Ch4z5ereactions 110115225323-phpapp01

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Chapter 4
Aqueous solutions
Types of reactions pp
http://www.youtube.com/watch?v=h2UczzWu-Qs&mode=related&search=

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4.1 Water, the Common Solvent - Parts of Solutions
Solution - homogeneous mixture.
Solute - what gets dissolved.
Solvent - what does the dissolving.
Soluble - Can be dissolved.
Miscible - liquids dissolve in each other.

3
Aqueous solutions
Dissolved in water.
Water is a good solvent
because the molecules are
polar.
The oxygen atoms have a
partial negative charge.
The hydrogen atoms have a
partial positive charge.
The angle is 105º.

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Figure 4.1
The Water
Molecule is
Polar

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Hydration
The process of breaking the ions of
salts apart.
Ions have charges and attract the
opposite charges on the water
molecules.

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Hydration
H H
O
H
H
O
H
HO
H H
O
HH
O
HH
O
HH
O
H
H
O
H
H
O

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Figure 4.2
Polar Water Molecules Interact with the
Positive and Negative Ions of a Salt

8
Solubility
How much of a substance will dissolve
in a given amount of water.
Usually in g/100 mL
Solubility varies greatly, but if they do
dissolve the ions are separated,
and they can move around.
Water can also dissolve non-ionic
compounds if they have polar bonds.

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4.2 Nature of Aqueous Solutions
Strong and Weak Electrolytes
Electricity is moving charges.
The ions that are dissolved can move.
Solutions of ionic compounds can
conduct electricity.
Called “electrolytes”.
Solutions are classified three ways.

11
Types of solutions
Strong electrolytes- completely
dissociate (fall apart into ions).
Many ions- Conduct well.
Weak electrolytes- Partially fall apart
into ions.
Few ions -Conduct electricity slightly.
Non-electrolytes- Don’t fall apart.
No ions- Don’t conduct.

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Figure 4.5
BaCI2 Dissolving

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Types of solutions
Acids - form H+
ions when dissolved.
Strong acids fall apart completely.
Many ions
H2SO4 HNO3 HCl HBr HI HClO4
Weak acids- don’t dissociate
completely.
Bases - form OH-
ions when dissolved.
Strong bases - many ions.
KOH, NaOH, Ca(OH)2

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Figure 4.6
HCI (aq) is
Completely
Ionized

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Figure 4.7
An Aqueous
Solution of
Sodium
Hydroxide

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Figure
4.8
Acetic
Acid in
Water

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Figure 4.9
The
Reaction
of NH3 in
Water

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4.3 Composition of Solutions
Concentration - how much is dissolved.
Molarity = Moles of solute
Liters of solution
abbreviated M
1 M = 1 mol solute / 1 liter solution
Calculate the molarity of a solution with
34.6 g of NaCl dissolved in 125 mL of
solution. Steps . . .

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4.3 Composition of Solutions pp
Calculate the molarity of a solution with 34.6 g
of NaCl dissolved in 125 mL of solution.
Convert grams to moles. How? Convert mL
to L. How? Divide moles by liters
(34.6 g)(1 mol)
(58.5 g)(0.125 L)
= 4.73 M

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Molarity
How many grams of HCl (M = 36.5)
would be required to make 50.0 mL of a
2.7 M solution? Ans. . .
(0.050 L)(2.7 moles/L)(36.5 g/mole) =
4.9 g HCl

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Molarity
What would the concentration of CaCl2 be
if you used 27g of CaCl2 (M = 111) to
make 500. mL of solution? Ans. . .
(27 g)(1 mole/110.986)(1/0.500L) = 0.49 M
(about 0.50 molarity)
What is the approximate concentration of
each ion?
Ca2+
= 0.50 M; Cl1-
= 1.00 M

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Molarity
Calculate the concentration of a
solution made by dissolving 45.6 g of
Fe2(SO4)3 to a total volume of 475 mL.
(M of Fe2(SO4)3 = 400. g/mol) Ans. . .
(45.6 g)(1 mole/400. g)/.475 L = 0.240 M
What is the concentration of each ion?
Fe = 0.48 M; SO4 = 0.72 M

