5.1.1 Updated 10/30/2018

5.1: THE NATURE OF AQUEOUS SOLUTIONS

SKILLS TO DEVELOP

To understand how and why solutions form

The solvent in aqueous solutions is water, which makes up about 70% of the mass of the human body and is essential for life. Many ofthe chemical reactions that keep us alive depend on the interaction of water molecules with dissolved compounds. Moreover, thepresence of large amounts of water on Earth’s surface helps maintain its surface temperature in a range suitable for life. In this section,we describe some of the interactions of water with various substances and introduce you to the characteristics of aqueous solutions.

POLAR SUBSTANCES

As shown in Figure 5.1.1, the individual water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. As is typical of group 16 elements, the oxygen atom in each O–H covalent bond attracts electrons more strongly thanthe hydrogen atom does. Consequently, the oxygen and hydrogen nuclei do not equally share electrons. Instead, hydrogen atoms areelectron poor compared with a neutral hydrogen atom and have a partial positive charge, which is indicated by δ . The oxygen atom, incontrast, is more electron rich than a neutral oxygen atom, so it has a partial negative charge. This charge must be twice as large as thepartial positive charge on each hydrogen for the molecule to have a net charge of zero. Thus its charge is indicated by 2δ . Thisunequal distribution of charge creates a polar bond in which one portion of the molecule carries a partial negative charge, while theother portion carries a partial positive charge (Figure 5.1.1). Because of the arrangement of polar bonds in a water molecule, water isdescribed as a polar substance.

Figure 5.1.1: The Polar Nature of Water. Each water molecule consists of two hydrogen atoms bonded to an oxygen atom in a bent (V-shaped) structure. Because the oxygen atom attracts electrons more strongly than the hydrogen atoms do, the oxygen atom is partially

negatively charged (2δ ; blue) and the hydrogen atoms are partially positively charged (δ ; red). For the molecule to have a netcharge of zero, the partial negative charge on oxygen must be twice as large as the partial positive charge on each hydrogen.

Because of the asymmetric charge distribution in the water molecule, adjacent water molecules are held together by attractiveelectrostatic (δ …δ ) interactions between the partially negatively charged oxygen atom of one molecule and the partially positivelycharged hydrogen atoms of adjacent molecules (Figure 5.1.2). Energy is needed to overcome these electrostatic attractions. In fact,without them, water would evaporate at a much lower temperature, and neither Earth’s oceans nor we would exist!

+

−

− +

+ −

5.1.2 Updated 10/30/2018

Figure 5.1.2: The Structure of Liquid Water. Two views of a water molecule are shown: (a) a ball-and-stick structure and (b) a space-filling model. Water molecules are held together by electrostatic attractions (dotted lines) between the partially negatively charged

oxygen atom of one molecule and the partially positively charged hydrogen atoms on adjacent molecules. As a result, the watermolecules in liquid water form transient networks with structures similar to that shown. Because the interactions between water

molecules are continually breaking and reforming, liquid water does not have a single fixed structure.

As you learned previously,, ionic compounds such as sodium chloride (NaCl) are also held together by electrostatic interactions—inthis case, between oppositely charged ions in the highly ordered solid, where each ion is surrounded by ions of the opposite charge in afixed arrangement. In contrast to an ionic solid, the structure of liquid water is not completely ordered because the interactions betweenmolecules in a liquid are constantly breaking and reforming.

The unequal charge distribution in polar liquids such as water makes them good solvents for ionic compounds. When an ionic soliddissolves in water, the ions dissociate. That is, the partially negatively charged oxygen atoms of the H O molecules surround thecations (Na in the case of NaCl), and the partially positively charged hydrogen atoms in H O surround the anions (Cl ; Figure 5.1.3).Individual cations and anions that are each surrounded by their own shell of water molecules are called hydrated ions. We can describethe dissolution of NaCl in water as

where (aq) indicates that Na and Cl are hydrated ions.

Figure 5.1.3: The Dissolution of Sodium Chloride in Water. An ionic solid such as sodium chloride dissolves in water because of theelectrostatic attraction between the cations (Na ) and the partially negatively charged oxygen atoms of water molecules, and between

the anions (Cl ) and the partially positively charged hydrogen atoms of water.

NOTE

Polar liquids are good solvents for ionic compounds.

ELECTROLYTES

When electricity, in the form of an electrical potential, is applied to a solution, ions in solution migrate toward the oppositely chargedrod or plate to complete an electrical circuit, whereas neutral molecules in solution do not (Figure 5.1.4). Thus solutions that containions conduct electricity, while solutions that contain only neutral molecules do not. Electrical current will flow through the circuitshown in Figure 5.1.4 and the bulb will glow only if ions are present. The lower the concentration of ions in solution, the weaker thecurrent and the dimmer the glow. Pure water, for example, contains only very low concentrations of ions, so it is a poor electricalconductor.

2+

2−

NaCl(s) N (aq) + C (aq)− →−−−O(l)H2

a+ l− (5.1.1)

+ −

+

−

5.1.3 Updated 10/30/2018

NOTE

Solutions that contain ions conduct electricity.

Figure 5.1.4: The conductivity of electrolyte solutions: (a) 0.1 M NaCl; (b) 0.05 M NaCl; (c) 0.1 M HgCl . An electrolyte solutionconducts electricity because of the movement of ions in the solution. The larger the concentration of ions, the better the solution

conducts. Weak electrolytes, such as HgCl , conduct badly because they produce very few ions when dissolved and exist mainly in theform of molecules. Figure used with permission from ChemePrime.

An electrolyte is any compound that can form ions when dissolved in water (c.f. nonelectrolytes). Electrolytes may be strong or weak.is any compound that can form ions when it dissolves in water. When strong electrolytes dissolve, the constituent ions dissociatecompletely due to strong electrostatic interactions with the solvent, producing aqueous solutions that conduct electricity very well(Figure 5.1.4). Examples include ionic compounds such as barium chloride ( ) and sodium hydroxide (NaOH), which are bothstrong electrolytes and dissociate as follows:

The single arrows from reactant to products in Equation 5.1.2 and Equation 5.1.3 indicate that dissociation is complete.

When weak electrolytes dissolve, they produce relatively few ions in solution. This does not mean that the compounds do not dissolvereadily in water; many weak electrolytes contain polar bonds and are therefore very soluble in a polar solvent such as water. They donot completely dissociate to form ions, however, because of their weaker electrostatic interactions with the solvent. Because very fewof the dissolved particles are ions, aqueous solutions of weak electrolytes do not conduct electricity as well as solutions of strongelectrolytes. One such compound is acetic acid (CH CO H), which contains the –CO H unit. Although it is soluble in water, it is aweak acid and therefore also a weak electrolyte. Similarly, ammonia (NH ) is a weak base and therefore a weak electrolyte. Thebehavior of weak acids and weak bases will be described in more detail elsewhere.

2

2

BaCl2

BaC (s) B (aq) + 2C (aq)l2 − →−−−O(l)H2

a2+ l− (5.1.2)

NaOH(s) N (aq) + O (aq)− →−−−O(l)H2

a+ H − (5.1.3)

3 2 2

3

5.1.4 Updated 10/30/2018

General structure of an aldehyde and a ketone. Notice that both contain the C=O group.

Nonelectrolytes (a substance that dissolves in water to form neutral molecules and has essentially no effect on electrical conductivity)that dissolve in water do so as neutral molecules and thus have essentially no effect on conductivity. Examples of nonelectrolytes thatare very soluble in water but that are essentially nonconductive are ethanol, ethylene glycol, glucose, and sucrose, all of which containthe –OH group that is characteristic of alcohols. The topic of why alcohols and carboxylic acids behave differently in aqueous solutionis for a different Module; for now, however, you can simply look for the presence of the –OH and –CO H groups when trying topredict whether a substance is a strong electrolyte, a weak electrolyte, or a nonelectrolyte. The distinctions between soluble andinsoluble substances and between strong, weak, and nonelectrolytes are illustrated in Figure 5.1.5.

Figure 5.1.5: The Difference between Soluble and Insoluble Compounds (a) and Strong, Weak, and Nonelectrolytes (b). When asoluble compound dissolves, its constituent atoms, molecules, or ions disperse throughout the solvent. In contrast, the constituents ofan insoluble compound remain associated with one another in the solid. A soluble compound is a strong electrolyte if it dissociatescompletely into ions, a weak electrolyte if it dissociates only slightly into ions, and a nonelectrolyte if it dissolves to produce only

neutral molecules.

NOTE

Ionic substances and carboxylic acids are electrolytes; alcohols, aldehydes, and ketones are nonelectrolytes.

EXAMPLE 5.1.1

Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.

a. formaldehyde

b. cesium chloride

Given: compound

Asked for: relative ability to form ions in water

Strategy:

A Classify the compound as ionic or covalent.

B If the compound is ionic and dissolves, it is a strong electrolyte that will dissociate in water completely to produce a solutionthat conducts electricity well. If the compound is covalent and organic, determine whether it contains the carboxylic acid group. Ifthe compound contains this group, it is a weak electrolyte. If not, it is a nonelectrolyte.

2

5.1.5 Updated 10/30/2018

Solution:

1. A Formaldehyde is an organic compound, so it is covalent. B It contains an aldehyde group, not a carboxylic acid group, so itshould be a nonelectrolyte.

2. A Cesium chloride (CsCl) is an ionic compound that consists of Cs and Cl ions. B Like virtually all other ionic compoundsthat are soluble in water, cesium chloride will dissociate completely into Cs (aq) and Cl (aq) ions. Hence it should be a strongelectrolyte.

EXERCISE 5.1.1

Predict whether each compound is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in water.

a. (CH ) CHOH (2-propanol)

b. ammonium sulfate

Answer:

a. nonelectrolyteb. strong electrolyte

SUMMARY

Aqueous solutions can be classified as polar or nonpolar depending on how well they conduct electricity.

Most chemical reactions are carried out in solutions, which are homogeneous mixtures of two or more substances. In a solution, asolute (the substance present in the lesser amount) is dispersed in a solvent (the substance present in the greater amount). Aqueoussolutions contain water as the solvent, whereas nonaqueous solutions have solvents other than water.

Polar substances, such as water, contain asymmetric arrangements of polar bonds, in which electrons are shared unequally betweenbonded atoms. Polar substances and ionic compounds tend to be most soluble in water because they interact favorably with itsstructure. In aqueous solution, dissolved ions become hydrated; that is, a shell of water molecules surrounds them.

