Chapter 4. Reduction/Oxidation Reactions (Redox)
4.2 Reduction/Oxidation Reaction
Learning Objectives
By the end of this section, you will be able to:
- Compute the oxidation states for elements in compounds and identify reducing agents and oxidizing agents.
Oxidation-Reduction Reactions
Earth’s atmosphere contains about 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 meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions. A few examples of such reactions will be used to develop a clear picture of this classification.
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:
It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction:
[latex]\text{Cl}_2(g) + 2\text{e}^{-} \longrightarrow 2\text{Cl}^{-}(s)[/latex]
These equations show that Na atoms lose electrons while Cl atoms (in the Cl2 molecule) gain electrons, the “s” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:
In this reaction, then, sodium is oxidized and chlorine undergoes reduction. Viewed from a more active perspective, sodium functions as a reducing agent (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.
Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:
- The oxidation number of an atom in an elemental substance is zero.
- The oxidation number of a monatomic ion is equal to the ion’s charge.
- Oxidation numbers for common nonmetals are usually assigned as follows:
- Hydrogen: +1 when combined with nonmetals, −1 when combined with metals
- Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, O22−), very rarely [latex]-\frac{1}{2}[/latex] (so-called superoxides, O2−), positive values when combined with F (values vary)
- Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)
- 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., 2+), 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 two related properties.
Example 3
Assigning Oxidation Numbers
Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:
(a) H2S
(b) SO32−
(c) Na2SO4
Solution
(a) According 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 calculate the oxidation number for sulfur:
(b) Guideline 3 suggests the oxidation number for oxygen is −2.
Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:
(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.
Assuming the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:
Check Your Learning
Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:
(a) KNO3
(b) AlH3
(c) NH4+
(d) H2PO4−
Answer:
(a) N, +5; (b) Al, +3; (c) N, −3; (d) P, +5
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 change 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 rule do exist Example 4.) Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:
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 +1 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 +1 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 (also called a fuel) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted in Figure 1 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:
Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at
3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.
Single-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 example of this type of reaction is the acid oxidation of certain metals:
Metallic elements may also be oxidized by solutions of other metal salts; for example:
This reaction may be observed by placing copper wire in a solution containing a dissolved silver 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 blue color (Figure 5).
Example 4
Describing Redox Reactions
Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.
(a) [latex]\text{ZnCO}_3(s) \longrightarrow \text{ZnO}(s) + \text{CO}_2(g)[/latex]
(b) [latex]2\text{Ga}(l) + 3\text{Br}_2(l) \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(g)[/latex]
(d) [latex]\text{BaCl}_2(aq) + \text{K}_2 \text{SO}_4(aq) \longrightarrow \text{BaSO}_4(s) + 2\text{KCl}(aq)[/latex]
(e) [latex]\text{C}_2 \text{H}_4(g) + 3\text{O}_2(g) \longrightarrow 2\text{CO}_2(g) + 2\text{H}_2 \text{O}(l)[/latex]
Solution
Redox reactions are identified per definition if one or more 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 +3 in GaBr3(s). The reducing agent is Ga(l). Bromine is reduced, its oxidation number decreasing from 0 in Br2(l) to −1 in GaBr3(s). The oxidizing agent is Br2(l).
(c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction). Oxygen is oxidized, its oxidation number increasing from −1 in H2O2(aq) to 0 in O2(g). Oxygen is also reduced, its oxidation number decreasing from −1 in H2O2(aq) to −2 in H2O(l). 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 −2 in C2H4(g) to +4 in CO2(g). The reducing agent (fuel) is C2H4(g). Oxygen is reduced, its oxidation number decreasing from 0 in O2(g) to −2 in H2O(l). The oxidizing agent is O2(g).
Check Your Learning
This equation describes the production of tin(II) chloride:
Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.
Answer:
Yes, a single-replacement reaction. Sn(s) is the reductant, HCl(g) is the oxidant.
Key Concepts and Summary
Redox reactions involve a change in oxidation number for one or more reactant elements. The Reducing Agent cause another reagent to become reduced. In turn the reducing agent becomes ‘oxidized’. The Oxidizing agent causes another agent to become oxidized. In turn the oxidizing agent becomes reduced. Analogy: A cleaning agent cleans something , but in turn beocmes ‘dirty’ as it picks up dirt from the thing it cleans
Chemistry End of Chapter Exercises
- Determine the oxidation states of the elements in the following compounds:
(a) NaI
(b) GdCl3
(c) LiNO3
(d) H2Se
(e) Mg2Si
(f) RbO2, rubidium superoxide
(g) HF
- Determine 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) SeO2
(d) KNO2
(e) In2S3
(f) P4O6
- Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.
