CHEMISTRY THE CENTRAL SCIENCE

4 REACTIONS IN AQUEOUS SOLUTION

4.4 OXIDATION-REDUCTION REACTIONS

In precipitation reactions, cations and anions come together to form an insoluble ionic compound. In neutralization reactions, H+ ions and OH ions come together to form H2O molecules. Now let's consider a third kind of reaction, one in which electrons are transferred from one reactant to another. Such reactions are called either oxidation-reduction reactions or redox reactions. In this chapter we concentrate on redox reactions where one of the reactants is a metal in its elemental form.

Oxidation and Reduction

One of the most familiar redox reactions is corrosion of a metal (FIGURE 4.11). In some instances corrosion is limited to the surface of the metal, with the green coating that forms on copper roofs and statues being one such case. In other instances the corrosion goes deeper, eventually compromising the structural integrity of the metal. Iron rusting is an important example.

Corrosion is the conversion of a metal into a metal compound by a reaction between the metal and some substance in its environment. When a metal corrodes, each metal atom loses electrons and so forms a cation, which can combine with an anion to form an ionic compound. The green coating on the Statue of Liberty contains Cu2+ combined with carbonate and hydroxide anions, rust contains Fe3+ combined with oxide and hydroxide anions, and silver tarnish contains Ag+ combined with sulfide anions.

When an atom, ion, or molecule becomes more positively charged (that is, when it loses electrons), we say that it has been oxidized. Loss of electrons by a substance is called oxidation. The term oxidation is used because the first reactions of this sort to be studied were reactions with oxygen. Many metals react directly with O2 in air to form metal oxides. In these reactions the metal loses electrons to oxygen, forming an ionic compound of the metal ion and oxide ion. The familiar example of rusting involves the reaction between iron metal and oxygen in the presence of water. In this process Fe is oxidized (loses electrons) to form Fe3+.

The reaction between iron and oxygen tends to be relatively slow, but other metals, such as the alkali and alkaline earth metals, react quickly upon exposure to air. FIGURE 4.12 shows how the bright metallic surface of calcium tarnishes as CaO forms in the reaction

In this reaction Ca is oxidized to Ca2+ and neutral O2 is transformed to O2− ions. When an atom, ion, or molecule becomes more negatively charged (gains electrons), we say that it is reducedThe gain of electrons by a substance is called reduction. When one reactant loses electrons (that is, when it is oxidized), another reactant must gain them. In other words, oxidation of one substance must be accompanied by reduction of some other substance.

FIGURE 4.11 Familiar corrosion products. (a) A green coating forms when copper is oxidized. (b) Rust forms when iron corrodes. (c) A black tarnish forms as silver corrodes.

FIGURE 4.12 Oxidation of calcium metal by molecular oxygen. The oxidation involves transfer of electrons from the calcium metal to the O2, leading to formation of CaO.

Oxidation Numbers

Before we can identify an oxidation-reduction reaction, we must have a bookkeeping system—a way of keeping track of electrons gained by the substance being reduced and electrons lost by the substance being oxidized. The concept of oxidation numbers (also called oxidation states) was devised as a way of doing this. Each atom in a neutral substance or ion is assigned an oxidation number. For monatomic ions the oxidation number is the same as the charge. For neutral molecules and polyatomic ions, the oxidation number of a given atom is a hypothetical charge. This charge is assigned by artificially dividing up the electrons among the atoms in the molecule or ion. We use the following rules for assigning oxidation numbers:

1. For an atom in its elemental form, the oxidation number is always zero. Thus, each H atom in the H2 molecule has an oxidation number of 0 and each P atom in the P4 molecule has an oxidation number of 0.

2. For any monatomic ion the oxidation number equals the ionic charge. Thus, K+ has an oxidation number of +1, S2− has an oxidation number of −2, and so forth. In ionic compounds the alkali metal ions (group 1A) always have a 1+ charge and therefore an oxidation number of +1. The alkaline earth metals (group 2A) are always +2, and aluminum (group 3A) is always +3 in ionic compounds. (In writing oxidation numbers we will write the sign before the number to distinguish them from the actual electronic charges, which we write with the number first.)

3. Nonmetals usually have negative oxidation numbers, although they can sometimes be positive:

(a) The oxidation number of oxygen is usually −2 in both ionic and molecular compounds. The major exception is in compounds called peroxides, which contain the O22− ion, giving each oxygen an oxidation number of −1.

(b) The oxidation number of hydrogen is usually +1 when bonded to nonmetals and −1 when bonded to metals.

(c) The oxidation number of fluorine is −1 in all compounds. The other halogens have an oxidation number of −1 in most binary compounds. When combined with oxygen, as in oxyanions, however, they have positive oxidation states.

