Chapter 8: Stoichiometry of Chemical Reactions

8.3 Other Types of Chemical Reactions

Learning Outcomes

  • Define two common types of chemical reactions (acid-base, and oxidation-reduction)
  • Classify chemical reactions as one of these two types given appropriate descriptions or chemical equations
  • Identify common acids and bases
  • Compute the oxidation states for elements in compounds

Acid-Base Reactions

An acid-base reaction is one in which a hydrogen ion, [latex]\ce{H+}[/latex], is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text.

For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, [latex]\ce{H3O+}[/latex]. As an example, consider the equation shown here:

[latex]\text{HCl(}aq\text{)}+{\text{H}}_{2}\text{O(}aq\text{)}\rightarrow{\text{Cl}}^{-}\text{(}aq\text{)}+{\text{H}}_{3}{\text{O}}^{\text{+}}\text{(}aq\text{)}[/latex]

The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, [latex]\ce{H3O+}[/latex] ions are produced by a chemical reaction in which [latex]\ce{H+}[/latex] ions are transferred from [latex]\ce{HCl}[/latex] molecules to [latex]\ce{H2O}[/latex] molecules (Figure 8.3.1).

This figure shows two flasks, labeled a and b. The flasks are both sealed with stoppers and are nearly three-quarters full of a liquid. Flask a is labeled H C l followed by g in parentheses. In the liquid there are approximately twenty space-filling molecular models composed of one red sphere and two smaller attached white spheres. The label H subscript 2 O followed by a q in parentheses is connected with a line to one of these models. In the space above the liquid in the flask, four space filling molecular models composed of one larger green sphere to which a smaller white sphere is bonded are shown. To one of these models, the label H C l followed by g in parentheses is attached with a line segment. An arrow is drawn from the space above the liquid pointing down into the liquid below. Flask b is labeled H subscript 3 O superscript positive sign followed by a q in parentheses. This is followed by a plus sign and C l superscript negative sign which is also followed by a q in parentheses. In this flask, no molecules are shown in the open space above the liquid. A label, C l superscript negative sign followed by a q in parentheses, is connected with a line segment to a green sphere. This sphere is surrounded by four molecules composed each of one red sphere and two white smaller spheres. A few of these same molecules appear separate from the green spheres in the liquid. A line segment connects one of them to the label H subscript 2 O which is followed by l in parentheses. There are a few molecules formed from one central larger red sphere to which three smaller white spheres are bonded. A line segment is drawn from one of these to the label H subscript 3 O superscript positive sign, followed by a q in parentheses.
Figure 8.3.1. When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).

This figure contains two images, each with an associated structural formula provided in the lower left corner of the image. The first image is a photograph of a variety of thinly sliced, circular cross sections of citrus fruits ranging in color for green to yellow, to orange and reddish-orange. The slices are closely packed on a white background. The structural formula with this picture shows a central chain of five C atoms. The leftmost C atom has an O atom double bonded above and to the left and a singly bonded O atom below and to the left. This single bonded O atom has an H atom indicated in red on its left side which is highlighted in pink. The second C atom moving to the right has H atoms bonded above and below. The third C atom has a single bonded O atom above which has an H atom on its right. This third C atom has a C atom bonded below it which has an O atom double bonded below and to the left and a singly bonded O atom below and to the right. An H atom appears in red and is highlighted in pink to the right of the singly bonded O atom. The fourth C atom has H atoms bonded above and below. The fifth C atom is at the right end of the structure. It has an O atom double bonded above and to the right and a singly bonded O atom below and to the right. This single bonded O atom has a red H atom on its right side which is highlighted in pink. The second image is a photograph of bottles of vinegar. The bottles are labeled, “Balsamic Vinegar,” and appear to be clear and colorless. The liquid in this bottle appears to be brown. The structural formula that appears with this image shows a chain of two C atoms. The leftmost C atom has H atoms bonded above, below, and to the left. The C atom on the right has a doubly bonded O atom above and to the right and a singly bonded O atom below and to the right. This O atom has an H atom bonded to its right which is highlighted in pink.
Figure 8.3.2. (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. The hydrogen atoms that may be transferred during an acid-base reaction are highlighted in the inset molecular structures. (credit a: modification of work by Scott Bauer; credit b: modification of work by Brücke-Osteuropa/Wikimedia Commons)

The nature of [latex]\ce{HCl}[/latex] is such that its reaction with water as just described is essentially 100% efficient: Virtually every [latex]\ce{HCl}[/latex] molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids, and [latex]\ce{HCl}[/latex] is one among just a handful of common acid compounds that are classified as strong (Table 8.3.1).

A far greater number of compounds behave as acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:

[latex]{\text{CH}}_{3}{\text{CO}}_{2}\text{H(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightleftharpoons{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\text{(}aq\text{)}+{\text{H}}_{3}{\text{O}}^{\text{+}}\text{(}aq\text{)}[/latex]

When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, [latex]\ce{CH3CO2-}[/latex] (Figure 8.3.2). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapter on chemical equilibrium.)