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Figure 4.10 Preparation of a Standard Solution
a.a. Weighed amount of solute is put into volumetric flaskWeighed amount of solute is put into volumetric flask
and small quantity of H2O addedand small quantity of H2O added
b.b. Solid dissolved by gently swirlingSolid dissolved by gently swirling with stopper inwith stopper in
placeplace
c.c. More water added until level just reaches mark onMore water added until level just reaches mark on
neck (d)neck (d)

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Making solutions
Describe how to make 100.0 mL of a 1.0 M
K2CrO4 solution.
(1.0 mole/L)(246 g/mole)(0.100 L) = dissolve
24.6 g in 85 ml distilled water; qsad to 100 ml

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Making solutions
Describe how to make 250. mL of an 2.0 M
copper(II) sulfate dihydrate solution.
What is formula for copper(II) sulfate dihydrate?
CuSO4•2H2O (M = 196)
What is the answer? . . .
(0.250L)(2.0 mole/L)(196 g/mole) = dissolve
98 g in 200 ml distilled water; qsad to 250 ml

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Dilution
Adding more solvent to a known solution.
The moles of solute stay the same.
moles = M x L
M1V1 = M2V2
moles = moles
Stock solution is a solution of known
concentration used to make more dilute
solutions

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Dilution
What volume of aWhat volume of a 1.71.7 MM solution is needed tosolution is needed to
make 250. mL of a 0.50make 250. mL of a 0.50 MM solution? Steps.solution? Steps.
VV11MM11 == VV22MM22
((xx)()(1.71.7) = () = (250.250.)()(0.500.50))
So, x = 74 mL, (rounded from 73.5 mL) of theSo, x = 74 mL, (rounded from 73.5 mL) of the
1.71.7 MM solution combined with enoughsolution combined with enough distilleddistilled
HH220 to make a0 to make a totaltotal volume of 250. mLvolume of 250. mL
Be careful! If you just add it to 250. mL then youBe careful! If you just add it to 250. mL then you
get a volume of 323 mL and this throws off yourget a volume of 323 mL and this throws off your
molarity.molarity.

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Dilution
18.5 mL of 2.3 M HCl is diluted to 250 mL
of water. What’s solution concentration?
Ans.
(18.5)(2.3) = 250(x) ; x = 0.170.17 MM
18.5 mL of 2.3 M HCl is added to 250 mL of
water. Solution concentration? (Tricky)
Find moles of HCl
(0.0185 L)(2.3 moles/L) = 0.0426 moles;
then M = 0.0426 moles/(0.250 + 0.0185 L)
= 0.160.16 MM (compare with above)

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Fig. 4.11 p. 145
(a)A Measuring Pipet is
graduated & can measure
various volumes accurately.
(b)A Volumetric (transfer) Pipet
measures one volume
accurately.

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More dilution (if time)
You have a 4.0 M stock solution. Describe
how to make 1.0L of a .75 M solution.
(1)(.75) = 4(x); x = 190 ml diluted with
distilled H20 to 1.0 L

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25 mL 0.67 M of H2SO4 added to 35 mL of
0.40 M CaCl2 . What mass CaSO4 is formed?
Steps. . .
1st get moles of both (moles = V • M)
0.025 L • 0.67 M = 0.017 mol H2SO4;
0.035 L • 0.40 M = 0.014 moles CaCl2
Then, need to find limiting reactant (2
stoichs)

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25 mL 0.67 M of H2SO4 added to 35 mL of
0.40 M CaCl2 . What mass CaSO4 formed?
0.01675 mol H2SO4 and 0.014 moles CaCl2
Then, need to find limiting reactant (2 stoichs)
H2SO4 + CaCl2 → CaSO4 + 2HCl
CaCl2 is LR and the mole ratio to CaSO4 = 1:1
(0.014 mol CaCl2)(1 mol CaSO4)(136 g CaSO4) = 1.9 g CaSO4
(1 mol CaCl2)(1 mol CaSO4)
IF this were NOT a ppt, then
conc = 0.014 moles/(0.025 + 0.035) = 0.23 M

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4.4 Types of Reactions
1 Precipitation reactions
When aqueous solutions of ionic
compounds are poured together a solid
forms.
A solid that forms from mixed solutions
is a precipitate
If you’re not a part of the solution, your
part of the precipitate