4. The sum of the oxidation numbers of all atoms in a neutral compound is zero. The sum of the oxidation numbers in a polyatomic ion equals the charge of the ion. For example, in the hydronium ion H3O+ the oxidation number of each hydrogen is +1 and that of oxygen is −2. Thus, the sum of the oxidation numbers is 3(+1) + (−2) = +1, which equals the net charge of the ion. This rule is useful in obtaining the oxidation number of one atom in a compound or ion if you know the oxidation numbers of the other atoms, as illustrated in Sample Exercise 4.8.

It's important to remember that in every oxidation-reduction reaction, the oxidation numbers of at least two atoms must change. The oxidation number increases for any atom that is oxidized and decreases for any atom that is reduced.

GIVE IT SOME THOUGHT

What is the oxidation number of nitrogen (a) in aluminum nitride, AlN, and (b) in nitric acid, HNO3?

SAMPLE EXERCISE 4.8 Determining Oxidation Numbers

Determine the oxidation number of sulfur in (a) H2S, (b) S8(c) SCl2(d) Na2SO3(e) SO42−.

SOLUTION

Analyze We are asked to determine the oxidation number of sulfur in two molecular species, in the elemental form, and in two substances containing ions.

Plan In each species the sum of oxidation numbers of all the atoms must equal the charge on the species. We will use the rules outlined previously to assign oxidation numbers.

Solve

(a) When bonded to a nonmetal, hydrogen has an oxidation number of +1 (rule 3b). Because the H2S molecule is neutral, the sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have 2(+1) + x = 0. Thus, S has an oxidation number of −2.

(b) Because this is an elemental form of sulfur, the oxidation number of S is 0 (rule 1).

(c) Because this is a binary compound, we expect chlorine to have an oxidation number of −1 (rule 3c). The sum of the oxidation numbers must equal zero (rule 4). Letting x equal the oxidation number of S, we have x + 2(−1) = 0. Consequently, the oxidation number of S must be +2.

(d) Sodium, an alkali metal, always has an oxidation number of +1 in its compounds (rule 2). Oxygen has a common oxidation state of −2 (rule 3a). Letting x equal the oxidation number of S, we have 2(+1) + x + 3(−2) = 0. Therefore, the oxidation number of S in this compound is +4.

(e) The oxidation state of O is −2 (rule 3a). The sum of the oxidation numbers equals −2, the net charge of the SO42− ion (rule 4). Thus, we have x + 4(−2) = −2. From this relation we conclude that the oxidation number of S in this ion is +6.

Comment These examples illustrate that the oxidation number of a given element depends on the compound in which it occurs. The oxidation numbers of sulfur, as seen in these examples, range from −2 to +6.

PRACTICE EXERCISE

What is the oxidation state of the boldfaced element in (a) P2O5(b) NaH, (c) Cr2O72−(d) SnBr4(e) BaO2?

Answers: (a) +5, (b) −1, (c) +6, (d) +4, (e) −1

Oxidation of Metals by Acids and Salts

The reaction between a metal and either an acid or a metal salt conforms to the general pattern

Examples:     

These reactions are called displacement reactions because the ion in solution is displaced (replaced) through oxidation of an element.

FIGURE 4.13 Reaction of magnesium metal with hydrochloric acid. The metal is readily oxidized by the acid, producing hydrogen gas, H2(g), and MgCl2(aq).

Many metals undergo displacement reactions with acids, producing salts and hydrogen gas. For example, magnesium metal reacts with hydrochloric acid to form magnesium chloride and hydrogen gas (FIGURE 4.13):

The oxidation number of Mg changes from 0 to +2, an increase that indicates the atom has lost electrons and has therefore been oxidized. The oxidation number of H+ in the acid decreases from +1 to 0, indicating that this ion has gained electrons and has therefore been reduced. Chlorine has an oxidation number of −1 both before and after the reaction, indicating that it is neither oxidized nor reduced. In fact the Cl ions are spectator ions, dropping out of the net ionic equation:

Metals can also be oxidized by aqueous solutions of various salts. Iron metal, for example, is oxidized to Fe2+ by aqueous solutions of Ni2+ such as Ni(NO3)2(aq):

The oxidation of Fe to Fe2+ in this reaction is accompanied by the reduction of Ni2+ to Ni. Remember: Whenever one substance is oxidized, another substance must be reduced.

SAMPLE EXERCISE 4.9 Writing Equations for Oxidation-Reduction Reactions

Write the balanced molecular and net ionic equations for the reaction of aluminum with hydrobromic acid.

SOLUTION

Analyze We must write two equations—molecular and net ionic—for the redox reaction between a metal and an acid.

Plan Metals react with acids to form salts and H2 gas. To write the balanced equations, we must write the chemical formulas for the two reactants and then determine the formula of the salt, which is composed of the cation formed by the metal and the anion of the acid.