Table 8.3.1. Common Strong Acids
Compound Formula Name in Aqueous Solution
[latex]\ce{HBr}[/latex] hydrobromic acid
[latex]\ce{HCl}[/latex] hydrochloric acid
[latex]\ce{HI}[/latex] hydroiodic acid
[latex]\ce{HNO3}[/latex] nitric acid
[latex]\ce{HClO4}[/latex] perchloric acid
[latex]\ce{H2SO4}[/latex] sulfuric acid

A base is a substance that will dissolve in water to yield hydroxide ions, [latex]\ce{OH-}[/latex]. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion—for example, [latex]\ce{NaOH}[/latex] and [latex]\ce{Ca(OH)2}[/latex]. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, [latex]\ce{KOH}[/latex] and [latex]\ce{Ba(OH)2}[/latex] dissolve in water and dissociate completely to produce cations ([latex]\ce{K+}[/latex] and [latex]\ce{Ba^{2+}}[/latex], respectively) and hydroxide ions, [latex]\ce{OH-}[/latex]. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases.

Consider as an example the dissolution of lye (sodium hydroxide) in water:

[latex]\text{NaOH(}s\text{)}\rightarrow{\text{Na}}^{\text{+}}\text{(}aq\text{)}+{\text{OH}}^{-}\text{(}aq\text{)}[/latex]

This equation confirms that sodium hydroxide is a base. When dissolved in water, [latex]\ce{NaOH}[/latex] dissociates to yield [latex]\ce{Na+}[/latex] and [latex]\ce{OH-}[/latex] ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, [latex]\ce{NaOH}[/latex] and other ionic hydroxides are all classified as strong bases.

Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure 8.3.3). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:

[latex]{\text{NH}}_{3}\text{(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightleftharpoons{\text{NH}}_{4}{}^{\text{+}}\text{(}aq\text{)}+{\text{OH}}^{-}\text{(}aq\text{)}[/latex]

This is, by definition, an acid-base reaction, in this case involving the transfer of [latex]\ce{H+}[/latex] ions from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as [latex]{\text{NH}}_{4}{}^{+}[/latex] ions.

This photograph shows a large agricultural tractor in a field pulling a field sprayer and a large, white cylindrical tank which is labeled “Caution Ammonia.”
Figure 8.3.3. Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)

A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base, and the products are often a salt and water:

[latex]\text{acid}+\text{base}\rightarrow\text{salt}+\text{water}[/latex]

To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid [latex]\ce{Mg(OH)2}[/latex]) is ingested to ease symptoms associated with excess stomach acid ([latex]\ce{HCl}[/latex]):

[latex]\text{Mg}{\text{(OH)}}_{2}\text{(}s\text{)}+2\text{HCl(}aq\text{)}\rightarrow{\text{MgCl}}_{2}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}[/latex]

Note that in addition to water, this reaction produces a salt, magnesium chloride.

Example 8.3.1: Writing Equations for Acid-Base Reactions

Write balanced chemical equations for the acid-base reactions described here:

  1. the weak acid hydrogen hypochlorite reacts with water
  2. a solution of barium hydroxide is neutralized with a solution of nitric acid
Show Solution
  1. The two reactants are provided, [latex]\ce{HOCl}[/latex] and [latex]\ce{H2O}[/latex]. Since the substance is reported to be an acid, its reaction with water will involve the transfer of [latex]\ce{H+}[/latex] from [latex]\ce{HOCl}[/latex] to [latex]\ce{H2O}[/latex] to generate hydronium ions, [latex]\ce{H3O+}[/latex] and hypochlorite ions, [latex]\ce{OCl-}[/latex]:
    • [latex]\text{HOCl(}aq\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightleftharpoons{\text{OCl}}^{-}\text{(}aq\text{)}+{\text{H}}_{3}{\text{O}}^{\text{+}}\text{(}aq\text{)}[/latex].
    • A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.
  2. The two reactants are provided, [latex]\ce{Ba(OH)2}[/latex] and [latex]\ce{HNO3}[/latex]. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide ([latex]\ce{Ba^2+}[/latex]) and the anion generated when the acid transfers its hydrogen ion [latex]\text{(}{\text{NO}}_{3}{}^{-}\text{)}[/latex]:
    • [latex]\text{Ba}{\text{(OH)}}_{2}\text{(}aq\text{)}+2{\text{HNO}}_{3}\text{(}aq\text{)}\rightarrow\text{Ba}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}[/latex]

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CHEMISTRY IN EVERYDAY LIFE

Stomach Antacids

Our stomachs contain a solution of roughly 0.03 M [latex]\ce{HCl}[/latex], which helps us digest the food we eat. The burning sensation associated with heartburn is a result of the acid of the stomach leaking through the muscular valve at the top of the stomach into the lower reaches of the esophagus. The lining of the esophagus is not protected from the corrosive effects of stomach acid the way the lining of the stomach is, and the results can be very painful. When we have heartburn, it feels better if we reduce the excess acid in the esophagus by taking an antacid. As you may have guessed, antacids are bases. One of the most common antacids is calcium carbonate, [latex]\ce{CaCO3}[/latex]. The reaction,

[latex]\text{CaCO}_{3}(\text{s})+2\text{HCl}(\text{aq})\rightleftharpoons\text{CaCl}_{2}(\text{aq})+\text{H}_{2}\text{O}(\text{l})+\text{CO}_{2}(\text{g})[/latex]

not only neutralizes stomach acid, it also produces [latex]\ce{CO2}[/latex](g), which may result in a belch.

