Oxidation–Reduction Reactions in Solution

4.8 Oxidation–Reduction Reactions in Solution

Learning Objective

To identify oxidation–reduction reactions in solution.

We described the defining characteristics of oxidation–reduction, or redox, reactions in Chapter 3 "Chemical Reactions". Most of the reactions we considered there 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 is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. Several mechanical routines can be applied to balance oxidation-reduction equations. We will use the "half-reaction" method..

Balancing Redox Equations Using Half Reactions

Table 4.4 Procedure for Balancing Oxidation–Reduction Reactions by the Half Reaction Method

Divide the unbalanced chemical equation for the reaction into an oxidation and a reduction.
Balance the number of atoms of all species except hydrogen and oxygen.
Balance the oxygen with H₂O.
Balance the hydrogen with H⁺.
Balance the charge with e^-.
Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.
Add the two equations and cancel the electrons.
Check to make sure that the equation is balanced in both atoms and total charges.
If the reaction is in acidic solution, your work is done. In basic solution, follow steps 10-13.
Add as many OH^- to the same side of the equation with H⁺ as there are H⁺.
H⁺ + OH^- → H₂O, so replace each combination of H⁺ and OH^- with an H₂O.
If H₂O appears on both sides of the arrow, cancel appropriately and leave the remainder.
Rewrite the balanced equation and check to make sure both atoms and total charges are balanced.

Consider, for example, the reaction of Cr²⁺(aq) with manganese dioxide (MnO₂) to make Cr³⁺(aq) and Mn²⁺(aq)in the presence of dilute acid. Equation 4.55 is a representation of the reaction before applying our balancing routine. Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the species that are oxidized and reduced:

Equation 4.55

Cr²⁺ + MnO₂ → Cr³⁺ + Mn²⁺

To balance a redox equation using the half reaction method , we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. The oxidation can be written as

Equation 4.56

Cr²⁺ → Cr³⁺

and the reduction as

Equation 4.57

MnO₂ → Mn²⁺

This completes step 1 from Table 4.4.

Step 2 is already complete since only oxygen remains unbalanced.

For step 3, the oxidation half reaction equation is done but we need to add two water molecules to the products side of the reduction half reaction equation (4.57) yielding equation 4.58.

Equation 4.58

MnO₂ → 2Mn²⁺ + 2H₂O

For step 4, again for this example, the oxidation half reaction equation is done as there are no hydrogen atoms to balance. For the reduction half reaction we need to add 4 H⁺ to the reactants side yielding equation 4.59.

Equation 4.59

MnO₂+ 4H⁺ → Mn²⁺ + 2H₂O

For step 5 we need to balance the charge on both sides of each equation. At this point the two equations may have different total charges; however, the charge needs to be the same on both sides of the arrow for each half reaction. Taking the oxidation half reaction (Equation 4.56) first, we need to add one electron to the products side of the equation. That will give us equation 4.60 showing a charge of +2 on both sides of the arrow which is a balanced charge.

Equation 4.60

Cr²⁺ → Cr³⁺ + e^-

Applying step 5 to the reduction half reaction requires two electrons be added to the reactants side giving us equation 4.61.

Equation 4.61

MnO₂+ 4H⁺ + 2e^- → Mn²⁺ + 2H₂O

For step 6 we compare the number of electrons in equation 4.60 to the number of electrons in equation 4.61. To make the number of electrons the same, we must multiply equation 4.60 by 2. This gives us equation 4.62.

Equation 4.62

2Cr²⁺ → 2Cr³⁺ + 2e^-

For step 7 we add equation 4.62 to equation 4.61 resulting in equation 4.63.

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

Equation 4.63

${2 Cr}^{2 +} \to {2 Cr}^{3 +} + {2 e}^{-}$

${MnO}_{2} + {4 H}^{+} + {2 e}^{-} \to {Mn}^{2 +} + {2 H}_{2} O$

———————————————————————

${2 Cr}^{2 +} + {MnO}_{2} + {4 H}^{+} \to {2 Cr}^{3 +} + {Mn}^{2 +} + {2 H}_{2} O$

Note the Pattern

In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. If electrons are ever "left over" in a half reaction equation, one of the steps in the balancing procedure has been incorrectly applied.

Step 9 is more of a direction than a "step." Since this reaction occurs in acidic solution, our work is done. Our balanced equation, including the physical states of the reactants and products, is shown in Equation 4.64.

Equation 4.64

2Cr²⁺(aq) + MnO₂(s) + 4H⁺(aq) → 2Cr³⁺(aq) + Mn²⁺(aq) + 2H₂O(l)

Example 17

Arsenic acid (H₃AsO₄) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH₃, a highly toxic and unstable gas) and Zn²⁺(aq). Balance the equation for this reaction using oxidation states:

H₃AsO₄(aq) + Zn(s) → AsH₃(g) + Zn²⁺(aq)

Given: reactants and products in acidic solution

Asked for: balanced chemical equation using half reactions

Strategy:

Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Half Reaction Method" for balancing a redox equation. When you are done, be certain to check that the equation is balanced.

