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Unformatted text preview: 5 Introduction to Reactions in Aqueous Solutions
Contents 51 The Nature of Aqueous Solutions 52 Precipitation Reactions 53 AcidBase Reactions 54 OxidationReduction:Some General Principles 55 Balancing OxidationReduction Equations 56 Oxidizing and Reducing Agents 57 Stoichiometry of Reactions in Aqueous Solutions:Titrations 5-1 The Nature of Aqueous Solutions
Reactions in aqueous(water) solution are important because :(1)water is inexpensive and able to dissolve a vast number of substances;(2)in aqueous solution,many substances are dissociated into ions,which can participate in chemical reactions;and (3) aqueous solutions are found everywhere,from seawater to living systems. Solutions are simply homogeneous mixtures. All solutions consist of a solvent and one or more solutes. Solutes may be electrolytes or nonelectrolytes. In this section you will see an animation of salt and sugar solutions used to complete an electrical circuit. You will also see an animation that illustrates the difference between strong and weak electrolytes. Solutes come in two varieties: el ectr ol ytes and nonel ectr ol ytes Solutions of electrolytes such as NaCl and HNO3 contain ions and thus conduct electricity. Solutions of nonelectrolytes such as sucrose (C12H22O11) and methanol (CH3OH) do not form ions in solution and thus do not conduct electricity. An electrolyte may be either a str ong el ectr ol yte or a wea k el ectr ol yte. A strong electrolyte exists in solution completely (or almost completely) as ions, while a weak electrolyte produces only a small concentration of ions when it dissolves in solution. Ionic compounds that are soluble in water di ssoci a te completely and exist in solution entirely as ions. Soluble ionic compounds are strong electrolytes. Molecular compounds such as sugar and alcohol are nonelectrolytes. They have no tendency to come apart, and they exist in solution entirely as aqueous molecules. Some molecular compounds, most notably acids and weak bases, are electrolytes The acids HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4 are molecular compounds that ionize in aqueous solution and exist completely as ions. For example, hydrogen chloride gas dissolves in water and ionizes in solution to give aqueous hydrogen ion and aqueous chloride ion. All of the seven acids listed above are strong Electrolytes.They are also called strong acids. In both instances the word strong denotes complete ionization. (Strong bases are ionic and are also strong electrolytes.) Weak acids and weak bases make up a group of molecular compounds that are weak electrolytes. Acetic acid (HC2H3O2) is a weak acid. Although it ionizes in water, the reverse process occurs more readily. At any given time most of the acetic acid exists as aqueous molecules in solution. Weak acids are weak electrolytes. To represent the equilibrium between the molecular acid and its ionized form, we use a double arrow in the equation. Weak bases are also weak electrolytes. Ammonia ionizes in water to produce aqueous ammonium ions and aqueous hydroxide ions. Example Calculating ion concentrations in a solution of a strong electrolyte.What are the aluminum and sulfate ion concentrations in 0.0165 M Al2(SO4)3 (aq)? Solution First ,identity the solute as a strong electrolyte,and write anequation to represent its dissocitation.The solute is an ionic compound consisting of the ions Al3+and SO42Al2(SO4)3 (s) 2 Al3+(aq) + 3 SO42-(aq) Now ,with stoichiometric factors from this equation,we can devise a conversion pathway to relate[Al3+] and [SO42-] to the given molarity of Al2(SO4)3 .The stoichiometric factors shown in blue in the following equations are derived from the fact that 1 mol Al2(SO4)3 consists of 2 mol Al3+and 3 mol SO42-. 