Milk of Magnesia is a suspension of the sparingly soluble base magnesium hydroxide, [latex]\ce{Mg(OH)2}[/latex]. It works according to the reaction:

[latex]\text{Mg(OH)}_{2}(\text{s})\rightleftharpoons\text{Mg}^{2+}(\text{aq})+2\text{OH}^{-}(\text{aq})[/latex]

The hydroxide ions generated in this equilibrium then go on to react with the hydronium ions from the stomach acid, so that:

[latex]\text{H}_{3}\text{O}^{+}+\text{OH}^{-}\rightleftharpoons{2}\text{H}_{2}\text{O}(\text{l})[/latex]

This reaction does not produce carbon dioxide, but magnesium-containing antacids can have a laxative effect. Several antacids have aluminum hydroxide, [latex]\ce{Al(OH)3}[/latex], as an active ingredient. The aluminum hydroxide tends to cause constipation, and some antacids use aluminum hydroxide in concert with magnesium hydroxide to balance the side effects of the two substances.

CHEMISTRY IN EVERYDAY LIFE

Culinary Aspects of Chemistry

Examples of acid-base chemistry are abundant in the culinary world. One example is the use of baking soda, or sodium bicarbonate in baking. [latex]\ce{NaHCO3}[/latex] is a base. When it reacts with an acid such as lemon juice, buttermilk, or sour cream in a batter, bubbles of carbon dioxide gas are formed from decomposition of the resulting carbonic acid, and the batter “rises.” Baking powder is a combination of sodium bicarbonate, and one or more acid salts that react when the two chemicals come in contact with water in the batter.

Many people like to put lemon juice or vinegar, both of which are acids, on cooked fish (Figure 8.3.4). It turns out that fish have volatile amines (bases) in their systems, which are neutralized by the acids to yield involatile ammonium salts. This reduces the odor of the fish, and also adds a “sour” taste that we seem to enjoy.

An image is shown of two fish with heads removed and skin on with lemon slices placed in the body cavity. The first line of an equation below the image reads C H subscript 3 C O O H plus N H subscript 2 C H subscript 2 C H subscript 2 C H subscript 2 C H subscript 2 N H subscript 2 arrow C H subscript 3 C O O superscript negative sign plus N H subscript 3 superscript positive sign C H subscript 2 C H subscript 2 C H subscript 2 C H subscript 2 N H subscript 2. The second line of the equation reads Acetic acid plus sign Putrescine arrow Acetate ion plus sign Putrescinium ion.
Figure 8.3.4. A neutralization reaction takes place between citric acid in lemons or acetic acid in vinegar, and the bases in the flesh of fish.

Pickling is a method used to preserve vegetables using a naturally produced acidic environment. The vegetable, such as a cucumber, is placed in a sealed jar submerged in a brine solution. The brine solution favors the growth of beneficial bacteria and suppresses the growth of harmful bacteria. The beneficial bacteria feed on starches in the cucumber and produce lactic acid as a waste product in a process called fermentation. The lactic acid eventually increases the acidity of the brine to a level that kills any harmful bacteria, which require a basic environment. Without the harmful bacteria consuming the cucumbers they are able to last much longer than if they were unprotected. A byproduct of the pickling process changes the flavor of the vegetables with the acid making them taste sour.

Oxidation-Reduction Reactions

Earth’s atmosphere contains about 20% molecular oxygen, [latex]\ce{O2}[/latex], 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 [latex]\ce{O2}[/latex], 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 classification.

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:

[latex]2\text{Na}(s)+\text{Cl}_2(g)\rightarrow{2}\text{NaCl}(s)[/latex]

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:

[latex]\begin{array}{l}2\text{Na(}s\text{)}\rightarrow 2{\text{Na}}^{\text{+}}\text{(}s\text{)}+2{\text{e}}^{-}\\ {\text{Cl}}_{2}\text{(}g\text{)}+2{\text{e}}^{-}\rightarrow 2{\text{Cl}}^{-}\text{(}s\text{)}\end{array}[/latex]

These equations show that Na atoms lose electrons while [latex]\ce{Cl}[/latex] atoms (in the [latex]\ce{Cl2}[/latex] 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 define the complementary processes that occur:

oxidation = loss of electrons

reduction = gain of electrons

In this reaction, then, sodium is oxidized and chlorine is undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (reductant), since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant), as it effectively removes electrons from (oxidizes) sodium.

