how to tell what element in a reaction is a base

Chapter 4. Stoichiometry of Chemic Reactions

4.ii Classifying Chemical Reactions

Learning Objectives

By the finish of this department, y'all will be able to:

• Ascertain three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction)
• Allocate chemical reactions as one of these three types given appropriate descriptions or chemical equations
• Identify common acids and bases
• Predict the solubility of common inorganic compounds by using solubility rules
• Compute the oxidation states for elements in compounds

Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When ii humans commutation data, 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 chemic substances, scientists have likewise establish information technology convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the near prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction.

Precipitation Reactions and Solubility Rules

A precipitation reaction is one in which dissolved substances react to form ane (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement, 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 product of a number of commodity and specialty chemicals. Precipitation reactions also play a fundamental function in many chemic assay techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of thing (run across the last module of this chapter).

The extent to which a substance may exist dissolved in h2o, or whatever solvent, is quantitatively expressed as its solubility, defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large 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 exist insoluble, and these are the substances that readily precipitate from solution. More data on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table 8).

Soluble compounds incorporate group i metallic cations (Li+, Na+, K+, Rb+, and Cs+) and ammonium ion (NHiv +) the halide ions (Cl−, Br−, and I−) the acetate (C2H3Otwo −), bicarbonate (HCOiii −), nitrate (NO3 −), and chlorate (ClOthree −) ions the sulfate (And soiv ii−) ion	Exceptions to these solubility rules include halides of Ag+, Hg2 two+, and Pb2+ sulfates of Ag+, Ba2+, Ca2+, Hg22+,Hg22+, Atomic number 822+, and Sr2+
Insoluble compounds contain carbonate (CO3 2−), chromate (CrOfour 2−), phosphate (POfour three−), and sulfide (Southwardtwo−) ions hydroxide ion (OH−)	Exceptions to these insolubility rules include compounds of these anions with grouping 1 metal cations and ammonium ion hydroxides of grouping one metallic cations and Ba2+
Table eight. Solubilities of Common Ionic Compounds in H2o

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

[latex]2\text{KI}(aq) + \text{Pb(NO}_3)_2(aq) \longrightarrow \text{PbI}_2(s) + 2\text{KNO}_3(aq)[/latex]

This ascertainment is consistent with the solubility guidelines: The just insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts.

The net ionic equation representing this reaction is:

[latex]\text{Pb}^{2+}(aq) + two\text{I}^{-}(aq) \longrightarrow \text{PbI}_2(s)[/latex]

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

Figure 1. A precipitate of PbI₂ forms when solutions containing Pb²⁺ and I⁻ are mixed. (credit: Der Kreole/Wikimedia Commons)

The solubility guidelines in Tabular array 9 may be used to predict whether a atmospheric precipitation reaction volition occur when solutions of soluble ionic compounds are mixed together. 1 merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For case, mixing solutions of silvery nitrate and sodium fluoride will yield a solution containing Ag⁺, NO₃ ⁻, Na⁺, and F⁻ ions. Aside from the two ionic compounds originally present in the solutions, AgNO₃ and NaF, two additional ionic compounds may exist derived from this collection of ions: NaNO₃ and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is i 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{(net ionic)}[/latex]

Instance 1

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

(a) potassium sulfate and barium nitrate

(b) lithium chloride and silver acetate

(c) lead nitrate and ammonium carbonate

Solution
(a) The two possible products for this combination are KNO_three and BaSO_four. The solubility guidelines indicate BaSO_four is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the style 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 LiC₂H₃O₂ 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 mode 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 NH₄NO₃. The solubility guidelines indicate PbCO₃ is insoluble, 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{Atomic number 82}^{2+}(aq) + {\text{CO}_3}^{2-}(aq) \longrightarrow \text{PbCO}_3(s)[/latex]

Cheque Your Learning
Which solution could exist used to precipitate the barium ion, Ba²⁺, in a h2o 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 of operations reaction is one in which a hydrogen ion, H⁺, is transferred from one chemic species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-calibration production of fertilizers, pharmaceuticals, and other substances essential to lodge. The subject of acid-base chemistry, therefore, is worthy of thorough give-and-take, and a full chapter is devoted to this topic later in the text.