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4.5 Precipitation reactions
NaOH(aq) + FeClNaOH(aq) + FeCl33(aq)(aq) →→ NaCl(aq) + Fe(OH)NaCl(aq) + Fe(OH)33(s)(s)
is really . . .is really . . .
NaNa++
(aq) + OH(aq) + OH--
(aq) + Fe(aq) + Fe+3+3
(aq) + Cl(aq) + Cl--
(aq)(aq) →→ NaNa++
(aq) +(aq) +
ClCl--
(aq) + Fe(OH)(aq) + Fe(OH)33(s)(s)
So all that really happens is . . .So all that really happens is . . .
NaNa++
(aq) + OH(aq) + OH--
(aq) + Fe(aq) + Fe+3+3
(aq) + Cl(aq) + Cl--
(aq)(aq) →→ NaNa++
(aq) +(aq) +
ClCl--
(aq) + Fe(OH)(aq) + Fe(OH)33(s)(s)
OHOH--
(aq) + Fe(aq) + Fe+3+3
(aq)(aq) →→ Fe(OH)Fe(OH)33(s)(s)
Then, balance:Then, balance: 33OHOH--
(aq) + Fe(aq) + Fe+3+3
(aq)(aq) →→ Fe(OH)Fe(OH)33(s)(s)
Double replacement reactionDouble replacement reaction

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Precipitation reaction
We can predict the products
Can only be certain by experimenting
The anion and cation switch partners
AgNO3(aq) + KCl(aq) → (molecularly) . . . ?
AgCl + KNO3
Zn(NO3)2(aq) + BaCr2O7(aq) → ?
ZnCr2O7 + Ba(NO3)2
CdCl2(aq) + Na2S(aq) → ?
CdS + 2NaCl

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Precipitations Reactions
Only happen if one of the products is
insoluble
Otherwise all the ions stay in solution-
nothing has happened.
Need to memorize the rules for
solubility (see, p. 150 or use solubility
poem)

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Let’s Sing!
Potassium, sodium and ammonium salts,
Whatever they may be,
Can always be depended on for solubility.
When asked about the nitrates
The answer is always clear,
They each and all are soluble,
Is all we want to hear.
Most every chloride’s soluble
At least we’ve always read
Save silver, mercurous mercury
And (slightly) chloride of lead

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Let’s Sing!
Every single sulfate
Is soluble, ‘Tis said
‘Cept barium and strontium
And calcium and lead.
Hydroxides of metals won’t dissolve
That is, all but three
Potassium, sodium and ammonium
Dissolve quite readily

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Let’s Sing!
And then you must remember
That you must not “forgit”
Calcium, barium, strontium
Dissolve a little bit.
The carbonates are insoluble,
It’s lucky that it’s so,
Or else, our marble buildings
Would melt away like snow.

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Repeat with feeling!!
Potassium, sodium, and ammonium salts
Whatever they may be
Can always be depended on
For solubility

41
Solubility Rules
1 All nitrates are soluble
2 Alkali metals ions and NH4
+
ions are
soluble
3 Halides are soluble except Ag+
, Pb+2
,
and Hg2
+2
4 Most sulfates are soluble, except Pb+2
,
Ba+2
, Hg+2
,and Ca+2

42
Solubility Rules
5 Most hydroxides are slightly soluble
(insoluble) except NaOH and KOH
6 Sulfides, carbonates, chromates, and
phosphates are insoluble
∗ Lower number rules supersede so Na2S
is soluble

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4.6 Describing Reactions in Solution
Molecular Equation - written as whole
formulas, not the ions.
K2CrO4(aq) + Ba(NO3)2(aq) → 2ΚNO3(aq) + BaCrO4 (s)
Complete Ionic equation - show dissolved
electrolytes as the ions.
2K+
+ CrO4
-2
+ Ba+2
+ 2 NO3
-
→ BaCrO4(s)+ 2K+
+ 2 NO3
-
Net Ionic equation - includes only solution
components undergoing a change. Spectator
ions are those that don’t react.
CrO4
-2
+ Ba+2
→ BaCrO4(s)

44
Three Type of Equations
Write the three types of equations for
the reactions when these solutions are
mixed. (Whiteboard)
iron (III) sulfate and potassium sulfide
lead (II) nitrate and sulfuric acid.