reducing agent = species that is oxidized

oxidizing agent = species that is reduced

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding [latex]\ce{NaCl}[/latex]:

[latex]{\text{H}}_{2}\text{(}g\text{)}+{\text{Cl}}_{2}\text{(}g\text{)}\rightarrow 2\text{HCl(}g\text{)}[/latex]

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 definition of redox reactions, a property called oxidation number has been defined. 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:
    • Hydrogen: +1 when combined with nonmetals, –1 when combined with metals
    • Oxygen: –2 in most compounds, sometimes –1 (so-called peroxides, [latex]{\text{O}}_{2}{}^{\text{2}-}\text{),}[/latex] very rarely [latex]-\frac{1}{2}[/latex] (so-called superoxides, [latex]{\text{O}}_{2}{}^{-}\text{),}[/latex] 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)
  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 first, 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 8.3.3: Assigning Oxidation Numbers

Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:

  1. [latex]\text{H}_{2}\text{S}[/latex]
  2. [latex]{\text{SO}}_{3}{}^{\text{2}-}[/latex]
  3. [latex]\text{Na}_{2}\text{SO}_{4}[/latex]
Show Solution
  1. According to guideline 1, the oxidation number for [latex]\ce{H}[/latex] is +1. Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:
    [latex]\begin{array}{l}\text{charge on }{\text{H}}_{2}\text{S}=0=\text{(}2\times +1\text{)}+\text{(}1\times x\text{)}\\ x=0-\text{(}2\times +1\text{)}=-2\end{array}[/latex]
  2. 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:
    [latex]\begin{array}{l}\text{charge on }{\text{SO}}_{3}{}^{\text{2-}}=-2=\text{(}3\times -1\text{)}+\text{(}1\times x\text{)}\\ x=-2-\text{(}3\times -2\text{)}=+4\end{array}[/latex]
  3. 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:
    [latex]\begin{array}{l}\text{charge on }{\text{SO}}_{4}{}^{2-}=-2=\text{(}4\times -2\text{)}+\text{(}1\times x\text{)}\\ x=-2-\text{(}4\times -2\text{)}=+6\end{array}[/latex]

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Using the oxidation number concept, an all-inclusive definition 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, as in Example 4.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

[latex]\begin{array}{lll}\hfill \mathbf{\text{oxidation}}& =& \text{increase in oxidation number}\hfill \\ \hfill \mathbf{\text{reduction}}& =& \text{decrease in oxidation number}\hfill \end{array}[/latex]

Returning to the reactions used to introduce this topic, they may now both be identified 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 [latex]\ce{NaCl}[/latex]) and chlorine is reduced (its oxidation number decreases from 0 in [latex]\ce{Cl2}[/latex] to –1 in [latex]\ce{NaCl}[/latex]). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in [latex]\ce{H2}[/latex] to +1 in [latex]\ce{HCl}[/latex]) and chlorine is reduced (its oxidation number decreases from 0 in [latex]\ce{Cl2}[/latex] to –1 in [latex]\ce{HCl}[/latex]).

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 significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:

[latex]10\text{Al(}s\text{)}+6{\text{NH}}_{4}{\text{ClO}}_{4}\text{(}s\text{)}\rightarrow 4{\text{Al}}_{2}{\text{O}}_{3}\text{(}s\text{)}+2{\text{AlCl}}_{3}\text{(}s\text{)}+12{\text{H}}_{2}\text{O(}g\text{)}+3{\text{N}}_{2}\text{(}g\text{)}[/latex]

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:

[latex]\text{Zn(}s\text{)}+2\text{HCl}\text{(}aq\text{)}\rightarrow{\text{ZnCl}}_{2}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}[/latex]

Metallic elements may also be oxidized by solutions of other metal salts; for example:

[latex]\text{Cu(}s\text{)}+2{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow\text{Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\text{(}aq\text{)}+2\text{Ag(}s\text{)}[/latex]

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 [latex]\ce{Cu^2+}[/latex] ions dissolve in the solution to yield a characteristic blue color (Figure 8.3.5).

This figure contains three photographs. In a, a coiled copper wire is shown beside a test tube filled with a clear, colorless liquid. In b, the wire has been inserted into the test tube with the clear, colorless liquid. In c, the test tube contains a light blue liquid and the coiled wire appears to have a fuzzy silver gray coating.
Figure 8.3.5. (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: modification of work by Mark Ott)

Example 8.3.4: Describing Redox Reactions

Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.