Substances that dissolve in water can be categorized according to whether the resulting aqueous solutions conduct electricity. Strongelectrolytes dissociate completely into ions to produce solutions that conduct electricity well. Weak electrolytes produce a relativelysmall number of ions, resulting in solutions that conduct electricity poorly. Nonelectrolytes dissolve as uncharged molecules and haveno effect on the electrical conductivity of water.

CONTRIBUTORS

Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC)

+ −

+ −

3 2

5.2.1 Updated 10/30/2018

5.2: PRECIPITATION REACTIONS

SKILLS TO DEVELOP

To identify a precipitation reaction and predict solubilities.

A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. We described a precipitation reaction inwhich a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silverdichromate:

This equation has the general form of an exchange reaction:

Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because bothcomponents of each compound change partners, such reactions are sometimes called double-displacement reactions. Two important uses of precipitationreactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling.

Video: Mixing Potassium Chromate and Silver Nitrate together to initiate a precipitation reaction (Equation ).

While chemical equations show the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little aboutwhat was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allowus to see similarities between reactions that might not otherwise be apparent.

Let’s consider the reaction of silver nitrate with potassium dichromate. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silverdichromate forms as a red solid. The overall balanced chemical equation for the reaction shows each reactant and product as undissociated, electricallyneutral compounds:

Although Equation gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionicsubstances such as AgNO and K Cr O are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag Cr O isnot very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as acomplete ionic equation showing which ions and molecules are hydrated and which are present in other forms and phases:

Note that K (aq) and NO (aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectatorions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species thatparticipate in the chemical reaction:

AgN (aq) + C (aq) → A C (s) + KN (aq)O3 K2 r2O7 g2 r2O7 O3 (5.2.1)

AC + BD → + BCADinsoluble

(5.2.2)

5.2.3

2AgN (aq) + C (aq) → A C (s) + 2KN (aq)O3 K2 r2O7 g2 r2O7 O3 (5.2.3)

5.2.3

3 2 2 7 2 2 7

2A (aq) + 2N (aq) + 2 (aq) + C (aq) → A C (s) + 2 (aq) + 2N (aq)g+ O−3 K+ r2O2−

7 g2 r2O7 K+ O−3 (5.2.4)

+3−

5.2.2 Updated 10/30/2018

Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, thesum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation , the charge on the left side is2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag Cr O formula unit.

By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reactionbetween silver fluoride and ammonium dichromate is as follows:

The complete ionic equation for this reaction is as follows:

Because two NH (aq) and two F (aq) ions appear on both sides of Equation , they are spectator ions. They can therefore be canceled to give the netionic equation (Equation ), which is identical to Equation :

If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example,we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.

EXAMPLE

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueoussodium phosphate to give solid barium phosphate and a solution of sodium nitrate.

Given: reactants and products

Asked for: overall, complete ionic, and net ionic equations

Strategy:

Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionicequation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.

Solution:

From the information given, we can write the unbalanced chemical equation for the reaction:

Because the product is Ba (PO ) , which contains three Ba ions and two PO ions per formula unit, we can balance the equation by inspection:

This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the completeionic equation, we write each soluble reactant and product in dissociated form:

The six NO (aq) ions and the six Na (aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionicequation:

2A (aq) + C (aq) → A C (s)g+ r2O2−7 g2 r2O7 (5.2.5)

5.2.5

2 2 7

2AgF(aq) + (N C (aq) → A C (s) + 2N F(aq)H4)2 r2O7 g2 r2O7 H4 (5.2.6)

2A (aq) + 2 (aq) + 2N (aq) + C (aq) → A C (s) + 2N (aq) + 2 (aq)g+ F − H +4 r2O2−

7 g2 r2O7 H +4 F − (5.2.7)

4+ − 5.2.7

5.2.8 5.2.5

2A (aq) + C (aq) → A C (s)g+ r2O2−7 g2 r2O7 (5.2.8)

5.2.1

Ba(N (aq) + N P (aq) → B (P (s) + NaN (aq)O3)2 a3 O4 a3 O4)2 O3

3 4 22+

43−

3Ba(N (aq) + 2N P (aq) → B (P (s) + 6NaN (aq)O3)2 a3 O4 a3 O4)2 O3

3B (aq) + 6N (aq) + 6N (aq) + 2P (aq) → B (P (s) + 6N (aq) + 6N (aq)a2+ O−3 a+ O3−

4 a3 O4)2 a+ O−3

3− +

3B (aq) + 2P (aq) → B (P (s)a2+ O3−4 a3 O4)2

5.2.3 Updated 10/30/2018

EXERCISE : MIXING SILVER FLUORIDE WITH SODIUM PHOSPHATE

Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueoussodium phosphate to give solid silver phosphate and a solution of sodium fluoride.

Answer

overall chemical equation:

complete ionic equation:

net ionic equation:

So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry,however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identitiesof the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could befurther from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, youmust begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.

The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually presentin the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make thatkind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions (acid–base,exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactionsthat occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.

PREDICTING SOLUBILITIES

Table gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, weidentify each species in the solution and then refer to Table to see which, if any, combination(s) of cation and anion are likely to produce an insolublesalt. In doing so, it is important to recognize that soluble and insoluble are relative terms that span a wide range of actual solubilities. We will discusssolubilities in more detail later, where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an“insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete.

Table : Guidelines for Predicting the Solubility of Ionic Compounds in Water

Soluble Exceptions

Rule 1 most salts that contain an alkali metal (Li , Na , K , Rb , and Cs ) and ammonium (NH )

Rule 2 most salts that contain the nitrate (NO ) anion

Rule 3 most salts of anions derived from monocarboxylic acids (e.g., CH CO ) but not silver acetate and salts of long-chaincarboxylates

Rule 4 most chloride, bromide, and iodide salts but not salts of metal ions located on the lowerright side of the periodic table (e.g., Cu ,Ag , Pb , and Hg ).

Insoluble Exceptions

Rule 5 most salts that contain the hydroxide (OH ) and sulfide (S ) anions but not salts of the alkali metals (group 1), theheavier alkaline earths (Ca , Sr , andBa in group 2), and the NH ion.

Rule 6 most carbonate (CO ) and phosphate (PO ) salts but not salts of the alkali metals or the NH ion.

Rule 7 most sulfate (SO ) salts that contain main group cations with a charge ≥ +2 but not salts of +1 cations, Mg , and dipositivetransition metal cations (e.g., Ni )

Just as important as predicting the product of a reaction is knowing when a chemical reaction will not occur. Simply mixing solutions of two differentchemical substances does not guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na (aq), 0.50 M Cl (aq), 0.50 M K (aq), and 0.50 M Br (aq). As

5.2.1

3AgF(aq) + N P (aq) → A P (s) + 3NaF(aq)a3 O4 g3 O4

3A (aq) + 3 (aq) + 3N (aq) + P (aq) → A P (s) + 3N (aq) + 3 (aq)g+ F − a+ O3−4 g3 O4 a+ F −

3A (aq) + P (aq) → A P (s)g+ O3−4 g3 O4

5.2.2

5.2.2

5.2.2

+ + + + +4+

3−

3 2−

++ 2+

22+

− 2−2+ 2+

2+4+

32−

43−

4+

42− 2+

2+

+ − + −

5.2.4 Updated 10/30/2018

you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute eachsolution with the other (Figure ).

Figure : The Effect of Mixing Aqueous KBr and NaCl Solutions. Because no net reaction occurs, the only effect is to dilute each solution with the other.(Water molecules are omitted from molecular views of the solutions for clarity.)

EXAMPLE

Using the information in Table , predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reactionthat occurs.

a. Aqueous solutions of barium chloride and lithium sulfate are mixed.b. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.c. Aqueous solutions of strontium bromide and aluminum nitrate are mixed.d. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.

Given: reactants

Asked for: reaction and net ionic equation

Strategy:

A. Identify the ions present in solution and write the products of each possible exchange reaction.B. Refer to Table to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net

ionic equation for the reaction.

Solution:

A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituentanions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba , Cl , Li , and SO ions. The only possible exchangereaction is to form LiCl and BaSO :

B We now need to decide whether either of these products is insoluble. Table shows that LiCl is soluble in water (rules 1 and 4), but BaSO is notsoluble in water (rule 5). Thus BaSO will precipitate according to the net ionic equation

Although soluble barium salts are toxic, BaSO is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbedinto tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—asuspension of very fine BaSO particles in water.

5.2.1

5.2.1

5.2.2

5.2.2

5.2.2

2+ − +42−

4

5.2.2 4

4

B (aq) + S (aq) → BaS (s)a2+ O2−4 O4 (5.2.9)

4

4

5.2.5 Updated 10/30/2018

An x-ray of the digestive organs of a patient who has swallowed a “barium milkshake.” A barium milkshake is a suspension of very fine BaSOparticles in water; the high atomic mass of barium makes it opaque to x-rays. Image used with permission from Wikipedia.

1. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resultingsolution initially contains Rb , OH , Co , and Cl ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II)hydroxide):

B According to Table , RbCl is soluble (rules 1 and 4), but Co(OH) is not soluble (rule 5). Hence Co(OH) will precipitate according to thefollowing net ionic equation:

2. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr , Br , Al , andNO ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:

B According to Table , both AlBr (rule 4) and Sr(NO ) (rule 2) are soluble. Thus no net reaction will occur.

1. A According to Table , lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb and CH CO ions. Because thesolution also contains NH and I ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:

B According to Table , ammonium acetate is soluble (rules 1 and 3), but PbI is insoluble (rule 4). Thus Pb(C H O ) will dissolve, and PbIwill precipitate. The net ionic equation is as follows:

4

+ − 2+ −

5.2.2 2 2

C (aq) + 2O (aq) → Co(OH (s)o2+ H − )2

2+ − 3+

3−

5.2.2 3 3 2

5.2.2 2+3 2

−

4+ −

5.2.2 2 2 3 2 2 2

P (aq) + 2 (aq) → Pb (s)b2+ I − I2

5.2.6 Updated 10/30/2018

EXERCISE

Using the information in Table , predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reactionthat occurs.

a. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.b. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.c. Solid sodium fluoride is added to an aqueous solution of ammonium formate.d. Aqueous solutions of calcium bromide and cesium carbonate are mixed.

Answer:

a. b.

c. dissolves; no net reactiond.

