Download Bis2A 06.Appendix A review of Red/Ox reactions

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1
Bis2A 06.Appendix A review of
∗
Red/Ox reactions
Mitch Singer
Based on Classifying Chemical Reactions† by
OpenStax College
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 4.0‡
Abstract
By the end of this section, you will be able to:
• Dene three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction)
• Classify chemical reactions as one of these three types given appropriate descriptions or chemical
equations
• Identify common acids and bases
• Predict the solubility of common inorganic compounds by using solubility rules
• Compute the oxidation states for elements in compounds
1 This module is a review and supplemental information
This module is meant as a review to oxidation-reduction reactions, balancing red/ox equations and calculating
red/ox states of atoms and simple molecules. In Bis2A, you will not need to balance equations or determine
(calculate) the red/ox state of an atom or molecule. However, given a pair of compounds you will have to
decide which one is the reduced form and which one is the oxidized form. This module may be of use and
is provided as an appendix of sorts to the modules in group 6: Energy.
Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their sts or feet, we say they are ghting. Faced with a wide
range of varied interactions between chemical substances, scientists have likewise found it convenient (or even
necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will
provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base,
and oxidation-reduction.
∗ Version 1.2: Jun 19, 2015 11:26 am +0000
† http://cnx.org/content/m51022/1.7/
‡ http://creativecommons.org/licenses/by/4.0/
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2 Oxidation-Reduction Reactions
Earth's atmosphere contains about 20% molecular oxygen, O2 , a chemically reactive gas that plays an
essential role in the metabolism of aerobic organisms and in many environmental processes that shape the
world. The term
oxidation was originally used to describe chemical reactions involving O2 , but its meaning
has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions.
A few examples of such reactions will be used to develop a clear picture of this classication.
Some redox reactions involve the transfer of electrons between reactant species to yield ionic products,
such as the reaction between sodium and chlorine to yield sodium chloride:
2Na(s)
+ Cl2 (g ) [U+27F6] 2NaCl(s)
(1)
It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of
each reactant in the form of an equation called a
half-reaction:
2Na(s) [U+27F6] 2Na+ (s) + 2e−
Cl2 (g )
(2)
+ 2e− [U+27F6] 2Cl− (s)
These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons, the
s subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For
redox reactions of this sort, the loss and gain of electrons dene the complementary processes that occur:
oxidation
=
loss of electrons
reduction
=
gain of electrons
(3)
reduction. Viewed from a more active
reducing agent (reductant), since it provides electrons to (or reduces)
Likewise, chlorine functions as an oxidizing agent (oxidant), as it eectively removes electrons
In this reaction, then, sodium is oxidized and chlorine undergoes
perspective, sodium functions as a
chlorine.
from (oxidizes) sodium.
reducing agent
=
species that is oxidized
oxidizing agent
=
species that is reduced
(4)
Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction
similar to the one yielding NaCl:
H2 (g )
+ Cl2 (g ) [U+27F6] 2HCl(g )
(5)
The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not
involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous denition
of redox reactions, a property called oxidation number has been dened. The oxidation number (or
oxidation state) of an element in a compound is the charge its atoms would possess if the compound was
ionic. The following guidelines are used to assign oxidation numbers to each element in a molecule or ion.
1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion's charge.
3. Oxidation numbers for common nonmetals are usually assigned as follows:
ˆ
ˆ
ˆ
Hydrogen: +1 when combined with nonmetals, 1 when combined with metals
Oxygen: 2 in most compounds, sometimes 1 (so-called peroxides, O2
called superoxides, O2
−
2− ), very rarely
− 21
(so-
), positive values when combined with F (values vary)
Halogens: 1 for F always, 1 for other halogens except when combined with oxygen or other
halogens (positive oxidation numbers in these cases, varying values)
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4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the
molecule or ion.
Note: The proper convention for reporting charge is to write the number rst, followed by the sign (e.g.,
2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This
convention aims to emphasize the distinction between these two related properties.
Example 1
Assigning Oxidation Numbers
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in
the following species:
(a) H2 S
(b) SO3
2−
(c) Na2 SO4
Solution
(a) According to guideline 1, the oxidation number for H is +1.
Using this oxidation number and the compound's formula, guideline 4 may then be used to
calculate the oxidation number for sulfur:
charge on H2 S
= 0 = (2 × + 1) + (1 × x)
x = 0 − (2 × + 1) = −2
(6)
(b) Guideline 3 suggests the oxidation number for oxygen is 2.
Using this oxidation number and the ion's formula, guideline 4 may then be used to calculate
the oxidation number for sulfur:
charge on
x
=
−2
2
= −2 = (3 × − 1) + (1 × x)
(
3
SO3
-
×
−2
)
=
(7)
+4
(c) For ionic compounds, it's convenient to assign oxidation numbers for the cation and anion
separately.
According to guideline 2, the oxidation number for sodium is +1.
Assuming the usual oxidation number for oxygen (2 per guideline 3), the oxidation number for
sulfur is calculated as directed by guideline 4:
charge on
x
=
−2
2−
= −2 = (4 × − 2) + (1 × x)
(
4
SO4
-
×
−2
)
=
+6
Check Your Learning
Assign oxidation states to the elements whose atoms are underlined in each of the following
compounds or ions:
(a) KNO3
(b) AlH3
+
Ψ
−
(d) H2 PO4
(c) N H4
_
note:
(a) N, +5; (b) Al, +3; (c) N, 3; (d) P, +5
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Using the oxidation number concept, an all-inclusive denition of redox reaction has been established.
Oxidation-reduction (redox) reactions
are those in which one or more elements involved undergo a
change in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist Example 2.) Denitions for
the complementary processes of this reaction class are correspondingly revised as shown here:
oxidation
=
increase in oxidation number
reduction
=
decrease in oxidation number
(9)
Returning to the reactions used to introduce this topic, they may now both be identied as redox processes.