  1. [latex]{\text{ZnCO}}_{3}\text{(}s\text{)}\rightarrow\text{ZnO(}s\text{)}+{\text{CO}}_{2}\text{(}g\text{)}[/latex]
  2. [latex]2\text{Ga}\text{(}l\text{)}+3{\text{Br}}_{2}\text{(}l\text{)}\rightarrow 2{\text{GaBr}}_{3}\text{(}s\text{)}[/latex]
  3. [latex]2{\text{H}}_{2}{\text{O}}_{2}\text{(}aq\text{)}\rightarrow 2{\text{H}}_{2}\text{O}\text{(}l\text{)}+{\text{O}}_{2}\text{(}g\text{)}[/latex]
  4. [latex]{\text{BaCl}}_{2}\text{(}aq\text{)}+{\text{K}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}\rightarrow{\text{BaSO}}_{4}\text{(}s\text{)}+2\text{KCl}\text{(}aq\text{)}[/latex]
  5. [latex]{\text{C}}_{2}{\text{H}}_{4}\text{(}g\text{)}+3{\text{O}}_{2}\text{(}g\text{)}\rightarrow 2{\text{CO}}_{2}\text{(}g\text{)}+2{\text{H}}_{2}\text{O}\text{(}l\text{)}[/latex]
Show Solution

Redox reactions are identified per definition if one or more elements undergo a change in oxidation number.

  1. This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
  2. This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in [latex]\ce{Ga}[/latex](l) to +3 in [latex]\ce{GaBr3}[/latex](s). The reducing agent is [latex]\ce{Ga}[/latex](l). Bromine is reduced, its oxidation number decreasing from 0 in [latex]\ce{Br2}[/latex](l) to –1 in [latex]\ce{GaBr3}[/latex](s). The oxidizing agent is [latex]\ce{Br2}[/latex](l).
  3. 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 reaction). Oxygen is oxidized, its oxidation number increasing from –1 in [latex]\ce{H2O2}[/latex](aq) to 0 in [latex]\ce{O2}[/latex](g). Oxygen is also reduced, its oxidation number decreasing from –1 in [latex]\ce{H2O2}[/latex](aq) to –2 in [latex]\ce{H2O}[/latex](l). For disproportionation reactions, the same substance functions as an oxidant and a reductant.
  4. This is not a redox reaction, since oxidation numbers remain unchanged for all elements.
  5. This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from –2 in [latex]\ce{C2H4}[/latex](g) to +4 in [latex]\ce{CO2}[/latex](g). The reducing agent (fuel) is [latex]\ce{C2H4}[/latex](g). Oxygen is reduced, its oxidation number decreasing from 0 in [latex]\ce{O2}[/latex](g) to –2 in [latex]\ce{H2O}[/latex](l). The oxidizing agent is [latex]\ce{O2}[/latex](g).

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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 difficult 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 [latex]\ce{H2O}[/latex] molecules.
  4. Balance hydrogen atoms by adding [latex]\ce{H+}[/latex] ions.
  5. Balance charge by adding electrons.
  6. If necessary, multiply each half-reaction’s coefficients 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:
    • Add [latex]\ce{OH-}[/latex] ions to both sides of the equation in numbers equal to the number of [latex]\ce{H+}[/latex] ions.
    • On the side of the equation containing both [latex]\ce{H+}[/latex] and [latex]\ce{OH-}[/latex] ions, combine these ions to yield water molecules.
    • Simplify the equation by removing any redundant water molecules.
  9. Finally, check to see that both the number of atoms and the total charges are balanced.

Example 8.3.5: 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.

[latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+{\text{Fe}}^{\text{2+}}\rightarrow{\text{Cr}}^{\text{3+}}+{\text{Fe}}^{\text{3+}}[/latex]

Show Solution

Step 1. Write the two half-reactions.

Each half-reaction will contain one reactant and one product with one element in common.

[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}[/latex]
[latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow{\text{Cr}}^{\text{3+}}[/latex]

Step 2. Balance all elements except oxygen and hydrogen.

The iron half-reaction is already balanced, but the chromium half-reaction shows two [latex]\ce{Cr}[/latex] atoms on the left and one [latex]\ce{Cr}[/latex] atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to [latex]\ce{Cr}[/latex] atoms.

[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}[/latex]
[latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow 2{\text{Cr}}^{\text{3+}}[/latex]

Step 3. Balance oxygen atoms by adding [latex]\ce{H2O}[/latex] molecules.

The iron half-reaction does not contain [latex]\ce{O}[/latex] atoms. The chromium half-reaction shows seven [latex]\ce{O}[/latex] atoms on the left and none on the right, so seven water molecules are added to the right side.

[latex]\begin{array}{c}{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}} \\ {\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}\end{array}[/latex]

Step 4. Balance hydrogen atoms by adding [latex]\ce{H+}[/latex] 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.

[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}[/latex]
[latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]

Step 5. Balance charge by adding electrons.

The iron half-reaction shows a total charge of 2+ on the left side (1 Fe2+ ion) and 3+ on the right side (1 [latex]\ce{Fe^3+}[/latex] 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-) + (14 × 1+) = 12+ on the left side [latex]{\text{(1 Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}[/latex] ion and 14 [latex]\ce{H+}[/latex] ions). The total charge on the right side is (2 × 3+) = 6 + (2 [latex]\ce{Cr^3+}[/latex] ions). Adding six electrons to the left side will bring that side’s total charge to (12+) + (6-) = 6+, and charge balance is achieved.