PRECIPITATION REACTIONS IN PHOTOGRAPHY

Precipitation reactions can be used to recover silver from solutions used to develop conventional photographic film. Although largely supplanted by digitalphotography, conventional methods are often used for artistic purposes. Silver bromide is an off-white solid that turns black when exposed to light, which isdue to the formation of small particles of silver metal. Black-and-white photography uses this reaction to capture images in shades of gray, with the darkestareas of the film corresponding to the areas that received the most light. The first step in film processing is to enhance the black/white contrast by using adeveloper to increase the amount of black. The developer is a reductant: because silver atoms catalyze the reduction reaction, grains of silver bromide thathave already been partially reduced by exposure to light react with the reductant much more rapidly than unexposed grains.

Figure : Outline of the Steps Involved in Producing a Black-and-White Photograph

After the film is developed, any unexposed silver bromide must be removed by a process called “fixing”; otherwise, the entire film would turn black withadditional exposure to light. Although silver bromide is insoluble in water, it is soluble in a dilute solution of sodium thiosulfate (Na S O ; photographer’shypo) because of the formation of [Ag(S O ) ] ions. Thus washing the film with thiosulfate solution dissolves unexposed silver bromide and leaves apattern of metallic silver granules that constitutes the negative. This procedure is summarized in Figure . The negative image is then projected ontopaper coated with silver halides, and the developing and fixing processes are repeated to give a positive image. (Color photography works in much the sameway, with a combination of silver halides and organic dyes superimposed in layers.) “Instant photo” operations can generate more than a hundred gallons ofdilute silver waste solution per day. Recovery of silver from thiosulfate fixing solutions involves first removing the thiosulfate by oxidation and thenprecipitating Ag ions with excess chloride ions.

EXAMPLE

A silver recovery unit can process 1500 L of photographic silver waste solution per day. Adding excess solid sodium chloride to a 500 mL sample of thewaste (after removing the thiosulfate as described previously) gives a white precipitate that, after filtration and drying, consists of 3.73 g of AgCl. Whatmass of NaCl must be added to the 1500 L of silver waste to ensure that all the Ag ions precipitate?

Given: volume of solution of one reactant and mass of product from a sample of reactant solution

Asked for: mass of second reactant needed for complete reaction

Strategy:

5.2.2

5.2.2

F (aq) + 2O (aq) → Fe(OH (s)e2+ H − )2

2P (aq) + 3H (aq) → H (P (s)O3−4 g2+ g3 O4)2

NaF(s)

C (aq) + C (aq) → CaC (s)a2+ O2−3 O3

5.2.2

2 2 3

2 3 23−

5.2.2

+

5.2.3

+

5.2.7 Updated 10/30/2018

A. Write the net ionic equation for the reaction. Calculate the number of moles of AgCl obtained from the 500 mL sample and then determine theconcentration of Ag in the sample by dividing the number of moles of AgCl formed by the volume of solution.

B. Determine the total number of moles of Ag in the 1500 L solution by multiplying the Ag concentration by the total volume.C. Use mole ratios to calculate the number of moles of chloride needed to react with Ag . Obtain the mass of NaCl by multiplying the number of moles

of NaCl needed by its molar mass.

Solution:

We can use the data provided to determine the concentration of Ag ions in the waste, from which the number of moles of Ag in the entire wastesolution can be calculated. From the net ionic equation, we can determine how many moles of Cl are needed, which in turn will give us the mass ofNaCl necessary.

A The first step is to write the net ionic equation for the reaction:

We know that 500 mL of solution produced 3.73 g of AgCl. We can convert this value to the number of moles of AgCl as follows:

Therefore, the 500 mL sample of the solution contained 0.0260 mol of Ag . The Ag concentration is determined as follows:

B The total number of moles of Ag present in 1500 L of solution is as follows:

C According to the net ionic equation, one Cl ion is required for each Ag ion. Thus 78.1 mol of NaCl are needed to precipitate the silver. Thecorresponding mass of NaCl is

Note that 78.1 mol of AgCl correspond to 8.43 kg of metallic silver, which is worth about $7983 at 2011 prices ($32.84 per troy ounce). Silver recoverymay be economically attractive as well as ecologically sound, although the procedure outlined is becoming nearly obsolete for all but artistic purposeswith the growth of digital photography.

EXERCISE

Because of its toxicity, arsenic is the active ingredient in many pesticides. The arsenic content of a pesticide can be measured by oxidizing arseniccompounds to the arsenate ion (AsO ), which forms an insoluble silver salt (Ag AsO ). Suppose you are asked to assess the purity of technical gradesodium arsenite (NaAsO ), the active ingredient in a pesticide used against termites. You dissolve a 10.00 g sample in water, oxidize it to arsenate, anddilute it with water to a final volume of 500 mL. You then add excess AgNO solution to a 50.0 mL sample of the arsenate solution. The resultingprecipitate of Ag AsO has a mass of 3.24 g after drying. What is the percentage by mass of NaAsO in the original sample?

Answer: 91.0%

NOTE

Precipitation reactions are a subclass of double displacement reactions.

SUMMARY

A complete ionic equation consists of the net ionic equation and spectator ions. Predicting the solubility of ionic compounds in water can give insight intowhether or not a reaction will occur. The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows allthe substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist insolution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equationwith the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.

+

+ +

+

+ +

−

C (aq) + A (aq) → AgCl(s)l− g+

moles AgCl = = 3.73 = 0.0260 mol AgClgrams AgCl

molar mass AgClg AgCl

⎛

⎝

1 mol AgCl

143.32 g AgCl

⎞

⎠(5.2.10)

+ +

[A ] = = = 0.0520 Mg+ moles Ag+

liters soln

0.0260 mol AgCl

0.500 L(5.2.11)

+

moles A = 1500 ( ) = 78.1 mol Ag+ L0.520 mol

1 Lg+ (5.2.12)

− +

mass NaCl = 78.1 ( ) = 4560 g NaCl = 4.56 kg NaClmol NaCl58.44 g NaCl

1 mol NaCl(5.2.13)

5.2.3

43−

3 4

2

3

3 4 2

5.2.8 Updated 10/30/2018

In a precipitation reaction, a subclass of exchange reactions, an insoluble material (a precipitate) forms when solutions of two substances are mixed. Topredict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce aninsoluble salt.

CONTRIBUTORS

Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC)

5.3.1 Updated 10/30/2018

5.3: ACID-BASE REACTIONS

SKILLS TO DEVELOP

To know the characteristic properties of acids and bases.

Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in ourhomes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids arebases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carriedout an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids andbases.

DEFINITIONS OF ACIDS AND BASES

We defined acids as substances that dissolve in water to produce H ions, whereas bases were defined as substances that dissolve inwater to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases havebeen known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned moreabout them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent colorchanges in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and asolution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO ) with the evolution of carbondioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyesthat differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions wereuseful, they were entirely descriptive.

Although we include their tastes among the common characteristics of acids and bases, we never advocate tasting an unknownchemical!

THE ARRHENIUS DEFINITION OF ACIDS AND BASES

The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry,1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H ions(protons; Equation 5.3.1), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH ) ions(Equation 5.3.2):

According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H and OH ions,respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations:

1. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substancesin aqueous solution.

+

−

3

+

−

+CHCl(g)an Arrhenius acid

− →−−−H2O(l)

H+(aq) l−(aq) (5.3.1)

N +ONaOH(s)an Arrhenius base

− →−−−H2O(l)

a+(aq) H−

(aq) (5.3.2)

+ −

5.3.2 Updated 10/30/2018

2. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce and ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the

reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride (Equation ) is not an acid–base reactionbecause it does not involve and :

THE BRØNSTED–LOWRY DEFINITION OF ACIDS AND BASES

Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposedindependently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), whodefined acid–base reactions in terms of the transfer of a proton (H ion) from one substance to another.

According to Brønsted and Lowry, an acid (A substance with at least one hydrogen atom that can dissociate to form an anion and an ion (a proton) in aqueous solution, thereby forming an acidic solution) is any substance that can donate a proton, and a base (a

substance that produces one or more hydroxide ions ( and a cation when dissolved in aqueous solution, thereby forming a basicsolution) is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrheniusdefinition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more generalbecause the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form

, so in Equation , is a Brønsted–Lowry base and is a Brønsted–Lowry acid. Because of its more general nature,the Brønsted–Lowry definition is used throughout this text unless otherwise specified.

POLYPROTIC ACIDS

Acids differ in the number of protons they can donate. For example, monoprotic acids (a compound that is capable of donating oneproton per molecule) are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl,HBr, HI, HNO , and HNO . All carboxylic acids that contain a single −CO H group, such as acetic acid (CH CO H), are monoproticacids, dissociating to form RCO and H . A compound that can donate more than one proton per molecule. can donate more than oneproton per molecule. For example, H SO can donate two H ions in separate steps, so it is a diprotic acid (a compound that can donatetwo protons per molecule in separate steps) and H PO , which is capable of donating three protons in successive steps, is a triproticacid (a compound that can donate three protons per molecule in separate steps), (Equations , , and ):

In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occursimultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactantsand products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. Thereaction is then said to be in equilibrium (the point at which the rates of the forward and reverse reactions become the same, so that thenet composition of the system no longer changes with time).

H+

OH−

5.3.3

H+ OH−

N +HC → N CH3 (g) l(g) H4 l(s) (5.3.3)

+

H+

OH−

NH+4 5.3.3 NH3 HCl

3 2 2 3 2

2− +

2 4+

3 45.3.4 5.3.5 5.3.6

P (l) (aq) + P (aq)H3 O4 ⇌

O(l)H2

H+

H2 O−4 (5.3.4)

P (aq) ⇌ (aq) +HP (aq)H2 O−4 H

+O2−

4 (5.3.5)

HP (aq) ⇌ (aq) +P (aq)O2−4 H

+O3−

4 (5.3.6)

5.3.3 Updated 10/30/2018

STRENGTHS OF ACIDS AND BASES

We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that strongacids (An acid that reacts essentially completely with water) to give and the corresponding anion. react essentially completely withwater to give and the corresponding anion. Similarly, strong bases (A base that dissociates essentially completely in water) to give

and the corresponding cation) dissociate essentially completely in water to give and the corresponding cation. Strongacids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids (An acid in which only afraction of the molecules react with water) to producee and the corresponding anion. and weak bases (A base in which only afraction of the molecules react with water to produce and the corresponding cation) react with water to produce ions, so weakacids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereasmore than 95% is present in undissociated form.

In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO , HClO , and H SO (H PO is only moderatelystrong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH,KOH, and Ca(OH) . Common weak acids include HCN, H S, HF, oxoacids such as HNO and HClO, and carboxylic acids such asacetic acid. The ionization reaction of acetic acid is as follows:

Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than1% dissociates), as we stated in section 8.1. Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation5.3.8) but a weak acid when it donates its second proton (Equation 5.3.9) as indicated by the single and double arrows, respectively:

Consequently, an aqueous solution of sulfuric acid contains ions and a mixture of and ions, but no

molecules.

