Download 3.Redox

SOLUTION REACTIONS
I. Solutions
A. Definitions
1. Solution = homogeneous mixture of two or more components.
2. Solvent = component present to largest extent. Phase of solution is same as
the phase of the solvent.
a. In some cases the solvent is not a pure substance . Mixed solvents are
used in many reactions .
b. In a few cases a minor component is chosen as the solvent for convenience.
For example, concentrated sulfuric acid is 98% H2SO4 and 2% water by
mass but water is still considered the “solvent”.
c. When the solvent is water the solute is said to be in an aqueous solution
3. Solute = minor component of the mixture.
a. The solute is dispersed throughout the solvent.
b. Concentration = a number giving the amount of solute dissolved in a given
amount solution or dispersed in a given amount of solvent.
B. Types of Solutes. Can classify solutes in terms of the properties of their solutions.
1. Electrolyte = substance that dissolve in water to yield a conducting solution.
Dissolves to produce ions.
2. Strong electrolytes = 100% ionized in water
a. Ionic compounds
NaCl, KNO3, MgSO4
b. Strong acids. Common strong acids are:
HCl, HI, HBr, HNO3,HClO4,H2SO4
HCl
H3O+ + Cl-
+ H2O
H3O+ = hydronium ion
3. Weak electrolytes = not 100% ionized in water.
a. Weak acids Example HC2H3O2
HC2H3O2
H3O+
+ H2O
+ C2H3O2–
b. Weak ammonia-type bases
NH3 = ammonia
NH3 +
H2O
NH4+ + OH4. Nonelectrolytes = solutes that do not ionize in water.
a. Exist in solution as neutral molecules
b. Examples: Sugar (C6H12O6), Ethanol (CH3CH2OH)
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B. Solubility - All ionic compounds are strong electrolytes in that they are 100%
ionized, however some are not particularly soluble in water. Qualitatively
the solubility of most common salts adhere to the following rules.
1. Most all Group 1 and NH4+ are soluble irrespective of the anion.
2. All nitrates (NO3- ), perchlorates (ClO4- ) , chlorates (ClO3- ) , and acetates
(C2H3O2- ) are soluble irrespective of the cation.
3. Except for the cations listed in Rule 1, all carbonates (CO32- ), phosphates (PO43-),
sulfides (S2- ), and sulfites (SO32- ) are insoluble.
4. Most hydroxides (OH- ) are insoluble except when the cation is a heavier Group 2
or one of those covered in Rule 1
5. Most chlorides (Cl- ), bromides (Br- ), and iodides (I- ) are soluble except when
the cation is Ag+, Hg22+, and Pb2+.
6. Most sulfates (SO42- ) are soluble except for the heavier Group 2 (Ba2+, Ca2+) ,
Pb2+, Hg2+, and Hg22+, with Ag2SO4 being moderately soluble.
II. Solution reactions.
A. Net Ionic Reactions.
1. Consider the following double displacement reaction run in aqueous solution.
NaCl(aq) + AgNO3(aq)
AgCl(s)
+ NaNO3(aq)
The subscript (aq) means dissolved in water. Since NaCl, AgNO3 and
NaNO3 are strong electrolytes, the individual ions are the species that exist
in solution. The reaction is better written as
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq)
AgCl(s) + Na+(aq) + NO3-(aq)
From the above equation, it is apparent that neither the Na+ nor the NO3- ions
are actively involved in the reaction. By convention these spectator ions are not
written in the net ionic reaction, as shown below
2
Ag+ + Cl-
AgCl(s)
Also the subscripts (aq) are many times deleted in writing the net reaction.
IN WRITING NET REACTIONS ONLY THOSE IONS AND MOLECULES
DIRECTLY INVOLVED IN THE REACTION ARE SHOWN.
2. Acid/base reactions.
a. Acid = any substance that dissociates in water to give H3O+ (hydronium ions)
1) Covalent compounds that react with H2O as follows:
H3O+ + X–
HX + H2O
2) Acidic protons are written first.
Monoprotic acids (one acidic proton) HCl, HNO3
Diprotic acids (two acidic protons) H2SO4, H2SO3
Triprotic acids (three acidic protons) H3PO4
b. Base = substance that dissociates in water to give OH– ions.
Ionic compounds in which the anion is OH– All are strong electrolytes but many
are not very soluble
c. Neutralization = reaction of acid and base to give a salt and water.