Solution:

Divide the unbalanced chemical equation for the reaction into an oxidation and a reduction. For no particular reason, let's tackle the oxidation half reaction first.
Zn →Zn²⁺

And now for the reduction half reaction.

H₃AsO₄ → AsH₃
Balance the number of atoms of all species except hydrogen and oxygen. An easy step since it's already done for this particular reaction.
Balance the oxygen with H₂O. Only the reduction half reaction contains oxygen. Since there are four oxygen atoms on the reactants' side, we add 4 H₂O to the product side.
H₃AsO₄→AsH₃ + 4H₂O
Balance the hydrogen with H⁺. Again, we only need to consider the reduction equation for this particular reaction since there is no hydrogen in the oxidation half. Since 3 hydrogen atoms appear on the reactants side and 11 hydrogen atoms appear on the products side, we add 8 H⁺ to the reactants side.
H₃AsO₄ + 8H⁺ → AsH₃ + 4H₂O
Balance the charge with e^-. Inserting the The oxidation half reaction has a 0 charge on the reactants side and a +2 on the products, so we add two electrons to the product side. actual chemical forms of arsenic and zinc and adjusting the coefficients gives
Oxidation: Zn → Zn²⁺ + 2e^-
And for the reduction half reaction we have a +8 charge on the reactants side and 0 charge on the products side, so we add eight electrons to the reactant side.
Reduction: H₃AsO₄ + 8H⁺ + 8e^-→ AsH₃ + 4H₂O
Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. For this reaction our lowest common multiple between 2 electrons and 8 electrons is 8, so we'll multiply the oxidation equation by 4.
4Zn → 4Zn²⁺ + 8e ^–
Add the two equations and cancel the electrons.
4Zn → 4Zn²⁺ + 8e ^–H₃AsO₄ + 8H⁺ + 8e^- → AsH₃ + 4H₂O

—————————————————————

4Zn + H₃AsO₄ + 8H⁺ →4Zn²⁺ + AsH₃ + 4H₂O
Check to make sure that the equation is balanced in both atoms and total charges. It's so easy to check and know that you have the correct "answer," so there's not much reason for "partial credit" on exam questions dealing with balancing equations. Most often when there's a snag, it's the charge. Make sure the charge is the same on both sides of the arrow. Atoms: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H Total charge: 8(+1) = 4(+2) = +8
If the reaction is in acidic solution, your work is done. In basic solution, follow steps 10-13. Since the problem stated the reaction occurs in acidic solution, we are done!

The balanced chemical equation for the reaction is therefore:

H₃AsO₄(aq) + 4Zn(s) + 8H⁺(aq) → AsH₃(g) + 4Zn²⁺(aq) + 4H₂O(l)

Exercise

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

CuS(s) + H⁺(aq) + NO₃⁻(aq) → Cu²⁺(aq) + NO(g) + SO₄²⁻(aq)

Answer: 3CuS(s) + 8H⁺(aq) + 8NO₃⁻(aq) → 3Cu²⁺(aq) + 8NO(g) + 3SO₄²⁻(aq) + 4H₂O(l)

Reactions in basic solutions are balanced in exactly the same manner through the first 9 steps of Table 4.4. Additionally steps 10 through 13 are employed. . To make sure you understand the procedure, consider Example 18.

Example 18

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

Al(s) + H₂O(aq) → [Al(OH)₄]⁻(aq) + H₂(g)

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 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Half Reaction Method" for balancing a redox reaction. When you are done, be certain to check that the equation is balanced.

Solution:

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

Al → [Al(OH)₄]^-

H₂O → H₂
Already done
Add four H₂O to the reactants side of the oxidation and one H₂O to the products side of the reduction.

Al + 4H₂O→ [Al(OH)₄]^-

H₂O → H₂ + H₂O
There are eight H's on the reactants side of the oxidation and four H's on the product side, so add four H⁺ to the reactant side. For the reduction there are two H's on the reactant side and four H's on the product side, so add two H⁺ to the reactant side.

Al + 4H₂O → [Al(OH)₄]^- + 4H⁺

H₂O + 2H⁺→ H₂ + H₂O
The charge for the oxidation equation is 0 on the reactants side and +3 on the products side, so add 3 electrons to the products side. The charge on the reactants side of the reduction equation is +2 and 0 on the products side, so add 2 electrons to the reactants side.

Al + 4H₂O → [Al(OH)₄]^- + 4H⁺ + 3e^-

H₂O + 2H⁺ + 2e^-→ H₂ + H₂O
The least common multiple of electrons between the two half reactions is 6 so multiply the oxidation by 2 and the reduction by 3.

