Classifing Chemical Reactions Review and Reinforcement on the Line at the Left Answer Key

Affiliate iv. Stoichiometry of Chemic Reactions

4.two Classifying Chemical Reactions

Learning Objectives

By the terminate of this section, you volition exist able to:

• Define three common types of chemical reactions (precipitation, acrid-base, and oxidation-reduction)
• Allocate chemical reactions equally i of these three types given appropriate descriptions or chemical equations
• Identify mutual acids and bases
• Predict the solubility of common inorganic compounds past using solubility rules
• Compute the oxidation states for elements in compounds

Humans interact with i another in various and complex means, and nosotros classify these interactions according to common patterns of behavior. When two humans substitution information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have also found it convenient (or even necessary) to classify chemical interactions past identifying mutual patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: atmospheric precipitation, acid-base, and oxidation-reduction.

Precipitation Reactions and Solubility Rules

A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the commutation of ions betwixt ionic compounds in aqueous solution and are sometimes referred to as double deportation, double replacement, or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central office in many chemic analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the limerick of affair (see the concluding module of this chapter).

The extent to which a substance may be dissolved in h2o, or any solvent, is quantitatively expressed as its solubility, defined every bit the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively big solubilities are said to be soluble. A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble, and these are the substances that readily precipitate from solution. More information on these of import concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may but refer to patterns of solubility that accept been observed for many ionic compounds (Table 8).

Soluble compounds contain group 1 metal cations (Li+, Na+, Yard+, Rb+, and Cs+) and ammonium ion (NHfour +) the halide ions (Cl−, Br−, and I−) the acetate (C2H3Otwo −), bicarbonate (HCO3 −), nitrate (NOiii −), and chlorate (ClO3 −) ions the sulfate (SOfour 2−) ion	Exceptions to these solubility rules include halides of Ag+, Hgtwo 2+, and Pb2+ sulfates of Ag+, Ba2+, Ca2+, Hg22+,Hg22+, Pb2+, and Srtwo+
Insoluble compounds comprise carbonate (CO3 2−), chromate (CrO4 2−), phosphate (POiv iii−), and sulfide (S2−) ions hydroxide ion (OH−)	Exceptions to these insolubility rules include compounds of these anions with group i metal cations and ammonium ion hydroxides of grouping 1 metallic cations and Batwo+
Table 8. Solubilities of Common Ionic Compounds in Water

A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide:

[latex]two\text{KI}(aq) + \text{Lead(NO}_3)_2(aq) \longrightarrow \text{PbI}_2(due south) + 2\text{KNO}_3(aq)[/latex]

This observation is consistent with the solubility guidelines: The only insoluble compound amidst all those involved is lead iodide, ane of the exceptions to the general solubility of iodide salts.

The net ionic equation representing this reaction is:

[latex]\text{Atomic number 82}^{2+}(aq) + 2\text{I}^{-}(aq) \longrightarrow \text{PbI}_2(south)[/latex]

Lead iodide is a bright xanthous solid that was formerly used as an creative person's pigment known as iodine yellow (Figure ane). The properties of pure PbI₂ crystals brand them useful for fabrication of X-ray and gamma ray detectors.

Effigy 1. A precipitate of PbI₂ forms when solutions containing Pb^two+ and I⁻ are mixed. (credit: Der Kreole/Wikimedia Eatables)

The solubility guidelines in Table nine may exist used to predict whether a atmospheric precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One only needs to identify all the ions present in the solution and and then consider if possible cation/anion pairing could event in an insoluble chemical compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag⁺, NO₃ ⁻, Na⁺, and F⁻ ions. Aside from the ii ionic compounds originally present in the solutions, AgNO_iii and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO₃ and AgF. The solubility guidelines bespeak all nitrate salts are soluble simply that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations:

[latex]\text{NaF}(aq) + \text{AgNO}_3(aq) \longrightarrow \text{AgF}(s) + \text{NaNO}_3(aq) \;\text{(molecular)}[/latex][latex]\text{Ag}^{+}(aq) + \text{F}^{-}(aq) \longrightarrow \text{AgF}(s) \;\text{(internet ionic)}[/latex]

Example ane

Predicting Precipitation Reactions
Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If atmospheric precipitation is expected, write a counterbalanced cyberspace ionic equation for the reaction.

(a) potassium sulfate and barium nitrate

(b) lithium chloride and silver acetate

(c) pb nitrate and ammonium carbonate

Solution
(a) The two possible products for this combination are KNO_three and BaSO₄. The solubility guidelines indicate BaSO₄ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

[latex]\text{Ba}^{2+}(aq) + {\text{SO}_4}^{2-}(aq) \longrightarrow \text{BaSO}_4(s)[/latex]

(b) The two possible products for this combination are LiC₂H₃O_two and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is

[latex]\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) \longrightarrow \text{AgCl}(s)[/latex]

(c) The two possible products for this combination are PbCO₃ and NH₄NO_iii. The solubility guidelines indicate PbCO_iii is insoluble, and and so a precipitation reaction is expected. The cyberspace ionic equation for this reaction, derived in the manner detailed in the previous module, is

[latex]\text{Pb}^{ii+}(aq) + {\text{CO}_3}^{ii-}(aq) \longrightarrow \text{PbCO}_3(south)[/latex]

Check Your Learning
Which solution could be used to precipitate the barium ion, Ba²⁺, in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate?

Answer:

sodium sulfate, BaSO_iv

Acid-Base of operations Reactions

An acid-base reaction is ane in which a hydrogen ion, H⁺, is transferred from ane chemic species to another. Such reactions are of cardinal importance to numerous natural and technological processes, ranging from the chemical transformations that take identify within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemical science, therefore, is worthy of thorough discussion, and a total chapter is devoted to this topic afterwards in the text.

For purposes of this cursory introduction, nosotros will consider only the more common types of acid-base of operations reactions that take place in aqueous solutions. In this context, an acid is a substance that volition deliquesce in h2o to yield hydronium ions, H₃O⁺. Every bit an example, consider the equation shown hither:

[latex]\text{HCl}(aq) + \text{H}_2 \text{O}(aq) \longrightarrow \text{Cl}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)[/latex]

The process represented by this equation confirms that hydrogen chloride is an acrid. When dissolved in h2o, H₃O⁺ ions are produced past a chemical reaction in which H⁺ ions are transferred from HCl molecules to H_iiO molecules (Figure 2).

Effigy two. When hydrogen chloride gas dissolves in water, (a) information technology reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).