45
4.7 Stoichiometry of Precipitationpp
Exactly the same, except you may have to
figure out what the pieces are.
What mass of solid is formed when 100.00
mL of 0.100 M Barium chloride is mixed with
100.00 mL of 0.100 M sodium hydroxide? . . .
Strategy: need to convert both reactants to
moles, then determine LR then use stoich.
Have 0.01 moles of both
BaCl2 + 2NaOH → Ba(OH)2 + 2NaCl

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BaCl2 continuedpp
What mass of solid is formed when 100.00
mL of 0.100 M Barium chloride is mixed with
100.00 mL of 0.100 M sodium hydroxide?
Have 0.010 moles of both
BaCl2 + 2NaOH → Ba(OH)2 + 2NaCl
NaOH is limiting & makes 0.005 mol Ba(OH)2
0.005 mol BaCl2 • 208 g/mol = 0.85 g formed

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Another Stoich of Precipitation problem
What volume of 0.204 M HCl is needed to
precipitate the silver from 50.ml of 0.0500 M
silver nitrate solution?
HCl + AgNO3 → AgCl + HNO3
50.ml of 0.0500 M AgNO3 = 0.0025 moles
Mole ratio = 1:1; need 0.0025 mol HCl also
Since M = mol/vol; vol = mol/M
Vol = 0.0025 ÷ 0.204 = 0.0122 L = 12.2 mL

48
Types of Reactions
4.8 Acid-Base
For our purposes an acid is a proton
donor.
A base is a proton acceptor (usually OH)
What is the net ionic equation for the
reaction of HCl(aq) and KOH(aq)?
Acid + Base → salt + water
H+
+ OH-
→ H2O

49
4.8 Acid - Base Reactions
Often called a neutralization reaction becauseOften called a neutralization reaction because
the acid neutralizes the base.the acid neutralizes the base.
OftenOften titratetitrate to determine concentrations.to determine concentrations.
Solution of known concentration (titrant) . . .Solution of known concentration (titrant) . . .
is added to the unknown (analyte) . . .is added to the unknown (analyte) . . .
until the equivalence point is reached whereuntil the equivalence point is reached where
enough titrant has been added to neutralize it.enough titrant has been added to neutralize it.

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Titrationpp
Where indicator changes color isWhere indicator changes color is endpointendpoint..
Not always at theNot always at the equivalenceequivalence point.point.
A 50.00 mL sample of aqueous Ca(OH)A 50.00 mL sample of aqueous Ca(OH)22
requires 34.66 mL of 0.0980requires 34.66 mL of 0.0980 MM Nitric acid forNitric acid for
neutralization. What is [Ca(OH)neutralization. What is [Ca(OH)22 ]?]?
# of H# of H++
xx MMAA x Vx VAA == # of OH# of OH--
xx MMBB x Vx VBB
(look at formulas, not at the balanced reaction)(look at formulas, not at the balanced reaction)
(1)(0.0980)(34.66) = (2)(50)(x) = 0.0340(1)(0.0980)(34.66) = (2)(50)(x) = 0.0340 MM
but look at Zumdahl method page 150!but look at Zumdahl method page 150!

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Acid-Base Reactionpp
75 mL of 0.2575 mL of 0.25 MM HCl is mixed with 225 mL of 0.055HCl is mixed with 225 mL of 0.055
MM Ba(OH)Ba(OH)22. What is the concentration of the. What is the concentration of the excessexcess
HH++
or OHor OH--
? We can do this in mmoles since the? We can do this in mmoles since the
units will cancel. Use mL and mmoles is same as Lunits will cancel. Use mL and mmoles is same as L
and moles. Steps. . .and moles. Steps. . .
Mmole HCl = 75 mL • 0.25Mmole HCl = 75 mL • 0.25 MM = 18.75 mmole;= 18.75 mmole;
Since 1:1 mole ratio of HSince 1:1 mole ratio of H++
: HCl; H: HCl; H1+1+
== 18.75 mmole18.75 mmole
Mmole Ba(OH)Mmole Ba(OH)22 = 225 mL • 0.055 M = 12.37 mmole= 225 mL • 0.055 M = 12.37 mmole
2:1 mole ratio OH2:1 mole ratio OH--
: Ba(OH): Ba(OH)22 ; OH; OH--
== 24.75 mmole24.75 mmole
H : OH mole ratio = 1:1; so 6.00 mmoles xs OH inH : OH mole ratio = 1:1; so 6.00 mmoles xs OH in
(75 + 225) ml = 0.0200(75 + 225) ml = 0.0200 MM OHOH1-1-

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Types of Reaction
4.9 Oxidation-Reduction called Redox
Ionic compounds are formed through
the transfer of electrons.
An Oxidation-reduction reaction
involves the transfer of electrons.
We need a way of keeping track.