2Al + 8H₂O → 2[Al(OH)₄]^- +8H⁺ + 6e^-

3H₂O + 6H⁺ + 6e^-→ 3H₂ + 3H₂O
Adding both half reactions, cancelling H₂O, H⁺ and e^- as appropriate.

2Al + 8H₂O → 2[Al(OH)₄]^- +8H⁺ + 6e^-

3H₂O + 6H⁺ + 6e^-→ 3H₂ + 3H₂O

———————————————————————

2Al + 8H₂O →2[Al(OH)₄]^- + 2H⁺ + 3H₂
Checking atoms and charge

2Al + 8O + 16H = 2Al + 8O + 16 H

Total charge 0 = -2 + 2
This reaction occurs in base, so we need to proceed to step 10.
Add as many OH^- to both sides of the equation as there are H⁺. In this case there are two H⁺ on the products side, so add two OH^- to the reactants side and two OH^- to the products side.

2Al + 8H₂O+ 2OH^- →2[Al(OH)₄]^- + 2H⁺ + 3H₂ + 2OH^-
H⁺ + OH^- → H₂O, so replace each combination of H⁺ and OH^- with an H₂O. In this case two H₂O molecules are formed on the products side.2Al + 8H₂O + 2OH^-→2[Al(OH)₄]^- + 2H₂O + 3H₂
If H₂O appears on both sides of the arrow, cancel appropriately and leave the remainder. And in this case, the two H₂O molecules on the product side are subtracted from the 8 H₂O on the reactant side leaving 6 H₂O on the reactant side. 2Al + 6H₂O + 2OH^- →2[Al(OH)₄]^- + 3H₂
Rewrite the balanced equation and check to make sure both atoms and total charges are balanced.
2Al(s) + 2OH^–(aq) + 6H₂O(l) → 2[Al(OH)₄]^–(aq) + 3H₂(g)

Thus 3 mol of H₂ 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 balanced chemical equation for the reaction.

Answer: 2MnO₄⁻(aq) + 3NO₂⁻(aq) + H₂O(l) → 2MnO₂(s) + 3NO₃⁻(aq) + 2OH⁻(aq)

As suggested in Example 17 and Example 18, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried 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 for predicting whether a redox reaction will occur and the identity of the products:

Compounds of elements in high oxidation states (such as ClO₄⁻, NO₃⁻, MnO₄⁻, Cr₂O₇²⁻, and UF₆) tend to act as oxidants and become reduced in chemical reactions.
Compounds of elements in low oxidation states (such as CH₄, NH₃, H₂S, and HI) tend to act as reductants and become oxidized in chemical reactions.

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 a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.

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 solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure 4.20 "Rust Formation"). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl⁻ ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:

Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)

In subsequent steps, FeCl₂ undergoes oxidation to form a reddish-brown precipitate of Fe(OH)₃.

Figure 4.20 Rust Formation

The corrosion process involves an oxidation–reduction reaction in which metallic iron is converted to Fe(OH)₃, a reddish-brown solid. Image Credit: By Messina, John, 1940-, Photographer (NARA record: 8464458) (U.S. National Archives and Records Administration) [Public domain], via Wikimedia Commons

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

metal + acid → salt + hydrogen

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

Equation 4.65

Pb(s) + 2H⁺(aq) → Pb²⁺(aq) + H₂(g)

Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite 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 of reactions are called single-displacement reactionsA chemical reaction in which an ion in solution is displaced through oxidation of a metal., in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 4.66) and the reduction of silver salts by copper (Equation 4.67 and Figure 4.21 "The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate"):

Equation 4.66

Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s)

Equation 4.67

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)

The reaction in Equation 4.66 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.

Figure 4.21 The Single-Displacement Reaction of Metallic Copper with a Solution of Silver Nitrate

When a copper coil is placed in a 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 of aqueous Cu²⁺ ions. Image Credit: By ΛΦΠ (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC BY-SA 4.0-3.0-2.5-2.0-1.0 (http://creativecommons.org/licenses/by-sa/4.0-3.0-2.5-2.0-1.0)], via Wikimedia Commons

The Activity Series

By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, we saw in Equation 4.66 and Equation 4.67 that metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn²⁺. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:

Equation 4.68

Zn (s) + {Mg}^{2 +} (aq) \to {Zn}^{2 +} (aq) + Mg (s)

Equation 4.69

Mg(s) + Zn²⁺(aq) → Mg²⁺(aq) + Zn(s)

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

Pairwise reactions of this sort are the basis of the activity seriesA list of metals and hydrogen in order of their relative tendency to be oxidized. (Figure 4.22 "The Activity Series"), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metalsThe metals at the top of the activity series, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metalsThe metals at the bottom of the activity series, which have the least tendency to be oxidized. (at the bottom of the series).