Acid
+ Base
Salt
HCl(aq) + KOH(aq)
+ Water
KCl(aq) + H2O(l)
Neutralization reactions also are usually written in the net ionic form.
net Ionic equation:
H3O+
+
OH-
2H2O
B. Oxidation-reduction ( REDOX) reactions.
1. Definitions.
a. Oxidation = loss of electrons.
b. Reduction = gain of electrons.
c. Oxidizing agent = reactant that causes oxidation by being reduced.
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( gains electrons)
d. Reducing agent = reactant that causes reduction by being oxidized.
( loses electrons)
2. Example
2Fe + 3Cl2
a. Fe is oxidized
b. Cl2 is reduced
2FeCl3
Half reaction: 2(Fe -------> Fe3+ + 3e- )
Half reaction: 3(Cl2 + 2e- -----> 2Cl-)
2 Fe + 3 Cl2
2FeCl3
c. Cl2 is the oxidizing agent and Fe is the reducing agent.
d. Note that the number of electrons is conserved, that is, the total number of
of electrons lost by the Fe is equal to the total number gained by the Cl2.
3. Oxidation numbers = number assigned to atoms in substances using a set of rules
that are useful in tracking electron changes.
a. The oxidation numbers (O.N.) of atoms of an element in its elemental form are
equal to zero.
Example : Fe (O.N. of Fe = 0); Cl2 (O.N. of Cl = 0)
b. The oxidation number of simple monatomic ions = charge on ion.
Example: Fe3+ ( O.N. of Fe = +3)
Cl- ( O.N. of Cl = -1)
c. In covalent compounds, the oxidation number of oxygen is -2 except in its
elemental form, in peroxides ( X-O-O-X is peroxide linkage, O22- is the peroxide
ion O.N. of O = -1), superoxides ( O2- is the superoxide ion, O.N. of O = -1/2)
and in binary compounds with fluorine ( where O.N. F is -1).
d. The oxidation number of hydrogen is +1 except in its elemental form and in
hydrides of Groups 1 and 2 metals and Al (where H = –1).
e. The sum of the oxidation numbers of all atoms in a formula = charge on formula.
1) Sum for neutral molecules = 0
2) Sum for complex ions = charge on ion.
4. Examples:
a. CO
b. CO2
c.. C2O42-
O = -2 , C = +2
O = -2 , C = +4
O = -2 , C = +3
d. CH2O
e. CO32f. Fe3O4
III. Balancing Redox reactions.
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O = -2, H = +1, C = 0
O = -2, C = +4
O = -2, Fe = +8/3
A. Using Oxidation numbers.
1. Loss of electrons ( oxidation ) = increase in O.N. ( becomes more positive)
Gain of electrons ( reduction ) = decrease in O.N. ( becomes more negative )
2. Since the number of electrons is conserved, O.N. is also conserved
Examples:
__3x(_+ 2_) = +6______
+4 -2
+2 -2
+1 -2
2NO2 + 3CO
|
2x (- 3 ) = -6
0
+4 -2
N2O + 3CO2
|
+1 -2
+1 -1
Cl2 + H2O
+
1 +5 -2
HCl + HClO3
Note: Cl2 is being both oxidized and reduced. Since it is doing two things, write it
down twice and treat as two separate reactants
_____5x( -2 ) = -10______________
0
0
+1 -2
+1 -1
+1 +5 -2
5Cl2 + Cl2 + 6H2O
10HCl + 2HClO3
|______1x( + 10 ) = +10____________|
Net 3Cl2 + 3H2O
5HCl + HClO3
8( - 7) = - 56
+6
4 H2SO4
+7
+ 8 KClO4
+1
+ 7N2O
+ 3 H2O
+5
4 Cl2 + 14 HNO3
+6
+ 4 K2SO4
7(+8) = + 56
Note that those coefficients in bold were obtained from oxidation number changes, the
others by mass balance.
The oxidation number changes are in terms of reactant
molecules, that is that in N2O two N’s are oxidized from +1 to +5, for a total increase of
+8 for each N2O that reacts.
B. Ion-electron method ( Half reaction method)
1. Consider the following aqueous solution reaction
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KMnO4(aq) + KNO2(aq) + H2SO4(aq)
MnSO4(aq) + KNO3(aq) + K2SO4(aq) + H2O(l)
Net ionic Reaction : (K+ and SO42– are spectator ions)
MnO4- + NO2- + H+
Mn2+ + NO3- + H2O
a. Write the half reactions
MnO4-
Mn2+
NO2-
NO3-
b. Balance each half reaction by imposing both a mass and a charge balance.
1) Mass balance = equal number of atoms of the same element of both sides of the equation.