In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation
number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from
0 in Cl2 to 1 in NaCl).
In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized
(its oxidation number increases from 0 in H2 to +1 in HCl) and chlorine is reduced (its oxidation number
decreases from 0 in Cl2 to 1 in HCl).
Several subclasses of redox reactions are recognized, including
combustion reactions
in which the
reductant (also called a fuel ) and oxidant (often, but not necessarily, molecular oxygen) react vigorously
and produce signicant amounts of heat, and often light, in the form of a ame. Solid rocket-fuel reactions
such as the one depicted in here
1 are combustion processes. A typical propellant reaction in which solid
aluminum is oxidized by ammonium perchlorate is represented by this equation:
10Al(s) + 6NH4 ClO4 (s) [U+27F6] 4Al2 O3 (s) + 2AlCl3 (s) + 12H2 O(g ) + 3N2 (g )
Watch a brief video
note:
(10)
2 showing the test ring of a small-scale, prototype,
hybrid rocket engine planned for use in the new Space Launch System being developed by NASA.
The rst engines ring at 3 s (green ame) use a liquid fuel/oxidant mixture, and the second, more
powerful engines ring at 4 s (yellow ame) use a solid mixture.
Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced
(or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the
acid oxidation of certain metals:
Zn(s)
+ 2HCl(aq ) [U+27F6] ZnCl2 (aq ) + H2 (g )
(11)
Metallic elements may also be oxidized by solutions of other metal salts; for example:
Cu(s)
+ 2AgNO3 (aq ) [U+27F6] Cu (NO3 )2 (aq ) + 2Ag(s)
(12)
This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver
ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu
dissolve in the solution to yield a characteristic blue color (Figure 1).
1 "Introduction", Figure 1 <http://cnx.org/content/m51020/latest/#CNX_Chem_04_00_Rocket>
2 http://openstaxcollege.org/l/16hybridrocket
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Figure 1: (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of
dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire
and development of a blue color in the solution, due to dissolved copper ions. (credit: modication of
work by Mark Ott)
Example 2
Describing Redox Reactions
Identify which equations represent redox reactions, providing a name for the reaction if appropriate.
For those reactions identied as redox, name the oxidant and reductant.
(a) ZnCO3 (s) [U+27F6] ZnO(s)
+ CO2 (g )
2Ga(l) + 3Br2 (l) [U+27F6] 2GaBr3 (s)
(c) 2H2 O2 (aq ) [U+27F6] 2H2 O(l) + O2 (g )
(d) BaCl2 (aq ) + K2 SO4 (aq ) [U+27F6] BaSO4 (s) + 2KCl(aq )
(e) C2 H4 (g ) + 3O2 (g ) [U+27F6] 2CO2 (g ) + 2H2 O(l)
(b)
Solution
Redox reactions are identied per denition if one or more elements undergo a change in oxidation
number.
(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
(b) This is a redox reaction.
Gallium is oxidized, its oxidation number increasing from 0 in
Ga(l ) to +3 in GaBr3 (s ). The reducing agent is Ga(l ). Bromine is reduced, its oxidation number
decreasing from 0 in Br2 (l ) to 1 in GaBr3 (s ). The oxidizing agent is Br2 (l ).
(c) This is a redox reaction.
It is a particularly interesting process, as it involves the same
element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reac-
tion). Oxygen is oxidized, its oxidation number increasing from 1 in H2 O2 (aq) to 0 in O2 (g ).
Oxygen is also reduced, its oxidation number decreasing from 1 in H2 O2 (aq ) to 2 in H2 O(l ). For
disproportionation reactions, the same substance functions as an oxidant and a reductant.
(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing
from 2 in C2 H4 (g ) to +4 in CO2 (g ). The reducing agent (fuel) is C2 H4 (g ). Oxygen is reduced,
its oxidation number decreasing from 0 in O2 (g ) to 2 in H2 O(l ). The oxidizing agent is O2 (g ).
Check Your Learning
This equation describes the production of tin(II) chloride:
Sn(s)
+ 2HCl(g ) [U+27F6] SnCl2 (s) + H2 (g )
Is this a redox reaction? If so, provide a more specic name for the reaction if appropriate, and
identify the oxidant and reductant.
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Yes, a single-replacement reaction. Sn(s )is the reductant, HCl(g ) is the oxidant.
2.1 Balancing Redox Reactions via the Half-Reaction Method
Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions
as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical
change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing
these reactions are sometimes very dicult to balance by inspection, so systematic approaches have been
developed to assist in the process. One very useful approach is to use the method of half-reactions, which
involves the following steps:
1. Write the two half-reactions representing the redox process.
2. Balance all elements except oxygen and hydrogen.
3. Balance oxygen atoms by adding H2 O molecules.
+ ions.
4. Balance hydrogen atoms by adding H
3
5. Balance charge by adding electrons.
6. If necessary, multiply each half-reaction's coecients by the smallest possible integers to yield equal
numbers of electrons in each.
7. Add the balanced half-reactions together and simplify by removing species that appear on both sides
of the equation.
8. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps:
ions to both sides of the equation in numbers equal to the number of H+ ions.
+ and OH ions, combine these ions to yield water
(b) On the side of the equation containing both H
(a) Add OH
molecules.
(c) Simplify the equation by removing any redundant water molecules.
4 are balanced.
9. Finally, check to see that both the number of atoms and the total charges
Example 3
Balancing Redox Reactions in Acidic Solution
Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III)
and chromium(III) in acidic solution.
Cr2 O7
2−
+ Fe2+ [U+27F6] Cr3+ + Fe3+
(14)
Solution
Step 1. Write the two half-reactions.
Each half-reaction will contain one reactant and one product with one element in common.
2+
Fe
Cr2 O7
3 The
[U+27F6] Fe3+
2−
[U+27F6] Cr3+
(15)
(16)
requirement of charge balance is just a specic type of mass balance in which the species in question are electrons.
An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must
be balanced.