(a) H2SO4
(b) Ca(OH)2
(c) BrOH
(d) ClNO2
(e) TiCl4
(f) NaH
- Identify the 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) + 2\text{HCl}(aq) \longrightarrow 2\text{NaCl}(aq) + \text{H}_2(g)[/latex]
(c) [latex]\text{Mg}(s) + \text{Cl}_2(g) \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}(l)[/latex]
(e) [latex]\text{K}_3 \text{P}(s) + 2\text{O}_2(g) \longrightarrow \text{K}_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]
- 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}(s) + \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(s)[/latex]
(c) [latex]\text{C}_2 \text{H}_4(g) + 3\text{O}_2(g) \longrightarrow 2\text{CO}_2(g) + 2\text{H}_2 \text{O}(g)[/latex]
(d) [latex]\text{Zn}(s) + \text{H}_2 \text{SO}_4(aq) \longrightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)[/latex]
(e) [latex]2\text{K}_2 \text{S}_2 \text{O}_3(s) + \text{I}_2(s) \longrightarrow 2\text{K}_2 \text{S}_4 \text{O}_6(s) + 2\text{KI}(s)[/latex]
(f) [latex]3 \text{Cu}(s) + 8\text{HNO}_3(aq) \longrightarrow 3 \text{Cu(NO}_3)_2(aq) + 2\text{NO}(g) + 4\text{H}_2 \text{O}(l)[/latex]
-
- Great Lakes Chemical Company produces bromine, Br2, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl2.
- In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of Mg3N2, a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.
- 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 CO2. (Hint: Water is one of the products.)
- Calcium propionate is sometimes added to bread to retard spoilage. This compound can be prepared by the reaction of calcium carbonate, CaCO3, with propionic acid, C2H5CO2H, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate.
- Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:
(a) [latex]\text{Ca(OH)}_2(s) + \text{H}_2 \text{S}(g) \longrightarrow[/latex]
(b) [latex]\text{Na}_2 \text{CO}_3(aq) + \text{H}_2 \text{S}(g) \longrightarrow[/latex]
- 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 water to produce liquid hydrogen sulfate as the only product. Write the two equations which represent these reactions.
- Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.
(a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)
(b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction
(c) gaseous H2S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)
- Calcium cyclamate Ca(C6H11NHSO3)2 is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid C6H11NHSO3H with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions.
- Complete and balance each of the following 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}(s) + \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(s)[/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]
- Complete and balance each of the following 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}^{2-}(aq)[/latex]
(c) [latex]\text{ZnS}(s) \longrightarrow \text{Zn}(s) + \text{S}^{2-}(aq)[/latex]
(d) [latex]\text{H}_2(g) \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(s) \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]
- Balance each of the following equations according to the 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}(g) + {\text{Hg}_2}^{2+}(aq) \longrightarrow \text{Hg}(l) + \text{S}(s) \;\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]
- 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}(s) + {\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}^{2+} + \text{S}(s) + \text{NO}(g) \;\text{(in acid)}[/latex]
(d) [latex]\text{NH}_3(aq) + \text{O}_2(g) \longrightarrow \text{NO}_2(g) \;\text{(gas phase)}[/latex]
(e) [latex]\text{Cl}_2(g) + \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(g) \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]
- 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(g) \longrightarrow \text{Br}^{-}(aq) + {\text{SO}_4}^{2-}(aq) \;\text{(in acid)}[/latex]
Glossary
- half-reaction
- an equation that shows whether each reactant loses or gains electrons in a reaction.
- oxidation
- process in which an element’s oxidation number is increased by loss of electrons
- oxidation-reduction reaction
- (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements
- oxidation number
- (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic
- oxidizing agent
- (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced
- reduction
- process in which an element’s oxidation number is decreased by gain of electrons
- reducing agent
- (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized
- s
Solutions
Answers to Chemistry End of Chapter Exercises
2. (a) oxidation-reduction (addition); (b) acid-base (neutralization); (c) oxidation-reduction (combustion)
4. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.
6. (a) H +1, P +5, O −2; (b) Al +3, H +1, O −2; (c) Se +4, O −2; (d) K +1, N +3, O −2; (e) In +3, S −2; (f) P +3, O −2
8. (a) acid-base; (b) oxidation-reduction: Na is oxidized, H+ is reduced; (c) oxidation-reduction: Mg is oxidized, Cl2 is reduced; (d) acid-base; (e) oxidation-reduction: P3− is oxidized, O2 is reduced; (f) acid-base
10.