0.0165 mol Al2(SO4)3 [Al3+] = 1L = 0.0330 M 0.0165 mol Al2(SO4)3 1L = 0.0495 M 2 mol Al3+ 1 mol Al2(SO4)3 0.0330 mol Al3+ 1L 0.0495 mol SO423 mol SO42 1 mol Al2(SO4)3 1L 52 Precipitation Reactions Many chemical reactions occur in aqueous solution. A precipitation reaction is one that occurs in solution and results in the formation of an insoluble product. For example, an aqueous solution of the soluble ionic compound, sodium sulfate, can be mixed with an aqueous solution of the soluble ionic compound, barium hydroxide. The result is the formation of a precipitate, the insoluble ionic compound barium sulfate. Note that only one of the products of the reaction is a solid. Sodium hydroxide is soluble. Many combinations of such solutions will not result in a precipitation because they produce no insoluble product. Whether or not an ionic product of a reaction in solution will precipitate from the solution can be predicted using a set of solubility guidelines To predict the outcome when combining aqueous solutions of ionic compounds: 1. Identify the cation and anion in each reactant. 2. Combine the cation from the first reactant with the anion of the second to get the first product. 3. Combine the anion from the first reactant with the cation of the second to get the second product. 4. Determine whether the products are aqueous or solid by consulting the solubility guidelines Having learned that substances such as sodium sulfate, barium hydroxide, and sodium hydroxide are strong electrolytes and exist entirely as ions in solution, we are equipped to write these chemical equations in a way that better represents what actually happens in solution. The equation above is a molecular equation in which none of the species is represented as ionized. A molecular equation shows the complete chemical formulas for the reactants and products. We know, however, that several of the species in the equation dissociate completely in solution. We convert the equation to a complete ionic equation by identifying the strong electrolytes and representing them as separated ions. The equation above becomes The ionic equation reveals that two of the species in solution (sodium ion and hydroxide ion) do not undergo any change in the course of the reaction. Ions that are present but play no role in the reaction are called spectator ions. Eliminating the spectator ions from both sides of the equation gives the net ionic reaction The net ionic equation for the combination of aqueous sodium sulfate andaqueous barium hydroxide is the same as the net ionic equation for any combination of a soluble sulfate and a soluble barium compound. To wr i te a net i oni c equati on: 1.Wr i te a bal anced mol ecul ar equati on. 2.Rewr i te the equati on to show the i ons that for m i n sol uti on when each sol ubl e str ong el ectr ol yte di ssoci ates or i oni zes i nto i ts component i ons. Onl y di ssol ved str ong el ectr ol ytes ar e wr i tten i n i oni c for m. 3.I denti fy and cancel spectator i ons that occur on both si des of the equati on. Example Using solubility rules to predict precipitation reactions.Predict whether a reaction will occur in each of the following cases.If so,write a net ionic equation for the reaction. (a) NaOH(aq) + MgCl2(aq) (b)BaS(aq) + CuSO4(aq) (c)(NH4)2SO4(aq) + ZnCl2(aq) Solution (a) Because all common Na compounds are water soluble,Na+ remains in solution.The combination of Mg2+ and OH-produces insoluble Mg(OH)2.with this information.we can write 2 Na+ (aq) + 2 OH-(aq) + Mg2+(aq) + 2 Cl-(aq) Mg(OH)2 (s) + 2 Na+ (aq) + 2 Cl-(aq) ? ? ? Write the elimination of spectator ions,we obtain 2 OH-(aq) + Mg2+(aq) Mg(OH)2 (s) (b) From the solubility rules,we conclude that insoluble combinations formed when Ba2+,S2-,Cu2+,and SO42- are found together in solution are BaSO4(s) and CuS(s).The net ionic equation is Ba2+(aq) + S2-(aq) + Cu2+(aq) + SO42-(aq) BaSO4(s) + CuS(s) (c) A careful review of the solubility rules shows that all the possible ion combinations lead to water-soluble compounds. 2 NH4+(aq) + SO42-(aq) + Zn2+(aq) + 2 Cl-(aq) no reaction Keep i n mi nd that a solution made by combining ZnSO4(aq) and NH4Cl(aq) is indistinguishable from the combination of ZnCl2(aq) and (NH4)2SO4(aq). 53 AcidBase Reactions