[latex]{\text{Fe}}^{\text{2+}}\rightarrow{\text{Fe}}^{\text{3+}}+{\text{e}}^{-}[/latex]
[latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+14{\text{H}}^{+}+6{\text{e}}^{-}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]

Step 6. Multiply the two half-reactions.

This step makes it 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 coefficient must be multiplied by 6.

[latex]{\text{6Fe}}^{\text{2+}}\rightarrow 6{\text{Fe}}^{\text{3+}}+6{\text{e}}^{-}[/latex]
[latex]{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+6{\text{e}}^{-}+14{\text{H}}^{+}\rightarrow 2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]

Step 7. Add the balanced half-reactions and cancel species that appear on both sides of the equation.

[latex]6{\text{Fe}}^{\text{2+}}+{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2}-}+6{\text{e}}^{-}+14{\text{H}}^{+}\rightarrow 6{\text{Fe}}^{\text{3+}}+6{\text{e}}^{-}+2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]

Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here:

[latex]6{\text{Fe}}^{\text{2+}}+{\text{Cr}}_{2}{\text{O}}_{7}{}^{\text{2-}}+14{\text{H}}^{+}\rightarrow 6{\text{Fe}}^{\text{3+}}+2{\text{Cr}}^{\text{3+}}+7{\text{H}}_{2}\text{O}[/latex]

A final check of atom and charge balance confirms the equation is balanced.

Reactants Products
[latex]\ce{Fe}[/latex] 6 6
[latex]\ce{Cr}[/latex] 2 2
[latex]\ce{O}[/latex] 7 7
[latex]\ce{H}[/latex] 14 14
charge 24+ 24+

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Key Concepts and Summary

Chemical reactions are classified 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 simplified by using a systematic approach called the half-reaction method.