4 The
requirement of charge balance is just a specic type of mass balance in which the species in question are electrons.
An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must
be balanced.
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Step 2. Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced,
but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right.
Changing the coecient on the right side of the equation to 2 achieves balance with regard
to Cr atoms.
2+
Fe
Cr2 O7
[U+27F6] Fe3+
2−
(17)
[U+27F6] 2Cr3+
(18)
Step 3. Balance oxygen atoms by adding H2 O molecules. The iron half-reaction does not contain O
atoms. The chromium half-reaction shows seven O atoms on the left and none on the right,
so seven water molecules are added to the right side.
2+
Fe
[U+27F6] Fe3+
(19)
Cr2 O7
[U+27F6] 2Cr3+ + 7H2 O
2−
+
Step 4. Balance hydrogen atoms by adding H ions. The iron half-reaction does not contain H atoms.
The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14
hydrogen ions are added to the left side.
2+
Fe
Cr2 O7
[U+27F6] Fe3+
(20)
[U+27F6] 2Cr3+ + 7H2 O
2−
(21)
Step 5. Balance charge by adding electrons. The iron half-reaction shows a total charge of 2+ on the
left side (1 Fe
2+ ion) and 3+ on the right side (1 Fe3+ ion). Adding one electron to the right
side bring that side's total charge to (3+) + (1) = 2+, and charge balance is achieved.
The chromium half-reaction shows a total charge of (1
2−
+
× 2−)
+ (14
×
1+) = 12+ on the left
side (1 Cr2 O7
ion and 14 H
ions). The total charge on the right side is (2 × 3+) = 6 +
3+ ions). Adding six electrons to the left side will bring that side's total charge to (12+
(2 Cr
+ 6−) = 6+, and charge balance is achieved.
2+
Fe
Cr2 O7
2−
[U+27F6] Fe3+ + e−
+ 14H+ + 6e− [U+27F6] 2Cr3+ + 7H2 O
(22)
(23)
Step 6. Multiply the two half-reactions so the number of electrons in one reaction equals the number
of electrons in the other reaction. To be consistent with mass conservation, and the idea
that redox reactions involve the transfer (not creation or destruction) of electrons, the iron
half-reaction's coecient must be multiplied by 6.
2+
6Fe
Cr2 O7
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2−
[U+27F6] 6Fe3+ + 6e−
+ 6e− + 14H+ [U+27F6] 2Cr3+ + 7H2 O
(24)
(25)
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Step 7. Add the balanced half-reactions and cancel species that appear on both sides of the equation.
6Fe2+ + Cr2 O7 2− + 6e− + 14H+ [U+27F6] 6Fe3+ + 6e− + 2Cr3+ + 7H2 O
(26)
Only the six electrons are redundant species. Removing them from each side of the equation
yields the simplied, balanced equation here:
6Fe2+ + Cr2 O7 2− + 14H+ [U+27F6] 6Fe3+ + 2Cr3+ + 7H2 O
(27)
A nal check of atom and charge balance conrms the equation is balanced.
Reactants
Products
Fe
6
6
Cr
2
2
O
7
7
H
14
14
charge
24+
24+
Table 1
Check Your Learning
2+ to produce Fe3+ and H O. Write a balanced
2
In acidic solution, hydrogen peroxide reacts with Fe
equation for this reaction.
note:
H2 O2 (aq )
+ 2H+ (aq ) + 2Fe2+ [U+27F6] 2H2 O(l) + 2Fe3+
3 Key Concepts and Summary
Chemical reactions are classied according to similar patterns of behavior.
A large number of important
reactions are included in three categories: precipitation, acid-base, and oxidation-reduction (redox). Precipitation reactions involve the formation of one or more insoluble products.
Acid-base reactions involve
the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for
one or more reactant elements. Writing balanced equations for some redox reactions that occur in aqueous
solutions is simplied by using a systematic approach called the half-reaction method.
4 Chemistry End of Chapter Exercises
Exercise 1
(Solution on p. 13.)
Use the following equations to answer the next ve questions:
i. H2 O(s)
[U+27F6] H2 O(l)
+ Cl− (aq ) Ag+ (aq ) + NO3 − (aq ) [U+27F6] AgCl(s) + Na+ (aq ) + NO3 − (aq )
iii. CH3 OH(g ) + O2 (g ) [U+27F6] CO2 (g ) + H2 O(g )
iv. 2H2 O(l) [U+27F6] 2H2 (g ) + O2 (g )
+
−
v. H (aq ) + OH (aq ) [U+27F6] H2 O(l)
+
ii. Na (aq )
(a) Which equation describes a physical change?
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(b) Which equation identies the reactants and products of a combustion reaction?
(c) Which equation is not balanced?
(d) Which is a net ionic equation?
Exercise 2
(Solution on p. 13.)
Indicate what type, or types, of reaction each of the following represents:
(a) Ca(s)
+ Br2 (l) [U+27F6] CaBr2 (s)
(b) Ca (OH)2 (aq ) + 2HBr(aq ) [U+27F6] CaBr2 (aq ) +
(c) C6 H12 (l) + 9O2 (g ) [U+27F6] 6CO2 (g ) + 6H2 O(g )
2H2 O(l)
Exercise 3
(Solution on p. 13.)
Indicate what type, or types, of reaction each of the following represents:
(a) H2 O(g )
+ C(s) [U+27F6] CO(g ) + H2 (g )
+ 3 O2 (g )
(c) Al(OH)3 (aq ) + 3HCl(aq ) [U+27F6] AlBr3 (aq ) + 3H2 O(l)
(d) Pb(NO3 )2 (aq ) + H2 SO4 (aq ) [U+27F6] PbSO4 (s) + 2HNO3 (aq )
(b) 2KClO3 (s) [U+27F6] 2KCl(s)
Exercise 4
(Solution on p. 13.)
Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the
dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain
your answer.
Exercise 5
(Solution on p. 13.)