I n thi s secti on you wi l l see two mor e ani mati ons that i ntr oduce aqueous aci ds and bases. Reacti ons between aci ds and bases ar e al so i ntr oduced. An acid is a molecular substance that ionizes to form a hydrogen ion (H+) and increases the concentration of aqueous H+ ions when it is dissolved in water. Because a hydrogen atom consists of a proton and an electron, H+ is simply a proton; therefore, an acid can also be thought of as a substance that can donate a proton. Hydrogen chloride ionizes in water, increasing the aqueous concentration of H +. Strong acids are those that ionize completely in water. Strong acids are also strong electrolytes. Hydrochloric acid is one of seven strong acids. strong acid
hydrochloric hydrobromic hydroiodic Nitric chloric perchloric sulfuric formula
HCl HBr HI HNO3 HClO3 HClO4 H2SO4 Sulfuric acid is the only strong acid that is diprotic, meaning it has two protons to donate. A base is a substance that increases the concentration of aqueous OH ions when it is dissolved in water. Bases can be either ionic or molecular substances. A base can be thought of as a substance that can accept a proton; in other words, bases are substances that react with H+ ions.Ammonia ionizes in water, accepting a proton from water and increasing the aqueous concentration of OH ion.A strong base is an ionic compound that dissociates completely in water. Group 1A hydroxides, such as sodium hydroxide,and Heavy Group 2A hydroxides, such as calcium hydroxide, are strong bases. Weak bases are molecular compounds that ionize only partially in water. Ammonia is an example of a weak base. An acid and a base can react with one another to form a molecular compound and a salt. The combination of hydrochloric acid and sodium hydroxide is a familiar neutralization reaction. The molecular compound produced by the reaction is water,and the salt is sodium chloride.The molecular Compound produced by the reaction of an acid and a base can also be a gas. Aqueous hydrochloric acid reacts with aqueous sodium sulfide (a weak base) to produce hydrogen sulfide (a gaseous molecular compound) and sodium chloride. More Acid-Base Reactions
Mg(OH)2 is a base because it contains OH-, but this compound is quite insoluble in water. Its finely divided solid particles form a suspension in water that is the familiar mill of magnesia, used as an antacid. In this suspension, Mg(OH)2(s) does dissolve very slightly, producing some OHin solution. If an acid is added, H+ from the acid combines with this OH- to form water--neutralization occurs. More Mg(OH)2(s) dissolves to produce more OH- in solution, which is neutralized by more H+, and so on. In this way, the neutralization reaction results in the dissolving of otherwise insoluble Mg(OH)2(s). The net ionic equation for the reaction of Mg(OH)2(s) with a strong acid is Mg(OH)2 + 2H+ Mg2+ + 2H2O Mg(OH)2(s) also reacts with a weak acid such as acetic acid. In the net ionic equation, acetic acid is written in its molecular form. But remember that some H+ and C2H3O2- ions are always present in an acetic acid solution. The H+ ions react with OH- ions, as in reaction (5.8), followed by further ionization of HC2H3O2, more neutralization, and so on. If enough acetic acid is present, the Mg(OH)2 will dissolve completely. Mg(OH)2(s) + 2 H+ Mg2+(aq) + 2 H2O Some Common Gas-Forming Reactions Ion
HSO3- + H+ SO32- + 2H+ HCO3- + H+ S2- + 2H+ SO2(g) + H2O(l) SO2(g) + H2O(l) CO2(g) + H2O(l) H2S(g) Example Writing equation(s) for acid-base reactions.Write a net ionic equation to represent the reaction of (a) aqueous strontium hydroxide with nitric acid;(b) solid aluminum hydroxide with hydrochloric acid. solution In each case we begin by writing the reactants in the whole-formula form.Then we substitute ionic forms,where appropriate.Finally we complete the equation as a net ionic equation. (a)Whole-formula form: HNO3(aq) + Sr(OH)2(aq) ? Ionic form: 2H+(aq) + 2NO3-(aq) + Sr2+(aq) + 2OH-(aq) Sr2+(aq) + 2NO3-(aq) + 2H2O(l) Net ionic equation:Remove the spectator ions(Sr2+and NO3-) 2H+(aq) + 2OH-(aq) 2H2O(l) H2O(l) ? Or,more simply,H+(aq) + OH-(aq) (b)Whole-formula