The nature of HCl is such that its reaction with water as only described is essentially 100% efficient: Virtually every HCl molecule that dissolves in h2o will undergo this reaction. Acids that completely react in this way are called strong acids, and HCl is one amidst just a handful of common acid compounds that are classified as strong (Table 9). A far greater number of compounds carry as weak acids and only partially react with water, leaving a large bulk of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy gustatory modality of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body scent. A familiar instance of a weak acrid is acetic acrid, the chief ingredient in food vinegars:

[latex]\text{CH}_3 \text{CO}_2 \text{H}(aq) + \text{H}_2 \text{O}(l) \leftrightharpoons \text{CH}_3 {\text{CO}_2}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)[/latex]

When dissolved in water under typical weather, only near 1% of acerb acid molecules are present in the ionized class, [latex]\text{CH}_3 {\text{CO}_2}^{-}[/latex](Effigy 3). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemic equilibrium.)

Figure 3. (a) Fruits such as oranges, lemons, and grapefruit comprise the weak acid citric acid. (b) Vinegars contain the weak acid acerb acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Brücke-Osteuropa/Wikimedia Commons)

Chemical compound Formula	Name in Aqueous Solution
HBr	hydrobromic acid
HCl	hydrochloric acrid
Hullo	hydroiodic acid
HNOiii	nitric acid
HClO4	perchloric acid
H2Thenfour	sulfuric acid
Tabular array ix. Common Strong Acids

A base is a substance that volition deliquesce in h2o to yield hydroxide ions, OH⁻. The nigh common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion—for case, NaOH and Ca(OH)₂. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)₂ dissolve in water and dissociate completely to produce cations (Yard⁺ and Ba²⁺, respectively) and hydroxide ions, OH⁻. These bases, forth with other hydroxides that completely dissociate in water, are considered strong bases.

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

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

This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na⁺ and OH⁻ ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water nether typical weather condition, NaOH and other ionic hydroxides are all classified as strong bases.

Different ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and and so are classified equally weak bases. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agronomical fertilizer, a raw cloth for chemic synthesis of other compounds, and an active ingredient in household cleaners (Figure 4). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:

[latex]\text{NH}_3(aq) + \text{H}_2 \text{O}(l) \rightleftharpoons {\text{NH}_4}^{+}(aq) + \text{OH}^{-}(aq)[/latex]

This is, by definition, an acid-base of operations reaction, in this case involving the transfer of H⁺ ions from h2o molecules to ammonia molecules. Under typical atmospheric condition, only nearly 1% of the dissolved ammonia is present equally NH₄ ⁺ ions.

Effigy four. Ammonia is a weak base used in a diverseness of applications. (a) Pure ammonia is normally practical as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)

The chemical reactions described in which acids and bases dissolved in water produce hydronium and hydroxide ions, respectively, are, by definition, acrid-base reactions. In these reactions, water serves as both a solvent and a reactant. A neutralization reaction is a specific type of acid-base of operations reaction in which the reactants are an acid and a base of operations, the products are often a salt and water, and neither reactant is the water itself:

[latex]\text{acid} + \text{base of operations} \longrightarrow \text{common salt} + \text{water}[/latex]

To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous interruption of solid Mg(OH)₂) is ingested to ease symptoms associated with excess breadbasket acid (HCl):

[latex]\text{Mg(OH)}_2(south) + two\text{HCl}(aq) \longrightarrow \text{MgCl}_2(aq) + 2\text{H}_2 \text{O}(l).[/latex]

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

Example 2

Writing Equations for Acid-Base of operations Reactions
Write balanced chemical equations for the acid-base of operations reactions described here:

(a) the weak acrid hydrogen hypochlorite reacts with water

(b) a solution of barium hydroxide is neutralized with a solution of nitric acid

Solution
(a) The two reactants are provided, HOCl and H₂O. Since the substance is reported to exist an acid, its reaction with water will involve the transfer of H⁺ from HOCl to H_twoO to generate hydronium ions, H_iiiO⁺ and hypochlorite ions, OCl⁻.

[latex]\text{HOCl}(aq) + \text{H}_2 \text{O}(fifty) \rightleftharpoons \text{OCl}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)[/latex]

A double-arrow is advisable in this equation because it indicates the HOCl is a weak acid that has not reacted completely.

(b) The two reactants are provided, Ba(OH)₂ and HNO_three. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba²⁺) and the anion generated when the acid transfers its hydrogen ion (NO3−).(NO3−).

[latex]\text{Ba(OH)}_2(aq) + 2\text{HNO}_3(aq) \longrightarrow \text{Ba(NO}_3)_2(aq) + ii\text{H}_2 \text{O}(fifty)[/latex]

Check Your Learning
Write the net ionic equation representing the neutralization of whatsoever strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acrid is dissolved in h2o.)

Answer:

[latex]\text{H}_3 \text{O}^{+}(aq) + \text{OH}^{-}(aq) \longrightarrow 2\text{H}_2 \text{O}(l)[/latex]

Explore the microscopic view of stiff and weak acids and bases.

Oxidation-Reduction Reactions

Earth's atmosphere contains virtually 20% molecular oxygen, O₂, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O₂, but its significant has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will exist used to develop a clear picture of this classification.

Some redox reactions involve the transfer of electrons betwixt reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:

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

Information technology is helpful to view the process with regard to each individual reactant, that is, to correspond the fate of each reactant in the course of an equation called a half-reaction:

[latex]2\text{Na}(due south) \longrightarrow 2\text{Na}^{+}(south) + 2\text{e}^{-}[/latex]
[latex]\text{Cl}_2(thou) + ii\text{e}^{-} \longrightarrow 2\text{Cl}^{-}(southward)[/latex]

These equations show that Na atoms lose electrons while Cl atoms (in the Cl_ii molecule) proceeds electrons, the "s" subscripts for the resulting ions signifying they are present in the form of a solid ionic chemical compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:

[latex]\begin{array}{r @ {{}={}} l} \pmb{\text{oxidation}} & \text{loss of electrons} \\[1em] \pmb{\text{reduction}} & \text{proceeds of electrons} \end{array}[/latex]

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

[latex]\begin{array}{r @ {{}={}} fifty} \pmb{\text{reducing amanuensis}} & \text{species that is oxidized} \\[1em] \pmb{\text{oxidizing agent}} & \text{species that is reduced} \end{array}[/latex]

Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction like to the one yielding NaCl:

[latex]\text{H}_2(thou) + \text{Cl}_2(yard) \longrightarrow two \text{HCl}(g)[/latex]

The product of this reaction is a covalent compound, then transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous ane and permit an unambiguous definition of redox reactions, a property called oxidation number has been divers. The oxidation number (or oxidation state) of an element in a compound is the charge its atoms would possess if the compound was ionic. The post-obit guidelines are used to assign oxidation numbers to each element in a molecule or ion.