(a) [latex]2\text{HCl}(g) + \text{Ca(OH)}_2(s) \longrightarrow \text{CaCl}_2(s) + 2\text{H}_2 \text{O}(l)[/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 2\text{AlF}_3(s)[/latex];
(b) [latex]2\text{Al}(s) + 3\text{CuBr}_2(aq) \longrightarrow 3\text{Cu}(s) + 2\text{AlBr}_3(aq)[/latex];
(c) [latex]\text{P}_4(s) + 5\text{O}_2(g) \longrightarrow \text{P}_4 \text{O}_{10}(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];
14.
(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{SO}_4);[/latex]
(c) [latex]\text{SrO}(s) + \text{H}_2 \text{SO}_4(l) \longrightarrow \text{SrSO}_4(s) + \text{H}_2 \text{O}[/latex]
16. [latex]\text{H}_2(g) + \text{F}_2(g) \longrightarrow 2\text{HF}(g)[/latex]
18. [latex]2\text{NaBr}(aq) + \text{Cl}_2(g) \longrightarrow 2\text{NaCl}(aq) + \text{Br}_2(l)[/latex]
20. [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(s) + \text{H}_2 \text{S}(g) \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{S}(g) \longrightarrow \text{Na}_2 \text{S}(aq) + \text{CO}_2(g) + \text{H}_2 \text{O}(l)[/latex]
24.
(a) step 1: [latex]\text{N}_2(g) + 3\text{H}_2(g) \longrightarrow 2\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(l) \longrightarrow 2\text{HBr}(g)[/latex];
(c) [latex]\text{Zn}(s) + \text{S}(s) \longrightarrow \text{ZnS}(s) \;\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) + 2\text{e}^{-} \longrightarrow \text{Sn}^{2+}(aq)[/latex]
(b) [latex][{\text{Ag(NH}_3)_2}]^{+}(aq) + \text{e}^{-} \longrightarrow \text{Ag}(s) + 2\text{NH}_3(aq)[/latex]
(c) [latex]\text{Hg}_2 \text{Cl}_2(s) + 2\text{e}^{-} \longrightarrow 2\text{Hg}(l) + 2\text{Cl}^{-}(aq)[/latex]
(d) [latex]2\text{H}_2 \text{O}(l) \longrightarrow \text{O}_2 + 4\text{H}^{+}(aq) + 4\text{e}^{-}[/latex]
(e) [latex]6 \text{H}_2 \text{O}(l) + 2{\text{IO}_3}^{-}(aq) + 10\text{e}^{-} \longrightarrow \text{I}_2(s) + 12 \text{OH}^{-}(aq)[/latex]
(f) [latex]\text{H}_2 \text{O}(l) + {\text{SO}_3}^{2-}(aq) \longrightarrow {\text{SO}_4}^{2-}(aq) + 2\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) + 6 \text{OH}^{-}(aq) \longrightarrow {\text{ClO}_3}^{-}(aq) + 3\text{H}_2 \text{O}(l) + 6\text{e}^{-}[/latex]
28.
(a) [latex]\text{Sn}^{2+}(aq) + 2\text{Cu}^{2+}(aq) \longrightarrow \text{Sn}^{4+}(aq) + 2\text{Cu}^{+}(aq)[/latex]
(b) [latex]\text{H}_2 \text{S}(g) + {\text{Hg}_2}^{2+}(aq) + 2\text{H}_2 \text{O}(l) \longrightarrow 2\text{Hg}(l) + \text{S}(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 5\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{Fe}^{3+}(aq) + \text{Ce}^{3+}(aq)[/latex]
(e) [latex]2\text{HBrO}(aq) + 2\text{H}_2 \text{O}(l) \longrightarrow 2\text{H}_3 \text{O}(aq) + 2\text{Br}^{-}(aq) + \text{O}_2(g)[/latex]
30.
(a) [latex]2\text{MnO}^{4-}(aq) + 3{\text{NO}_2}^{-}(aq) + \text{H}_2 \text{O}(l) \longrightarrow 2\text{MnO}_{2}(s) + 3{\text{NO}_3}^{-}(aq) + 2\text{OH}^{-}(aq)[/latex]
(b) [latex]3{\text{MnO}_4}^{2-}(aq) + 2\text{H}_2 \text{O}(l) \longrightarrow 2{\text{MnO}_4}^{-}(aq) + 4\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) + 2\text{Br}^{-}(aq) + {\text{SO}_4}^{2-}(aq)[/latex]