For purposes of this brief introduction, we will consider only the more than mutual types of acid-base reactions that take identify in aqueous solutions. In this context, an acid is a substance that volition dissolve in water to yield hydronium ions, H_iiiO⁺. As an example, consider the equation shown here:

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

The process represented past this equation confirms that hydrogen chloride is an acid. When dissolved in water, H₃O⁺ ions are produced by a chemical reaction in which H⁺ ions are transferred from HCl molecules to H₂O molecules (Effigy 2).

Figure 2. When hydrogen chloride gas dissolves in h2o, (a) it reacts as an acrid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).

The nature of HCl is such that its reaction with h2o every bit but described is essentially 100% efficient: Virtually every HCl molecule that dissolves in h2o will undergo this reaction. Acids that completely react in this style are chosen strong acids, and HCl is 1 among simply a handful of common acid compounds that are classified every bit strong (Table nine). A far greater number of compounds behave equally weak acids and but partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body aroma. A familiar instance of a weak acid is acetic acrid, the primary ingredient in food vinegars:

[latex]\text{CH}_3 \text{CO}_2 \text{H}(aq) + \text{H}_2 \text{O}(50) \leftrightharpoons \text{CH}_3 {\text{CO}_2}^{-}(aq) + \text{H}_3 \text{O}^{+}(aq)[/latex]

When dissolved in water under typical conditions, only almost 1% of acetic acrid molecules are present in the ionized form, [latex]\text{CH}_3 {\text{CO}_2}^{-}[/latex](Figure three). (The apply of a double-arrow in the equation in a higher place denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemic equilibrium.)

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

Chemical compound Formula	Proper noun in Aqueous Solution
HBr	hydrobromic acid
HCl	hydrochloric acid
HI	hydroiodic acid
HNO3	nitric acid
HClO4	perchloric acrid
HtwoSOiv	sulfuric acid
Table 9. Common Strong Acids

A base of operations is a substance that will deliquesce in water to yield hydroxide ions, OH⁻. The well-nigh mutual bases are ionic compounds composed of alkali or element of group ii cations (groups 1 and two) combined with the hydroxide ion—for example, NaOH and Ca(OH)_two. When these compounds dissolve in h2o, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)₂ deliquesce in water and dissociate completely to produce cations (Yard⁺ and Ba²⁺, respectively) and hydroxide ions, OH⁻. These bases, along with other hydroxides that completely dissociate in water, are considered strong bases.

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

[latex]\text{NaOH}(southward) \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 whatsoever other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds deliquesce in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.

Dissimilar ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases. These types of compounds are as well abundant in nature and important commodities in diverse technologies. For case, global production of the weak base ammonia is typically well over 100 metric tons annually, existence widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an agile ingredient in household cleaners (Figure 4). When dissolved in h2o, ammonia reacts partially to yield hydroxide ions, as shown here:

[latex]\text{NH}_3(aq) + \text{H}_2 \text{O}(50) \rightleftharpoons {\text{NH}_4}^{+}(aq) + \text{OH}^{-}(aq)[/latex]

This is, by definition, an acid-base reaction, in this case involving the transfer of H⁺ ions from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as NH₄ ⁺ ions.

Figure 4. Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied every bit an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of piece of work by National Resources Conservation Service; credit b: modification of piece of work by pat00139)

The chemic reactions described in which acids and bases dissolved in water produce hydronium and hydroxide ions, respectively, are, by definition, acid-base reactions. In these reactions, water serves as both a solvent and a reactant. A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acrid 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{salt} + \text{h2o}[/latex]

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

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

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

Example ii

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

(a) the weak acid hydrogen hypochlorite reacts with h2o

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

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 H₂O to generate hydronium ions, H_iiiO⁺ and hypochlorite ions, OCl⁻.

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

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

(b) The ii reactants are provided, Ba(OH)₂ and HNO₃. Since this is a neutralization reaction, the two products will be h2o 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) + 2\text{H}_2 \text{O}(l)[/latex]

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

Answer:

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

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

Oxidation-Reduction Reactions

Earth'south atmosphere contains nigh 20% molecular oxygen, O₂, a chemically reactive gas that plays an essential office in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemic reactions involving O₂, but its meaning has evolved to refer to a broad and important reaction course known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear motion picture of this nomenclature.