53
Oxidation States - Memorize!
A way of keeping track of the electrons.
Not necessarily true of what is in nature,
but it works (i.e., a model only).
Need the rules for assigning (memorize).

54
Oxidation States - Memorize!
Elements & Ions
1The oxidation state of elements in their
standard states is zero.
Na(s) = 0; Cl2(g) = 0, but O2(s) ≠ 0!
2Oxidation states for monoatomic ions are
the same as their charge.
Na1+
= 1; Cl1-
= 1-
; N3-
= 3-

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Oxidation states - memorize!
•Oxygen & Hydrogen
3Oxygen is assigned an oxidation state of
-2 in its covalent compounds (H2O)
except as peroxide (H2O2) where = -1
or with fluorine (OF2) where = +2
4Hydrogen in compounds with:
nonmetals is assigned +1 (H2O)
with metals it is -1 (CaH2)

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Oxidation states - memorize!
5fluorine is always -1 in compounds (HF).
6The sum of the oxidation states must be:
zero in compounds (Na2SO4) or
equal the charge of the ion (SO4
2-
)

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Tr103 Assigning Oxidation Numbers
What is the oxidation # of Mn in
KMnO4?
1•1 + 1•Mn + 4•-2 = 0; Mn = +7
What is the oxidation # of Cr in
BaCr2O7?
1•2 + 2•Cr + 7•-2 = 0; Cr = +6
What is the oxidation # of Cl in
Sr(ClO4)2?
1•2 + 2*Cl + 8•-2 = 0; Cl = +7!
Even though as an ion is would
be Cl1-

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Assign the oxidation states to each element
CO2
NO3
-1
H2SO4
Fe2O3
Fe3O4
C = +4; O = -2
N = +5; O = -2
H = +1; S = +6; O = -2
Fe = +3; O = -2
Fe = +2.67; O = -2
Remember: This isn’t real.
It’s only a model to help
us.

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Oxidation-Reduction
Electrons transfer, so oxidation states change.
2Na + Cl2 → 2NaCl
CH4 + 2O2 → CO2 + 2H2O
Oxidation is the loss of electrons (an increase
in the oxidation state).
Reduction is the gain of electrons (a decrease
in the oxidation state).
OIL RIG
LEO GER
Coefficients or subscripts do NOT change the
oxidation values.

60
Oxidation-Reduction
Oxidation means an increase in oxidation
state - lose electrons.
Na → Na1+
+ 1 e-
Reduction means a decrease in oxidation
state - gain electrons.
2 e-
+ Cl2 → 2Cl1-
The substance that is oxidized is called
the reducing agent.
The substance that is reduced is called
the oxidizing agent.

61
A Summary
of an
Oxidation-
Reduction
Process

62
Agents
Oxidizing agent - has the potential to
cause another substance to get oxidized.
Other substance loses electrons, but . . .
In causing that other substance to get
reduced the oxidizing agent gets reduced.
The oxidizing agent gains electrons (gets
a more negative oxidation state).
MnO4
1-
will oxidize Fe2+
to Fe3+
so here it is
an oxidizing agent.
But the Mn in MnO4
1-
gets reduced from +7
to +2 (MnO4
1-
+ 5e-
→ Mn2+
)
+7 +2

63
Agents
Reducing agent - has the potential to
cause another substance to get reduced.
Other substance gains electrons, but . . .
In causing that other substance to get
reduced the reducing agent gets oxidized.
The reducing agent loses electrons (gets
a more positive oxidation state).
Na will reduce Zn2+
to Zn so here it is a
reducing agent.
But the Na gets oxidized from 0 to +1
(Na → Na1+
+ 1e-
)