Determine the oxidation states of the elements in the following compounds:
(a) NaI
(b) GdCl3
(c) LiNO3
(d) H2 Se
(e) Mg2 Si
(f ) RbO2 , rubidium superoxide
(g) HF
Exercise 6
Determine the oxidation states of the elements in the compounds listed.
(Solution on p. 13.)
None of the oxygen-
containing compounds are peroxides or superoxides.
(a) H3 PO4
(b) Al(OH)3
(c) SeO2
(d) KNO2
(e) In2 S3
(f ) P4 O6
Exercise 7
Determine the oxidation states of the elements in the compounds listed.
(Solution on p. 13.)
None of the oxygen-
containing compounds are peroxides or superoxides.
(a) H2 SO4
(b) Ca(OH)2
(c) BrOH
(d) ClNO2
(e) TiCl4
(f ) NaH
Exercise 8
(Solution on p. 13.)
Classify the following as acid-base reactions or oxidation-reduction reactions:
(a) Na2 S(aq )
(b)
+ 2HCl(aq ) [U+27F6] 2NaCl(aq ) + H2 S(g )
2Na(s) + 2HCl(aq ) [U+27F6] 2NaCl(aq ) + H2 (g )
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(c) Mg(s)
+ Cl2 (g ) [U+27F6] MgCl2 (s)
+ 2HCl(aq ) [U+27F6] MgCl2 (aq ) + H2 O(l)
(e) K3 P(s) + 2O2 (g ) [U+27F6] K3 PO4 (s)
(f ) 3KOH(aq ) + H3 PO4 (aq ) [U+27F6] K3 PO4 (aq ) + 3H2 O(l)
(d) MgO(s)
Exercise 9
(Solution on p. 13.)
Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the
oxidizing and reducing agents in each of the following equations:
(a) Mg(s)
+ NiCl2 (aq ) [U+27F6] MgCl2 (aq ) + Ni(s)
+ Cl2 (g ) [U+27F6] PCl5 (s)
(c) C2 H4 (g ) + 3O2 (g ) [U+27F6] 2CO2 (g ) + 2H2 O(g )
(d) Zn(s) + H2 SO4 (aq ) [U+27F6] ZnSO4 (aq ) + H2 (g )
(e) 2K2 S2 O3 (s) + I2 (s) [U+27F6] K2 S4 O6 (s) + 2KI(s)
(f ) 3Cu(s) + 8HNO3 (aq ) [U+27F6] 3Cu(NO3 )2 (aq ) + 2NO(g ) + 4H2 O(l)
(b) PCl3 (l)
Exercise 10
(Solution on p. 13.)
Complete and balance the following acid-base equations:
(a) HCl gas reacts with solid Ca(OH)2 (s ).
(b) A solution of Sr(OH)2 is added to a solution of HNO3 .
Exercise 11
(Solution on p. 13.)
Complete and balance the following acid-base equations:
(a) A solution of HClO4 is added to a solution of LiOH.
(b) Aqueous H2 SO4 reacts with NaOH.
(c) Ba(OH)2 reacts with HF gas.
Exercise 12
(Solution on p. 13.)
Complete and balance the following oxidation-reduction reactions, which give the highest possible
oxidation state for the oxidized atoms.
(a) Al(s)
+ F2 (g ) [U+27F6]
+ CuBr2 (aq ) [U+27F6] (single
(c) P4 (s) + O2 (g ) [U+27F6]
(d) Ca(s) + H2 O(l) [U+27F6] (products
(b) Al(s)
displacement)
are a strong base and a diatomic gas)
Exercise 13
(Solution on p. 13.)
Complete and balance the following oxidation-reduction reactions, which give the highest possible
oxidation state for the oxidized atoms.
(a) K(s)
+ H2 O(l) [U+27F6]
+ HBr(aq ) [U+27F6]
Sn(s) + I2 (s) [U+27F6]
(b) Ba(s)
(c)
Exercise 14
(Solution on p. 13.)
Complete and balance the equations for the following acid-base neutralization reactions. If water
is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be
more than one correct answer, depending on the amounts of reactants used.
(a) Mg(OH)2 (s) + HClO4 (aq ) [U+27F6]
+ H2 O(l) [U+27F6] (assume an excess of water and that the product dissolves)
(b) SO3 (g )
(c) SrO(s)
+ H2 SO4 (l) [U+27F6]
Exercise 15
When heated to 700800
(Solution on p. 13.)
◦
C, diamonds, which are pure carbon, are oxidized by atmospheric
oxygen. (They burn!) Write the balanced equation for this reaction.
Exercise 16
(Solution on p. 13.)
The military has experimented with lasers that produce very intense light when uorine combines
explosively with hydrogen. What is the balanced equation for this reaction?
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Exercise 17
11
(Solution on p. 13.)
Write the molecular, total ionic, and net ionic equations for the following reactions:
(a) Ca(OH)2 (aq )
(b) H3 PO4 (aq )
+ HC2 H3 O2 (aq ) [U+27F6]
+ CaCl2 (aq ) [U+27F6]
Exercise 18
(Solution on p. 14.)
Great Lakes Chemical Company produces bromine, Br2 , from bromide salts such as NaBr, in
Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction
of NaBr with Cl2 .
Exercise 19
(Solution on p. 14.)
In a common experiment in the general chemistry laboratory, magnesium metal is heated in air
to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small
amounts of Mg3 N2 , a compound formed as some of the magnesium reacts with nitrogen. Write a
balanced equation for each reaction.
Exercise 20
(Solution on p. 14.)
Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned
spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1
mol of CO2 . (Hint: Water is one of the products.)
Exercise 21
(Solution on p. 14.)
Calcium propionate is sometimes added to bread to retard spoilage.
This compound can be
prepared by the reaction of calcium carbonate, CaCO3 , with propionic acid, C2 H5 CO2 H, which has
properties similar to those of acetic acid. Write the balanced equation for the formation of calcium
propionate.