Try It

  1. Indicate what type, or types, of reaction each of the following represents:
    1. [latex]\text{Ca(}s\text{)}+{\text{Br}}_{2}\text{(}l\text{)}\rightarrow{\text{CaBr}}_{2}\text{(}s\text{)}[/latex]
    2. [latex]\text{Ca}{\text{(OH)}}_{2}\text{(}aq\text{)}+2\text{HBr(}aq\text{)}\rightarrow{\text{CaBr}}_{2}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}[/latex]
    3. [latex]{\text{C}}_{6}{\text{H}}_{12}\text{(}l\text{)}+9{\text{O}}_{2}\text{(}g\text{)}\rightarrow 6{\text{CO}}_{2}\text{(}g\text{)}+6{\text{H}}_{2}\text{O(}g\text{)}[/latex]
  2. 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.
  3. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
    1. [latex]\ce{H3PO4}[/latex]
    2. [latex]\ce{Al(OH)3}[/latex]
    3. [latex]\ce{SeO2}[/latex]
    4. [latex]\ce{KNO2}[/latex]
    5. [latex]\ce{In2S3}[/latex]
    6. [latex]\ce{P4O6}[/latex]
  4. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.
    1. [latex]\text{Al(}s\text{)}+{\text{F}}_{2}\text{(}g\text{)}\rightarrow[/latex]
    2. [latex]\text{Al(}s\text{)}+{\text{CuBr}}_{2}\text{(}aq\text{)}\rightarrow[/latex] (single displacement)
    3. [latex]{\text{P}}_{4}\text{(}s\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow[/latex]
    4. [latex]\text{Ca(}s\text{)}+{\text{H}}_{2}\text{O(}l\text{)}\rightarrow[/latex] (products are a strong base and a diatomic gas)
  5. Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method):
    1. [latex]{\text{Sn}}^{\text{4+}}\text{(}aq\text{)}\rightarrow{\text{Sn}}^{\text{2+}}\text{(}aq\text{)}[/latex]
    2. [latex]{\left[\text{Ag}{\text{(}{\text{NH}}_{3}\text{)}}_{2}\right]}^{+}\text{(}aq\text{)}\rightarrow\text{Ag}\text{(}s\text{)}+{\text{NH}}_{3}\text{(}aq\text{)}[/latex]
    3. [latex]{\text{Hg}}_{2}{\text{Cl}}_{2}\text{(}s\text{)}\rightarrow\text{Hg}\text{(}l\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}[/latex]
    4. [latex]{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{O}}_{2}\text{(}\text{in acidic solution}\text{)}[/latex]
    5. [latex]{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}[/latex]
    6. [latex]{\text{SO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{) (in acidic solution)}[/latex]
    7. [latex]{\text{MnO}}_{4}{}^{-}\text{(}aq\text{)}\rightarrow{\text{Mn}}^{\text{2+}}\text{(}aq\text{) (in acidic solution)}[/latex]
    8. [latex]{\text{Cl}}^{-}\text{(}aq\text{)}\rightarrow{\text{ClO}}_{3}{}^{-}\text{(}aq\text{) (in basic solution)}[/latex]
Show Selected Solutions
  1. (a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)
  2. An oxidation-reduction reaction, because the oxidation state of the silver changes during the reaction.
  3. (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
  4. The answers are as follows:
    1. [latex]2\text{Al(}s\text{)}+3{\text{F}}_{2}\text{(}g\text{)}\rightarrow 2{\text{AlF}}_{3}\text{(}s\text{)};[/latex]
    2.  [latex]2\text{Al(}s\text{)}+3{\text{CuBr}}_{2}\text{(}aq\text{)}\rightarrow 3\text{Cu(}s\text{)}+2{\text{AlB}}_{3}\text{(}aq\text{)};[/latex]
    3.  [latex]{\text{P}}_{4}\text{(}s\text{)}+5{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{P}}_{4}{\text{O}}_{10}\text{(}s\text{)};[/latex]
    4.  [latex]\text{Ca(}s\text{)}+2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow\text{Ca}{\text{(OH)}}_{2}\text{(}aq\text{)}+{\text{H}}_{2}\text{(}g\text{)}[/latex]
  5. The answers are as follows:
    1. [latex]\begin{array}{l}\\ S{\text{n}}^{\text{4+}}\text{(}aq\text{)}\rightarrow{\text{Sn}}^{\text{2+}}\text{(}aq\text{)}\\ {\text{Sn}}^{\text{4+}}\text{(}aq\text{)}+{\text{Se}}^{-}\rightarrow{\text{Sn}}^{\text{2+}}\text{(}aq\text{)}\end{array}[/latex],
    2.  [latex]\begin{array}{l}{\left[\text{Ag}{\text{(}{\text{NH}}_{3}\text{)}}_{2}\right]}^{+}\text{(}aq\text{)}\rightarrow\text{Ag(}s\text{)}+2{\text{NH}}_{3}\text{(}aq\text{)}\\ {\left[\text{Ag}{\text{(}{\text{NH}}_{3}\text{)}}_{2}\right]}^{+}\text{(}aq\text{)}+{\text{e}}^{-}\rightarrow\text{Ag(}s\text{)}+2{\text{NH}}_{3}\text{(}aq\text{)}\end{array};[/latex]
    3.  [latex]\begin{array}{l}{\text{Hg}}_{2}{\text{Cl}}_{2}\text{(}s\text{)}\rightarrow\text{Hg(}l\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}\\ {\text{Hg}}_{2}{\text{Cl}}_{2}\text{(}s\text{)}\rightarrow 2\text{Hg(}l\text{)}+2{\text{Cl}}^{-}\text{(}aq\text{)}\\ {\text{Hg}}_{2}{\text{Cl}}_{2}\text{(}s\text{)}+2{\text{e}}^{-}\rightarrow 2\text{Hg(}l\text{)}+2{\text{Cl}}^{-}\text{(}aq\text{)}\end{array};[/latex]
    4.  [latex]\begin{array}{l}2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{O}}_{2}\text{(}g\text{)}\\ 2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{O}}_{2}\text{(}g\text{)}+4{\text{H}}^{\text{+}}\text{(}aq\text{)}\\ 2{\text{H}}_{2}\text{O(}l\text{)}\rightarrow{\text{O}}_{2}\text{(}g\text{)}+4{\text{H}}^{\text{+}}\text{(}aq\text{)}+4{\text{e}}^{-}\end{array};[/latex]
    5.  [latex]\begin{array}{l}{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}\\ 2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}\\ 2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}+6{\text{H}}_{2}\text{O(}l\text{)}\\ 12{\text{H}}^{\text{+}}\text{(}aq\text{)}+2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}+6{\text{H}}_{2}\text{O(}l\text{)}\\ 12{\text{H}}^{\text{+}}\text{(}aq\text{)}+\mathbf{12}\mathbf{\text{O}}{\mathbf{\text{H}}}^{\mathbf{-}}\left(\text{aq}\right)+2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}+6{\text{H}}_{2}\text{O(}l\text{)}+\mathbf{12}\mathbf{\text{O}}{\mathbf{\text{H}}}^{\mathbf{-}}\left(\text{aq}\right)\\ 12{\text{H}}_{2}\text{O(}l\text{)}+2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}+6{\text{H}}_{2}\text{O(}l\text{)}+12{\text{OH}}^{-}\text{(}aq\text{)}\\ 6{\text{H}}_{2}\text{O(}l\text{)}+2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}\rightarrow{\text{I}}_{2}\text{(}s\text{)}+12{\text{OH}}^{-}\text{(}aq\text{)}\\ 6{\text{H}}_{2}\text{O(}l\text{)}+2{\text{IO}}_{3}{}^{-}\text{(}aq\text{)}10{\text{e}}^{-}\rightarrow{\text{I}}_{2}\text{(}s\text{)}+12{\text{OH}}^{-}\text{(}aq\text{)}\end{array};[/latex]
    6.  [latex]\begin{array}{l}\\ {\text{SO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\\ {\text{H}}_{2}\text{O(}l\text{)}+{\text{SO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}\\ {\text{H}}_{2}\text{O(}l\text{)}+{\text{SO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}+2{\text{H}}^{\text{+}}\text{(}aq\text{)}\\ {\text{H}}_{2}\text{O(}l\text{)}+{\text{SO}}_{3}{}^{\text{2-}}\text{(}aq\text{)}\rightarrow{\text{SO}}_{4}{}^{\text{2-}}\text{(}aq\text{)}+2{\text{H}}^{\text{+}}\text{(}aq\text{)}+2{\text{e}}^{-}\end{array};[/latex]
    7.  [latex]\begin{array}{l}\\ {\text{MnO}}_{4}{}^{-}\text{(}aq\text{)}\rightarrow{\text{Mn}}^{\text{2+}}\text{(}aq\text{)}\\ {\text{MnO}}_{4}{}^{-}\text{(}aq\text{)}\rightarrow{\text{Mn}}^{\text{2+}}\text{(}aq\text{)}+4{\text{H}}_{2}\text{O(}l\text{)}\\ 8{\text{H}}^{\text{+}}\text{(}aq\text{)}+{\text{MnO}}_{4}{}^{-}\text{(}aq\text{)}\rightarrow{\text{Mn}}^{\text{2+}}\text{(}aq\text{)}+4{\text{H}}_{2}\text{O(}l\text{)}\\ 8{\text{H}}^{\text{+}}\text{(}aq\text{)}+{\text{MnO}}_{4}{}^{-}\text{(}aq\text{)}+5{\text{e}}^{-}\rightarrow{\text{Mn}}^{\text{2+}}\text{(}aq\text{)}+4{\text{H}}_{2}\text{O(}l\text{)}\end{array};[/latex]
    8.  [latex]\begin{array}{l}{\text{Cl}}^{-}\text{(}aq\text{)}\rightarrow{\text{ClO}}_{3}{}^{-}\text{(}aq\text{)}\\ 3{\text{H}}_{2}\text{O(}l\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}\rightarrow{\text{ClO}}_{3}{}^{-}\text{(}aq\text{)}\\ 3{\text{H}}_{2}\text{O(}l\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}\rightarrow{\text{ClO}}_{3}{}^{-}\text{(}aq\text{)}+6{\text{H}}^{\text{+}}\text{(}aq\text{)}\\ 3{\text{H}}_{2}\text{O(}l\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}+\mathbf{6}{\mathbf{\text{OH}}}^{\mathbf{-}}\left(\text{a}\text{q}\right)\rightarrow{\text{ClO}}_{3}{}^{-}\text{(}aq\text{)}+6{\text{H}}^{\text{+}}\text{(}aq\text{)}+\mathbf{6}{\mathbf{\text{OH}}}^{\mathbf{-}}\left(\text{aq}\right)\\ 3{\text{H}}_{2}\text{O(}l\text{)}+{\text{Cl}}^{-}\text{(}aq\text{)}+6{\text{OH}}^{-}\text{(}aq\text{)}\rightarrow{\text{ClO}}_{3}{}^{-}\text{(}aq\text{)}+6{\text{H}}_{2}\text{O(}l\text{)}\\ {\text{Cl}}^{-}\text{(}aq\text{)}+6{\text{OH}}^{-}\text{(}aq\text{)}\rightarrow\text{Cl}{\text{O}}_{3}{}^{-}\text{(}aq\text{)}+3{\text{H}}_{2}\text{O(}l\text{)}+6{\text{e}}^{-}\end{array}[/latex]