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

[latex]ii\text{Na}(due south) + \text{Cl}_2(thousand) \longrightarrow 2\text{NaCl}(south)[/latex]

It is helpful to view the procedure with regard to each individual reactant, that is, to stand for the fate of each reactant in the form of an equation called a half-reaction:

[latex]two\text{Na}(s) \longrightarrow two\text{Na}^{+}(southward) + 2\text{eastward}^{-}[/latex]
[latex]\text{Cl}_2(grand) + 2\text{east}^{-} \longrightarrow 2\text{Cl}^{-}(s)[/latex]

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

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

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

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

Some redox processes, however, practice non involve the transfer of electrons. Consider, for instance, a reaction like to the one yielding NaCl:

[latex]\text{H}_2(grand) + \text{Cl}_2(chiliad) \longrightarrow 2 \text{HCl}(g)[/latex]

The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and allow an unambiguous definition of redox reactions, a belongings 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 following guidelines are used to assign oxidation numbers to each element in a molecule or ion.

1. The oxidation number of an atom in an elemental substance is zero.
2. The oxidation number of a monatomic ion is equal to the ion'southward charge.
3. Oxidation numbers for common nonmetals are usually assigned as follows:
   • Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
   • Oxygen: −ii in most compounds, sometimes −1 (so-called peroxides, O_ii ²⁻), very rarely [latex]-\frac{one}{2}[/latex] (so-called superoxides, O_ii ⁻), positive values when combined with F (values vary)
   • Halogens: −1 for F always, −one 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 accuse on the molecule or ion.

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

Case 3

Assigning Oxidation Numbers
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the post-obit species:

(a) H_iiSouth

(b) SO₃ ⁱⁱ⁻

(c) Na_twoSO_iv

Solution
(a) According to guideline 1, the oxidation number for H is +i.

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 +1) + (1 \times x)[/latex]

[latex]x = 0 = - (ii \times +ane) = -two[/latex]

(b) Guideline iii suggests the oxidation number for oxygen is −2.

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

[latex]{\text{charge on So}_3}^{2-} = -two = (3 \times -2) + (i \times x)[/latex]

[latex]x = -ii - (iii \times -2) = +four[/latex]

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

According to guideline 2, the oxidation number for sodium is +1.

Bold the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:

[latex]{\text{charge on And so}_4}^{2-} = -ii = (iv \times -2) + (1 \times x)[/latex]

[latex]ten = -2 -(four \times -2) = +6[/latex]

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

(a) GrandNO₃

(b) AlH₃

(c) NH₄ ⁺

(d) H₂ PO_four ⁻

Respond:

(a) North, +five; (b) Al, +three; (c) N, −3; (d) P, +v

Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a modify in oxidation number. (While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this dominion do be Example 4.) Definitions for the complementary processes of this reaction class are correspondingly revised every bit shown hither:

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

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

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

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

[latex]ten\text{Al}(s) + half dozen\text{NH}_4 \text{ClO}_4(due south) \longrightarrow 4\text{Al}_2 \text{O}_3(s) + 2\text{AlCl}_3(south) + 12\text{H}_2 \text{O}(g) + 3\text{N}_2(m)[/latex]

Scout a brief video showing the test firing of a minor-calibration, prototype, hybrid rocket engine planned for utilise in the new Space Launch System being developed by NASA. The first engines firing at

3 south (green flame) apply a liquid fuel/oxidant mixture, and the second, more than 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 metal chemical element. One mutual example of this blazon of reaction is the acid oxidation of certain metals:

[latex]\text{Zn}(s) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)[/latex]

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

[latex]\text{Cu}(s) + 2 \text{AgNO}_3(aq) \longrightarrow \text{Cu(NO}_3)_2(aq) + 2 \text{Ag}(southward)[/latex]

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

Effigy 5. (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Deportation of dissolved silver ions by copper ions results in (c) aggregating of gray-colored silver metal on the wire and development of a blue colour in the solution, due to dissolved copper ions. (credit: modification of work past Mark Ott)

Example 4

Describing Redox Reactions
Place which equations stand for redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, proper name the oxidant and reductant.