form: Al(OH)3(s) + HCl(aq) Ionic form: Al(OH)3(s) + 3H+(aq) + 3Cl-(aq) Al3+(aq) + 3Cl-(aq) + 3H2O(l) Net ionic equation:Remove the spectator ion(Cl-) Al(OH)3(s) + 3H+ (aq) Al3+(aq) + 3H2O(l) 54 OxidationReduction:Some General Principles Oxi dati on-r educti on r eacti ons i nvol ve the tr ansfer of el ectr ons fr om one speci es to another . Oxi dati on Number s ar e i ntr oduced.You wi l l see an ani mati on And a demonstr ati on of oxi dati on-r educti on r eacti ons, And you wi l l l ear n to pr edi ct what r eacti ons wi l l occur , based on the acti vi ty ser i es In addition to precipitation and neutralization reactions, aqueous ions can participate in oxidation-reduction reactions. Oxidation-reduction reactions involve the transfer of electrons from one chemical species to another. A piece of calcium metal, for example, dissolves in aqueous acid. Electrons from the calcium metal have been transferred to the hydrogen ions. The calcium has undergone oxidation (loss of electrons), and the hydrogen ion has undergone reduction (gain of electrons). In order to determine whether a reaction is an oxidation-reduction reaction, we compare the oxidation numbers(or oxidation states) of each Element on both sides of the equation. Oxidation numbers can be assigned using the following rules. 1. Any a tom i n i ts el ementa l for m ha s a n oxi da ti on sta te of zer o. N o excepti ons. 2. Any mona tomi c i on ha s a n oxi da ti on sta te equa l to i ts cha r ge. N o excepti ons. 3. Oxygen a l most a l wa ys ha s a n oxi da ti on sta te of mi nus 2. Excepti ons i ncl ude the el ementa l for ms of oxygen (O2 a nd O3) i n whi ch i ts oxi da ti on sta te i s zer o, the per oxi de i on (O22) i n whi ch i ts oxi da ti on sta te i s mi nus 1, a nd the super oxi de i on (O2) i n whi ch i ts oxi da ti on sta te i s mi nus . 4. H ydr ogen a l most a l wa ys ha s a n oxi da ti on sta te of pl us 1. Excepti ons i ncl ude el ementa l hydr ogen (H 2) i n whi ch i ts oxi da ti on sta te i s zer o a nd meta l hydr i des (such a s N a H ) i n whi ch i ts oxi da ti on sta te i s mi nus 1. 5. Fl uor i ne a l most a l wa ys ha s a n oxi da ti on sta te of mi nus 1. T he onl y excepti on i s tha t of el ementa l fl uor i ne (F2), whi ch ha s a n oxi da ti on sta te of zer o. T he other ha l ogens usua l l y ha ve oxi da ti on sta tes of mi nus 1 a l so. T he excepti ons a r e the el ementa l ha l ogens, i n whi ch thei r oxi da ti on sta tes a r e zer o, a nd compounds or pol ya tomi c i ons i n whi ch chl or i ne, br omi ne, or i odi ne a r e combi ned wi th fl uor i ne or oxygen. I n these ca ses the ha l ogens (other tha n fl uor i ne) ca n ha ve posi ti ve oxi da ti on sta tes. 6. T he sum of oxi da ti on sta tes must equa l the over a l l cha r ge on the speci es; i .e., the sum of oxi da ti on sta tes i n a mol ecul e must equa l zer o, a nd the sum of oxi da ti on sta tes i n a pol ya tomi c i on must equa l the cha r ge on the pol ya tomi c i on. Some common oxidation-reduction reactions are those between metals and acids or metals and salts. Calcium being oxidized by acid is one example. Magnesium is also oxidized in acidic solution. Metals can also be oxidized by dissolved salts. Nickel metal is oxidized by a solution of copper sulfate. Nickel metal, however, is not oxidized by a solution of calcium sulfate. Whether or not such a reaction will occur can be predicted using the activity series. A metal solid will be oxidized (dissolved) by the ions of any metal that occurs below it on the series. Oxidation and Reduction Half-Reaction The reaction Zn(s) + Cu2+(aq) reduction reaction. Zn2+(aq) + Cu(s) is an oxidation- We can show this by evaluating changes in oxidation state, but there is another especially useful way to establish that it is an oxidation-reduction reaction. Think of the reaction as involving two half-reactions occurring at the same time. The overall reaction is the sum of the two half-reactions. We can represent the half-reactions by half-equations and the overall reaction by an overall equation.