Glossary

acid: substance that produces [latex]\ce{H3O+}[/latex] when dissolved in water

acid-base reaction: reaction involving the transfer of a hydrogen ion between reactant species

base: substance that produces [latex]\ce{​OH-}[/latex] when dissolved in water

complete ionic equation: chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions

combustion reaction: vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light

half-reaction: an equation that shows whether each reactant loses or gains electrons in a reaction.

insoluble: of relatively low solubility; dissolving only to a slight extent

molecular equation: chemical equation in which all reactants and products are represented as neutral substances

neutralization reaction: reaction between an acid and a base to produce salt and water

net ionic equation: chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)

oxidation: process in which an element’s oxidation number is increased by loss of electrons

oxidation-reduction reaction: (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements

oxidation number: (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic

oxidizing agent: (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced

precipitate: insoluble product that forms from reaction of soluble reactants

precipitation reaction: reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double-displacement or metathesis

reactant: substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation

reduction: process in which an element’s oxidation number is decreased by gain of electrons

reducing agent: (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized

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

single-displacement reaction: (also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species

soluble: of relatively high solubility; dissolving to a relatively large extent

solubility: the extent to which a substance may be dissolved in water, or any solvent

spectator ion: ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality

strong acid: acid that reacts completely when dissolved in water to yield hydronium ions

strong base: base that reacts completely when dissolved in water to yield hydroxide ions

weak acid: acid that reacts only to a slight extent when dissolved in water to yield hydronium ions

weak base: base that reacts only to a slight extent when dissolved in water to yield hydroxide ions

definition

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Chemistry Fundamentals Copyright © by Dr. Julie Donnelly, Dr. Nicole Lapeyrouse, and Dr. Matthew Rex is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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