Figure 4.22 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 the elements below it in the series. Because magnesium is above zinc in Figure 4.22 "The Activity Series", magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H₂. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example 19 demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. We will return to the activity series when we discuss oxidation–reduction reactions in future chapters.

Example 19

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

A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
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 4.22 "The Activity Series" and from their relative positions, predict whether a reaction will occur. 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:

A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:
Al(s) + 3Ag⁺(aq) → Al³⁺(aq) + 3Ag(s)
Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.
A Mercury lies below lead in the activity series, so no reaction will occur.
A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H₂. B From our discussion of solubilities, recall that Pb²⁺ and SO₄²⁻ form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:
Pb(s) + 2H⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) + H₂(g)
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 sulfuric acid. Image Credit: Highland Community College, Freeport, IL

Exercise

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

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

Answer:

no reaction
3Zn(s) + 2Cr³⁺(aq) → 3Zn²⁺(aq) + 2Cr(s)
2Al(s) + 6CH₃CO₂H(aq) → 2Al³⁺(aq) + 6CH₃CO₂⁻(aq) + 3H₂(g)

Summary

In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method (Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method"), in which the overall reaction is separated into an oxidation equation and a reduction equation. Single-displacement reactions are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the activity series (Figure 4.22 "The Activity Series"), which arranges metals and H₂ in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. Active metals lie at the top of the activity series, whereas inert metals are at the bottom of the activity series.

Key Takeaway

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

Conceptual Problems

Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants?
If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why?
In each redox reaction, determine which species is oxidized and which is reduced:
1. Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
2. Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O(l)
3. BrO₃⁻(aq) + 2MnO₂(s) + H₂O(l) → Br⁻(aq) + 2MnO₄⁻(aq) + 2H⁺(aq)
Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is not a single-displacement reaction.
Of the following elements, which would you expect to have the greatest tendency to be oxidized: Zn, Li, or S? Explain your reasoning.
Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning.
Which of these metals produces H₂ in acidic solution?
1. Ag
2. Cd
3. Ca
4. Cu
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. Mg(s) + Cu²⁺(aq) →
2. Au(s) + Ag⁺(aq) →
3. Cr(s) + Pb²⁺(aq) →
4. K(s) + H₂O(l) →
5. Hg(l) + Pb²⁺(aq) →

Numerical Problems

Balance each redox reaction under the conditions indicated.
1. CuS(s) + NO₃⁻(aq) → Cu²⁺(aq) + SO₄²⁻(aq) + NO(g); acidic solution
2. Ag(s) + HS⁻(aq) + CrO₄²⁻(aq) → Ag₂S(s) + Cr(OH)₃(s); basic solution
3. Zn(s) + H₂O(l) → Zn²⁺(aq) + H₂(g); acidic solution
4. O₂(g) + Sb(s) → H₂O₂(aq) + SbO₂⁻(aq); basic solution
5. UO₂²⁺(aq) + Te(s) → U⁴⁺(aq) + TeO₄²⁻(aq); acidic solution
Balance each redox reaction under the conditions indicated.
1. MnO₄⁻(aq) + S₂O₃²⁻(aq) → Mn²⁺(aq) + SO₄²⁻(aq); acidic solution
2. Fe²⁺(aq) + Cr₂O₇²⁻(aq) → Fe³⁺(aq) + Cr³⁺(aq); acidic solution
3. Fe(s) + CrO₄²⁻(aq) → Fe₂O₃(s) + Cr₂O₃(s); basic solution
4. Cl₂(aq) → ClO₃⁻(aq) + Cl⁻(aq); acidic solution
5. CO₃²⁻(aq) + N₂H₄(aq) → CO(g) + N₂(g); basic solution
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.
1. Platinum wire is dipped in hydrochloric acid.
2. Manganese metal is added to a solution of iron(II) chloride.
3. Tin is heated with steam.
4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.
1. A few drops of NiBr₂ are dropped onto a piece of iron.
2. A strip of zinc is placed into a solution of HCl.
3. Copper is dipped into a solution of ZnCl₂.
4. A solution of silver nitrate is dropped onto an aluminum plate.
Dentists occasionally use metallic mixtures called amalgams for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction.
Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu²⁺(aq) and nitric oxide gas.
1. What has been oxidized? What has been reduced?
2. Balance the chemical equation.
Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:
1. Pt²⁺(aq) + Ag(s) →
2. HCN(aq) + NaOH(aq) →
3. Fe(NO₃)₃(aq) + NaOH(aq) →
4. CH₄(g) + O₂(g) →
Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation:
1. Zn(s) + HCl(aq) →
2. HNO₃(aq) + AlCl₃(aq) →
3. K₂CrO₄(aq) + Ba(NO₃)₂(aq) →
4. Zn(s) + Ni²⁺(aq) → Zn²⁺(aq) + Ni(s)