MnO4- +8H+
Mn2+ + 4H2O
Note: When O is lost, H2O is formed. Get the needed H's as H+
NO2- + H2O
NO3- + 2H+
Note: When O is needed, obtain from H2O. Will have H+ 's as products
2) Charge balance --- add electrons to more positive side to equalize the ion charges
MnO4- + 8H+
[
+7
]
Mn2+ + 4H2O
[
+2
]
must add 5e- to reactant side to equalize charge
MnO4- + 8H+ + 5eNO2- +
[
-1
H2O
]
Mn2+ + 4H2O
NO3- + 2H+
[
+1
]
must add 2e- to product side to equalize charge
NO2- +
H2O
NO3- + 2H+ + 2e-
c. Add two half reactions so that electrons cancel
2x( MnO4- + 8H+ + 5e–
Mn2+ + 4H2O)
5x(NO2- + H2O
NO3- + 2H+ + 2e-)
__________________________________________
2MnO4- + 5NO2- + 6H+
2Mn2+
___
+ 5NO3 + 3H2O
NOTE: MnO4- is the oxidizing agent
NO2- is the reducing agent
2. Balance the following assuming an acidic aqueous solution
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-
Cr2O72- + SO32Cr3+ + SO42a. balanced half reactions
Cr2O72- + 14H+ + 6e
2Cr3+ + 7H2O
SO32- + H2O
SO42- + 2H+ + 2eb. add and cancel electrons
Cr2O72- + 14H+ + 6e-
2Cr3+ + 7H2O
3x(SO32- + H2O
SO42- + 2H+ + 2e-)
_______________________________________________
Cr2O72- + 3SO32- + 8H+
2Cr3+ + 3SO42-
+ 4H2O
3. Basic solutions
a. In acidic solutions oxygen balance was obtained using a H+/H2O couple.
In basic solutions protons are not available so a H2O/OH- couple is used, that is,
oxygens are supplied by ( or go to form) OH- and H2O supplies the needed H+'s.
b. Balance the following redox reaction
MnO4- + SO32- + H2O
1) MnO4- half reaction
MnO4- + 2H2O
MnO2(s) + SO42- + OHMnO2(s) + 4OH-
Note : If lose O's form OH- and obtain H+ 's from H2O.
The net result is that two OH-'s are formed for every O lost.
Charge balance in same way as in acidic solutions
MnO4- + 2H2O + 3eMnO2(s) + 4OH2) SO32- half reaction
SO32- + 2OH-
SO42- + H2O + 2eNote: If need O's obtain from OH-'s with H2O as a product.
The net result is that two OH-'s are used for every O needed
3) Net reaction
2(MnO4- + 2H2O + 3e-
MnO2(s) + 4OH-)
3(SO32- + 2OH-
SO42- + H2O + 2e-)
_____________________________________________
2MnO4- + 3SO32- + H2O
2MnO2(s)
7
+
_____________
3SO42- + 2OH-
IV. Concentration.
A. Concentration Units
1. Molarity (M) = moles of solute per liter of solution = mmol solute per mL of solution
a. A solution was prepared by dissolving 10.0 g of NaOH (FM = 40.0) in
enough water to give 350 mL of solution. Calculate the Molarity of the solution .
10.0 g
mol of NaOH = 40 g/mol = 0.250 mol
0.250 mol
Molarity = 0.350 L = 0.714 M ( or 0.714 molar)
In terms of mmols: mmol of NaOH = 250 mmol, and mL of solution = 350 mL
Molarity = 250 mmol / 350 mL = 0.714 M
b. What volume of a 6.0M stock solution of HCl is required to prepare
700 mL of a 0.83 M HCl solution?
mmoles of HCl required = (0.83 mmol/mL)x(700 mL) = 581 mmol
581 mmol
mL of 6.0 M HCl required to obtain 581 mmol = 6.0 mmol/mL = 96.8 mL
Note: Let Mf = final Molarity = 0.83M
Vf = final volume = 700mL
Mi = initial Molarity = 6.0M
Vi = initial volume = ?
Then for dilutions (Mf)( Vf) = (Mi)(Vi)
Mf
in the above Vi = Vf M = 700 mL (0.83M / 6.0M) = 96.8 mL
i
2. Mole fraction (X)
a. The mole fraction of a component , i , in a solution , Xi =
ni
n total
where ni = moles of i
ntotal = total number of moles of all components of the solution.
b. Note that solvent and solute are not used in this expression of concentration.