Solve The reactants are Al and HBr. The cation formed by Al is Al3+, and the anion from hydrobromic acid is Br. Thus, the salt formed in the reaction is AlBr3. Writing the reactants and products and then balancing the equation gives the molecular equation:

2 Al(s) + 6 HBr(aq) → 2 AlBr3(aq) + 3 H2(g)

Both HBr and AlBr3 are soluble strong electrolytes. Thus, the complete ionic equation is

2 Al(s) + 6H+(aq) + 6Br(aq) → 2 Al3+(aq) + 6 Br(aq) + 3 H2(g)

Because Br is a spectator ion, the net ionic equation is

2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g)

Comment The substance oxidized is the aluminum metal because its oxidation state changes from 0 in the metal to +3 in the cation, thereby increasing in oxidation number. The H+ is reduced because its oxidation state changes from +1 in the acid to 0 in H2.

PRACTICE EXERCISE

(a) Write the balanced molecular and net ionic equations for the reaction between magnesium and cobalt(II) sulfate. (b) What is oxidized and what is reduced in the reaction?

Answers: (a) Mg(s) + CoSO4(aq) → MgSO4(aq) + Co(s); Mg(s) + Co2+(aq) → Mg2+(aq) + Co(s), (b) Mg is oxidized and Co2+ is reduced.

The Activity Series

Can we predict whether a certain metal will be oxidized either by an acid or by a particular salt? This question is of practical importance as well as chemical interest. According to Equation 4.27, for example, it would be unwise to store a solution of nickel nitrate in an iron container because the solution would dissolve the container. When a metal is oxidized, it forms various compounds. Extensive oxidation can lead to the failure of metal machinery parts or the deterioration of metal structures.

Different metals vary in the ease with which they are oxidized. Zn is oxidized by aqueous solutions of Cu2+, for example, but Ag is not. Zn, therefore, loses electrons more readily than Ag; that is, Zn is easier to oxidize than Ag.

A list of metals arranged in order of decreasing ease of oxidation, such as TABLE 4.5, is called an activity series. The metals at the top of the table, such as the alkali metals and the alkaline earth metals, are most easily oxidized; that is, they react most readily to form compounds. They are called the active metals. The metals at the bottom of the activity series, such as the transition elements from groups 8B and 1B, are very stable and form compounds less readily. These metals, which are used to make coins and jewelry, are called noble metals because of their low reactivity.

The activity series can be used to predict the outcome of reactions between metals and either metal salts or acids. Any metal on the list can be oxidized by the ions of elements below it. For example, copper is above silver in the series. Thus, copper metal is oxidized by silver ions:

TABLE 4.5 • Activity Series of Metals in Aqueous Solution

The oxidation of copper to copper ions is accompanied by the reduction of silver ions to silver metal. The silver metal is evident on the surface of the copper wire in FIGURE 4.14. The copper(II) nitrate produces a blue color in the solution, as can be seen most clearly in the photograph on the right of Figure 4.14.

GIVE IT SOME THOUGHT

Does a reaction occur (a) when an aqueous solution of NiCl2(aq) is added to a test tube containing strips of metallic zinc, and (b) when NiCl2(aq) is added to a test tube containing Zn(NO3)2(aq)?

Only metals above hydrogen in the activity series are able to react with acids to form H2. For example, Ni reacts with HCl(aq) to form H2:

Because elements below hydrogen in the activity series are not oxidized by H+, Cu does not react with HCl(aq). Interestingly, copper does react with nitric acid, as shown in Figure 1.11, but the reaction is not oxidation of Cu by H+ ions. Instead, the metal is oxidized to Cu2+ by the nitrate ion, accompanied by the formation of brown nitrogen dioxide, NO2(g):

As the copper is oxidized in this reaction, NO3, where the oxidation number of nitrogen is +5, is reduced to NO2, where the oxidation number of nitrogen is +4. We will examine reactions of this type in Chapter 20.

FIGURE 4.14 Reaction of copper metal with silver ion. When copper metal is placed in a solution of silver nitrate, a redox reaction forms silver metal and a blue solution of copper(II) nitrate.

SAMPLE EXERCISE 4.10 Determining When an Oxidation-Reduction Reaction Can Occur

Will an aqueous solution of iron(II) chloride oxidize magnesium metal? If so, write the balanced molecular and net ionic equations for the reaction.

SOLUTION

Analyze We are given two substances—an aqueous salt, FeCl2, and a metal, Mg—and asked if they react with each other.

Plan A reaction occurs if the reactant that is a metal in its elemental form (Mg) is located above the reactant that is a metal in its oxidized form (Fe2+) in Table 4.5. If the reaction occurs, the Fe2+ ion in FeCl2 is reduced to Fe, and the Mg is oxidized to Mg2+.