64
Identify the
Substance oxidized
Substance reduced
Oxidizing agent
Reducing agent
in the following reactions:
Fe(s) + O2(g) → Fe2O3(s) oxidation #s?…
0 0 +3 -2 O, R, OA, RA?
O R
RA OA

65
Identify the
O, R, OA, RA in the following reactions:
Fe2O3(s)+ 3 CO(g) → 2 Fe(s) + 3 CO2(g)
Fe3+
reduced to Feo
so Fe2O3 is the OA C2+
oxidized to C4+
so CO is the RA.
SO3
2-
+ H+
+ MnO4
1-
→ SO4
2-
+ H2O + Mn+2
Sulfur goes from +4 to +6 (from SO3
2-
to SO4
2-
)
so it is oxidized and its compound, SO3
2-
is the
reducing agent which reduces Mn in MnO4
1-
from +7 to +2.
Mn is reduced & MnO4
1-
is the Oxidizing
agent (it oxidizes S from +4 to +6).

66
Half-Reactions
All redox reactions can be thought of as
happening in two halves.
One produces electrons - Oxidation half.
Other requires electrons - Reduction half.
We write the half-reactions to help us
balance the equation.
Then we add the half-reactions to cancel
out the electrons.
Every redox reaction has an oxidation
half-reaction & a reduction half-reaction.

67
Half-Reactions
Write the half reactions for the following.
Na + Cl2 → Na+
+ Cl-
Steps. . .
Na → Na1+
+ 1e-
2e- + Cl2 → 2Cl1-

68
Half-Reactions (more steps)
Write the half reactions for: SO3
-
2
+ H+
+ MnO4
-
→ SO4
-2
+ H2O + Mn+2
(steps …)
1st, identify the elements that are changing their
oxidation numbers, they are . . .
S from +4 to +6 so it is losing 2 e- (it is being
oxidized so it is the reducing agent)
Mn from +7 to +2 so it is gaining 5 e- (being
reduced so it is the oxidizing agent).
SO3
-2
+ H+
+ MnO4
-
→ SO4
-2
+ H2O + Mn+2
+4
+7 +6 +2
So, the half reactions are the stuff with sulfur
and manganese.

69
Half-Reactions (more steps)
SO3
-2
+ H+
+ MnO4
-
→ SO4
-2
+ H2O + Mn+2
+4
+7 +6 +7 S
from +4 to +6 (losing) and . . . Mn
from +7 to +2 (gaining)
The half-reactions are . . .
SO3
-2
→ SO4
-2
+ 2 e-
+4 +6
5 e- + MnO4
-
→ Mn+2
+7 +2

70
Not every reaction is redox
Only reactions that change oxidation state (gain
or lose electrons) are redox.
Is SO2 + H2O → H2SO3 a redox?
+4 -2 +1 -2 +1+4-2 No!
Is H2 + CuO → Cu + H2O a redox?
0 +2 -2 0 +1-2 Yes!
(Discuss Rainbow Matrix)

71
4.10 Balancing Redox Equations
In aqueous solutions the key is the
number of electrons produced must be
the same as those required.
For reactions in acidic solution: 8-step
procedure.
1Write separate half-reactions (assign
oxidation numbers, temporarily delete
substances that don’t change)
2For each half reaction balance all
reactants except H and O
3Balance O using H2O

72
Acidic Solution
4Balance H using H+
5Balance charge using e-
6Multiply equations to make electrons
equal
7Add equations and cancel identical
species (cancel coefficients as needed)
8Check that charges and elements are
balanced.

73
Examplepp
A breathalyzer uses potassium
dichromate to test for ethanol because
the orange potassium dichromate
changes to the green chromium 3+
ion in
the presence of alcohol.
Write and balance the Breathalyzer
reaction given the following:
The reactants are K2Cr2O7, HCl and
C2H5OH.
The products are CrCl3, CO2, KCl & H2O.