The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:

Most of the ammonia (>99%) is present in the form of NH (g). Amines, which are organic analogues of ammonia, are also weak bases,as are ionic compounds that contain anions derived from weak acids (such as S ).

There is no correlation between the solubility of a substance and whether it is a strongelectrolyte, a weak electrolyte, or a nonelectrolyte.

Table lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids.The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other basesyou encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or aweak electrolyte! Many weak acids and bases are extremely soluble in water.

H+

H+

OH− OH−

H+

OH−

3 4 2 4 3 4

2 2 2

C C H(l) (aq) +C C (aq)H3 O2 ⇌

O(l)H2

H+ H3 O−2 (5.3.7)

(aq) +HS (aq)S (l)H2 O4strong acid

− →−−−O(l)H2

H+ O−4 (5.3.8)

⇌ (aq) +S (aq)HS (aq)O−4

weak acid

H+ O2−4 (5.3.9)

H+(aq)

HSO−4 (aq)

SO2−4 (aq)

SH2 O4

N (g) + O(l) ⇌ N (aq) +O (aq)H3 H2 H+4 H− (5.3.10)

32−

5.3.1

5.3.4 Updated 10/30/2018

Table : Common Strong Acids and Bases

Strong Acids Strong Bases

HydrogenHalides

Oxoacids Group 1Hydroxides

Hydroxides of theHeavier Group 2

Elements

HCl HNO LiOH Ca(OH)

HBr H SO NaOH Sr(OH)

HI HClO KOH Ba(OH)

RbOH

CsOH

EXAMPLE

Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.

a. CH CH CO Hb. CH OHc. Sr(OH)d. CH CH NHe. HBrO

Given: compound

Asked for: acid or base strength

Strategy:

A Determine whether the compound is organic or inorganic.

B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H or OH ions, respectively. Iforganic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group,respectively. Recall that all polyprotic acids except H SO are weak acids.

Solution:

a. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, soit must be a weak acid.

b. A CH OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociateto form the OH ion. Because it does not contain a carboxylic acid (−CO H) group, methanol also cannot dissociate to form H(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base.

c. A Sr(OH) is an inorganic compound that contains one Sr and two OH ions per formula unit. B We therefore expect it to bea strong base, similar to Ca(OH) .

d. A CH CH NH is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an Rgroup. B Consequently, we expect it to behave similarly to ammonia (Equation 5.3.7), reacting with water to produce smallamounts of the OH ion. Ethylamine is therefore a weak base.

e. A HBrO is perbromic acid, an inorganic compound. B It is not listed in Table 5.3.1 as one of the common strong acids, butthat does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly belowCl in group 17. We might therefore expect that HBrO is chemically similar to HClO , a strong acid—and, in fact, it is.

5.3.1

3 2

2 4 2

4 2

5.3.1

3 2 2

3

2

3 2 2

4

+ −

2 4

3−

2+

22+ −

2

3 2 2

−

4

4 4

5.3.5 Updated 10/30/2018

EXERCISE

Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.

a. Ba(OH)b. HIOc. CH CH CH CO Hd. (CH ) NHe. CH O

Answer:

a. strong baseb. strong acidc. weak acidd. weak basee. none of these; formaldehyde is a neutral molecule

THE HYDRONIUM ION

Because isolated protons are very unstable and hence very reactive, an acid never simply “loses” an H ion. Instead, the proton isalways transferred to another substance, which acts as a base in the Brønsted–Lowry definition. Thus in every acid–base reaction, onespecies acts as an acid and one species acts as a base. Occasionally, the same substance performs both roles, as you will see later. Whena strong acid dissolves in water, the proton that is released is transferred to a water molecule that acts as a proton acceptor or base, asshown for the dissociation of sulfuric acid:

Technically, therefore, it is imprecise to describe the dissociation of a strong acid as producing ions, as we have been doing. The

resulting ion, called the hydronium ion and is a more accurate representation of . For the sake of brevity, however, indiscussing acid dissociation reactions, we often show the product as (as in Equation 5.3.7) with the understanding that the

product is actually the ion.

Conversely, bases that do not contain the hydroxide ion accept a proton from water, so small amounts of OH are produced, as in thefollowing:

Again, the double arrow indicates that the reaction does not go to completion but rather reaches a state of equilibrium. In this reaction,water acts as an acid by donating a proton to ammonia, and ammonia acts as a base by accepting a proton from water. Thus water canact as either an acid or a base by donating a proton to a base or by accepting a proton from an acid. Substances that can behave as bothan acid and a base are said to be amphotericWhen substances can behave as both an acid and a base..

5.3.1

2

4

3 2 2 2

3 2

2

+

+ → +S (l)H2 O4acid (proton donor)

O(l)H2base (proton acceptor)

(aq)H3O+

acid

HS (aq)O−4

base

(5.3.11)

H+(aq)

H3O+ H+

(aq)

H+(aq)

H3O+(aq)

−

+ ⇌ +N (g)H3base

O(l)H2acid

N (aq)H+4acid

O (aq)H−

base

(5.3.12)

5.3.6 Updated 10/30/2018

The products of an acid–base reaction are also an acid and a base. In Equation , for example, the products of the reaction are thehydronium ion, here an acid, and the hydrogen sulfate ion, here a weak base. In Equation , the products are NH , an acid, andOH , a base. The product NH is called the conjugate acid of the base NH , and the product OH is called the conjugate baseThesubstance formed when a Brønsted–Lowry acid donates a proton. of the acid H O. Thus all acid–base reactions actually involve twoconjugate acid–base pairs. All acid–base reactions involve two conjugate acid–base pairs, the Brønsted–Lowry acid and the base itforms after donating its proton, and the Brønsted–Lowry base and the acid it forms after accepting a proton.; in Equation , theyare NH /NH and H O/OH .

NEUTRALIZATION REACTIONS

A neutralization reaction (a chemical reaction in which an acid and a base react in stoichiometric amounts to produce water and a salt)is one in which an acid and a base react in stoichiometric amounts to produce water and a salt (the general term for any ionic substancethat does not have OH− as the anion or H+ as the cation), the general term for any ionic substance that does not have OH as the anionor H as the cation. If the base is a metal hydroxide, then the general formula for the reaction of an acid with a base is described asfollows: Acid plus base yields water plus salt. For example, the reaction of equimolar amounts of HBr and NaOH to give water and asalt (NaBr) is a neutralization reaction:

Acid plus base yields water plus salt.

If we write the complete ionic equation for the reaction in Equation , we see that and are spectator ions and are

not involved in the reaction:

The overall reaction is therefore simply the combination of H (aq) and OH (aq) to produce H O, as shown in the net ionic equation:

The net ionic equation for the reaction of any strong acid with any strong base is identical to Equation .

The strengths of the acid and the base generally determine whether the reaction goes to completion. The reaction of any strong acidwith any strong base goes essentially to completion, as does the reaction of a strong acid with a weak base, and a weak acid with astrong base. Examples of the last two are as follows:

Sodium acetate is written with the organic component first followed by the cation, as is usual for organic salts. Most reactions of aweak acid with a weak base also go essentially to completion. One example is the reaction of acetic acid with ammonia:

An example of an acid–base reaction that does not go to completion is the reaction of a weak acid or a weak base with water, which isboth an extremely weak acid and an extremely weak base.

NOTE

Except for the reaction of a weak acid or a weak base with water, acid–base reactions essentially go to completion.

5.3.11

5.3.12 4+

−4+

3−

2

5.3.12

4+

3 2−

−

+

+ → +HBr(aq)acid

NaOH(aq)base

O(l)H2water

NaBr(aq)salt

(5.3.13)

5.3.13 Na+(aq)

Br−(aq)

(aq) + + +O (aq) → O(l) + +H+ B (aq)r− N (aq)a+ H− H2 N (aq)a+ B (aq)r− (5.3.14)

+ −2

(aq) +O (aq) → O(l)H+ H− H2 (5.3.15)

5.3.15

+ →HCl(aq)strong acid

N (aq)H3weak base

N Cl(aq)H4salt

(5.3.16)

+ → + O(l)C C H(aq)H3 O2weak acid

NaOH(aq)strong base

C C Na(aq)H3 O2salt

H2 (5.3.17)

+ →C C H(aq)H3 O2weak acid

N (aq)H3weak base

C C N (aq)H3 O2 H4salt

(5.3.18)

5.3.7 Updated 10/30/2018

In some cases, the reaction of an acid with an anion derived from a weak acid (such as HS ) produces a gas (in this case, H S).Because the gaseous product escapes from solution in the form of bubbles, the reverse reaction cannot occur. Therefore, these reactionstend to be forced, or driven, to completion. Examples include reactions in which an acid is added to ionic compounds that contain theHCO , CN , or S anions, all of which are driven to completion:

The reactions in Equations - are responsible for the rotten egg smell that is produced when metal sulfides come in contactwith acids.

EXAMPLE : CALCIUM PROPIONATE

Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemicalequation for the reaction of aqueous propionic acid (CH CH CO H) with aqueous calcium hydroxide [Ca(OH) ] to give calciumpropionate. Do you expect this reaction to go to completion, making it a feasible method for the preparation of calciumpropionate?

Given: reactants and product

Asked for: balanced chemical equation and whether the reaction will go to completion

Strategy

Write the balanced chemical equation for the reaction of propionic acid with calcium hydroxide. Based on their acid and basestrengths, predict whether the reaction will go to completion.

Solution

Propionic acid is an organic compound that is a weak acid, and calcium hydroxide is an inorganic compound that is a strong base.The balanced chemical equation is as follows:

The reaction of a weak acid and a strong base will go to completion, so it is reasonable to prepare calcium propionate by mixingsolutions of propionic acid and calcium hydroxide in a 2:1 mole ratio.

EXERCISE

Write a balanced chemical equation for the reaction of solid sodium acetate with dilute sulfuric acid to give sodium sulfate.