Solve Because Mg is above Fe in the table, the reaction occurs. To write the formula for the salt produced in the reaction, we must remember the charges on common ions. Magnesium is always present in compounds as Mg2+; the chloride ion is Cl. The magnesium salt formed in the reaction is MgCl2, meaning the balanced molecular equation is

Mg(s) + FeCl2(aq) → MgCl2(aq) + Fe(s)

Both FeCl2 and MgCl2 are soluble strong electrolytes and can be written in ionic form, which shows us that Cl is a spectator ion in the reaction. The net ionic equation is

Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)

The net ionic equation shows that Mg is oxidized and Fe2+ is reduced in this reaction.

Check Note that the net ionic equation is balanced with respect to both charge and mass.

PRACTICE EXERCISE

Which of the following metals will be oxidized by Pb(NO3)2: Zn, Cu, Fe?

Answer: Zn and Fe

A CLOSER LOOK
THE AURA OF GOLD

Throughout history, people have cherished gold, fought for it, and even died for it.

The physical and chemical properties of gold make it special. First, its intrinsic beauty and rarity make it precious. Second, it is soft and can be easily formed into jewelry, coins, and other objects. Third, it is one of the least active metals (Table 4.5). It is not oxidized in air and does not react with water (FIGURE 4.15), with basic solutions, or with most acidic solutions.

FIGURE 4.15 The chemical inertness of gold. Two, stamped, weighed, and priced (in the 1800's) gold bars recently recovered from a steamship that sank off the coast of Central America in 1857.

Many of the early studies on gold arose from alchemy, in which people attempted to turn cheap metals, such as lead, into gold. Alchemists discovered that gold can be dissolved in a 3:1 mixture of concentrated hydrochloric and nitric acids, known as aqua regia (“royal water”). The action of the nitric acid on gold is similar to that on copper (Equation 4.31) in that the nitrate ion, rather than H+, oxidizes the metal to Au3+. The Cl ions interact with Au3+ to form highly stable AuCl4 ions. The net ionic equation is

All the gold ever mined would fit in a cube 21 m on a side and weighing about 1.6 × 108 kg. More than 90% of this amount has been produced since the 1848 California gold rush. Annual worldwide production of gold is about 2.4 × 106 kg. By contrast 16,000 times more aluminum, over 3.97 × 1010 kg, are produced annually.

Roughly three-quarters of gold production goes to make jewelry, where it is often alloyed with other metals. Approximately 12% of gold production is used to meet a variety of industrial applications, most notably in electronic devices where its excellent conductivity and corrosion resistance make it a valuable component. A typical touch-tone telephone contains 33 gold-plated contacts. Gold is also used in computers and other microelectronic devices where fine gold wire is used to link components.

Because of its resistance to corrosion, gold is an ideal metal for dental crowns and caps, which accounts for about 3% of the annual use of the element. The pure metal is too soft to use in dentistry, so it is combined with other metals to form alloys.

RELATED EXERCISE: 4.91

STRATEGIES IN CHEMISTRY
ANALYZING CHEMICAL REACTIONS

In this chapter you have been introduced to a great number of chemical reactions. A major difficulty students face in trying to master material of this sort is gaining a “feel” for what happens when chemicals react. In fact, you might marvel at the ease with which your professor or teaching assistant can figure out the results of a chemical reaction. One of our goals in this textbook is to help you become more adept at predicting the outcomes of reactions. The key to gaining this “chemical intuition” is understanding how to categorize reactions.

Attempting to memorize individual reactions would be a futile task. It is far more fruitful to recognize patterns to determine the general category of a reaction, such as metathesis or oxidation-reduction. Thus, when you are faced with the challenge of predicting the outcome of a chemical reaction, ask yourself the following questions:

• What are the reactants?

• Are they electrolytes or nonelectrolytes?

• Are they acids and bases?

• If the reactants are electrolytes, will metathesis produce a precipitate? Water? A gas?

• If metathesis cannot occur, can the reactants engage in an oxidation-reduction reaction? This requires that there be both a reactant that can be oxidized and a reactant that can be reduced.

By asking questions such as these, you should be able to predict what happens during the reaction. You might not always be entirely correct, but if you keep your wits about you, you will not be far off. As you gain experience, you will begin to look for reactants that might not be immediately obvious, such as water from the solution or oxygen from the atmosphere.

One of the greatest tools available to chemists is experimentation. If you perform an experiment in which two solutions are mixed, you can make observations that help you understand what is happening. For example, using Table 4.1 to predict whether a precipitate will form is not nearly as exciting as seeing the precipitate form, as in Figure 4.4. Careful observations in the laboratory portion of the course will make your lecture material easier to master.