(a) [latex]\text{ZnCO}_3(due south) \longrightarrow \text{ZnO}(s) + \text{CO}_2(grand)[/latex]

(b) [latex]2\text{Ga}(l) + 3\text{Br}_2(fifty) \longrightarrow 2\text{GaBr}_3(due south)[/latex]

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

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

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

Solution
Redox reactions are identified per definition if i or more than elements undergo a change 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(l) to +iii in GaBr_three(s). The reducing amanuensis is Ga(50). Bromine is reduced, its oxidation number decreasing from 0 in Br₂(l) to −1 in GaBr_iii(s). The oxidizing agent is Br₂(l).

(c) This is a redox reaction. It is a specially interesting process, equally it involves the same element, oxygen, undergoing both oxidation and reduction (a so-chosen disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H_twoO₂(aq) to 0 in O₂(chiliad). Oxygen is as well reduced, its oxidation number decreasing from −one in H₂O₂(aq) to −2 in H₂O(50). For disproportionation reactions, the same 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_four(thou) to +4 in CO₂(g). The reducing agent (fuel) is C₂H_four(g). Oxygen is reduced, its oxidation number decreasing from 0 in O₂(thou) to −2 in H₂O(l). The oxidizing agent is O_two(yard).

Check Your Learning
This equation describes the product of tin(II) chloride:

[latex]\text{Sn}(southward) + 2\text{HCl}(chiliad) \longrightarrow \text{SnCl}_2(s) + \text{H}_2(thousand)[/latex]

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

Respond:

Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(m) is the oxidant.

Balancing Redox Reactions via the Half-Reaction Method

Redox reactions that accept place in aqueous media often involve water, hydronium ions, and hydroxide ions equally reactants or products. Although these species are non oxidized or reduced, they do participate in chemic change in other ways (eastward.k., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very hard to balance by inspection, so systematic approaches have been developed to assist in the process. 1 very useful approach is to utilize the method of one-half-reactions, which involves the post-obit steps:

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

2. Balance all elements except oxygen and hydrogen.

3. Balance oxygen atoms by calculation H₂O molecules.

4. Balance hydrogen atoms by adding H⁺ ions.

v. Balance charge^[one] by adding electrons.

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

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

8. For reactions occurring in basic media (excess hydroxide ions), bear 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 water molecules.
3. Simplify the equation by removing any redundant water molecules.

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

Instance 5

Balancing Redox Reactions in Acidic Solution
Write a counterbalanced equation for the reaction between dichromate ion and iron(Two) to yield atomic number 26(III) and chromium(3) in acidic solution.

[latex]\text{Cr}_2 {\text{O}_7}^{2-} + \text{Fe}^{2+} \longrightarrow \text{Cr}^{3+} + \text{Atomic number 26}^{3+}[/latex]

Solution

1. Write the two half-reactions.

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

   [latex]\text{Atomic number 26}^{two+} \longrightarrow \text{Fe}^{3+}[/latex]

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

2. Balance all elements except oxygen and hydrogen. The atomic number 26 half-reaction is already balanced, merely the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the correct. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms.

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

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

3. Balance oxygen atoms past calculation H_iiO molecules. The fe half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven h2o molecules are added to the right side.

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

4. Balance hydrogen atoms by calculation H⁺ ions. The iron half-reaction does not incorporate H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, then 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}^{two-} + xiv\text{H}^{+} \longrightarrow two\text{Cr}^{3+} + 7 \text{H}_2 \text{O}[/latex]

5. Rest accuse by calculation electrons. The atomic number 26 half-reaction shows a total charge of two+ on the left side (1 Fe²⁺ ion) and iii+ on the right side (1 Fe³⁺ ion). Calculation i electron to the right side bring that side'south total charge to (3+) + (one−) = two+, and charge balance is achieved.

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

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

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

6. Multiply the 2 half-reactions so the number of electrons in 1 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 cosmos or destruction) of electrons, the atomic number 26 half-reaction's coefficient must be multiplied by 6.