Since the sum of all mole fractions must equal unity, the following holds
!X
i
= 1.000
i
A solution prepared by dissolving 10.0 g of K2Cr2O7 (FM = 294) in 90.0 g of
water has a density of 1.075 g/mL. Calculate
1. the mole fraction of K2Cr2O7 X K 2 Cr2 O 7 =
8
# moles K 2 Cr2 O7
Total moles in Solution
10.0 g
moles of K2Cr2O7 = 294 g/mol = 0.0340 mol
90.0 g
moles of H2O = 18.0 g/mol = 5.00 mol
0.0340
X = 5.00 + 0.034 = .00675 = 6.75x10-3
# moles os solute
# Liters of solution
total mass of solution = 10.0 g + 90.0g = 100.0 g
100.0 g
volume of solution = 1.075 g/mL = 93.5 mL = 0.0935 L
0.0340 mol
Molarity = 0.935 L = 0.364 molar
2. the Molarity of the solution M =
3. The number of mL of this solution that would be required to react with 8.30 g of
Na2SO3 (FM = 126) according to the equation:
Cr2O72- + 3SO32- + 8H+
2Cr3+ + 3SO42- + 4H2O
8.30 g
= 0.0659 mol
mol SO32– = mol Na2SO3 =
126 g/mol
! 1 Cr2 O 2–
7 $
2–
2– #
& = 0.0219 mol of Cr2O72–
mol Cr2O7 required = (0.0659 mol SO3 )
2–
" 3 SO 3 %
or 21.9 mmol of Cr2O72–
∴ mL of 0.364 M K2Cr2O7 =
21.9 mmol
= 60.3 mL
0.364 mmol/mL
B. Titration.
1. Titration = the controlled addition, from a buret, of the solution of one reactant
(titrant) to a solution of another reactant until an “end point” is reached.
a. At the end point, one has mixed stoichiometric amounts of the two reactants.
b. The end point is signaled by using an indicator. This is a substance that will
react with one of the reactants, usually the titrant, to produce a change in color.
When the solution changes color, the end point has been reached.
2. The titration reaction can be any of the solution reactions discussed in the Unit-- that
is, precipitation, neutralization and redox.
For Example in the problem in B.4. above, the reaction could be done by
dissolving the 8.30 g of Na2SO3 in water, adding some acid, a few drops of an
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indicator and titrating with the 0.364 M K2Cr2O7 solution. The end point would
occur after 60.3 mL of the K2Cr2O7 solution had been added.
3. The Neutralization reactions.
a. Neutralization = reaction of an acid with a base to give a salt and water.
Examples
Acid + Base
----> Salt + water
HCl + NaOH -----> NaCl + H2O
H2SO4 + 2NaOH -----> Na2SO4 + 2 H2O
2 HCl + Ba(OH)2 ----> BaCl2 + 2 H2O
b. Note that ordinarily all the acidic protons and OH–’s are assumed to react.
c. Numerical example: How many mL of a 0.30 M NaOH solution would be
required to neutralize 35.0 mL of a 0.20 M H2SO4 solution.
Note the balanced equation is: H2SO4 + 2NaOH -----> Na2SO4 + 2 H2O
Working in mmoles, mmol of H2SO4 = (0.20 mmol/mL)(35.0 mL) = 7.00 mmol
2 NaOH
= 14.00 mmol NaOH
mmol NaOH needed = 7.00 mmol H 2 SO 4
1 H2 SO 4
14.00 mmol
= 46.7 mL
mL NaOH needed =
0.30 mmol/mL
4. Standardization = Process by which the concentration of a solution in determined
by titration with a solution of known concentration (a standard solution) or a very
pure substance that has been accurately weighed out (a primary standard).
a. Example. A 10.00 mL sample of an HCl solution required 31.4 mL of a
0.250 M solution of KOH for neutralization. Calculate the concentration of the HCl.
Equation:
HCl + KOH -----> KCl + H2O
mmol of KOH used = (0.250 mmol/mL)(31.4 mL) = 7.85 mmol
Since the reaction ratio of HCl-to-KOH is one-to-one,
mmol of HCl present = mmol KOH = 7.85 mmol
7.85 mmol
M=
= 0.785M
10.00 mL
We say that the HCl solution has been standardized.
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Problems
1. Balance the following ionic redox equations using the half-reaction method. Show both
balanced half reactions as well as the over-all balanced equations. Acidic or basic
solutions are indicated for each reaction.