Answer:

One of the most familiar and most heavily advertised applications of acid–base chemistry is antacids, which are bases that neutralizestomach acid. The human stomach contains an approximately 0.1 M solution of hydrochloric acid that helps digest foods. If theprotective lining of the stomach breaks down, this acid can attack the stomach tissue, resulting in the formation of an ulcer. Becauseone factor that is believed to contribute to the formation of stomach ulcers is the production of excess acid in the stomach, manyindividuals routinely consume large quantities of antacids. The active ingredients in antacids include sodium bicarbonate andpotassium bicarbonate (NaHCO and KHCO ; Alka-Seltzer); a mixture of magnesium hydroxide and aluminum hydroxide [Mg(OH)and Al(OH) ; Maalox, Mylanta]; calcium carbonate (CaCO ; Tums); and a complex salt, dihydroxyaluminum sodium carbonate

−2

3− − 2−

HC (aq) + (aq) → C (aq)O−3 H+ H2 O3 (5.3.19)

C (aq) → C (g) + O(l)H2 O3 O2 H2 (5.3.20)

C (aq) + (aq) → HCN(g)N− H+ (5.3.21)

(aq) + (aq) → H (aq)S2− H+ S− (5.3.22)

H (aq) + (aq) → S(g)S− H+ H2 (5.3.23)

5.3.19 5.3.22

5.3.2

3 2 2 2

2C C C H(aq) +Ca(OH (aq) → (C C C Ca(aq) + 2 O(l)H3 H2 O2 )2 H3 H2 O2)2 H2

5.3.2

2C C Na(s) + S (aq) → N S (aq) + 2C C H(aq)H3 O2 H2 O4 a2 O4 H3 O2

3 3 2

3 3

5.3.8 Updated 10/30/2018

[NaAl(OH) CO ; original Rolaids]. Each has certain advantages and disadvantages. For example, Mg(OH) is a powerful laxative (it isthe active ingredient in milk of magnesia), whereas Al(OH) causes constipation. When mixed, each tends to counteract the unwantedeffects of the other. Although all antacids contain both an anionic base (OH , CO , or HCO ) and an appropriate cation, they differsubstantially in the amount of active ingredient in a given mass of product.

Stomach acid. An antacid tablet reacts with 0.1 M HCl (the approximate concentration found in the human stomach).

EXAMPLE

Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets arerequired to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO ? (Neutralizing all of the stomach acid isnot desirable because that would completely shut down digestion.)

Given: volume and molarity of acid and mass of base in an antacid tablet

Asked for: number of tablets required for 90% neutralization

Strategy:

A. Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion.B. Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid

that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid.C. Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass.

Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet.

Solution:

A We first write the balanced chemical equation for the reaction:

Each carbonate ion can react with 2 mol of H to produce H CO , which rapidly decomposes to H O and CO . Because HCl is astrong acid and CO is a weak base, the reaction will go to completion.

B Next we need to determine the number of moles of HCl present:

Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90:

We know from the stoichiometry of the reaction that each mole of CaCO reacts with 2 mol of HCl, so we need

2 3 2

3−

32−

3−

5.3.3

3

2HCl(aq) +CaC (s) → CaC (aq) + C (aq)O3 l2 H2 O3

+2 3 2 2

32−

75 ( )( ) = 0.015 mol HClmL1 L

1000 mL

0.20 mol HCl

L

(0.015 mol HCl)(0.90) = 0.014 mol HCl

3

moles CaC = 0.014 ( ) = 0.0070 mol CaCO3 mol HCl1 mol CaCO3

2 mol HClO3

5.3.9 Updated 10/30/2018

C Each Tums tablet contains

Thus we need Tums tablets.

EXERCISE

Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must beconsumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH) CO ? The neutralization reaction can bewritten as follows:

Answer: 6.4 tablets

THE PH SCALE

One of the key factors affecting reactions that occur in dilute solutions of acids and bases is the concentration of H and OH ions. ThepH scale provides a convenient way of expressing the hydrogen ion (H ) concentration of a solution and enables us to describe acidityor basicity in quantitative terms.

Pure liquid water contains extremely low, but measurable concentrations of H O (aq) and OH (aq) ions produced via an autoionizationreaction, in which water acts simultaneously as an acid and as a base:

The concentration of hydrogen ions in pure water is only 1.0 × 10 M at 25°C. Because the autoionization reaction produces both aproton and a hydroxide ion, the OH concentration in pure water is also 1.0 × 10 M. Pure water is a neutral solution in which thetotal positive charge from all the cations is matched by an identical total negative charge from all the anions. At room temperature thisis [H ] = [OH ] = 1.0 × 10 M.

The pH scale describes the hydrogen ion concentration of a solution in a way that avoids the use of exponential notation. The negativebase-10 logarithm of the hydrogen ion concentration: pH is often defined as the negative base-10 logarithm of the hydrogen ionconcentration; pH is actually defined as the negative base-10 logarithm of hydrogen ion activity. As you will learn in a more advancedcourse, the activity of a substance in solution is related to its concentration. For dilute solutions such as those we are discussing, theactivity and the concentration are approximately the same.

Conversely,

(If you are not familiar with logarithms or using a calculator to obtain logarithms and antilogarithms)

Because the hydrogen ion concentration is 1.0 × 10 M in pure water at 25°C, the pH of pure liquid water (and, by extension, of anyneutral solution) is

Adding an acid to pure water increases the hydrogen ion concentration and decreases the hydroxide ion concentration because aneutralization reaction occurs, such as that shown in Equation 5.3.15. Because the negative exponent of [H ] becomes smaller as [H ]

( ) = 0.00500 mol CaC⎛

⎝

500 mg CaCO3

1 Tums tablet

⎞

⎠

⎛

⎝

1 g

1000 mg CaCO3

⎞

⎠

1 mol CaCO3

100.1 gO3

= 1.40.0070 mol CaCO3

0.00500 mol CaCO3

5.3.3

2 3

NaAl(OH C (s) + 4HCl(aq) → AlC (aq) +NaCl(aq) +C (g) + 3 O(l))2 O3 l3 O2 H2

+ −

+

3+ −

O(l) + O(l) ⇌ (aq) +O (aq)H2 H2 H3O+ H− (5.3.24)

−7

− −7

+ − −7

pH = −log[ ]H+ (5.3.25)

[ ] =H+ 10−pH (5.3.26)

−7

pH = − log[1.0 × ] = 7.0010−7 (5.3.27)

+ +

5.3.10 Updated 10/30/2018

increases, the pH decreases with increasing [H ]. For example, a 1.0 M solution of a strong monoprotic acid such as HCl or HNO hasa pH of 0.00:

NOTE

pH decreases with increasing [H ].

Conversely, adding a base to pure water increases the hydroxide ion concentration and decreases the hydrogen ion concentration.Because the autoionization reaction of water does not go to completion, neither does the neutralization reaction. Even a strongly basicsolution contains a detectable amount of H ions. For example, a 1.0 M OH solution has [H ] = 1.0 × 10 M. The pH of a 1.0 MNaOH solution is therefore

For practical purposes, the pH scale runs from pH = 0 (corresponding to 1 M H ) to pH 14 (corresponding to 1 M OH ), although pHvalues less than 0 or greater than 14 are possible.

We can summarize the relationships between acidity, basicity, and pH as follows:

If pH = 7.0, the solution is neutral.If pH < 7.0, the solution is acidic.If pH > 7.0, the solution is basic.

Keep in mind that the pH scale is logarithmic, so a change of 1.0 in the pH of a solution corresponds to a tenfold change in thehydrogen ion concentration. The foods and consumer products we encounter daily represent a wide range of pH values, as shown inFigure 5.3.2.

Figure 5.3.2 A Plot of pH versus [H ] for Some Common Aqueous Solutions. Although many substances exist in a range of pH values(indicated in parentheses), they are plotted using typical values.

EXAMPLE 5.3.4

+3

pH = − log[1.0] = 0.00 (5.3.28)

+

+ − + −14

pH = −log[1.0 × ] = 14.0010−14 (5.3.29)

+ −

+

5.3.11 Updated 10/30/2018

a. What is the pH of a 2.1 × 10 M aqueous solution of HClO ?b. The pH of a vinegar sample is 3.80. What is its hydrogen ion concentration?

Given: molarity of acid or pH

Asked for: pH or [H ]

Strategy:

Using the balanced chemical equation for the acid dissociation reaction and Equation 5.3.25 or 5.3.24, determine [H ] and convertit to pH or vice versa.

Solution:

a. HClO (perchloric acid) is a strong acid, so it dissociates completely into H ions and ClO ions:

The H ion concentration is therefore the same as the perchloric acid concentration. The pH of the perchloric acid solution isthus

The result makes sense: the H ion concentration is between 10 M and 10 M, so the pH must be between 1 and 2.

Note: The assumption that [H ] is the same as the concentration of the acid is valid for only strong acids. Because weak acidsdo not dissociate completely in aqueous solution, a more complex procedure is needed to calculate the pH of their solutions.

b. We are given the pH and asked to calculate the hydrogen ion concentration. From Equation 5.3.24,

Thus .

EXERCISE 5.3.4

a. What is the pH of a 3.0 × 10 M aqueous solution of HNO ?b. What is the hydrogen ion concentration of turnip juice, which has a pH of 5.41?

Answer:

a. b.

Tools have been developed that make the measurement of pH simple and convenient (Figures 5.3.3 and 5.3.4). For example, pH paperconsists of strips of paper impregnated with one or more acid–base indicators, which are intensely colored organic molecules whosecolors change dramatically depending on the pH of the solution. Placing a drop of a solution on a strip of pH paper and comparing itscolor with standards give the solution’s approximate pH. A more accurate tool, the pH meter, uses a glass electrode, a device whosevoltage depends on the H ion concentration (Figure 5.3.4).

−24

+

+

4+

4−

HCl (l) → (aq) +Cl (aq)O4 H+ O−4

+

pH = −log[ ] = −log(2.1 × ) = 1.68H+ 10−2

+ −1 −2

+

= [ ]10−pH H+

[ ] = = 1.6 × MH+ 10−3.80 10−4

−53

pH = 4.52

[ ] = 3.9 × MH+ 10−6

+

5.3.12 Updated 10/30/2018

Figure 5.3.3: These pH strips can measure pH in a series of ranges by putting sample on a strip and comparing its color change withcolors on the box that correspond to a certain pH. The strip on the left measures pH 0-7 and shows results of a strong acid sample; thecenter strip is pH range 5-10 and shows results of a 6.97 buffer solution sample; the strip on the right measures a broad range (pH 1-

14) and shows results of a 10% bleach water solution sample. This image can be enlarged by clicking on it. Photo by Monica Bruckner(courtesy of the Science Education Resource Center (SERC) at Carleton College).

Figure 5.3.4: These three meters and probes can measure pH (left) and electrical conductivity (center and right). The image can beenlarged if you click on it. Photo by Monica Bruckner (courtesy of the Science Education Resource Center (SERC) at Carleton

College).