1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion's charge.
3. Oxidation numbers for common nonmetals are ordinarily assigned as follows:
   • Hydrogen: +i when combined with nonmetals, −1 when combined with metals
   • Oxygen: −2 in near compounds, sometimes −1 (so-called peroxides, O₂ ⁱⁱ⁻), very rarely [latex]-\frac{1}{2}[/latex] (then-called superoxides, O_ii ⁻), positive values when combined with F (values vary)
   • Halogens: −1 for F always, −ane for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.

Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., two+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these 2 related properties.

Example 3

Assigning Oxidation Numbers
Follow the guidelines in this department of the text to assign oxidation numbers to all the elements in the following species:

(a) H_twoS

(b) And so_iii ²⁻

(c) Na_twoSo₄

Solution
(a) Co-ordinate to guideline 1, the oxidation number for H is +1.

Using this oxidation number and the compound's formula, guideline 4 may then be used to summate the oxidation number for sulfur:

[latex]\text{accuse on H}_2 \text{Southward} = 0 = (2 \times +ane) + (one \times 10)[/latex]

[latex]10 = 0 = - (two \times +one) = -2[/latex]

(b) Guideline 3 suggests the oxidation number for oxygen is −ii.

Using this oxidation number and the ion's formula, guideline 4 may then be used to calculate the oxidation number for sulfur:

[latex]{\text{charge on SO}_3}^{2-} = -ii = (3 \times -2) + (1 \times x)[/latex]

[latex]x = -2 - (3 \times -ii) = +iv[/latex]

(c) For ionic compounds, it's convenient to assign oxidation numbers for the cation and anion separately.

According to guideline ii, the oxidation number for sodium is +1.

Assuming the usual oxidation number for oxygen (-2 per guideline three), the oxidation number for sulfur is calculated as directed past guideline 4:

[latex]{\text{charge on SO}_4}^{two-} = -2 = (4 \times -2) + (1 \times x)[/latex]

[latex]x = -2 -(iv \times -2) = +vi[/latex]

Cheque Your Learning
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:

(a) GDue northO_iii

(b) AlH₃

(c) NH₄ ⁺

(d) H₂ PO_four ⁻

Answer:

(a) N, +5; (b) Al, +3; (c) Northward, −3; (d) P, +5

Using the oxidation number concept, an spread-out definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which ane or more elements involved undergo a change in oxidation number. (While the vast bulk of redox reactions involve changes in oxidation number for 2 or more elements, a few interesting exceptions to this rule practice exist Case 4.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:

[latex]\pmb{\text{oxidation}} = \text{increase in oxidation number}[/latex]

[latex]\pmb{\text{reduction}} = \text{decrease in oxidation number}[/latex]

Returning to the reactions used to introduce this topic, they may now both exist identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +ane in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H₂ to +ane in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl₂ to −1 in HCl).

Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (likewise called a fuel) and oxidant (oft, simply not necessarily, molecular oxygen) react vigorously and produce pregnant amounts of oestrus, and ofttimes low-cal, in the form of a flame. Solid rocket-fuel reactions such every bit the 1 depicted in Figure ane in Chapter 4 Introduction are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation:

[latex]x\text{Al}(south) + 6\text{NH}_4 \text{ClO}_4(s) \longrightarrow 4\text{Al}_2 \text{O}_3(s) + 2\text{AlCl}_3(s) + 12\text{H}_2 \text{O}(chiliad) + 3\text{N}_2(thou)[/latex]

Watch a brief video showing the examination firing of a small-scale, paradigm, hybrid rocket engine planned for apply in the new Infinite Launch Arrangement being developed by NASA. The get-go engines firing at

3 south (green flame) utilise a liquid fuel/oxidant mixture, and the 2nd, more powerful engines firing at 4 s (yellow flame) utilise a solid mixture.

Unmarried-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common case of this blazon of reaction is the acid oxidation of certain metals:

[latex]\text{Zn}(south) + ii\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2(grand)[/latex]

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

[latex]\text{Cu}(s) + two \text{AgNO}_3(aq) \longrightarrow \text{Cu(NO}_3)_2(aq) + ii \text{Ag}(s)[/latex]

This reaction may be observed past placing copper wire in a solution containing a dissolved silver salt. Argent ions in solution are reduced to elemental argent at the surface of the copper wire, and the resulting Cu²⁺ ions dissolve in the solution to yield a characteristic blue color (Figure v).

Figure 5. (a) A copper wire is shown adjacent to a solution containing silver(I) ions. (b) Deportation of dissolved silvery ions by copper ions results in (c) accumulation of grayness-colored silver metal on the wire and evolution of a bluish colour in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)

Instance iv

Describing Redox Reactions
Identify which equations represent redox reactions, providing a name for the reaction if advisable. For those reactions identified every bit redox, name the oxidant and reductant.

(a) [latex]\text{ZnCO}_3(s) \longrightarrow \text{ZnO}(southward) + \text{CO}_2(g)[/latex]

(b) [latex]2\text{Ga}(l) + three\text{Br}_2(fifty) \longrightarrow 2\text{GaBr}_3(s)[/latex]

(c) [latex]2\text{H}_2 \text{O}_2(aq) \longrightarrow 2\text{H}_2 \text{O}(l) + \text{O}_2(k)[/latex]

(d) [latex]\text{BaCl}_2(aq) + \text{Chiliad}_2 \text{SO}_4(aq) \longrightarrow \text{BaSO}_4(s) + 2\text{KCl}(aq)[/latex]

(e) [latex]\text{C}_2 \text{H}_4(k) + 3\text{O}_2(thousand) \longrightarrow 2\text{CO}_2(m) + 2\text{H}_2 \text{O}(l)[/latex]

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

(a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(fifty) to +3 in GaBr_three(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br₂(l) to −1 in GaBr₃(due south). The oxidizing amanuensis is Br₂(l).

(c) This is a redox reaction. It is a particularly interesting procedure, every bit it involves the same element, oxygen, undergoing both oxidation and reduction (a so-chosen disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −ane in H₂O_two(aq) to 0 in O₂(m). Oxygen is also reduced, its oxidation number decreasing from −ane in H_twoO_ii(aq) to −2 in H₂O(fifty). For disproportionation reactions, the aforementioned substance functions as an oxidant and a reductant.

(d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.

(e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −ii in C₂H_iv(g) to +4 in CO₂(g). The reducing agent (fuel) is C₂H₄(g). Oxygen is reduced, its oxidation number decreasing from 0 in O₂(yard) to −2 in H₂O(50). The oxidizing amanuensis is O₂(chiliad).