-
a. Cu + NO3 ----> Cu2+ + NO2
b. Cu +
c. Zn +
d. Zn +
-
e. MnO4
NO3
NO3
NO3
(acidic)
---> Cu2+ + NO
(acidic)
---> Zn2+ + N2
(acidic)
---> Zn2+ + N2O
(acidic)
+ H2S ---> Mn2+ + S(s)
(acidic)
f. MnO4 + S2- ---> MnO2(s) + S(s)
g. MnO4 + NO2 ---> MnO2(s) + NO3
2h. SO4 + Br- ---> SO2 + Br2
2i. SO4 + I- ---> H2S + I2
j. NO3 + I2 ---> IO3 + NO2
k. CuS + NO3 ---> Cu2+ + S(s) + NO
l. H2O2 + MnO4 ---> Mn2+ + O2
2m. C2O4 + MnO4 ---> MnO2(s) + CO2
n. Al + H+ ---> Al3+ H2
2o. Br2 + SO2 ---> Br- + SO4
p. ClO3 + I2 ---> IO3 + Clq. HClO3 ---> ClO4 + ClO2
r. Cl2 ---> ClO3 + Cl2s. MnO4 + Cl2 ---> MnO4 + Cl2t. MnO4 ---> MnO4 + MnO2(s)
u. NO2 ---> NO3 + NO2
2v. C + NO3 ---> NO2 + CO3
(acidic)
(acidic)
(acidic)
(acidic)
(acidic)
(acidic)
(acidic)
(basic)
(acidic)
(basic)
(basic)
(basic)
(basic)
(basic)
(acidic)
(basic)
(basic)
2. Balance the following redox reactions using oxidation number changes.
a. N2H2 + KMnO4 + H2SO4 ---> NO + MnSO4 + K2SO4 + H2O
b. Cl2 + H2O ---> HClO3 + HCl
c. Cu + HNO3 ---> NO + Cu(NO3)2 + H2O
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d. As2O3 + HClO4 + H2O ---> H3AsO4 + HClO
e. Br2 + HClO3 + H2O ---> HBrO + HCl
f. I2O5 + CO ---> I2 + CO2
g. HI + HNO3 ---> I2 + NO + H2O
h. Ag + H2SO4 ---> Ag2SO4 + SO2 + H2O
-
3. Consider the reaction NO3
+ S2- ---> NO + SO2 that takes place in an acidic solution.
a. How many grams of Na2S would be required to react with 8.00 g of Ca(NO3)2?
b. How many grams of Na2S would be required to react with 25 mL of a 2M Ca(NO3)2 solution?
4. To standardize an acidified KMnO4 solution, 0.9812 g of FeSO4 is weighed out, dissolved in
water, and titrated by the addition of the KMnO4 solution. To reach the end point, 25.61 mL of
KMnO4 solution is required. Write the balanced equation for this reaction assuming the the
products are Fe3+ and Mn2+. What is the molarity of the KMnO4 solution?
5. A 0.3500 g sample of an iron ore was dissolved in acid and all the iron was reduced to
Fe2+ . All the iron was then reoxidized to Fe3+ by titration with a 0.0950 M KMnO4
solution. If it required 5.50 mL of the KMnO4 to reach the end point in the titration,
what was the percent iron in the ore sample?
6. Iodate ion will oxidize sulfite to sulfate in acidic solution. Suppose that a 1.390 g
sample
of KIO3 was required to react with 23.21 mL of 0.700 M Na2SO3. What is
the
oxidation number of iodine after the reaction has occurred?
7. Consider the following redox equation (unbalanced)
2-
Cr2O7
2-
+ S2- ---> Cr3+ + SO4
a. Calculate the Molarity of a solution prepared by dissolving 0.624 g of Na2S in
enough water to give 250 mL of solution.
b How many mL of this solution will be required to react with 0.512 g of
Al2(Cr2O7)3?
c. Suppose that 24.86 mL of this Na2S solution was required to react with 0.377 g
of a dichromate salt. Which one of the following could be this salt?
Fe2(Cr2O7)3
CaCr2O7
PbCr2O7
BaCr2O7
Bi2(Cr2O7)3.
d. How many mL of the Na2S solution would be required to react with 50.00 mL of
0.300 M K2Cr2O7 solution?
8. Consider the following redox reaction (unbalanced)
-
ClO3 + Br- ---> Br2 + ClHow many mL of a 0.350 M Fe(ClO3)3 solution would be required to react with 2.00 g of AlBr3?
9. At 20° C the solubility of KCl in water is 34.0 g KCl per 100 g water. The density of
the
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resulting saturated solution is 1.17 g / mL. Calculate (a) the mole fraction of KCl in
the solution, (b) the Molarity of the solution.