SUMMARY

An acidic solution and a basic solution react together in a neutralization reaction that also forms a salt. Acid–base reactions requireboth an acid and a base. In Brønsted–Lowry terms, an acid is a substance that can donate a proton (H ), and a base is a substance thatcan accept a proton. All acid–base reactions contain two acid–base pairs: the reactants and the products. Acids can donate one proton(monoprotic acids), two protons (diprotic acids), or three protons (triprotic acids). Compounds that are capable of donating morethan one proton are generally called polyprotic acids. Acids also differ in their tendency to donate a proton, a measure of their acidstrength. Strong acids react completely with water to produce H O (aq) (the hydronium ion), whereas weak acids dissociate onlypartially in water. Conversely, strong bases react completely with water to produce the hydroxide ion, whereas weak bases react onlypartially with water to form hydroxide ions. The reaction of a strong acid with a strong base is a neutralization reaction, whichproduces water plus a salt.

The acidity or basicity of an aqueous solution is described quantitatively using the pH scale. The pH of a solution is the negativelogarithm of the H ion concentration and typically ranges from 0 for strongly acidic solutions to 14 for strongly basic ones. Becauseof the autoionization reaction of water, which produces small amounts of hydronium ions and hydroxide ions, a neutral solution of

+

3+

+

5.3.13 Updated 10/30/2018

water contains 1 × 10 M H ions and has a pH of 7.0. An indicator is an intensely colored organic substance whose color is pHdependent; it is used to determine the pH of a solution.

CONTRIBUTORS

Modified by Joshua Halpern (Howard University), Scott Sinex, and Scott Johnson (PGCC)

−7 +

5.4.1 Updated 10/30/2018

5.4: OXIDATION-REDUCTION: SOME GENERAL PRINCIPLES

SKILLS TO DEVELOP

To identify oxidation–reduction reactions in solution.

The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron isexposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metaldevelops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive chargeby transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative chargeand form oxide ions (O ). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss ofelectrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons.For every oxidation, there must be an associated reduction.

NOTE

Any oxidation must ALWAYS be accompanied by a reduction and vice versa.

Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, areaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, itsmass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). Thereaction is as follows:

Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements inthe reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state of each atom in acompound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction forelectrons. Atoms in their elemental form, such as O or H , are assigned an oxidation state of zero. For example, the reaction ofaluminum with oxygen to produce aluminum oxide is

Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidationstate of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with anoxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al O , electrons are transferred as follows(the superscript 0 emphasizes the oxidation state of the elements):

Equation 5.4.1 and Equation 5.4.2 are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer ofelectrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gainedto preserve electrical neutrality. In Equation 5.4.3, for example, the total number of electrons lost by aluminum is equal to the totalnumber gained by oxygen:

The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained.

An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in Figure 5.4.1.

2−

C O(s) + (g) → 2Cu(s) + O(g)u2 H2 H2 (5.4.1)

2 2

4Al(s) + 3 → 2A (s)O2 l2O3 (5.4.2)

2 3

4A + 3 → 4A + 6l0

O02 l

3+O

2− (5.4.3)

electrons lost = 4Al atoms× = 12 lost3 loste−

Al atome− (5.4.4a)

electrons gained = 6O atoms× = 12 gained2 gainede−

O atome− (5.4.4a)

5.4.2 Updated 10/30/2018

NOTE

In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.

ASSIGNING OXIDATION STATES

Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements areidentical to the charges on the monatomic ions. Previosuly, you learned how to predict the formulas of simple ionic compounds basedon the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodiumchloride (NaCl; Figure 5.4.1), magnesium oxide (MgO), and calcium chloride (CaCl ). In covalent compounds, in contrast, atomsshare electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help youunderstand and predict many reactions.

Figure 5.4.1: The Reaction of a Neutral Sodium Atom with a Neutral Chlorine Atom. The result is the transfer of one electron fromsodium to chlorine, forming the ionic compound NaCl.

A set of rules for assigning oxidation states to atoms in chemical compounds follows.

Rules for Assigning Oxidation States

1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.2. The oxidation state of a monatomic ion is the same as its charge—for example, Na = +1, Cl = −1.3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well,

except when combined with oxygen or other halogens.4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–

fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.

Nonintegral oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of thesame element with different oxidation states.

In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant,just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in amolecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1.The net charge is zero, as it must be for any compound.

2

+ −

5.4.3 Updated 10/30/2018

Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6"The Structure of Atoms". Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemicalcompounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) asopposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H andCl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rulealso prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF but −½ in KO . Note that an oxidation state of −½ forO in KO is perfectly acceptable.

The reduction of copper(I) oxide shown in Equation 5.4.5 demonstrates how to apply these rules. Rule 1 states that atoms in theirelemental form have an oxidation state of zero, which applies to H and Cu. From rule 4, hydrogen in H O has an oxidation state of +1,and from rule 5, oxygen in both Cu O and H O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in amolecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu O must have a charge of+1: 2(+1) + (−2) = 0. So the oxidation states are as follows:

Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0).So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a netconservation of charge:

Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidationstate of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way ofassigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.

EXAMPLE 5.4.1

Assign oxidation states to all atoms in each compound.

a. sulfur hexafluoride (SF )b. methanol (CH OH)c. ammonium sulfate [(NH )2SO ]d. magnetite (Fe O )e. ethanoic (acetic) acid (CH CO H)

Given: molecular or empirical formula

Asked for: oxidation states

Strategy:

Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, otherhalogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.

Solution:

a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfurhexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero ina neutral molecule (here SF ), the oxidation state of sulfur must be +6:

[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0

− +

−

2 2

2

2 2

2 2

2

(s) + (g) → 2 (s) + (g)Cu2

+1O−2

H2

0Cu

0H2

+1O−2

(5.4.5)

electrons lost = 2Hatoms× = 2 lost1 loste−

Hatome− (5.4.6a)

electrons gained = 2Cuatoms× = 2 gained1 gainede−

Cuatome

− (5.4.6b)

6

3

4 4

3 4

3 2

6

5.4.4 Updated 10/30/2018

b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has nonet charge, carbon must have an oxidation state of −2:

[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0

c. Note that (NH ) SO is an ionic compound that consists of both a polyatomic cation (NH ) and a polyatomic anion (SO )(see Table 2.4 "Common Polyatomic Ions and Their Names"). We assign oxidation states to the atoms in each polyatomic ionseparately. For NH , hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:

[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH ion

For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:

[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion

d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by thepositive charge on three iron atoms, giving an oxidation state of +8/3 for iron:

[(4 O atoms)(−2)]+[(3 Fe atoms) ]= 0

Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. Infact, Fe O can be viewed as having two Fe ions and one Fe ion per formula unit, giving a net positive charge of +8 performula unit. Fe O is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestonebecause it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the“lodestar.”

e. Initially, we assign oxidation states to the components of CH CO H in the same way as any other compound. Hydrogen andoxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of

[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0

So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atomspresent. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. Todetermine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumptionthat bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methylgroup (−CH ) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation stateof +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. Forthe methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of thecarboxylic acid group (−CO H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, weassign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of

[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3

To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individualatoms in acetic acid are thus

Thus the sum of the oxidation states of the two carbon atoms is indeed zero.

4 2 4 4+

42−

4+

4+

(+ )83

3 43+ 2+

3 4

3 2

3

2

C−3H3

+1C+3O2−2

H+1

(5.4.1)

5.4.5 Updated 10/30/2018

EXERCISE 5.4.1

Assign oxidation states to all atoms in each compound.

a. barium fluoride (BaF )b. formaldehyde (CH O)c. potassium dichromate (K Cr O )d. cesium oxide (CsO )e. ethanol (CH CH OH)

Answer:

a. Ba, +2; F, −1b. C, 0; H, +1; O, −2c. K, +1; Cr, +6; O, −2d. Cs, +1; O, −½e. C, −3; H, +1; C, −1; H, +1; O, −2; H, +1

REDOX REACTIONS OF SOLID METALS IN AQUEOUS SOLUTION

A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solidmetals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure 5.4.1). Rust is formed from acomplex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, andoxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:

In subsequent steps, FeCl undergoes oxidation to form a reddish-brown precipitate of Fe(OH) .

Figure 5.4.2: Rust Formation. The corrosion process involves an oxidation–reduction reaction in which metallic iron is converted toFe(OH) , a reddish-brown solid.

Many metals dissolve through reactions of this type, which have the general form

Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall ofthe Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, isslightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:

2

2

2 2 7

2

3 2

−

Fe(s) + 2HCl(aq) → FeC (aq) + (g)l2 H2 (5.4.5)

2 3

3

metal+ acid → salt + hydrogen (5.4.6)

Pb(s) + 2 (aq) → P (aq) + (g)H+

b2+

H2 (5.4.7)

5.4.6 Updated 10/30/2018

Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, whichresulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed hisfavorite horse as consul!

SINGLE-DISPLACEMENT REACTIONS

Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types ofreactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Twoexamples of single-displacement reactions are the reduction of iron salts by zinc (Equation 5.4.8) and the reduction of silver salts bycopper (Equation 5.4.9 and Figure 5.4.2):

The reaction in Equation 5.4.8 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails andsheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it fromoxidation as long as zinc remains on the object.

Figure 5.4.3: The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate. When a copper coil is placed ina solution of silver nitrate, silver ions are reduced to metallic silver on the copper surface, and some of the copper metal dissolves.

Note the formation of a metallic silver precipitate on the copper coil and a blue color in the surrounding solution due to the presence ofaqueous Cu ions. Figure used with permission of Wikipedia

THE ACTIVITY SERIES

By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists havearranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. Forexample, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reactswith both copper salts and silver salts, producing Zn . Zinc therefore has a greater tendency to be oxidized than does iron, copper, orsilver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to givezinc metal:

Magnesium has a greater tendency to be oxidized than zinc does.

Pairwise reactions of this sort are the basis of the activity series (Figure 5.4.4), which lists metals and hydrogen in order of theirrelative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkalimetals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which havethe lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which

Zn(s) +F (aq) → Z (aq) +Fe(s)e2+ n2+ (5.4.8)

Cu(s) + 2A (aq) → C (aq) + 2Ag(s)g+ u2+ (5.4.9)

2+

2+

Zn(s) +M (aq) Z (aq) +Mg(s)g2+ → n2+ (5.4.10)

Mg(s) +Z (aq) → M (aq) +Zn(s)n2+ g2+ (5.4.11)

5.4.7 Updated 10/30/2018

are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metalsare active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, whichhave the least tendency to be oxidized. (at the bottom of the series).