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

   [latex]\text{Cr}_2 {\text{O}_7}^{two-} + 14\text{H}^{+} + 6\text{due east}^{-} \longrightarrow 2\text{Cr}^{three+} + vii \text{H}_2 \text{O}[/latex]

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

   [latex]6\text{Fe}^{2+} + \text{Cr}_2 {\text{O}_7}^{2-} + 6\text{east}^{-} + 14\text{H}^{+} \longrightarrow 6\text{Iron}^{3+} + six\text{e}^{-} + 2\text{Cr}^{3+} + 7 \text{H}_2 \text{O}[/latex]

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

   [latex]6\text{Fe}^{2+} + \text{Cr}_2 {\text{O}_7}^{2-} + xiv\text{H}^{+} \longrightarrow 6\text{Fe}^{3+} + 2\text{Cr}^{three+} + 7 \text{H}_2 \text{O}[/latex]

A concluding check of cantlet and charge balance confirms the equation is counterbalanced.

	Reactants	Products
Fe	half dozen	6
Cr	2	two
O	7	7
H	14	xiv
accuse	24+	24+
Tabular array 10.

Check Your Learning
In acidic solution, hydrogen peroxide reacts with Fe²⁺ to produce Iron³⁺ and H₂O. Write a counterbalanced equation for this reaction.

Answer:

[latex]\text{H}_2 \text{O}_2(aq) + 2\text{H}^{+}(aq) + 2\text{Fe}^{two+} \longrightarrow 2\text{H}_2 \text{O}(50) + 2\text{Fe}^{3+}[/latex]

Key Concepts and Summary

Chemical reactions are classified according to similar patterns of behavior. A large number of of import reactions are included in three categories: precipitation, acid-base, and oxidation-reduction (redox). Precipitation reactions involve the formation of i or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for 1 or more than 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.

Chemical science End of Chapter Exercises

1. Employ the following equations to respond the next 5 questions:

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

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

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

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

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

   (a) Which equation describes a physical change?

   (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 type, or types, of reaction each of the post-obit represents:

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

   (b) [latex]\text{Ca(OH)}_2 (aq) + 2\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) + 9\text{O}_2(one thousand) \longrightarrow 6\text{CO}_2(yard) + six\text{H}_2 \text{O}(grand)[/latex]

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

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

   (b) [latex]ii\text{KClO}_3(southward) \longrightarrow 2\text{KCl}(s) + 3\text{O}_2(yard)[/latex]

   (c) [latex]\text{Al(OH)}_3(aq) + 3\text{HCl}(aq) \longrightarrow \text{AlCl}_3(aq) + 3\text{H}_2 \text{O}(50)[/latex]

   (d) [latex]\text{Pb(NO}_3)_2(aq) + \text{H}_2 \text{SO}_4(sq) \longrightarrow \text{PbSO}_4(s) + 2\text{HNO}_3(aq)[/latex]

4. Silverish can be separated from gold because silver dissolves in nitric acid while gold does non. Is the dissolution of silver in nitric acrid an acrid-base reaction or an oxidation-reduction reaction? Explain your answer.
5. Determine the oxidation states of the elements in the following compounds:

   (a) NaI

   (b) GdCl₃

   (c) LiNO₃

   (d) H_twoSe

   (e) Mg₂Si

   (f) RbO_two, rubidium superoxide

   (g) HF

6. Make up one's mind the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.

   (a) H₃PO₄

   (b) Al(OH)₃

   (c) SeO_two

   (d) KNO_two

   (e) In_iiS_three

   (f) P_ivO_{half dozen}

7. Make up one's mind 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) ClNO₂

   (eastward) TiCl_iv

   (f) NaH

8. Classify the following as acid-base of operations reactions or oxidation-reduction reactions:

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

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

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

   (d) [latex]\text{MgO}(s) + 2\text{HCl}(aq) \longrightarrow \text{MgCl}_2(s) + \text{H}_2 \text{O}(50)[/latex]

   (e) [latex]\text{K}_3 \text{P}(s) + ii\text{O}_2(g) \longrightarrow \text{Chiliad}_3 \text{PO}_4(s)[/latex]

   (f) [latex]3\text{KOH}(aq) + \text{H}_3 \text{PO}_4(aq) \longrightarrow \text{K}_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 state for each, and the oxidizing and reducing agents in each of the following equations:

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

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

   (c) [latex]\text{C}_2 \text{H}_4(g) + iii\text{O}_2(g) \longrightarrow two\text{CO}_2(g) + ii\text{H}_2 \text{O}(g)[/latex]

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

   (e) [latex]2\text{K}_2 \text{S}_2 \text{O}_3(s) + \text{I}_2(due south) \longrightarrow two\text{1000}_2 \text{Southward}_4 \text{O}_6(southward) + 2\text{KI}(s)[/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 rest the post-obit acid-base equations:

    (a) HCl gas reacts with solid Ca(OH)₂(southward).