Oxidation: Zn(s) Zn2+(aq) + 2eReduction: Cu2+(aq) + 2eCu(s) Overall: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Oxidation is a process in which The O.S.of some element increa -ses and in which electrons appe -ar on the right side of a halfequation. Reduction is a process in which The O.S.of some element decreases and in which electrons appear on The left side of a half-equation. Example Expressing an oxidation-reduction reaction through half-equations and Solution an overall equation.show the oxidation and reduction that Oxidation: Fe(s) Fe2+(aq) + 2eoccur,and write an overall ionic equation for the reaction of iron Reduction: 2H+(aq) + 2eH2(g) which hydrochloric acid solution to Overall: Fe(s) + 2 H+(aq) Fe2+(aq) produce H2(g)and Fe2+(aq). 55 Balancing OxidationReduction Equations
The Half-Reaction (Ion-Electron) Method The basic steps in this method of balancing a redox equation are as follows : Write and balance separate half-equations for oxidation and reduction. Adjust coefficients in the two half-equations so that the same number of electrons appears in each half-equation. Add together the two half-equations (canceling out electrons)to obtain the balanced overall equation. Example Balancing the Equation for a Redox Reaction in Acidic Solution. The reaction described by expression is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution SO32-(aq) + MnO4-(aq)
Solution Step 1. Write skeleton half-equations based on the species undergoing oxidation and reduction. The O.S. of sulfur increases from +4 in SO32- to +6 in SO42-. The O.S. 0f Mn decreases from +7 in MnO4to +2 in Mn2+.The skeleton half-equations are SO3 2-(aq) MnO4-(aq) SO42-(aq) Mn2+(aq) SO42- (aq) + Mn2+(aq) Step 2. Balance each half-equation for number of atoms, in this order. Atoms other than H and O O atoms, by adding H2O with the appropriate coefficient H atoms, by adding H+ with the appropriate coefficient The other atoms (S and Mn) are already balanced in the skeleton half-equations. To balance O atoms, we add one H2O molecule to the left side of the first half-equation and four to the fight side of the second.
SO32 -(aq) + H2O(l) SO42-(aq) MnO4-(aq) Mn2+(aq) + 4 H2O(l) To balance H atoms, we ami two H+ ions to the right side of the first half-equation and eight to the left side of thc second. SO3 2-(aq) + H2O(l) SO42-(aq) + 2H+(aq) MnO4-(aq) +8H+(aq) Mn2+(aq) + 4 H2O(l) Step 3. Balance each half-equation for electric charge. Add the number of electrons necessary to get the same electric charge on both sides of each half-equation , By doing this, you will see that the half-equation in which electrons appear on the tight side is the oxidation halfequation . The other half-equation , with electrons on the left side, is the reduction half-equation. Oxidation: SO3 2-(aq) + H2O(l) SO42-(aq) + 2H+(aq) + 2e(net charge on each side. -2) MnO4-(aq) +8H+(aq) + 5 eMn2+(aq) + 4 H2O(l) (net charge on each side, +2) Step 4. Obtain the overall redox equation by combining the half-equations. Multiply through the oxidation halfequation by 5 and through the reduction half-equation by 2.This results in 10 e- on each side of the overall equation. These terms cancel out. Electrons must not appear in the final equation. 5SO3 2-(aq) + 5H2O(l) 5 SO42-(aq) + 10H+(aq) + 10e2MnO4-(aq) +16H+(aq) +10 e2Mn2+(aq) + 8 H2O(l) 5SO3 2-(aq) + 2MnO4-(aq) +5H2O(l) + 16H+(aq) 5 SO42-(aq) +2Mn2+(aq) + 8 H2O(l) + 10H+(aq)
Step 5. Simplify. The overall equation should not contain the same species on both sides.subtract five H2O from each side of the equation in step 4. This leaves three H2O on the right. Also subtract ten H+ from each side, leaving six on the left.