10. Suppose a 8.75 M aqueous CH3OH solution has a density of 0.789 g / mL. Calculate
the mole fraction of CH3OH in the solution..
11. A solution prepared by dissolving 6.0 g of NaCl in enough water to give 500 mL of
solution has a density of 1.05 g / mL. Calculate (a) the Molarity of the solution, (b) the
mole fraction of NaCl in the solution.
12. Dilute HNO3 can oxidize Cu to give NO and Cu2+. How many mL of a 0.25 M HNO3
solution would be required to react with 6.00 g of Cu? How many mL of a 6.0M HNO3
stock solution would be required to prepare this solution?
13. How many mL of a 0.532 M H2SO4 solution would be required to titrate 62.0 mL of
a 0.784 M KOH solution.
14. A 35.0 g sample of a mixture of KCl and NaCl required 199 mL of a 2.54 M AgNO3
solution to precipitate all the chloride as AgCl. Calculate the grams of KCl in the mixture.
15. Aluminum reacts with HCl to give aluminum chloride and hydrogen . Suppose that 0.500 g
of Al is placed in 200 mL of 0.30 M HCl and hydrogen is generated. What reactant will be
in excess and how many moles of this reactant will remain? What volume of hydrogen could
be collected over water at 27° C and 755 Torr atmospheric pressure in this reaction? ( vapor
pressure of H2O at 27° C = 27 Torr.)
16. Calculate the molar concentration of He in a mixture of 0.250 g of He and 20.0 g of CH4
when measured at 250° C and a total pressure of 1.13 atm. Calculate the partial pressure of
He in the mixture.
17. Potassium permanganate, KMnO4, reacts with Na2C2O4 in acidic solution to give
Mn2+ and CO2. How many mL of a 0.513 M KMnO4 solution would be required to
react with 357 mL of a 0.135 M Na2C2O4 solution?
18. Write net ionic equations for the following. (The equations need not be balanced.)
a. ZnSO4(aq) + KOH(aq) -----> K2SO4(aq) + ZnO(s) + H2O(l)
b. KMnO4(aq) + KCl(aq) + H2SO4(aq) -----> MnSO4(aq) + Cl2(g) + K2SO4(aq) + H2O(l)
c. H3AsO4(aq) + KOH(aq) -----> K3AsO4(aq) + H2O(l)
d. Cu(s) + KNO3(aq) + H2SO4(aq) -----> CuSO4(aq) + K2SO4(aq) + NO(g)
e. Ca(NO3)2(aq) + K3PO4(aq) -----> Ca3(PO4)2(s) + KNO3(aq)
f. Na2SO3(aq) + Na2Cr2O7(aq) + HClO4(aq) ---> Cr(ClO4)3(aq) + Na2SO4(aq) + NaClO4(aq)
g. Al (s) + HCl(aq) -----> AlCl3(aq) + H2(g)
h. CO(g) + HNO3(aq) -----> CO2(g) + NO2(g) + H2O(l)
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i. BaO(s) + HClO4(aq) -----> Ba(ClO4)2(aq) + H2O(l)
j. Na2S(aq) + FeCl3(aq) ----> Fe2S3(s) + NaCl(aq)
19. In each of the following reactions identify the substances, if any, that are oxidized
or reduced. Give the oxidation numbers of the atoms in each reaction.
a. N2O + O2
b. Al + I2
-----> N2O4
-----> AlI3
c. P2O5 + NaOH
d. CuCl + Cl2
-----> Na3PO4 + H2O
-----> CuCl2
e. CuSO4 + Zn
-----> ZnSO4 + Cu
20. Perchlorate ions react with sulfide ions in acidic solution to give sulfate ions and
gaseous chlorine.
a. Balance the equation for this reaction using the half reaction method.
b. How many mL of 0.150 M sodium perchlorate would be required to react with
1.25 g of potassium sulfide?
c. Calculate how many grams of calcium perchlorate would be required to react with 2.50 g of
potassium sufide.
d. How many mL of 0.261 M potassium sulfide would be required to react with 32.6 mL of
0.424 M calcium perchlorate ?
21. Consider a 0.250 M solution of KOH.
a. How many grams of KOH would have to be dissolved in a total volume of 300 mL
to make 0.250 M KOH?
b. How many mL of a 18.0 M KOH stock solution would have to be diluted to make
1200 mL of 0.250 M KOH?
c. How many moles of KOH would be present in 452 mL of 0.250 M KOH?
d. How many mmoles of KOH would be present in 62.1 mL of 0.250 M KOH?
e. How many mL of 0.150 M HCl would be required to neutralize 26.2 mL of
0.250 M KOH?
14
f. How many mL of 0.150 M H2SO4 would be required to neutralize 26.2 mL of
0.250 M KOH?
g. Suppose that 41.2 mL of 0.250 M KOH was required to neutralize 1.162 g of a
solid monoprotic acid. Calculate the molar mass of the acid.