Figure 5.4.4: The Activity Series

When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of theelements below it in the series. Because magnesium is above zinc in Figure 5.4.4, magnesium metal will reduce zinc salts but not viceversa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal saltsto the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by itsposition relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids toproduce H . Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily.Example 5.4.2 demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacementreactions.

EXAMPLE 5.4.2

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.

Given: reactants

Asked for: overall reaction and net ionic equation

Strategy:

A. Locate the reactants in the activity series in Figure .5.4.4 and from their relative positions, predict whether a reaction willoccur. If a reaction does occur, identify which metal is oxidized and which is reduced.

B. Write the net ionic equation for the redox reaction.

Solution:

1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to theirrelative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionicequation is as follows:

2

5.4.8 Updated 10/30/2018

Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions andare not involved in the reaction.

2. A Mercury lies below lead in the activity series, so no reaction will occur.

3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H .B From our discussion of solubilities, recall that Pb and SO form insoluble lead(II) sulfate. In this case, the sulfate ionsare not spectator ions, and the reaction is as follows:

Lead(II) sulfate is the white solid that forms on corroded battery terminals.

Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuricacid.

EXERCISE 5.4.2

Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.

1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).

Answer:

1. 2. 3.

SUMMARY

Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reductionequation.

In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reductionreactions in solution using the oxidation state method (Table 5.4.1), in which the overall reaction is separated into an oxidationequation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt thatresult in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions canbe predicted using the activity series (Figure 5.4.4), which arranges metals and H in decreasing order of their tendency to be oxidized.

Al(s) + 3A (aq) → A (aq) + 3Ag(s)g+ l3+ (5.4.2)

22+

42−

Pb(s) + 2 (aq) +S (aq) → PbS (s) + (g)H+ O2−4 O4 H2 (5.4.3)

no reaction

3Zn(s) + 2C (aq) → 3Z (aq) + 2Cr(s)r3+ n2+

2Al(s) + 6C C H(aq) → 2A (aq) + 6C C (aq) + 3 (g)H3 O2 l3+ H3 O−2 H2

2

5.4.9 Updated 10/30/2018

Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metalsare at the bottom of the activity series.

5.5.1 Updated 10/30/2018

5.5: BALANCING OXIDATION-REDUCTION EQUATIONSThe Learning Objectives of this Module are:

1. To identify oxidation–reduction reactions in solution.

We described the defining characteristics of oxidation–reduction, or redox, reactions already. Most of the reactions we considered therewere relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution,however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation isthe most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactionsin aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reactionmethod. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is discussed inthe second semester

BALANCING REDOX EQUATIONS USING OXIDATION STATESTo balance a redox equation using the oxidation state method, we conceptually separate the overall reaction into two parts: anoxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons.Consider, for example, the reaction of Cr (aq) with manganese dioxide (MnO ) in the presence of dilute acid. Equation 8.8.1 is the netionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using theprocedure described in Section 7.5:

Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidationstate. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced:

The oxidation can be written as

and the reduction as

For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by theoxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same numberof electrons in both. In this example, we must multiply the oxidation equation by 2 to give

Note the Patterns

In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.

The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction:

2+2

+2

C (aq)r2+

+4

+ Mn (aq)O2

−2

+1

+ (aq)H+

+3

→ C (aq)r3+

+2

+ M (aq)n2+

+1

+ O (l)H2

−2

(8.8.1)

C +M → C +Mr2+ n4+ r3+ n2+ (8.8.2)

C → C +r2+ r3+ e− (8.8.3)

M + 2 → Mn4+ e− n2+ (8.8.4)

2C → 2C + 2r2+ r3+ e− (8.8.5)

2C → 2C + 2r2+ r3+ e− (8.8.6)

M + 2 → Mn4+ e− n2+

5.5.2 Updated 10/30/2018

We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actualchemical forms of the reactants and products:

Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of the equation (+4) does not equal thecharge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add H as necessary to eitherside of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, wecould balance the charge by adding OH as necessary to either side of the equation to balance the charges. In this case, adding four Hions to the left side of the equation gives

Although the charges are now balanced, we have two oxygen atoms on the left side of the equation and none on the right. We canbalance the oxygen atoms without affecting the overall charge balance by adding H O as necessary to either side of the equation. Here,we need to add two H O molecules to the right side:

Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is nowbalanced. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions issummarized in Table 8.8.1 and illustrated in Example 17.

Table 8.8.1 Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method

1. Write the unbalanced chemical equation for the reaction, showing the reactants and the products.2. Assign oxidation states to all atoms in the reactants and the products (see Section 3.5 ) and determine which atoms change oxidation state.

3. Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number ofelectrons accepted or donated by each.

4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.

5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients asnecessary to give the numbers of atoms in step 4.

6. Add the two equations and cancel the electrons.

7. Balance the charge by adding H or OH ions as necessary for reactions in acidic or basic solution, respectively.

8. Balance the oxygen atoms by adding H O molecules to one side of the equation.9. Check to make sure that the equation is balanced in both atoms and total charges.

EXAMPLE 17

Arsenic acid (H AsO ) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenicacid in acidic solution yields arsine (AsH , a highly toxic and unstable gas) and Zn (aq). Balance the equation for this reaction usingoxidation states:

Given: reactants and products in acidic solution

Asked for: balanced chemical equation using oxidation states

Strategy:

2C (aq)r2+

2C (aq) +r2+ 2e−

2C (aq) + 2C (aq)r2+ r2+

→

→

→

2C (aq) +r3+ 2e−

M (aq)n2+

2C (aq) +M (aq)r3+ n2+

(8.8.7)

+

− +

2C (aq) +Mn (s) + 4 (aq) → 2C (aq) +M (aq)r2+ O2 H+ r3+ n2+ (8.8.8)

2

2

2C (aq) +Mn (s) + 4 (aq) → 2C (aq) +M (aq) + 2 O(l)r2+ O2 H+ r3+ n2+ H2 (8.8.9)

+ −

2

3 4

32+

As (aq) +Zn(s) → As (g) +Z (aq)H3 O4 H3 n2+

5.5.3 Updated 10/30/2018

Follow the procedure given in Table 8.8.1 for balancing a redox equation using oxidation states. When you are done, be certain tocheck that the equation is balanced.

Solution:

1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip thisstep.

2. Assign oxidation states using the procedure described in Section 3.5 and determine which atoms change oxidation state. Theoxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state ofzinc in elemental zinc is 0, and the oxidation state of zinc in Zn (aq) is +2:

3. Write separate equations for oxidation and reduction. The arsenic atom in H AsO is reduced from the +5 to the −3 oxidationstate, which requires the addition of eight electrons:

Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:

4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number ofelectrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply theoxidation equation by 4 to obtain

5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjustingcoefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zincand adjusting the coefficients gives

6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is

which then yields

7. Balance the charge by adding H or OH ions as necessary for reactions in acidic or basic solution, respectively. Because thereaction is carried out in acidic solution, we can add H ions to whichever side of the equation requires them to balance the charge.The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H ions to the leftside gives a charge of +8 on both sides of the equation:

8. Balance the oxygen atoms by adding H O molecules to one side of the equation. There are 4 O atoms on the left side of theequation. Adding 4 H O molecules to the right side balances the O atoms:

Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms.

2+

3 4

Reduction: A + 8 →s5+ e− As3−

−3

Oxidation: Zn → Z + 2n2+ e−

Reduction (×1): A + 8 → As5+ e− s3−

Oxidation (×4): 4Z → 4Z + 8n0 n2+ e−

(5.5.1)

(5.5.2)

Reduction: As (aq) + 8 → As (g)H3 O4 e−

H3

Oxidation: 4Zn(s) → 4Z (aq) + 8n2+

e−

(5.5.3)

(5.5.4)

As (aq) + 4Zn(s) + → As (g) + 4Z (aq) +H3 O4 8e− H3 n2+ 8e−

As (aq) + 4Zn(s) → As (g) + 4Z (aq)H3 O4 H3 n2+

+ −

+

+

As (aq) + 4Zn(s) + 8 (aq) → As (g) + 4Z (aq)H3 O4 H+ H3 n2+

2

2

As (aq) + 4Zn(s) + 8 (aq) → As (g) + 4Z (aq) + 4 O(l)H3 O4 H+ H3 n2+ H2

5.5.4 Updated 10/30/2018

9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is importantto check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:

The balanced chemical equation for the reaction is therefore:

Exercise

Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitricacid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states:

Answer:

Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example18.

EXAMPLE 18

The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxidedissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may beclogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solutionto help break up the clogs. The reaction is as follows:

Balance this equation using oxidation states.

Given: reactants and products in a basic solution

Asked for: balanced chemical equation

Strategy:

Follow the procedure given in Table 8.8.1 for balancing a redox reaction using oxidation states. When you are done, be certain tocheck that the equation is balanced.

Solution:

We will apply the same procedure used in Example 17 but in a more abbreviated form.

1. The equation for the reaction is given, so we can skip this step.

2. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH) ] . The oxidation state of hydrogen changes from+1 in H O to 0 in H . Aluminum is oxidized, while hydrogen is reduced:

3.

4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:

Atoms: 1As+ 4Zn + 4O+ 11H = 1As+ 4Zn + 4O+ 11H

Total charge: 8(+1) = 4(+2) = +8

(5.5.5)

(5.5.6)

As (aq) + 4Zn(s) + 8 (aq) → As (g) + 4Z (aq) + 4 O(l)H3 O4 H+ H3 n2+ H2

CuS(s) + (aq) +N (aq) → C (aq) +NO(g) +S (aq)H+ O−3 u2+ O2−

4

3CuS(s) + 8 (aq) + 8N (aq) → 3C (aq) + 8NO(g) + 3S (aq) + 4 O(l)H+ O−3 u2+ O2−

4 H2

Al(s) + O(aq) → [Al(OH (aq) + (g)H2 )4]− H2

4−

2 2

(s) + O(aq) → [ (OH (aq) + (g)Al0

H+1

2 Al+3

)4]− H2

0

Reduction: A → A + 3l0 l3+ e−

Oxidation: + → (in )H+ e− H 0 H2

(5.5.7)

(5.5.8)

5.5.5 Updated 10/30/2018

5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbersof atoms as in step 4. Because a molecule of H O contains two protons, in this case, 3H corresponds to 3/2H O. Similarly, eachmolecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H .