Exercise 22
(Solution on p. 14.)
Complete and balance the equations of the following reactions, each of which could be used to
remove hydrogen sulde from natural gas:
(a) Ca(OH)2 (s)
+ H2 S(g ) [U+27F6]
+ H2 S(g ) [U+27F6]
(b) Na2 CO3 (aq )
Exercise 23
(Solution on p. 14.)
Copper(II) sulde is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid
copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid hydrogen
sulfate as the only product. Write the two equations which represent these reactions.
Exercise 24
(Solution on p. 14.)
Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.
(a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (rst reduce
the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)
(b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction
(c) gaseous H2 S from solid Zn and S via a two-step process (rst a redox reaction between the
starting materials, then reaction of the product with a strong acid)
Exercise 25
(Solution on p. 14.)
Calcium cyclamate Ca(C6 H11 NHSO3 )2 is an articial sweetener used in many countries around
the world but is banned in the United States. It can be puried industrially by converting it to the
barium salt through reaction of the acid C6 H11 NHSO3 H with barium carbonate, treatment with
sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide.
Write the balanced equations for these reactions.
Exercise 26
(Solution on p. 14.)
Complete and balance each of the following half-reactions (steps 25 in half-reaction method):
[U+27F6] Sn2+ (aq )
+
[Ag(NH3 )2 ] (aq ) [U+27F6] Ag(s) + NH3 (aq )
(a) Sn
(b)
4+
(aq )
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−
(c) Hg2 Cl2 (s) [U+27F6] Hg(l) + Cl (aq )
(d) H2 O(l) [U+27F6] O2 (in acidic solution)
(e) IO3
−
(f ) SO3
(aq ) [U+27F6] I2 (s)
2− (aq ) [U+27F6] SO 2− (aq ) (in acidic solution)
4
2+
−
(aq )
(g) MnO4
(h) Cl
−
[U+27F6] Mn
(aq ) [U+27F6] ClO3
−
(aq ) (in acidic solution)
(aq ) (in basic solution)
Exercise 27
(Solution on p. 15.)
Complete and balance each of the following half-reactions (steps 25 in half-reaction method):
2+
(aq ) [U+27F6] Cr
3+
(aq )
−
2− (aq )
(b) Hg(l) + Br (aq ) [U+27F6] HgBr4
2−
(c) ZnS(s) [U+27F6] Zn(s) + S
(aq )
(a) Cr
(d) H2 (g ) [U+27F6] H2 O(l) (in basic solution)
+
(e) H2 (g ) [U+27F6] H3 O (aq ) (in acidic solution)
(f ) NO3
−
(aq ) [U+27F6] HNO2 (aq ) (in acidic solution)
[U+27F6] MnO4 − (aq ) (in basic solution)
−
(aq ) [U+27F6] ClO3 (aq ) (in acidic solution)
(g) MnO2 (s)
(h) Cl
−
Exercise 28
(Solution on p. 15.)
Balance each of the following equations according to the half-reaction method:
+ Cu2+ (aq ) [U+27F6] Sn4+ (aq ) + Cu+ (aq )
2+ (aq ) [U+27F6] Hg (l) + S(s) (in acid)
(b) H2 S(g ) + Hg2
−
−
−
(c) CN (aq ) + ClO2 (aq ) [U+27F6] CNO (aq ) + Cl (aq )
2+
4+
3+
3+
(d) Fe
(aq ) + Ce
(aq ) [U+27F6] Fe
(aq ) + Ce
(aq )
−
(e) HBrO(aq ) [U+27F6] Br (aq ) + O2 (g ) (in acid)
(a) Sn
2+
(aq )
(in acid)
Exercise 29
(Solution on p. 15.)
Balance each of the following equations according to the half-reaction method:
+ NO3 − (aq ) [U+27F6] Zn2+ (aq ) + N2 (g ) (in acid)
2+
−
(b) Zn(s) + NO3 (aq ) [U+27F6] Zn
(aq ) + NH3 (aq ) (in base)
2+
−
(c) CuS(s) + NO3 (aq ) [U+27F6] Cu
(aq ) + S(s) + NO(g ) (in acid)
(d) NH3 (aq ) + O2 (g ) [U+27F6] NO2 (g ) (gas phase)
−
−
−
(e) Cl2 (g ) + OH (aq ) [U+27F6] Cl (aq ) + ClO3 (aq ) (in base)
2+
−
(f ) H2 O2 (aq ) + MnO4 (aq ) [U+27F6] Mn
(aq ) + O2 (g ) (in acid)
−
−
(g) NO2 (g ) [U+27F6] NO3 (aq ) + NO2 (aq ) (in base)
3+
−
2+
(h) Fe
(aq ) + I (aq ) [U+27F6] Fe
(aq ) + I2 (aq )
(a) Zn(s)
Exercise 30
(Solution on p. 15.)
Balance each of the following equations according to the half-reaction method:
(a) MnO4
−
(aq )
+ NO2 − (aq ) [U+27F6] MnO2 (s) + NO3 − (aq )
2− (aq ) [U+27F6] MnO − (aq ) + MnO (s) (in base)
(b) MnO4
4
2
−
2− (aq ) (in acid)
(c) Br (l) + SO (g ) [U+27F6] Br (aq ) + SO
2
2
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4
(in base)
OpenStax-CNX module: m56014
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Solutions to Exercises in this Module
Solution to Exercise (p. 8)
(a) i. The transition is from ice to liquid water. (b) iii. Combustion with oxygen generally produces both
CO2 and H2 O. (c) iii. The balanced equation is
2CH3 OH(g ) + 3O2 (g ) [U+27F6] 2CO2 (g ) + 4H2 O(g )
(d) v.
Only reacting ionic species are present.
Solution to Exercise (p. 9)
(a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)
Solution to Exercise (p. 9)
(a) oxidation-reduction (combustion); (b) oxidation-reduction; (c) acid-base (neutralization); (d) precipitation
Solution to Exercise (p. 9)
An oxidation-reduction reaction, because the oxidation state of the silver changes during the reaction.