22. Potassium acid phthalate is a monoprotic acid with a molar mass of 204.2 g. If in a
titration 24.62 mL of a solution of NaOH is required to neutralize 2.9462 g of
potassium acid phthalate, calculate the Molarity of the NaOH.
15
SOLUTION REACTIONS
(Answers)
Problems
1. Balance the following ionic redox equations using the half-reaction method. Show both
balanced half reactions as well as the over-all balanced equations. Acidic or basic
solutions are indicated for each reaction.
-
a) Cu + 4 H+ + 2 NO3 ----> Cu2+ + 2 NO2 + 2 H2O
(acidic)
-
b) 3 Cu + 8 H+ + 2 NO3 ---> 3 Cu2+ + 2 NO + 4 H2O
c) 5 Zn +
d) 4 Zn +
-
e) 2 MnO4
-
f) 2 MnO4
g) 2 MnO4
-2
h) SO4 +
2i) SO4 +
j) 10 NO3
12 H+ + 2 NO3
10 H+ + 2 NO3
---> 5 Zn2+ + N2 + 6 H2O
(acidic)
---> 4 Zn2+ + N2O + 5 H2O
(acidic)
+ 6 H+ + 5 H2S ---> 2 Mn2+ + 5 S(s) + 8 H2O
(acidic)
+ 8 H+ + 3 S2- ---> 2 MnO2(s) + 3 S(s) + 4 H2O
(acidic)
+ 2 H+ +
3 NO2
3 NO3
---> 2 MnO2(s) +
+ H2 O
(acidic)
10 H+ + 8 I- ---> H2S + 4 I2 + 4 H2O
(acidic)
-
+ 8 H+ + I2 ---> 2 IO3
-
l) 5 H2O2 + 6 H+ +
+ 10 NO2 + 4 H2O
---> 3 Cu2+ + 3 S(s) + 2 NO+ 4 H2O
2 MnO4
2-
---> 2 Mn2+ + 5 O2 + 8 H2O
-
m) 3 C2O4 + 4 H2O + 2 MnO4 ---> 2 MnO2(s) + 6 CO2 + 8 OHn) 2 Al + 6 H+ ---> 2 Al3+ + 3 H2
2-
o) Br2 + 4 OH- + SO2 ---> 2 Br- + SO4
-
-
+ 6 OH- + 3 I2 ---> 6 IO3
-
r) 6 OH- + 3 Cl2 --->
2-
2-
t) 3 MnO4
ClO3
+ 5 Cl- + 3 H2O
-
-
+ 4 H+ ---> 2 MnO4
-
v) C + H2O +
4 NO3
+ 5 Cl- + 3 H2O
+ ClO2 + 2 H2O
+ Cl2 ---> 2 MnO4
u) 2 NO2 + 2OH- ---> NO3
+ 2 H2 O
-
q) 2 HClO3 + 2 OH- ---> ClO4
s) 2 MnO4
(acidic)
4 H+ + 2 Br- ---> SO2 + Br2 + 2 H2O
k) 3 CuS + 8 H+ + 2 NO3
p) 5 ClO3
(acidic)
---> 4 NO2 +
16
+ H2 O
2CO3
(acidic)
(basic)
(acidic)
(basic)
(basic)
(basic)
(basic)
+ MnO2(s) + 2 H2O
-
(acidic)
(basic)
+ 2 Cl-
+ NO2
(acidic)
+
(acidic)
(basic)
2 OH-
(basic)