    (b) A solution of Sr(OH)₂ is added to a solution of HNO₃.

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

    (a) A solution of HClO_four 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 following oxidation-reduction reactions, which give 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{(single displacement)}[/latex]

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

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

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

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

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

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

14. Complete and rest the equations for the post-obit acrid-base neutralization reactions. If water is used every bit a solvent, write the reactants and products equally aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.

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

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

    (c) [latex]\text{SrO}(southward) + \text{H}_2 \text{And then}_4(50) \longrightarrow[/latex]

15. When heated to 700–800 °C, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn down!) Write the balanced equation for this reaction.
16. The military has experimented with lasers that produce very intense low-cal when fluorine combines explosively with hydrogen. What is the counterbalanced equation for this reaction?
17. Write the molecular, total ionic, and internet 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. Slap-up Lakes Chemic Company produces bromine, Br₂, from bromide salts such equally NaBr, in Arkansas alkali by treating the brine 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, merely in these experiments information technology often looks greyness, due to small amounts of Mg₃N₂, a compound formed as 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 two mol of LiOH per 1 mol of CO_two. (Hint: Water is one of the products.)
21. Calcium propionate is sometimes added to bread to retard spoilage. This chemical compound tin be prepared past the reaction of calcium carbonate, CaCO₃, with propionic acid, C₂H_vCO₂H, which has properties similar to those of acetic acid. Write the counterbalanced equation for the germination of calcium propionate.
22. Complete and balance the equations of the post-obit reactions, each of which could be used to remove hydrogen sulfide from natural gas:

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

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

23. Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product and so reacts with liquid water to produce liquid hydrogen sulfate every bit the only product. Write the ii equations which correspond these reactions.
24. Write counterbalanced chemical equations for the reactions used to prepare each of the post-obit compounds from the given starting material(southward). In some cases, additional reactants may be required.

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

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

    (c) gaseous H_iiSouth from solid Zn and S via a two-step process (first a redox reaction between the starting materials, so reaction of the product with a strong acid)

25. Calcium cyclamate Ca(C₆H_xiNHSO₃)₂ is an artificial sweetener used in many countries effectually the earth only is banned in the United States. It can be purified industrially by converting it to the barium table salt through reaction of the acid C₆H_elevenNHSO_threeH 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 post-obit half-reactions (steps 2–5 in half-reaction method):

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

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

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

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

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

27. Consummate and residual each of the following half-reactions (steps two–5 in one-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}^{2-}(aq)[/latex]

    (c) [latex]\text{ZnS}(s) \longrightarrow \text{Zn}(s) + \text{S}^{ii-}(aq)[/latex]

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

    (e) [latex]\text{H}_2(g) \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(due south) \longrightarrow {\text{MnO}_4}^{-}(aq) \;\text{(in basic solution)}[/latex]

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

28. Residue each of the post-obit equations co-ordinate to the half-reaction method:

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

    (b) [latex]\text{H}_2 \text{S}(m) + {\text{Hg}_2}^{2+}(aq) \longrightarrow \text{Hg}(l) + \text{Due south}(southward) \;\text{(in acrid)}[/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{Atomic number 26}^{3+}(aq) + \text{Ce}^{3+}(aq)[/latex]

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

29. Balance 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(g) \;\text{(in acid)}[/latex]

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

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

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

    (due east) [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(one thousand) \longrightarrow {\text{NO}_3}^{-}(aq) + {\text{NO}_2}^{-}(aq) \;\text{(in base)}[/latex]

    (h) [latex]\text{Fe}^{3+}(aq) + \text{I}^{-}(aq) \longrightarrow \text{Fe}^{2+}(aq) + \text{I}_2(aq)[/latex]

30. Balance each of the following equations according to the 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)}[/latex]