Check Your Learning
This equation describes the production of can(Two) chloride:

[latex]\text{Sn}(s) + 2\text{HCl}(g) \longrightarrow \text{SnCl}_2(s) + \text{H}_2(k)[/latex]

Is this a redox reaction? If so, provide a more specific name for the reaction if advisable, and identify the oxidant and reductant.

Respond:

Yeah, a single-replacement reaction. Sn(southward) is the reductant, HCl(g) is the oxidant.

Balancing Redox Reactions via the Half-Reaction Method

Redox reactions that accept identify in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical alter in other ways (east.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to aid in the procedure. One very useful arroyo is to use the method of one-half-reactions, which involves the post-obit steps:

1. Write the two half-reactions representing the redox process.

2. Balance all elements except oxygen and hydrogen.

iii. Balance oxygen atoms by calculation H_iiO molecules.

4. Balance hydrogen atoms by adding H⁺ ions.

5. Rest charge^[1] past calculation electrons.

6. If necessary, multiply each half-reaction'due south coefficients by the smallest possible integers to yield equal numbers of electrons in each.

seven. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation.

8. For reactions occurring in bones media (excess hydroxide ions), carry out these boosted steps:

1. Add OH⁻ ions to both sides of the equation in numbers equal to the number of H⁺ ions.
2. On the side of the equation containing both H⁺ and OH⁻ ions, combine these ions to yield h2o molecules.
3. Simplify the equation past removing any redundant water molecules.

9. Finally, check to see that both the number of atoms and the total charges^[2] are balanced.

Example 5

Balancing Redox Reactions in Acidic Solution
Write a balanced equation for the reaction between dichromate ion and iron(2) to yield iron(Iii) and chromium(III) in acidic solution.

[latex]\text{Cr}_2 {\text{O}_7}^{2-} + \text{Fe}^{2+} \longrightarrow \text{Cr}^{3+} + \text{Fe}^{3+}[/latex]

Solution

1. Write the ii half-reactions.

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

   [latex]\text{Fe}^{two+} \longrightarrow \text{Iron}^{three+}[/latex]

   [latex]\text{Cr}_2 {\text{O}_7}^{2-} \longrightarrow \text{Cr}^{3+}[/latex]

2. Balance all elements except oxygen and hydrogen. The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and i Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves residuum with regard to Cr atoms.

   [latex]\text{Fe}^{2+} \longrightarrow \text{Fe}^{3+}[/latex]

   [latex]\text{Cr}_2 {\text{O}_7}^{2-} \longrightarrow two\text{Cr}^{3+}[/latex]

3. Balance oxygen atoms by adding H₂O molecules. The iron one-half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, and so 7 water molecules are added to the right side.

   [latex]\text{Fe}^{2+} \longrightarrow \text{Fe}^{three+}[/latex]
   [latex]\text{Cr}_2 {\text{O}_7}^{two-} \longrightarrow 2\text{Cr}^{3+} + vii \text{H}_2 \text{O}[/latex]

4. Balance hydrogen atoms by adding H⁺ ions. The iron one-half-reaction does not incorporate H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side.

   [latex]\text{Fe}^{two+} \longrightarrow \text{Fe}^{3+}[/latex]

   [latex]\text{Cr}_2 {\text{O}_7}^{2-} + 14\text{H}^{+} \longrightarrow 2\text{Cr}^{3+} + 7 \text{H}_2 \text{O}[/latex]

5. Rest charge by adding electrons. The fe half-reaction shows a total charge of 2+ on the left side (1 Ironⁱⁱ⁺ ion) and 3+ on the right side (1 Feⁱⁱⁱ⁺ ion). Calculation ane electron to the right side bring that side's total accuse to (3+) + (1−) = ii+, and charge balance is accomplished.

   The chromium half-reaction shows a total accuse of (ane × 2−) + (14 × 1+) = 12+ on the left side (1 Cr2O72−(i Cr2O72− ion and 14 H⁺ ions). The total charge on the right side is (2 × 3+) = vi + (2 Cr³⁺ ions). Calculation vi electrons to the left side will bring that side's total charge to (12+ + half dozen−) = 6+, and charge remainder is achieved.

   [latex]\text{Fe}^{2+} \longrightarrow \text{Fe}^{3+} + \text{e}^{-}[/latex]

   [latex]\text{Cr}_2 {\text{O}_7}^{2-} + 14\text{H}^{+} + six\text{due east}^{-} \longrightarrow 2\text{Cr}^{3+} + 7 \text{H}_2 \text{O}[/latex]

6. Multiply the two half-reactions so the number of electrons in i reaction equals the number of electrons in the other reaction. To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or devastation) of electrons, the iron half-reaction's coefficient must be multiplied by half-dozen.

   [latex]6\text{Iron}^{2+} \longrightarrow six\text{Fe}^{3+} + 6\text{e}^{-}[/latex]

   [latex]\text{Cr}_2 {\text{O}_7}^{2-} + 14\text{H}^{+} + 6\text{east}^{-} \longrightarrow 2\text{Cr}^{iii+} + 7 \text{H}_2 \text{O}[/latex]

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

   [latex]6\text{Atomic number 26}^{2+} + \text{Cr}_2 {\text{O}_7}^{two-} + 6\text{e}^{-} + 14\text{H}^{+} \longrightarrow 6\text{Fe}^{three+} + 6\text{e}^{-} + 2\text{Cr}^{3+} + 7 \text{H}_2 \text{O}[/latex]

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

   [latex]half-dozen\text{Iron}^{2+} + \text{Cr}_2 {\text{O}_7}^{two-} + fourteen\text{H}^{+} \longrightarrow 6\text{Iron}^{3+} + 2\text{Cr}^{3+} + vii \text{H}_2 \text{O}[/latex]

A last check of cantlet and charge residue confirms the equation is balanced.

	Reactants	Products
Iron	6	6
Cr	ii	2
O	7	7
H	14	14
charge	24+	24+
Table 10.

Check Your Learning
In acidic solution, hydrogen peroxide reacts with Iron²⁺ to produce Iron³⁺ and H₂O. Write a balanced equation for this reaction.

Reply:

[latex]\text{H}_2 \text{O}_2(aq) + 2\text{H}^{+}(aq) + two\text{Iron}^{two+} \longrightarrow ii\text{H}_2 \text{O}(50) + two\text{Fe}^{iii+}[/latex]

Key Concepts and Summary

Chemical reactions are classified according to similar patterns of beliefs. A large number of important reactions are included in iii categories: precipitation, acid-base of operations, and oxidation-reduction (redox). Atmospheric precipitation reactions involve the formation of i or more than insoluble products. Acrid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for i or more reactant elements. Writing counterbalanced equations for some redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method.