Solution to Exercise (p. 9)
(a) Na +1, I 1; (b) Gd +3, Cl 1; (c) Li +1, N +5, O 2; (d) H +1, Se 2; (e) Mg +2, Si 4; (f ) Rb +1;
O
− 12 ;
(g) H +1, F 1
Solution to Exercise (p. 9)
(a) H +1, P +5, O 2; (b) Al +3, H +1, O 2; (c) Se +4, O 2; (d) K +1, N +3, O 2; (e) In +3, S 2; (f )
P +3, O 2
Solution to Exercise (p. 9)
(a) H +1, S +6, O 2; (b) Ca +2, O 2, H +1; (c) Br +1, O 2, H +1; (d) Cl +1, N +3, O 2; (e) Ti +4,
Cl 1; (f ) Na +1, H 1
Solution to Exercise (p. 9)
(a) acid-base; (b) oxidation-reduction: Na is oxidized, H
+ is reduced; (c) oxidation-reduction: Mg is oxidized,
3 is oxidized, O is reduced; (f ) acid-base
2
Cl2 is reduced; (d) acid-base; (e) oxidation-reduction: P
Solution to Exercise (p. 10)
(a) Mg is oxidized from 0 to +2 and is the reducing agent, Ni is reduced from +2 to 0 and is the oxidizing
agent; (b) P is oxidized from +3 to +5 and is the reducing agent, Cl is reduced from 0 to 1 and is the
oxidizing agent; (c) C is oxidized from +2 to +4 and is the reducing agent, O is reduced from 0 to 2 and
is the oxidizing agent; (d) Zn is oxidized from 0 to +2 and is the reducing agent, H is reduced from +1 to 0
and is the oxidizing agent; (e) S is oxidized from +2 to +2.5 and is the reducing agent, I2 is reduced from 0
to 1 and is the oxidizing agent; (f ) Cu is oxidized from O to +2, N is reduced from +5 to +2
Solution to Exercise (p. 10)
(a) 2HCl(g )+Ca(OH)2 (s) [U+27F6] CaCl2 (s)+2H2 O(l); (b) Sr(OH)2 (aq )+2HNO3 (aq ) [U+27F6] Sr(NO3 )2 (aq )+
2H2 O(l)
Solution to Exercise (p. 10)
(a) HClO4 (aq )+LiOH(aq )
2H2 O(l);
(c) Ba(OH)2 (aq )
[U+27F6] H2 O(l)+LiClO4 (aq ); (b) H2 SO4 (aq )+2NaOH(aq ) [U+27F6] Na2 SO4 (aq )+
+ 2HF(g ) [U+27F6] BaF2 (aq ) + 2H2 O(l)
Solution to Exercise (p. 10)
(a) 2Al(s) + 3F2 (g ) [U+27F6] 2AlF3 (s); (b) 2Al(s) + 3CuBr2 (aq ) [U+27F6] 3Cu(s) + 2AlB3 (aq );
5O2 (g ) [U+27F6] P4 O10 (s); (d) Ca(s) + 2H2 O(l) [U+27F6] Ca(OH)2 (aq ) + H2 (g )
(c) P4 (s) +
Solution to Exercise (p. 10)
(a)
2K(s) + 2H2 O(l) [U+27F6] 2KOH(aq ) + H2 (g );
+ 2I2 (s) [U+27F6] SnI4 (s)
(b) Ba(s)
+ 2HBr(aq ) [U+27F6] BaBr2 (aq ) + H2 (g );
(c)
Sn(s)
Solution to Exercise (p. 10)
2+
−
+
(a) Mg(OH)2 (s)+2HClO4 (aq ) [U+27F6] Mg
(aq )+2ClO4 (aq )+2H2 O(l); (b) SO3 (g )+2H2 O(l) [U+27F6] H3 O
−
HSO4 (aq ), (a solution of H2 SO4 ; (c) SrO(s) + H2 SO4 (l) [U+27F6] SrSO4 (s) + H2 O
Solution to Exercise (p. 10)
C(s)
+ O2 (g ) [U+27F6] CO2 (g )
Solution to Exercise (p. 10)
H2 (g )
+ F2 (g ) [U+27F6] 2HF(g )
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(aq )+
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14
Solution to Exercise (p. 11)
(a) Ca(OH)2 (aq ) + 2HC2 H3 O2 (aq ) [U+27F6] Ca(C2 H3 O2 )2 (aq ) + 2H2 O(l),
2+
−
2+
−
(aq ) + 2OH (aq ) + 2HC2 H3 O2 (aq ) [U+27F6] Ca
(aq ) + 2C2 H3 O2 (aq )
+ 2H2 O(l),
+ HC2 H3 O2 (aq ) [U+27F6] C2 H3 O2 − (aq ) + H2 O(l)
(b) 2H3 PO4 (aq ) + 3CaCl2 (aq ) [U+27F6] Ca3 (PO4 )2 (s) + 6HCl(aq ),
2H3 PO4 (aq ) + 3Ca2+ (aq ) + 6Cl− (aq ) [U+27F6] Ca3 (PO4 )2 (s) + 6H+ (aq ) + 6Cl− (aq ),