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

Glossary

acid

substance that produces H₃O⁺ 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 water

combustion reaction

vigorous redox reaction producing pregnant amounts of energy in the form of heat and, sometimes, light

one-half-reaction

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

insoluble

of relatively low solubility; dissolving only to a slight extent

neutralization reaction

reaction between an acid and a base to produce salt and h2o

oxidation

procedure in which an chemical element's oxidation number is increased by loss of electrons

oxidation-reduction reaction

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

oxidation number

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

oxidizing agent

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

precipitate

insoluble production that forms from reaction of soluble reactants

precipitation reaction

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

reduction

process in which an element's oxidation number is decreased by proceeds of electrons

reducing amanuensis

(also, reductant) substance that brings most the reduction of another substance, and in the process becomes oxidized

salt

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

single-displacement reaction

(as well, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species

soluble

of relatively high solubility; dissolving to a relatively big extent

solubility

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

potent acid

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

strong base

base of operations that reacts completely when dissolved in h2o to yield hydroxide ions

weak acid

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

weak base

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

Solutions

Answers to Chemistry Terminate of Affiliate Exercises

two. (a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)

four. It is an oxidation-reduction reaction considering the oxidation country of the silvery changes during the reaction.

6. (a) H +1, P +five, O −ii; (b) Al +3, H +1, O −2; (c) Se +four, O −2; (d) Chiliad +1, N +3, O −2; (due east) In +iii, S −2; (f) P +3, O −two

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

x.
(a) [latex]2\text{HCl}(yard) + \text{Ca(OH)}_2(s) \longrightarrow \text{CaCl}_2(s) + 2\text{H}_2 \text{O}(50)[/latex];
(b) [latex]\text{Sr(OH)}_2(aq) + 2\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) + 3\text{F}_2 \longrightarrow two\text{AlF}_3(s)[/latex];
(b) [latex]two\text{Al}(s) + 3\text{CuBr}_2(aq) \longrightarrow three\text{Cu}(due south) + 2\text{AlBr}_3(aq)[/latex];
(c) [latex]\text{P}_4(due south) + 5\text{O}_2(1000) \longrightarrow \text{P}_4 \text{O}_{x}(s)[/latex];
(d) [latex]\text{Ca}(s) + 2\text{H}_2 \text{O}(l) \longrightarrow \text{Ca(OH)}_2(aq) + \text{H}_2(g)[/latex];

xiv.
(a) [latex]\text{Mg(OH)}_2(s) + 2\text{HClO}_4(aq) \longrightarrow \text{Mg}^{2+}(aq) + 2{\text{ClO}_4}^{-}(aq) + 2\text{H}_2 \text{O}(l);[/latex]
(b) [latex]\text{SO}_3(g) + 2\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{And so}_4);[/latex]
(c) [latex]\text{SrO}(s) + \text{H}_2 \text{And so}_4(l) \longrightarrow \text{SrSO}_4(s) + \text{H}_2 \text{O}[/latex]

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

eighteen. [latex]2\text{NaBr}(aq) + \text{Cl}_2(g) \longrightarrow 2\text{NaCl}(aq) + \text{Br}_2(l)[/latex]

twenty. [latex]2\text{LiOH}(aq) + \text{CO}_2(g) \longrightarrow \text{Li}_2 \text{CO}_3(aq) + \text{H}_2 \text{O}(l)[/latex]

22.
(a) [latex]\text{Ca(OH)}_2(south) + \text{H}_2 \text{S}(thou) \longrightarrow \text{CaS}(southward) + 2\text{H}_2\text{O}(50);[/latex]
(b) [latex]\text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{South}(thousand) \longrightarrow \text{Na}_2 \text{S}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(fifty)[/latex]

24.
(a) step ane: [latex]\text{N}_2(g) + three\text{H}_2(chiliad) \longrightarrow 2\text{NH}_3(m)[/latex], pace 2: [latex]\text{NH}_3(m) + \text{HNO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3(aq) \longrightarrow \text{NH}_4 \text{NO}_3 \;\text{(later drying)}[/latex] ;
(b) [latex]\text{H}_2(grand) + \text{Br}_2(l) \longrightarrow 2\text{HBr}(g)[/latex];
(c) [latex]\text{Zn}(s) + \text{S}(s) \longrightarrow \text{ZnS}(south) \;\text{and} \;\text{ZnS}(s) + 2\text{HCl}(aq) \longrightarrow \text{ZnCl}_2(aq) + \text{H}_2 \text{S}(g)[/latex];