Chemistry Terminate of Chapter Exercises

1. Utilise the following equations to answer the side by side five questions:

   i. [latex]\text{H}_2 \text{O}(due south) \longrightarrow \text{H}_2 \text{O}(l)[/latex]

   ii. [latex]\text{Na}^{+}(aq) + \text{Cl}^{-}(aq) + \text{Ag}^{+}(aq) + {\text{NO}_3}^{-}(aq) \longrightarrow \text{Ag} \text{Cl}(southward) + \text{Na}^{+}(aq) + {\text{NO}_3}^{-}(aq)[/latex]

   iii. [latex]\text{CH}_3 \text{OH}(aq) + \text{O}_2(g) \longrightarrow \text{CO}_2(1000) + \text{H}_2 \text{O}(yard)[/latex]

   iv. [latex]2\text{H}_2 \text{O}(l) \longrightarrow 2 \text{H}_2(g) + \text{O}_2(thou)[/latex]

   v. [latex]\text{H}^{+}(aq) + \text{OH}^{-}(aq) \longrightarrow \text{H}_2 \text{O}(l)[/latex]

   (a) Which equation describes a physical modify?

   (b) Which equation identifies the reactants and products of a combustion reaction?

   (c) Which equation is not balanced?

   (d) Which is a net ionic equation?

2. Indicate what blazon, or types, of reaction each of the following represents:

   (a) [latex]\text{Ca}(s) + \text{Br}_2(l) \longrightarrow \text{CaBr}_2(due south)[/latex]

   (b) [latex]\text{Ca(OH)}_2 (aq) + two\text{HBr}(aq) \longrightarrow \text{CaBr}_2(aq) + 2\text{H}_2 \text{O}(l)[/latex]

   (c) [latex]\text{C}_6 \text{H}_{12}(l) + ix\text{O}_2(g) \longrightarrow half dozen\text{CO}_2(one thousand) + 6\text{H}_2 \text{O}(g)[/latex]

3. Indicate what type, or types, of reaction each of the following represents:

   (a) [latex]\text{H}_2 \text{O}(grand) + \text{C}(s) \longrightarrow \text{CO}(k) + \text{H}_2(1000)[/latex]

   (b) [latex]two\text{KClO}_3(s) \longrightarrow 2\text{KCl}(s) + three\text{O}_2(g)[/latex]

   (c) [latex]\text{Al(OH)}_3(aq) + 3\text{HCl}(aq) \longrightarrow \text{AlCl}_3(aq) + three\text{H}_2 \text{O}(l)[/latex]

   (d) [latex]\text{Pb(NO}_3)_2(aq) + \text{H}_2 \text{Then}_4(sq) \longrightarrow \text{PbSO}_4(s) + ii\text{HNO}_3(aq)[/latex]

4. Argent can exist separated from aureate because silvery dissolves in nitric acid while gold does non. Is the dissolution of silverish in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explicate your answer.
5. Decide the oxidation states of the elements in the following compounds:

   (a) NaI

   (b) GdCl_three

   (c) LiNO_iii

   (d) H₂Se

   (e) Mg₂Si

   (f) RbO₂, rubidium superoxide

   (g) HF

6. Decide the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

   (a) H_iiiPO_iv

   (b) Al(OH)₃

   (c) SeO_ii

   (d) KNO₂

   (due east) In_iiSouth₃

   (f) P₄O₆

7. Decide the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

   (a) H₂And so₄

   (b) Ca(OH)₂

   (c) BrOH

   (d) ClNO_two

   (e) TiCl₄

   (f) NaH

8. Classify the post-obit as acrid-base reactions or oxidation-reduction reactions:

   (a) [latex]\text{Na}_2 \text{S}(aq) + 2 \text{HCl}(aq) \longrightarrow 2 \text{NaCl}(aq) + \text{H}_2 \text{S}(g)[/latex]

   (b) [latex]2\text{Na}(s) + ii\text{HCl}(aq) \longrightarrow ii\text{NaCl}(aq) + \text{H}_2(yard)[/latex]

   (c) [latex]\text{Mg}(southward) + \text{Cl}_2(g) \longrightarrow \text{MgCl}_2(aq)[/latex]

   (d) [latex]\text{MgO}(due south) + 2\text{HCl}(aq) \longrightarrow \text{MgCl}_2(s) + \text{H}_2 \text{O}(l)[/latex]

   (east) [latex]\text{K}_3 \text{P}(s) + 2\text{O}_2(1000) \longrightarrow \text{Chiliad}_3 \text{PO}_4(s)[/latex]

   (f) [latex]3\text{KOH}(aq) + \text{H}_3 \text{PO}_4(aq) \longrightarrow \text{One thousand}_3\text{PO}_4(aq) + 3 \text{H}_2 \text{O}(l)[/latex]

9. Identify the atoms that are oxidized and reduced, the change in oxidation country for each, and the oxidizing and reducing agents in each of the post-obit equations:

   (a) [latex]\text{Mg}(south) + \text{NiCl}_2(aq) \longrightarrow \text{MgCl}_2(aq) + \text{Ni}(due south)[/latex]

   (b) [latex]\text{PCl}_3(fifty) + \text{Cl}_2(g) \longrightarrow \text{PCl}_5(south)[/latex]

   (c) [latex]\text{C}_2 \text{H}_4(m) + 3\text{O}_2(g) \longrightarrow two\text{CO}_2(m) + 2\text{H}_2 \text{O}(chiliad)[/latex]

   (d) [latex]\text{Zn}(southward) + \text{H}_2 \text{And so}_4(aq) \longrightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)[/latex]

   (e) [latex]ii\text{K}_2 \text{S}_2 \text{O}_3(s) + \text{I}_2(s) \longrightarrow 2\text{G}_2 \text{S}_4 \text{O}_6(southward) + 2\text{KI}(south)[/latex]

   (f) [latex]3 \text{Cu}(s) + 8\text{HNO}_3(aq) \longrightarrow 3 \text{Cu(NO}_3)_2(aq) + two\text{NO}(g) + 4\text{H}_2 \text{O}(l)[/latex]

10. Complete and remainder the following acrid-base equations:

    (a) HCl gas reacts with solid Ca(OH)_ii(s).

    (b) A solution of Sr(OH)₂ is added to a solution of HNO_three.

11. Complete and balance the following acrid-base equations:

    (a) A solution of HClO₄ is added to a solution of LiOH.