2H3 PO4 (aq ) + 3Ca2+ (aq ) [U+27F6] Ca3 (PO4 )2 (s) + 6H+ (aq )
Ca
OH
−
(aq )
Solution to Exercise (p. 11)
2NaBr(aq )
+ Cl2 (g ) [U+27F6] 2NaCl(aq ) + Br2 (l)
Solution to Exercise (p. 11)
2Mg(s) + O2 (g ) [U+27F6] 2MgO(s); 3Mg(s) + N2 (g ) [U+27F6] Mg3 N2 (s)
Solution to Exercise (p. 11)
2LiOH(aq ) + CO2 (g ) [U+27F6] Li2 CO3 (aq ) + H2 O(l)
Solution to Exercise (p. 11)
CaCO3 (s)
+ 2C2 H5 CO2 H(aq ) [U+27F6] Ca(C2 H5 CO2 )2 (aq ) + CO2 (g ) + H2 O(l)
Solution to Exercise (p. 11)
(a) Ca(OH)2 (s)
CO2 (g )
+
H2 S(g ) [U+27F6] CaS(s)
+ 2H2 O(l);
(b) Na2 CO3 (aq )
+
H2 S(g ) [U+27F6] Na2 S(aq )
+
+ H2 O(l)
Solution to Exercise (p. 11)
CuS(s)
+ 2O2 (g ) [U+27F6] SO2 (g ) + CuO(s)
SO3 (g )
+ H2 O(l) [U+27F6] H2 SO4 (l)
Solution to Exercise (p. 11)
(a) step 1: N2 (g )+3H2 (g ) [U+27F6] 2NH3 (g ), step 2: NH3 (g )+HNO3 (aq ) [U+27F6] NH4 NO3 (aq )
(b) H2 (g )+Br2 (l) [U+27F6] 2HBr(g ); (c) Zn(s)+S(s)
[U+27F6] NH4 NO3 (s)
[U+27F6] ZnS(s) and ZnS(s)+2HCl(aq ) [U+27F6] ZnCl2 (aq )+
(after
H2 S(g )
Solution to Exercise (p. 11)
2C6 H11 NHSO3 H(aq ) + BaCO3 (aq ) [U+27F6] (C6 H11 NHSO3 )2 Ba(s) + H2 CO3 (aq )
+ H2 SO4 (aq ) [U+27F6] BaSO4 (s) + 2C6 H11 NHSO3 H(aq )
2C6 H11 NHSO3 H(aq ) + Ca(OH)2 (aq ) [U+27F6] Ca(C6 H11 NHSO3 )2 (s) + 2H2 O(l)
(C6 H11 NHSO3 )2 Ba(s)
Solution to Exercise (p. 11)
2+
(a) S n4+ (aq ) [U+27F6] Sn
(aq )
Sn
4+
(aq )
+
−
Se
[U+27F6] Sn
2+
, (b)
(aq )
[Ag(NH3 )2 ]
+
(aq )
[U+27F6] Ag(s) + 2NH3 (aq )
[Ag(NH3 )2 ]
+
(aq )
+ e− [U+27F6] Ag(s) + 2NH3 (aq )
IO3
2H2 O(l) [U+27F6] O2 (g )
(d)
2H2 O(l) [U+27F6] O2 (g ) + 4H+ (aq )
2H2 O(l) [U+27F6] O2 (g ) + 4H+ (aq ) + 4e−
−
2IO3
(aq ) [U+27F6] I2 (s)
2IO3
−
(aq ) [U+27F6] I2 (s)
12H
; (c)
Hg2 Cl2 (s) [U+27F6
Hg2 Cl2 (s)
(aq )
+ 6H2 O(l)
+ 2IO3 − (aq ) [U+27F6] I2 (s) + 6H2 O(l)
12H+ (aq ) + 12OH− (aq) + 2IO3 − (aq ) [U+27F6] I2 (s) + 6H2 O(l) + 12O
12H2 O(l) + 2IO3 − (aq ) [U+27F6] I2 (s) + 6H2 O(l) + 12OH− (aq )
6H2 O(l) + 2IO3 − (aq ) [U+27F6] I2 (s) + 12OH− (aq )
6H2 O(l) + 2IO3 − (aq )10e− [U+27F6] I2 (s) + 12OH− (aq )
http://cnx.org/content/m56014/1.2/
+ 2e− [U
(aq ) [U+27F6] I2 (s)
−
+
; (e)
Hg2 Cl2 (s) [U+27F6
OpenStax-CNX module: m56014
SO3
15
2− (aq ) [U+27F6] SO 2− (aq )
4
(f ) H2 O(l)
H2 O(l)
+
+
SO3
MnO4
−
(aq )
(aq )
[U+27F6] Mn2+ (aq )
2− (aq ) [U+27F6] SO 2− (aq ); (g) MnO − (aq ) [U+27F6] Mn2+ (aq )
4
4
2− (aq ) [U+27F6] SO 2− (aq )
SO3
4
+
+ 2H
(aq )
2− (aq ) [U+27F6] SO 2− (aq ) + 2H+ (aq ) + 2e−
H2 O(l) + SO3
4
Cl
−
+
8H (aq )
+
MnO4
−
(aq )
+
[U+27F6] Mn
+
−
−
8H (aq ) + MnO4 (aq ) + 5e
2+
4H2 O
(aq )
[U+27F6] Mn
2+
+ 4H2 O
(aq )
+ 4H2 O
[U+27F6] ClO3 − (aq )
3H2 O(l) + Cl− (aq ) [U+27F6] ClO3 − (aq )
3H2 O(l) + Cl− (aq ) [U+27F6] ClO3 − (aq ) + 6H+ (aq )
(h)
3H2 O(l) + Cl− (aq ) + 6OH− (aq) [U+27F6] ClO3 − (aq ) + 6H+ (aq ) + 6OH− (aq)
3H2 O(l) + Cl− (aq ) + 6OH− (aq ) [U+27F6] ClO3 − (aq ) + 6H2 O(l)
−
Cl
(aq )
+ 6OH− (aq ) [U+27F6] ClO3 − (aq ) + 3H2 O(l) + 6e−
Solution to Exercise (p. 12)
For an example of the fully worked out solution, see the solution to Solution to Exercise (p.
2+
3+
14).
(a)
14).