26.
(a) [latex]\text{Sn}^{4+}(aq) + ii\text{e}^{-} \longrightarrow \text{Sn}^{two+}(aq)[/latex]
(b) [latex][{\text{Ag(NH}_3)_2}]^{+}(aq) + \text{e}^{-} \longrightarrow \text{Ag}(due south) + two\text{NH}_3(aq)[/latex]
(c) [latex]\text{Hg}_2 \text{Cl}_2(s) + two\text{e}^{-} \longrightarrow ii\text{Hg}(l) + 2\text{Cl}^{-}(aq)[/latex]
(d) [latex]2\text{H}_2 \text{O}(fifty) \longrightarrow \text{O}_2 + 4\text{H}^{+}(aq) + iv\text{due east}^{-}[/latex]
(eastward) [latex]half dozen \text{H}_2 \text{O}(l) + 2{\text{IO}_3}^{-}(aq) + x\text{eastward}^{-} \longrightarrow \text{I}_2(south) + 12 \text{OH}^{-}(aq)[/latex]
(f) [latex]\text{H}_2 \text{O}(l) + {\text{And so}_3}^{ii-}(aq) \longrightarrow {\text{And then}_4}^{ii-}(aq) + 2\text{H}^{+}(aq) + 2\text{due east}^{-}[/latex]
(g) [latex]8\text{H}^{+}(aq) + {\text{MnO}_4}^{-}(aq) + five\text{due east}^{-} \longrightarrow \text{Mn}^{two+}(aq) + four\text{H}_2 \text{O}(50)[/latex]
(h) [latex]\text{Cl}^{-}(aq) + six \text{OH}^{-}(aq) \longrightarrow {\text{ClO}_3}^{-}(aq) + 3\text{H}_2 \text{O}(l) + 6\text{due east}^{-}[/latex]

28.
(a) [latex]\text{Sn}^{2+}(aq) + 2\text{Cu}^{2+}(aq) \longrightarrow \text{Sn}^{4+}(aq) + two\text{Cu}^{+}(aq)[/latex]
(b) [latex]\text{H}_2 \text{Southward}(m) + {\text{Hg}_2}^{2+}(aq) + 2\text{H}_2 \text{O}(fifty) \longrightarrow two\text{Hg}(50) + \text{Due south}(due south) + 2\text{H}_3 \text{O}^{+}(aq)[/latex]
(c) [latex]5\text{CN}^{-}(aq) + 2\text{ClO}_2(aq) + iii\text{H}_2 \text{O}(fifty) \longrightarrow 5\text{CNO}^{-}(aq) + 2\text{Cl}^{-}(aq) + 2\text{H}_3 \text{O}^{+}(aq)[/latex]
(d) [latex]\text{Atomic number 26}^{ii+}(aq) + \text{Ce}^{iv+}(aq) \longrightarrow \text{Iron}^{3+}(aq) + \text{Ce}^{3+}(aq)[/latex]
(e) [latex]two\text{HBrO}(aq) + 2\text{H}_2 \text{O}(50) \longrightarrow two\text{H}_3 \text{O}(aq) + ii\text{Br}^{-}(aq) + \text{O}_2(thousand)[/latex]

thirty.
(a) [latex]2\text{MnO}^{4-}(aq) + 3{\text{NO}_2}^{-}(aq) + \text{H}_2 \text{O}(l) \longrightarrow 2\text{MnO}_{two}(s) + three{\text{NO}_3}^{-}(aq) + 2\text{OH}^{-}(aq)[/latex]
(b) [latex]iii{\text{MnO}_4}^{2-}(aq) + 2\text{H}_2 \text{O}(50) \longrightarrow 2{\text{MnO}_4}^{-}(aq) + four\text{OH}^{-}(aq) + {\text{MnO}_2}(southward) \;\text{(in base)}[/latex]
(c) [latex]\text{Br}_2(50) + \text{And then}_2(chiliad) + two\text{H}_2 \text{O}(l) \longrightarrow four\text{H}^{+}(aq) + 2\text{Br}^{-}(aq) + {\text{And then}_4}^{2-}(aq)[/latex]

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

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