    (b) Aqueous H_iiSo₄ reacts with NaOH.

    (c) Ba(OH)_two reacts with HF gas.

12. Complete and balance the post-obit oxidation-reduction reactions, which requite the highest possible oxidation state for the oxidized atoms.

    (a) [latex]\text{Al}(s) + \text{F}_2(g) \longrightarrow[/latex]

    (b) [latex]\text{Al}(s) + \text{CuBr}_2(aq) \longrightarrow \;\text{(unmarried deportation)}[/latex]

    (c) [latex]\text{P}_4(s) + \text{O}_2(g) \longrightarrow[/latex]

    (d) [latex]\text{Ca}(due south) + \text{H}_2 \text{O}(50) \longrightarrow \;\text{(products are a strong base of operations and a diatomic gas)}[/latex]

13. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation country for the oxidized atoms.

    (a) [latex]\text{K}(south) + \text{H}_2 \text{O}(l) \longrightarrow[/latex]

    (b) [latex]\text{Ba}(s) + \text{HBr}(aq) \longrightarrow[/latex]

    (c) [latex]\text{Sn}(s) + \text{I}_2(s) \longrightarrow[/latex]

14. Complete and rest the equations for the following acid-base neutralization reactions. If water is used every bit a solvent, write the reactants and products as aqueous ions. In some cases, in that location may be more than one correct respond, depending on the amounts of reactants used.

    (a) [latex]\text{Mg(OH)}_2(s) + \text{HClO}_4(aq) \longrightarrow[/latex]

    (b) [latex]\text{And so}_3(m) + \text{H}_2 \text{O}(l) \longrightarrow \;\text{(assume an excess of water and that the production dissolves)}[/latex]

    (c) [latex]\text{SrO}(southward) + \text{H}_2 \text{SO}_4(l) \longrightarrow[/latex]

15. When heated to 700–800 °C, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the counterbalanced equation for this reaction.
16. The armed forces has experimented with lasers that produce very intense low-cal when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?
17. Write the molecular, total ionic, and net ionic equations for the following reactions:

    (a) [latex]\text{Ca(OH)}_2 + \text{HC}_2 \text{H}_3 \text{O}_2(aq) \longrightarrow[/latex]

    (b) [latex]\text{H}_3 \text{PO}_4(aq) + \text{CaCl}_2(aq) \longrightarrow[/latex]

18. Keen Lakes Chemic Visitor produces bromine, Br₂, from bromide salts such as NaBr, in Arkansas brine by treating the alkali with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl₂.
19. In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, just in these experiments it oftentimes looks gray, due to pocket-size amounts of Mg_threeN₂, a chemical compound formed equally some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.
20. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO_ii. (Hint: H2o is one of the products.)
21. Calcium propionate is sometimes added to staff of life to retard spoilage. This compound tin can be prepared past the reaction of calcium carbonate, CaCO_iii, with propionic acid, C_twoH₅CO₂H, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate.
22. Complete and balance the equations of the following reactions, each of which could exist used to remove hydrogen sulfide from natural gas:

    (a) [latex]\text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(1000) \longrightarrow[/latex]

    (b) [latex]\text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(m) \longrightarrow[/latex]

23. Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid h2o to produce liquid hydrogen sulfate as the only product. Write the two equations which represent these reactions.
24. Write balanced chemic equations for the reactions used to ready each of the post-obit compounds from the given starting material(south). In some cases, additional reactants may be required.

    (a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (kickoff reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acrid)

    (b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction

    (c) gaseous H₂Due south from solid Zn and S via a ii-step process (starting time a redox reaction between the starting materials, and so reaction of the product with a strong acid)

25. Calcium cyclamate Ca(C_{half dozen}H₁₁NHSO_three)_ii is an artificial sweetener used in many countries around the globe simply is banned in the United states. It can exist purified industrially by converting it to the barium table salt through reaction of the acid C₆H₁₁NHSO₃H with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and so neutralization with calcium hydroxide. Write the balanced equations for these reactions.
26. Complete and balance each of the following one-half-reactions (steps ii–5 in half-reaction method):

    (a) [latex]\text{Sn}^{iv+}(aq) \longrightarrow \text{Sn}^{ii+}(aq)[/latex]

    (b) [latex][{\text{Ag(NH}_3)_2}]^{+}(aq) \longrightarrow \text{Ag}(due south) + \text{NH}_3(aq)[/latex]

    (c) [latex]\text{Hg}_2 \text{Cl}_2(s) \longrightarrow \text{Hg}(50) + \text{Cl}^{-}(aq)[/latex]

    (d) [latex]\text{H}_2 \text{O}(l) \longrightarrow \text{O}_2(1000) \;\text{(in acidic solution)}[/latex]

    (e) [latex]{\text{IO}_3}^{-}(aq) \longrightarrow \text{I}_2(south)[/latex]

    (f) [latex]{\text{SO}_3}^{2-}(aq) \longrightarrow {\text{SO}_4}^{2-}(aq) \text{(in acidic solution)}[/latex]

    (g) [latex]{\text{MnO}_4}^{-}(aq) \longrightarrow \text{Mn}^{2+}(aq) \;\text{(in acidic solution)}[/latex]

    (h) [latex]\text{Cl}^{-}(aq) \longrightarrow {\text{ClO}_3}^{-}(aq) \;\text{(in bones solution)}[/latex]

27. Consummate and rest each of the following one-half-reactions (steps 2–5 in half-reaction method):

    (a) [latex]\text{Cr}^{2+}(aq) \longrightarrow \text{Cr}^{3+}(aq)[/latex]

    (b) [latex]\text{Hg}(l) + \text{Br}^{-}(aq) \longrightarrow {\text{HgBr}_4}^{ii-}(aq)[/latex]

    (c) [latex]\text{ZnS}(s) \longrightarrow \text{Zn}(southward) + \text{Southward}^{2-}(aq)[/latex]

    (d) [latex]\text{H}_2(k) \longrightarrow \text{H}_2 \text{O}(fifty) \text{(in basic solution)}[/latex]

    (e) [latex]\text{H}_2(chiliad) \longrightarrow \text{H}_3 \text{O}^{+}(aq) \text{(in acidic solution)}[/latex]

    (f) [latex]{\text{NO}_3}^{-}(aq) \longrightarrow \text{HNO}_2(aq) \;\text{(in acidic solution)}[/latex]

    (g) [latex]\text{MnO}_2(s) \longrightarrow {\text{MnO}_4}^{-}(aq) \;\text{(in bones solution)}[/latex]