(a)
; (b) Hg(l)+4Br− (aq ) [U+27F6] HgBr4 2− (aq )+2e− ; (c) ZnS(s)+2e− [U+27F6] Zn(s)+
2−
−
+
−
−
S
(aq ); (d) H2 (g ) +2OH (aq ) [U+27F6] 2H2 O(l) +2e ; (e) H2 (g ) +2H2 O(l) [U+27F6] 2H3 O (aq ) +2e ; (f )
+
−
−
−
−
NO3 (aq )+3H3 O (aq )+2e [U+27F6] HNO2 (aq )+4H2 O(l); (g) MnO2 (s)+4OH (aq ) [U+27F6] MnO4 (aq )+
2H2 O(l) + 3e− ; (h) Cl− (aq ) + 3H2 O(l) [U+27F6] ClO3 − (aq ) + 6H3 O+ (aq ) + 6e−
Cr
(aq ) [U+27F6] Cr
−
(aq )+e
Solution to Exercise (p. 12)
For an example of the fully worked out solution, see the solution to Solution to Exercise (p.
Sn
2+
(aq ) + 2Cu
2+
(aq ) [U+27F6] Sn
4+
+
(aq ) + 2Cu (aq ); (b) H2 S(g ) + Hg2
2+ (aq ) + 2H O(l) [U+27F6] 2Hg(l) +
2
−
−
+
+ 2H3 O+ (aq ); (c) 5CN− (aq ) + 2ClO2 (aq ) + 3H2 O(l) [U+27F6] 5CNO (aq ) + 2Cl (aq ) + 2H3 O (aq );
2+
4+
3+
3+
+
(d) Fe
(aq ) + Ce
(aq ) [U+27F6] Fe
(aq ) + Ce
(aq ); (e) 2HBrO(aq ) + 2H2 O(l) [U+27F6] 2H3 O (aq ) +
−
2Br (aq ) + O2 (g )
S(s)
Solution to Exercise (p. 12)
For an example of the fully worked out solution, see the solution to Solution to Exercise (p.
14).
(a)
14).
(a)
5Zn(s) + 12H3 O+ (aq ) + 2NO3 − (aq ) [U+27F6] 5Zn2+ (aq ) + N2 (g ) + 18H2 O(l); (b) 4Zn(s) + NO3 − (aq ) +
6H2 O(l) [U+27F6] 4Zn2+ (aq )+NH3 (aq )+9OH− (aq ); (c) 3CuS(s)+8H3 O+ (aq )+2NO3 − (aq ) [U+27F6] 3Cu2+ (aq )+
3S(s)+2NO(g )+12H2 O(l); (d) 4NH3 (aq )+7O2 (g ) [U+27F6] 4NO2 (g )+6H2 O(l); (e) 3Cl2 (g )+6OH− (aq ) [U+27F6] 5Cl− (aq )+
+
2+
−
−
ClO3 (aq ) +3H2 O(l); (f ) 5H2 O2 (aq ) +2MnO4 (aq ) +6H3 O (aq ) [U+27F6] 2Mn
(aq ) +5O2 (g ) +14H2 O(l);
−
3+
−
2+
−
−
(g) 2NO2 (g )+2OH (aq ) [U+27F6] NO3 (aq )+NO2 (aq )+H2 O(l); (h) 2Fe
(aq )+2I (aq ) [U+27F6] 2Fe
(aq )+
I2 (aq )
Solution to Exercise (p. 12)
For an example of the fully worked out solution, see the solution to Solution to Exercise (p.
2MnO4 − (aq ) + 3NO2 − (aq ) + H2 O(l) [U+27F6] 2MnO2 (s) + 3NO3 − (aq ) + 2OH− (aq ); (b) 3MnO4 2− (aq ) +
2H2 O(l) [U+27F6] 2MnO4 − (aq )+4OH− (aq )+MnO2 (s) (in base); (c) Br2 (l)+SO2 (g )+2H2 O(l) [U+27F6] 4H+ (aq )+
2Br− (aq ) + SO4 2− (aq )
Glossary
Denition 1: acid
substance that produces H3 O
+ when dissolved in water
Denition 2: acid-base reaction
reaction involving the transfer of a hydrogen ion between reactant species
Denition 3: base
when dissolved in water
substance that produces OH
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OpenStax-CNX module: m56014
Denition 4: combustion reaction
vigorous redox reaction producing signicant amounts of energy in the form of heat and, sometimes,
light
Denition 5: half-reaction
an equation that shows whether each reactant loses or gains electrons in a reaction.
Denition 6: insoluble
of relatively low solubility; dissolving only to a slight extent
Denition 7: neutralization reaction
reaction between an acid and a base to produce salt and water
Denition 8: oxidation
process in which an element's oxidation number is increased by loss of electrons
Denition 9: oxidation-reduction reaction
(also, redox reaction) reaction involving a change in oxidation number for one or more reactant
elements
Denition 10: oxidation number
(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic
Denition 11: oxidizing agent
(also, oxidant) substance that brings about the oxidation of another substance, and in the process
becomes reduced
Denition 12: precipitate
insoluble product that forms from reaction of soluble reactants
Denition 13: precipitation reaction
reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double-displacement or metathesis
Denition 14: reduction
process in which an element's oxidation number is decreased by gain of electrons
Denition 15: reducing agent
(also, reductant) substance that brings about the reduction of another substance, and in the process
becomes oxidized
Denition 16: salt
ionic compound that can be formed by the reaction of an acid with a base that contains a cation
and an anion other than hydroxide or oxide
Denition 17: single-displacement reaction
(also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic
species
Denition 18: soluble
of relatively high solubility; dissolving to a relatively large extent
Denition 19: solubility
the extent to which a substance may be dissolved in water, or any solvent
Denition 20: strong acid
acid that reacts completely when dissolved in water to yield hydronium ions
Denition 21: strong base
base that reacts completely when dissolved in water to yield hydroxide ions
Denition 22: weak acid
acid that reacts only to a slight extent when dissolved in water to yield hydronium ions
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16
OpenStax-CNX module: m56014
Denition 23: weak base
base that reacts only to a slight extent when dissolved in water to yield hydroxide ions
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17