2. Balance the following redox reactions using oxidation number changes.
a) 5 N2H2 + 6 KMnO4 + 9 H2SO4 ---> 10 NO + 6 MnSO4 + 3 K2SO4 +14 H2O
b) 3 Cl2 + 3 H2O -----> HClO3 + 5 HCl
c) 3 Cu + 8 HNO3
-----> 2 NO + 3 Cu(NO3)2 + 4 H2O
d) 3 As2O3 + 2 HClO4 + 9 H2O -----> 6 H3AsO4 + 2 HClO
e) 3 Br2 + HClO3 + 3 H2O -----> 6 HBrO + HCl
f) I2O5 + 5 CO -----> I2 + 5 CO2
g) 6 HI + 2 HNO3
-----> 3 I2 + 2 NO + 4 H2O
h) 2 Ag + 2 H2SO4
-----> Ag2SO4 + SO2 + 2 H2O
3. a. 3.81g b. 3.90
4. MnO4– + 5Fe2+ + 8H+ -----> Mn2+ + 5Fe3+ + 4H2O
0.05046M
5. 41.7%
6. zero
7. a. 0.032 M b. 51.3 c. Bi2(Cr2O7)3 d. 351 mL
+
–
–
Cl– + 3 Br2 + 3 H2O
8. Balanced: 6 H + ClO + 6 Br
3
3.57 mL
9. a. 0.076
b. 3.98 M
10. 0.236
11. a. 0.205 M
12. a. 0.252
b. 3.54x10–3
b. 10.5
13. 45.7
14. 18.9
15 HCl excess, 4x10–3mol 0.714 L O2
16. 1.25x10–3
17. 37.6 mL
18. Write net ionic equations for the following. (The equations need not be balanced.)
a. ZnSO4(aq) + KOH(aq) -----> K2SO4(aq) + ZnO(s) + H2O
-----> ZnO + H2O
Zn2+ + OHb. KMnO4(aq) + KCl(aq) + H2SO4(aq) -----> MnSO4(aq) + Cl2(g) + K2SO4
MnO4- + Cl- + H+ -----> Mn2+ + Cl2
c. H3AsO4(aq) + KOH(aq) -----> K3AsO4(aq) + H2O(l)
H+ + OH- ----> H2O
d. Cu(s) + KNO3(aq) + H2SO4(aq) ----> CuSO4(aq) + K2SO4(aq) + NO(g)
17
Cu + NO3- + H+ -----> Cu2+ + NO
e. Ca(NO3)2 + K3PO4(aq) -----> Ca3(PO4)2 (s) + KNO3(aq)
Ca2+ + PO43- ------> Ca3(PO4)2(s)
f. Na2SO3(aq) + Na2Cr2O7(aq) + HClO4(aq) ---->
Cr(ClO4)3(aq) + Na2SO4(aq) + NaClO4(aq)
SO32- + Cr2O72- + H+ ------> Cr3+ + SO42g. Al (s) + HCl(aq) -----> AlCl3(aq) + H2(g)
Al + H+ -----> Al3+ + H2
h. CO(g) + HNO3(aq) -----> CO2(g) + NO2(g) + H2O(l)
CO + NO3- + H+ -----> CO2 + NO2 + H2O
i. BaO(s) + HClO4(aq) -----> Ba(ClO4)2(aq) + H2O(l)
BaO + H+ ----> Ba2+ + H2O
j. Na2S(aq) + FeCl3(aq) ----> Fe2S3(s) + NaCl(aq)
S2- + Fe3+ -----> Fe2S3
19.
In each of the following reactions identify the substances, if any, that are oxidized or reduced.
Give the oxidation numbers of the atoms in each reaction.( oxidation nos. written over atoms)
+1 -2
0
+4 -2
a. 2N2O + 3O2 -----> 2N2O4
0
0
N oxidized
+3 -1
b. 2Al + 3I2 -----> 2AlI3
+5 -2
+1 -2 +1
0
Al oxidized
+1 +5 -2
c. P2O5 + 6NaOH ----->
+1 -1
0
e. CuSO4 + Zn
I2 reduced
+1 -2
2Na3PO4 + 3H2O
not redox
+2 -1
d. 2CuCl + Cl2 -----> 2CuCl2
+2 +6 -2
O reduced
+2 +6 -2
Cu+ oxidized
Cl2 reduced
0
-----> ZnSO4 + Cu
20. a ( 2 ClO4- + 16 H+ + 14 e- -----> Cl2 +
Zn oxidized
Cu2+ reduced
8 H2O ) X4
( S2- + 4 H2O ------> SO42- + 8 H+ + 8 e- ) X7
________________________________________________________
8 ClO4- + 7 S2- + 8 H+ ------> 4 Cl2 7 SO42- + 4 H2O
b. 86.3 mL c. 3.10 g d. 92.7 mL
21..a. 4.21 g b. 16.7 mL c. 0.113 mol d. 15.5 mmol e. 43.7 mL f. 21.8 mL
g. 112.8
22. 0.5860 M
23. 0.145 M
18