    (h) [latex]\text{Cl}^{-}(aq) \longrightarrow {\text{ClO}_3}^{-}(aq) \;\text{(in acidic solution)}[/latex]

28. Balance each of the following equations according to the one-half-reaction method:

    (a) [latex]\text{Sn}^{2+}(aq) + \text{Cu}^{2+}(aq) \longrightarrow \text{Sn}^{4+}(aq) + \text{Cu}^{+}(aq)[/latex]

    (b) [latex]\text{H}_2 \text{S}(chiliad) + {\text{Hg}_2}^{2+}(aq) \longrightarrow \text{Hg}(l) + \text{S}(due south) \;\text{(in acid)}[/latex]

    (c) [latex]\text{CN}^{-}(aq) + \text{ClO}_2(aq) \longrightarrow \text{CNO}^{-}(aq) + \text{Cl}^{-}(aq) \text{(in acid)}[/latex]

    (d) [latex]\text{Fe}^{2+}(aq) + \text{Ce}^{4+}(aq) \longrightarrow \text{Fe}^{3+}(aq) + \text{Ce}^{3+}(aq)[/latex]

    (e) [latex]\text{HBrO}(aq) \longrightarrow \text{Br}^{-}(aq) + \text{O}_2(g) \;\text{(in acid)}[/latex]

29. Remainder each of the following equations according to the half-reaction method:

    (a) [latex]\text{Zn}(s) + {\text{NO}_3}^{-}(aq) \longrightarrow \text{Zn}^{2+}(aq) + \text{N}_2(thousand) \;\text{(in acrid)}[/latex]

    (b) [latex]\text{Zn}(s) + {\text{NO}_3}^{-}(aq) \longrightarrow \text{Zn}^{2+}(aq) + \text{NH}_3(aq) \;\text{(in base of operations)}[/latex]

    (c) [latex]\text{CuS}(s) + {\text{NO}_3}^{-}(aq) \longrightarrow \text{Cu}^{2+} + \text{S}(s) + \text{NO}(chiliad) \;\text{(in acid)}[/latex]

    (d) [latex]\text{NH}_3(aq) + \text{O}_2(chiliad) \longrightarrow \text{NO}_2(yard) \;\text{(gas phase)}[/latex]

    (eastward) [latex]\text{Cl}_2(grand) + \text{OH}^{-}(aq) \longrightarrow \text{Cl}^{-}(aq) + {\text{ClO}_3}^{-}(aq) \;\text{(in base)}[/latex]

    (f) [latex]\text{H}_2 \text{O}_2(aq) + {\text{MnO}_4}^{-}(aq) \longrightarrow \text{Mn}^{2+}(aq) + \text{O}_2(g) \;\text{(in acid)}[/latex]

    (g) [latex]\text{NO}_2(1000) \longrightarrow {\text{NO}_3}^{-}(aq) + {\text{NO}_2}^{-}(aq) \;\text{(in base)}[/latex]

    (h) [latex]\text{Atomic number 26}^{iii+}(aq) + \text{I}^{-}(aq) \longrightarrow \text{Iron}^{two+}(aq) + \text{I}_2(aq)[/latex]

30. Remainder each of the post-obit equations according to the one-half-reaction method:

    (a) [latex]{\text{MnO}_4}^{-}(aq) + {\text{NO}_2}^{-}(aq) \longrightarrow \text{MnO}_{2}(s) + {\text{NO}_3}^{-}(aq) \;\text{(in base)}[/latex]

    (b) [latex]{\text{MnO}_4}^{2-}(aq) \longrightarrow {\text{MnO}_4}^{-}(aq) + {\text{MnO}_2}(s) \;\text{(in base of operations)}[/latex]

    (c) [latex]\text{Br}_2(l) + \text{SO}_2(g) \longrightarrow \text{Br}^{-}(aq) + {\text{SO}_4}^{2-}(aq) \;\text{(in acid)}[/latex]

Glossary

acid

substance that produces H_threeO⁺ when dissolved in h2o

acid-base reaction

reaction involving the transfer of a hydrogen ion between reactant species

base of operations

substance that produces OH⁻ when dissolved in h2o

combustion reaction

vigorous redox reaction producing significant amounts of energy in the form of estrus and, sometimes, lite

half-reaction

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

insoluble

of relatively depression solubility; dissolving only to a slight extent

neutralization reaction

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

oxidation

process in which an chemical element'south oxidation number is increased past loss of electrons

oxidation-reduction reaction

(besides, redox reaction) reaction involving a change in oxidation number for one or more reactant elements

oxidation number

(as well, oxidation state) the accuse each atom of an chemical element would have in a compound if the chemical compound were ionic

oxidizing agent

(also, oxidant) substance that brings virtually the oxidation of another substance, and in the process becomes reduced

precipitate

insoluble product that forms from reaction of soluble reactants

precipitation reaction

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

reduction

process in which an element'south oxidation number is decreased by gain of electrons

reducing agent

(likewise, reductant) substance that brings about the reduction of some other substance, and in the procedure becomes oxidized

salt

ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide

single-deportation reaction

(also, replacement) redox reaction involving the oxidation of an elemental substance past an ionic species

soluble

of relatively loftier solubility; dissolving to a relatively big extent

solubility

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

stiff acid

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

strong base

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

weak acid

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

weak base of operations

base that reacts but to a slight extent when dissolved in water to yield hydroxide ions

Solutions

Answers to Chemistry End of Affiliate Exercises

two. (a) oxidation-reduction (add-on); (b) acrid-base (neutralization); (c) oxidation-reduction (combustion)

iv. It is an oxidation-reduction reaction considering the oxidation state of the argent changes during the reaction.

6. (a) H +one, P +5, O −ii; (b) Al +3, H +one, O −two; (c) Se +four, O −ii; (d) K +i, N +iii, O −two; (e) In +3, S −two; (f) P +three, O −2

8. (a) acid-base; (b) oxidation-reduction: Na is oxidized, H⁺ is reduced; (c) oxidation-reduction: Mg is oxidized, Cl_two is reduced; (d) acid-base of operations; (e) oxidation-reduction: P³⁻ is oxidized, O₂ is reduced; (f) acrid-base

10.
(a) [latex]2\text{HCl}(g) + \text{Ca(OH)}_2(s) \longrightarrow \text{CaCl}_2(due south) + 2\text{H}_2 \text{O}(l)[/latex];
(b) [latex]\text{Sr(OH)}_2(aq) + ii\text{HNO}_3(aq) \longrightarrow \text{Sr(NO}_3)_2(aq) + 2\text{H}_2 \text{O}(l)[/latex];

