
-2
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)
+3
0
→ +3
2 x (- 3e- per Al)
0
2 x (+3e- per Fe)
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REDOX Balancing by Oxidation Number Method
+2
-1e- lost per Fe
+3
+1
-2
Fe+2(aq) + MnO4-(aq) + H+(aq)
+7
+1 -2
Fe+3(aq) + Mn+2(aq) + H2O(l)
+5 e- gained 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 Fe+2(aq) + MnO4-(aq) + H+(aq)
5 Fe+3(aq) + Mn+2(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 Fe+2(aq) + MnO4-(aq) +8 H+(aq)
5 Fe+3(aq) + Mn+2(aq) +4 H2O(l)
REDOX Balancing by Half-Reaction Method
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)
Mn+2(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 in the
water, add electrons to balance the charge.
5e- + MnO4-(aq) + 8H+(aq)
Mn+2(aq) + 4H2O(l)
Sum the two half-reactions (multiply Fe rxn by 5 to cancel electrons).
{ Fe+2(aq)
MnO4-(aq) + 8H+(aq) +5eMnO4-(aq)+ 8H+(aq)+5e- +5Fe+2(aq)
Fe+3(aq) +e- } x5
Mn+2(aq) + 4H2O(l)
5Fe+3(aq)+5e- + Mn+2(aq)+ 4H2O(l)
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