6. Adding the equations and canceling the electrons gives

To remove the fractional coefficients, multiply both sides of the equation by 2:

7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carriedout in basic solution, we can balance the charge by adding two OH ions to the left side:

8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms byadding three H O molecules to the left side:

9. Be sure the equation is balanced:

10. The balanced chemical equation is therefore

Thus 3 mol of H gas are produced for every 2 mol of Al.

Exercise

The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balancedchemical equation for the reaction.

Answer:

As suggested in Example 17 and Example 18, a wide variety of redox reactions are possible in aqueous solutions. The identity of theproducts obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction iscarried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines forpredicting whether a redox reaction will occur and the identity of the products:

1. Compounds of elements in high oxidation states (such as ClO , NO , MnO , Cr O , and UF ) tend to act as oxidants andbecome reduced in chemical reactions.

2. Compounds of elements in low oxidation states (such as CH , NH , H S, and HI) tend to act as reductants and become oxidized inchemical reactions.

Reduction: 3 + 3 → 3 (in )H+ e− H 0 H2

Oxidation: A → A + 3l0 l3+ e−

(5.5.9)

(5.5.10)

2+

2

2

Reduction: O+ 3 →3

2H2 e− 3

2H2

Oxidation: Al → [Al(OH + 3)4]− e−

(5.5.11)

(5.5.12)

Al+ O+ → [Al(OH + +3

2H2 3e− )4]−

3

2H2 3e−

Al+ O → [Al(OH +3

2H2 )4]−

3

2H2

2Al+ 3 O → 2[Al(OH + 3H2 )4]− H2

−

2Al+ 2O + 3 O → 2[Al(OH + 3H− H2 )4]− H2

2

2Al+ 2O + 6 O → 2[Al(OH + 3H− H2 )4]− H2

Atoms: 2Al+ 8O+ 14H = 2Al+ 8O+ 14H

Total charge: (2)(0) + (2)(−1) + (6)(0) = (2)(−1) + (3)(0)

(5.5.13)

(5.5.14)

−2 = −2

2Al(s) + 2O (aq) + 6 O(l) → 2[Al(OH (aq) + 3 (g)H− H2 )4]− H2

2

2Mn (aq) + 3N (aq) + O(l) → 2Mn (s) + 3N (aq) + 2O (aq)O−4 O−

2 H2 O2 O−3 H−

4−

3−

4−

2 72−

6

4 3 2

5.5.6 Updated 10/30/2018

Note the Pattern

Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.

When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of acompound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.

REDOX REACTIONS OF SOLID METALS IN AQUEOUS SOLUTIONA widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solidmetals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure 8.8.1). Rust is formed from acomplex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, andoxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:

In subsequent steps, FeCl undergoes oxidation to form a reddish-brown precipitate of Fe(OH) .

Figure 8.8.1 Rust Formation. The corrosion process involves an oxidation–reduction reaction in which metallic iron is converted toFe(OH) , a reddish-brown solid.

Many metals dissolve through reactions of this type, which have the general form

Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall ofthe Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, isslightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:

Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, whichresulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed hisfavorite horse as consul!

−

Fe(s) + 2HCl(aq) → FeC (aq) + (g)l2 H2 (8.8.10)

2 3

3

metal+ acid → salt + hydrogen (8.8.11)

Pb(s) + 2 (aq) → P (aq) + (g)H+ b2+ H2 (8.8.12)

5.5.7 Updated 10/30/2018

SINGLE-DISPLACEMENT REACTIONSCertain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types ofreactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Twoexamples of single-displacement reactions are the reduction of iron salts by zinc (Equation 8.8.13) and the reduction of silver salts bycopper (Equation 8.8.14 and Figure 8.8.2):

The reaction in Equation 8.8.13 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails andsheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it fromoxidation as long as zinc remains on the object.

Figure 8.8.2 The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate. When a copper coil is placed in asolution of silver nitrate, silver ions are reduced to metallic silver on the copper surface, and some of the copper metal dissolves. Note

the formation of a metallic silver precipitate on the copper coil and a blue color in the surrounding solution due to the presence ofaqueous Cu ions. Figure used with permission of Wikipedia

SUMMARYIn oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reductionreactions in solution using the oxidation state method (Table 8.8.1), in which the overall reaction is separated into an oxidationequation and a reduction equation.

KEY TAKEAWAY

Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reductionequation.

CONCEPTUAL PROBLEMS

1. Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?

2. If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, doyou expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weakreductant? Why?

Zn(s) +F (aq) → Z (aq) +Fe(s)e2+ n2+ (8.8.13)

Cu(s) + 2A (aq) → C (aq) + 2Ag(s)g+ u2+ (8.8.14)

2+

5.5.8 Updated 10/30/2018

3. In each redox reaction, determine which species is oxidized and which is reduced:

a. Zn(s) + H SO (aq) → ZnSO (aq) + H (g)b. Cu(s) + 4HNO (aq) → Cu(NO ) (aq) + 2NO (g) + 2H O(l)c. BrO (aq) + 2MnO (s) + H O(l) → Br (aq) + 2MnO (aq) + 2H (aq)

4. Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give anexample of a redox reaction that is not a single-displacement reaction.

NUMERICAL PROBLEMS

1. Balance each redox reaction under the conditions indicated.

a. CuS(s) + NO (aq) → Cu (aq) + SO (aq) + NO(g); acidic solutionb. Ag(s) + HS (aq) + CrO (aq) → Ag S(s) + Cr(OH) (s); basic solutionc. Zn(s) + H O(l) → Zn (aq) + H (g); acidic solutiond. O (g) + Sb(s) → H O (aq) + SbO (aq); basic solutione. UO (aq) + Te(s) → U (aq) + TeO (aq); acidic solution

2. Balance each redox reaction under the conditions indicated.

a. MnO (aq) + S O (aq) → Mn (aq) + SO (aq); acidic solutionb. Fe (aq) + Cr O (aq) → Fe (aq) + Cr (aq); acidic solutionc. Fe(s) + CrO (aq) → Fe O (s) + Cr O (s); basic solutiond. Cl (aq) → ClO (aq) + Cl (aq); acidic solutione. CO (aq) + N H (aq) → CO(g) + N (g); basic solution

3. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write thecomplete ionic equation for the reaction.

a. Platinum wire is dipped in hydrochloric acid.b. Manganese metal is added to a solution of iron(II) chloride.c. Tin is heated with steam.d. Hydrogen gas is bubbled through a solution of lead(II) nitrate.

4. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write thecomplete ionic equation for the reaction.

a. A few drops of NiBr are dropped onto a piece of iron.b. A strip of zinc is placed into a solution of HCl.c. Copper is dipped into a solution of ZnCl .d. A solution of silver nitrate is dropped onto an aluminum plate.

5. Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water cancontaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gascan be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.

6. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu (aq) and nitric oxide gas.

a. What has been oxidized? What has been reduced?b. Balance the chemical equation.

7. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; thencomplete and balance the chemical equation:

2 4 4 2

3 3 2 2 2

3−

2 2−

4− +

3− 2+

42−

−42−

2 3

22+

2

2 2 2 2−

22+ 4+

42−

4−

2 32− 2+

42−

2+2 7

2− 3+ 3+

42−

2 3 2 3

2 3− −

32−

2 4 2

2

2

2+

5.5.9 Updated 10/30/2018

a. Pt (aq) + Ag(s) →b. HCN(aq) + NaOH(aq) →c. Fe(NO ) (aq) + NaOH(aq) →d. CH (g) + O (g) →

8. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; thencomplete and balance the chemical equation:

a. Zn(s) + HCl(aq) →b. HNO (aq) + AlCl (aq) →c. K CrO (aq) + Ba(NO ) (aq) →d. Zn(s) + Ni (aq) → Zn (aq) + Ni(s)

CONTRIBUTORS

Anonymous

2+

3 3

4 2

3 3

2 4 3 22+ 2+

5.6.1 Updated 10/30/2018

5.6: OXIDIZING AND REDUCING AGENTSCompounds that are capable of accepting electrons, such as O or F , are called oxidants (or oxidizing agents) because they can oxidizeother compounds. In the process of accepting electrons, an oxidant is reduced. Compounds that are capable of donating electrons, suchas sodium metal or cyclohexane (C H ), are called reductants (or reducing agents) because they can cause the reduction of anothercompound. In the process of donating electrons, a reductant is oxidized. These relationships are summarized in Equation 3.30:

oxidant + reductant → oxidation−reduction

Some oxidants have a greater ability than others to remove electrons from other compounds. Oxidants can range from very powerful,capable of oxidizing most compounds with which they come in contact, to rather weak. Both F and Cl are powerful oxidants: forexample, F will oxidize H O in a vigorous, potentially explosive reaction. In contrast, S is a rather weak oxidant, and O fallssomewhere in between. Conversely, reductants vary in their tendency to donate electrons to other compounds. Reductants can alsorange from very powerful, capable of giving up electrons to almost anything, to weak. The alkali metals are powerful reductants, sothey must be kept away from atmospheric oxygen to avoid a potentially hazardous redox reaction.

A combustion reaction, first introduced in Section 3.2 "Determining Empirical and Molecular Formulas", is an oxidation–reductionreaction in which the oxidant is O . One example of a combustion reaction is the burning of a candle, shown in Figure 3.9 "AnExample of a Combustion Reaction". Consider, for example, the combustion of cyclohexane, a typical hydrocarbon, in excess oxygen.The balanced chemical equation for the reaction, with the oxidation state shown for each atom, is as follows:

If we compare the oxidation state of each element in the products and the reactants, we see that hydrogen is the only element whoseoxidation state does not change; it remains +1. Carbon, however, has an oxidation state of −2 in cyclohexane and +4 in CO ; that is,each carbon atom changes its oxidation state by six electrons during the reaction. Oxygen has an oxidation state of 0 in the reactants,but it gains electrons to have an oxidation state of −2 in CO and H O. Because carbon has been oxidized, cyclohexane is thereductant; because oxygen has been reduced, it is the oxidant. All combustion reactions are therefore oxidation–reduction reactions.

We described the defining characteristics of oxidation–reduction, or redox, reactions in Chapter 7. Most of the reactions we consideredthere were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution,however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation isthe most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactionsin aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reactionmethod. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is discussed inthe second semester

Note the Pattern

Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.

When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of acompound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.

CONTRIBUTORS

Anonymous

2 2

6 12

+ → 2(g)O2gains e (is reduced)

4Na(s)loses e (is oxidized)

N O(s)a2redox reaction

(3.30)

2 2

2 2 8 2

2

+ 9 → 6 + 6C6−2H12

+1O2

0C+4O2−2

H2

+1O−2

(3.31)

2

2 2

5.6.2 Updated 10/30/2018