12.
(a) [latex]2\text{Al}(s) + iii\text{F}_2 \longrightarrow 2\text{AlF}_3(s)[/latex];
(b) [latex]2\text{Al}(s) + 3\text{CuBr}_2(aq) \longrightarrow 3\text{Cu}(southward) + two\text{AlBr}_3(aq)[/latex];
(c) [latex]\text{P}_4(s) + 5\text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(southward)[/latex];
(d) [latex]\text{Ca}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)[/latex];

fourteen.
(a) [latex]\text{Mg(OH)}_2(southward) + 2\text{HClO}_4(aq) \longrightarrow \text{Mg}^{2+}(aq) + two{\text{ClO}_4}^{-}(aq) + 2\text{H}_2 \text{O}(fifty);[/latex]
(b) [latex]\text{SO}_3(g) + ii\text{H}_2 \text{O}(l) \longrightarrow \text{H}_3 \text{O}^{+}(aq) + {\text{HSO}_4}^{-}(aq), \text{(a solution of)} \; \text{H}_2 \text{SO}_4);[/latex]
(c) [latex]\text{SrO}(s) + \text{H}_2 \text{Then}_4(fifty) \longrightarrow \text{SrSO}_4(s) + \text{H}_2 \text{O}[/latex]

16. [latex]\text{H}_2(m) + \text{F}_2(1000) \longrightarrow 2\text{HF}(g)[/latex]

xviii. [latex]two\text{NaBr}(aq) + \text{Cl}_2(g) \longrightarrow ii\text{NaCl}(aq) + \text{Br}_2(l)[/latex]

20. [latex]ii\text{LiOH}(aq) + \text{CO}_2(chiliad) \longrightarrow \text{Li}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(fifty)[/latex]

22.
(a) [latex]\text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(thousand) \longrightarrow \text{CaS}(s) + 2\text{H}_2\text{O}(l);[/latex]
(b) [latex]\text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{Due south}(thousand) \longrightarrow \text{Na}_2 \text{S}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(fifty)[/latex]

24.
(a) pace 1: [latex]\text{N}_2(m) + three\text{H}_2(g) \longrightarrow ii\text{NH}_3(g)[/latex], step 2: [latex]\text{NH}_3(g) + \text{HNO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3 \;\text{(after drying)}[/latex] ;
(b) [latex]\text{H}_2(g) + \text{Br}_2(50) \longrightarrow 2\text{HBr}(g)[/latex];
(c) [latex]\text{Zn}(due south) + \text{S}(s) \longrightarrow \text{ZnS}(southward) \;\text{and} \;\text{ZnS}(south) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2 \text{S}(k)[/latex];

26.
(a) [latex]\text{Sn}^{four+}(aq) + ii\text{e}^{-} \longrightarrow \text{Sn}^{2+}(aq)[/latex]
(b) [latex][{\text{Ag(NH}_3)_2}]^{+}(aq) + \text{e}^{-} \longrightarrow \text{Ag}(due south) + 2\text{NH}_3(aq)[/latex]
(c) [latex]\text{Hg}_2 \text{Cl}_2(s) + 2\text{e}^{-} \longrightarrow two\text{Hg}(fifty) + 2\text{Cl}^{-}(aq)[/latex]
(d) [latex]2\text{H}_2 \text{O}(50) \longrightarrow \text{O}_2 + iv\text{H}^{+}(aq) + 4\text{e}^{-}[/latex]
(e) [latex]6 \text{H}_2 \text{O}(50) + 2{\text{IO}_3}^{-}(aq) + x\text{e}^{-} \longrightarrow \text{I}_2(southward) + 12 \text{OH}^{-}(aq)[/latex]
(f) [latex]\text{H}_2 \text{O}(50) + {\text{So}_3}^{2-}(aq) \longrightarrow {\text{And so}_4}^{2-}(aq) + ii\text{H}^{+}(aq) + 2\text{e}^{-}[/latex]
(g) [latex]8\text{H}^{+}(aq) + {\text{MnO}_4}^{-}(aq) + 5\text{e}^{-} \longrightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2 \text{O}(l)[/latex]
(h) [latex]\text{Cl}^{-}(aq) + half-dozen \text{OH}^{-}(aq) \longrightarrow {\text{ClO}_3}^{-}(aq) + iii\text{H}_2 \text{O}(l) + 6\text{e}^{-}[/latex]

28.
(a) [latex]\text{Sn}^{2+}(aq) + ii\text{Cu}^{2+}(aq) \longrightarrow \text{Sn}^{4+}(aq) + 2\text{Cu}^{+}(aq)[/latex]
(b) [latex]\text{H}_2 \text{South}(g) + {\text{Hg}_2}^{two+}(aq) + 2\text{H}_2 \text{O}(l) \longrightarrow ii\text{Hg}(l) + \text{Southward}(s) + 2\text{H}_3 \text{O}^{+}(aq)[/latex]
(c) [latex]5\text{CN}^{-}(aq) + 2\text{ClO}_2(aq) + 3\text{H}_2 \text{O}(l) \longrightarrow v\text{CNO}^{-}(aq) + 2\text{Cl}^{-}(aq) + 2\text{H}_3 \text{O}^{+}(aq)[/latex]
(d) [latex]\text{Fe}^{2+}(aq) + \text{Ce}^{4+}(aq) \longrightarrow \text{Atomic number 26}^{3+}(aq) + \text{Ce}^{three+}(aq)[/latex]
(due east) [latex]2\text{HBrO}(aq) + 2\text{H}_2 \text{O}(l) \longrightarrow two\text{H}_3 \text{O}(aq) + ii\text{Br}^{-}(aq) + \text{O}_2(g)[/latex]

xxx.
(a) [latex]2\text{MnO}^{4-}(aq) + three{\text{NO}_2}^{-}(aq) + \text{H}_2 \text{O}(fifty) \longrightarrow 2\text{MnO}_{2}(s) + 3{\text{NO}_3}^{-}(aq) + ii\text{OH}^{-}(aq)[/latex]
(b) [latex]3{\text{MnO}_4}^{two-}(aq) + ii\text{H}_2 \text{O}(l) \longrightarrow 2{\text{MnO}_4}^{-}(aq) + four\text{OH}^{-}(aq) + {\text{MnO}_2}(s) \;\text{(in base)}[/latex]
(c) [latex]\text{Br}_2(l) + \text{SO}_2(g) + 2\text{H}_2 \text{O}(l) \longrightarrow 4\text{H}^{+}(aq) + two\text{Br}^{-}(aq) + {\text{So}_4}^{two-}(aq)[/latex]

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Source: https://opentextbc.ca/chemistry/chapter/4-2-classifying-chemical-reactions/

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