74
Examplepp
Write and balance the Breathalyzer equation
given the following:
Reactants: K2Cr2O7, HCl and C2H5OH. Products:
CrCl3, CO2, KCl & H2O.
Write the formula equation:
K2Cr2O7 + HCl + C2H5OH → CrCl3 + CO2 + KCl + H2O
Write the ionic equation (solubility poem):
2K1+
Cr2O7
2-
+ H1+
+ Cl1-
+ C2H5OH → Cr3+
3Cl1-
+ CO2 + K1+
+ Cl1-
+ H2O
Assign oxidation numbers
2K1+
Cr2O7
2-
+ H1+
+ Cl1-
+ C2H5OH → Cr3+
3Cl1-
+ CO2 + K1+
+ Cl1-
+ H2O +1
+6 -2 +1 -1 -2+1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2

75
Examplepp
Write the formula equation: K2Cr2O7 + HCl +
C2H5OH → CrCl3 + CO2 + KCl + H2O Write the ionic
equation (solubility poem): 2K1+
Cr2O7
2-
+ H1+
+ Cl1-
+
C2H5OH → Cr3+
3Cl1-
+ CO2 + K1+
+ Cl1-
+ H2O Assign oxidation
numbers 2K1+
Cr2O7
2-
+ H1+
+ Cl1-
+ C2H5OH → Cr3+
+ 3Cl1-
+ CO2 +
K1+
+ Cl1-
+ H2O +1 +6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1
-1 +1 -2
Delete substances where no element changes its
oxidation number:
2K1+
Cr2O7
2-
+ H1+
Cl1-
+ C2H5OH → Cr3+
+ 3Cl1-
+ CO2 + K1+
+ Cl1-
+ H2O +1
+6 -2 +1 -1 -2 +1 -2+1 +3 -1 +4 -2 +1 -1 +1 -2
Only those where oxidation number changes :
Cr2O7
2-
+ C2H5OH → Cr3+
+ CO2 +6 -2
+3 +4

76
Balancing Redox Equationspp
Only those where oxidation number changes :
Cr2O7
2-
+ C2H5OH → Cr3+
+ CO2 +6
-2 +3 +4
For reactions in acidic solution: 8-step
procedure.
1Write separate half-reactions
Cr2O7
2-
→ Cr3+
C2H5OH → CO2

77
Cr2O7
2-
→ Cr3+ -
C2H5OH
→ CO2
2For each half-reaction balance all
reactants except H and O
Cr2O7
2-
→ 2Cr3+
C2H5OH
→ 2CO2

78
Cr2O7
2-
→ 2Cr3+ -
C2H5OH
→ 2CO2
3Balance O using H2O
Cr2O7
2-
→ 2Cr3+
+ 7H2O
3H2O + C2H5OH → 2CO2
4 Balance H using H1+
14H1+
+ Cr2O7
2-
→ 2Cr3+
+ 7H2O 3H2O +
C2H5OH → 2CO2 + 12H1+

79
Examplepp
14H1+
+ Cr2O7
2-
→ 2Cr3+
+ 7H2O 3H2O +
C2H5OH → 2CO2 + 12H1+
5Balance the charge using e-
6e-
+ 14H1+
+ Cr2O7
2-
→ 2Cr3+
+ 7H2O 3H2O +
C2H5OH → 2CO2 + 12H1+
+ 12e-
7 Multiply equations to equalize the charges
2(6e-
+ 14H1+
+ Cr2O7
2-
→ 2Cr3+
+ 7H2O) 3H2O +
C2H5OH → 2CO2 + 12H1+
+ 12e-
Get:
12e-
+ 28H1+
+ 2Cr2O7
2-
→ 4Cr3+
+ 14H2O
3H2O + C2H5OH → 2CO2 + 12H1+
+ 12e-

80
Examplepp
12e-
+ 28H1+
+ 2Cr2O7
2-
→ 4Cr3+
+ 14H2O 3H2O +
C2H5OH → 2CO2 + 12H1+
+ 12e-
7Add equations and cancel identical
species (cancel coefficients if needed)
12e-
+ 1628H1+
+ 2Cr2O7
2-
→ 4Cr3+
+ 1114H2O 3H2O
+ C2H5OH → 2CO2 + 12H1+
+ 12e-
• 16H1+
+ 2Cr2O7
2-
+ C2H5OH → 2CO2 + 4Cr3+
+ 11H2O

81
Acidic Solutionpp
16H1+
+ 2Cr2O7
2-
+ C2H5OH → 2CO2 + 4Cr3+
+ 11H2O
8Combine the net ions to form the original
compounds and check that everything is
balanced.
The original compounds were:
So the balanced equation is . . .(write it
out)
2K2Cr2O7 + 16HCl + C2H5OH → 4CrCl3 + 2CO2 + 4KCl + 11H2O