3 24+ 422+ 2 5SO (aq) + 2MnO (aq) + 6H (aq) 5 SO (aq) +2Mn (aq) +3 H O(l) Step 6. Verify. Check the overall equation m ensure that it is balanced both for numbers of atoms and electric charge. For example, show that in the balanced equation from Step5. the net charge on each side of the equation is minus six: (5 2-) + (2 1-) +(6 1+) - (5 2-) + (2 2+) = -6. Balancing Equations for Redox Reactions in Acidic Aqueous Solutions by the Half-Reaction Method: A Summary Write the equations for the oxidation and reduction half-reactions. In each half-equation: (1) Balance atoms of all the elements except H and O. (2) Balance oxygen using H2O. (3) Balance hydrogen using H+. (4) Balance charge, using electrons. If necessary, equalize the number of electrons in the oxidation and reduction half-equations by multiplying one or both half-equations by appropriate integers. Add the half-equations, then cancel species common to both sides of the over-all equation. Check that numbers of atoms and charges balance. Balancing Redox Equations in Basic Solution To balance equations for redox reactions in basic solution,the roblem is this:In basic solution,OH-,not H+,must appear in the final balanced equation.OneSimple approach is to treat the reaction a if it Were occurring in an acidic Solution,and balance it.Then,add to each side of the overall redox equation A number of OH- ions equal to the number of H+ ions.Where H+ and OH-Appear on the same side of the equation,combine them to produce H2O Molecules.If H2O now appears on both sides of the equation,subtract the same Number of H2O molecules from each side,leaving a remainder of H2O on just One side. Example Balancing the Equation for a Redox Reaction in Basic Solution. Balance the equation for the reaction in which permanganate ion oxidizes cyanide ion to cyanate ion in basic solution and is itself reduced to MnO2(s). MnO4-(aq) + CN - (aq)
Solution MnO2(s) + OCN - (aq) Initially, we treat the half-reactions and the overall reaction as if they were occurring in an acidic solution, and finally, we adjust the overall equation to a basic solution. Step 1. Write two skeleton half-equations and balance thern for Mn,C,and N atoms. MnO4-(aq) CN-(aq) MnO2(s) OCN-(aq) Note that in this case,the skeleton half-equations as initially written are balanced for Mn,C,and N atoms.
Step 2. Balance the half-equations for 0 and H atorns. Add H20 and/or H+ as required. MnO4-(aq) + 4 H+(aq) MnO2(s) + 2 H2O(I) CN-(aq) + H2O(1) OCN-(aq) + 2 H+ (aq) Step 3. Balance the half-equations for electric charge by adding the appropriate numbers of electrons. Reduction: Step 4. Combine the half-equations to obtain an overall redox equation. Multiply the reduction haft-equation by 2 and the oxidation half-equation by 3. Make the appropriate
4- Step 5. Change from an acidic to a basic medium by adding 2 OH- to both sides of the overall equation; combine 2H+ and 2OH- to form 2 Step 6. Verify. Check the final overall equation to ensure that it is balanced both for number of atoms and for electric charge. For example, show that in the balanced equation from Step. 5, the net charge on each side of the equation is 5-. Balancing Equations for Redox Reactions in Basic Aqueous Solutions by the Half-Reaction Method: A Summary the equation as ifthe reaction were occurring in acidic medium, by using the method for acidic aqueous solutions. ~ To both sides of the overall equation obtained, add a number of OH- that is equal to the number of H+ ions. ~ On the side of the overall equation containing both H+ and OHions, combine them to form H2O molecules. If H2O molecules now appear on both sides of the overall equation, cancel the same number from each side, leaving a remainder of H2Oon just one side. ~ Check that numbers of atoms and charges balance.