82
Balance the following in acidic solution
MnO4
1-
+ Fe+2
→ Mn+2
+ Fe+3
Ans. . .
8H1+
+ MnO4
-
+ 5Fe+2
→ Mn+2
+ 5Fe+3
+ 4H2O
Cu + NO3
1-
→ Cu+2
+ NO(g) Ans. . .
3Cu + 2NO3
1-
+ 8H1+
→ 3Cu+2
+ 2NO(g) + 4H2O

83
More Practice if time
The following reactions occur in aqueous
solution. Balance them
Pb + PbO2 + SO4
-2
→ PbSO4 (not done)
Mn+2
+ NaBiO3 → Bi+3
+ MnO4
-
(not done)

84
Now for a tough one
Fe(CN)6
-4
+ MnO4
-
→ Mn+2
+ Fe+3 +
CO2 + NO3
-
. . .
188 H+
+ 5 Fe(CN)6
4-
+ 61 MnO4
1-
→ 94 H2O + 61 Mn2+
+
5 Fe3+
+ 30 CO2 + 30 NO3
1-

85
Basic Solutionpp
Do everything you would with acid, but
we have to add a step because . . .
Because there is no H1+
in basic solution.
So, add OH1-
sufficient to convert the H1+
to
water (OH1-
+ H1+
→ H2O)
Be sure to add the OH1-
to both sides of
the reaction (conservation of mass law)
Cr(OH)3 + OCl-
+ OH-
→ CrO4
-2
+ Cl-
+ H2O
we’ll do this on the board to get . . .
2Cr(OH)3 + 3OCl-
+ 4OH-
→ 2CrO4
-2
+ 3Cl-
+ 5H2O

86
Basic Solution
After balancing like it was in acidic solution, add
OH1-
(to both sides) to cancel all the H1+
in basic
solution.
You try CrO2
1-
+ ClO1-
→ CrO4
1-
+ Cl1-
in basic
solution to get . . .
CrO2
1-
+ 2ClO1-
→ CrO4
1-
+ 2Cl1-
Try NO2
1-
+ Al → NH3 + AlO2
1-
in basic solution to
get . . .
OH1-
+ H2O + NO2
1-
+ 2Al → NH3 + 2AlO2
1-

87
Basic Solution - more problems if time
Do everything you would with acid, but
add one more step.
Add enough OH-
to both sides to
neutralize the H+
(by forming water)
CrI3 + Cl2 → CrO4
-
+ IO4
-
+ Cl-
Fe(OH)2 + H2O2 → Fe(OH)-

88
Redox Titrations
Same as any other titration.
The permanganate ion is used often
because it is its own indicator. MnO4
-
is
purple, Mn+2
is colorless. When reaction
solution remains clear, MnO4
-
is gone.
(Know for AP lab portion)
Chromate ion (CrO4
1-
) is also useful, but
color change, orangish yellow to green,
is harder to detect.

89
Redox Titration Examplepp
The iron content of iron ore can be
determined by titration with standard
KMnO4 solution. The iron ore is dissolved
in excess HCl, and the iron reduced to Fe+2
ions. This solution is then titrated with
KMnO4 solution, producing Fe+3
and Mn+2
ions in acidic solution.
If it requires 41.95 mL of 0.205 M KMnO4
to titrate a solution made with 6.128 g of
iron ore, what percent of the ore was iron?
Steps. . .

90
Examplepp
Iron ore + xs HCl → iron reduced to Fe+2
. Then titrated w/ KMnO4
solution → Fe+3
+ Mn+2
ions in acidic solution.
41.95 mL of 0.205 M KMnO4 titrates 6.128 g of iron ore, what percent is
iron?
Write NR Fe2+
reacting with MnO4
1-
8H1+
+ 5Fe2+
+ MnO4
1-
→ 5Fe3+
+ Mn2+
+ 4H20
Do stoich (calc moles MnO4
1-
1st) to get
0.0086 moles MnO4
1-
and 0.043 moles Fe2+
Find g of Fe2+
= 0.043 mol x 55.847 g/mol = 2.40 g
(2.40 g Fe/6.128 g Fe ore) x 100% = 39.2%

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