~ Balance 56 Oxidizing and Reducing Agents
In a redox reaction, the substance that makes it possible for some other substance to be oxidized is called the oxidizing agent, or oxidant. In doing so, the oxidizing agent is itself reduced. Similarly, the substance that causes some other substance to be reduced is called the reducing agent, or reductant. In the reaction, the reducing agent is itself oxidized. Or, stated in other ways,
An oxidizing agent (Oxidant) : contains an element whose oxidation state decreases in a redox reaction gains electrons (electrons are found on the left side of its half-equation) A ducing agent (reductant) : contains an element whose oxidation state increases in a redox reaction loses electrons (electrons are found on the right side of its half-equation) Example Identifying oxidizing and reducing agents.Hydrogen peroxide,H2O2,is a versatile chemical.Its uses include bleaching wood pulp and substituting for chlorine in water purification.One reason for its versatility is that it can be either an oxidizing or a reducing agent.For the following reactions,identify whether hydrogen peroxide is an oxidizing or reducing agent. (a) H2O2(aq) + 2Fe 2+(aq) + 2H+(aq) (b)5H2O2(aq) + 2MnO4-(aq) + 6H+(aq) 5O2(g) Solution 2H2O(l) + 2Fe 3+(aq) 8H2O(l) + 2Mn 2+(aq) + (a) Fe 2+ is oxidized to Fe 3+,and because H2O2 makes this possible ,it is an oxidizing agent.looking at the matter another way,we see that the Oxidation states of oxygen in H2O2is 1.In H2O,it is 2.Hydrogen Peroxide is reduced and thereby acts as an oxidizing agent. (b) MnO4- is reduced to Mn 2+,and H2O2 makes this possible.In this situation,hydrogen peroxide is a reducing agent.Or,the oxidation state of Oxygen increases from 1in H2O2 to 0 in O2.Hydrogen peroxide is oxidized and thereby acts as a reducing agent. When H2O2 acts as an oxidizing agent,it is reduced to H2O in acidic solution or to OH- in basic solution.When it acts as a reducing agent,it is oxidized to O2(g). Pamianganate ion, MnO4-, is a versatile oxidizing agent that has many uses in the chemical laboratory. ln the next section, we describe its use in the quantitative analysis of iron-that is, the determination of the exact (quantitative) amount of iron in an iron containing material. Ozone, O3(g), a triatomJc form of oxygen,is an oxidizing agent used in water purification, as in the oxidation of the organic compound phenol, C6H5OH.
C6H5OH(aq) + 14O3(g) 6CO2(g) + 3H2O(l) + 14O2(g) Thiosulfate ion, S2O32- , is an important reducing agent, One of its industrial uses is as an antichlor to destroy residual chlorine from the bleaching of fibers.
S2O32- (aq) + 4Cl2(aq) + 5H2O 8Cl2HSO4- (aq) + 8 H+(aq) + 57 Stoichiometry of Reactions in Aqueous Solutions:Titrations
Titration is a reaction carried out by the carefully controlled addition Of one solution to another.The trick is to stop the titration at the point Where both reactants have reacted completely,a condition called the equivalence point of the titration. Still widely used,though,is a technique in which a very small quantity of a Substance added to the reaction mixture changes color at or very near the Equivalence point,such substances are called indicators. Example Standardizing a solution for use in redox titrations.Apiece of iron wire weighing 0.1568g is converted to Fe 2+(aq) and requires 26.24mL of a KMnO4(aq) solution for its titration.What is the molarity of the KMnO4(aq)? 5Fe 2+(aq) + MnO4-(aq) + 8H+(aq) 5Fe 3+(aq) + Mn 2+(aq) +4H2O(l) Solution First ,determine the amount of KMnO4 consumed in the titration. ?mol KMnO4 = 0.1568g Fe 1 mol Fe 1 mol Fe 2+ 55.847g Fe 1 mol Fe 1 mol KMnO4 1 mol MnO4- 1 mol MnO45 mol Fe 2+ = 5.615 10-4 mol KMnO4 The volume of solution containing the 5.615 10-4 mol KMnO4 is 26.24 mL=0.02646L,which means that Concn KMnO4(aq) = 5.615 10-4 mol KMnO4 0.02624 L = 0.02140 M KMnO4(aq) ...
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