18 A C I D S A N D B A S E S ( A H L ) Introduction extend our understanding of K and develop c As our understanding of the reactions of acids and bases has increased, theories have evolved our understanding of the acid and base and the range of reactions considered as acid and base reactions has broadened. In this topic, dissociation constants, K and K respectively. As we dene Lewis acids and bases and examine their reactions. In topic 7, the equilibrium a b law described how the equilibrium constant can be determined for a specic chemical a quantitative analytical technique, the acid- reaction at equilibrium. Weak acids and bases partially ionize in water with the reactants and base titration has wide-ranging applications in products being in a state of equilibrium. We scientic research and industry. Increased power of instrumentation has improved the reliability of this technique. This chapter concludes with an in-depth analysis of the pH curve, its features and the chemistry of buffer solutions; a product of specic types of acid-base reactions. 18.1 lews cs ses Understandings Applications and skills ➔ A Lewis acid is a lone pair acceptor and a Lewis ➔ Application of Lewis acid–base theory to base is a lone pair donor. inorganic and organic chemistry to identify the ➔ When a Lewis base reacts with a Lewis acid a role of the reacting species. coordinate bond is formed. ➔ A nucleophile is a Lewis base and an electrophile is a Lewis acid. Nature of science ➔ Theories can be suppor ted, falsied, or replaced considering lone pairs of electrons. Lewis theory doesn’t falsify Brønsted–Lowry but extends it. by new theories – acid–base theories can be extended to a wider eld of applications by 395
18 ACIDS AND BA SE S (AHL) Acid–base theories have resulted Ex te ou uest from collaboration and competition within the global scientific community. In developing acid–base theories chemists collected evidence through Brønsted (Danish), Lowry (British), observation and experimentation and used it to support, refute, or and Lewis (American) were chemists replace existing theories. Rather than falsifying the Brønsted–Lowry who lived and worked during the theory, the Lewis theory of acids and bases extends our understanding late nineteenth and early twentieth of acid–base reactions, enabling further applications in this eld. centuries, before computers, the internet , or high-speed communication Dening Lewis acids and bases and transpor tation. Their endeavours built on the work of scientists before In sub-topic 8.1 a Brønsted–Lowry base was dened as a substance that can them in moving our understanding of accept a proton. It is the presence of at least one lone pair of electrons that acid–base theory forward. allows a Brønsted–Lowry base to form a coordinate bond with a proton. The hydroxide ion and ammonia are good examples of Brønsted–Lowry bases: A straight line between bonded atoms represents a covalent bond in which O each atom contributes an equal number of electrons. An arrow between bonded + OH atoms represents a coote o H in which one atom contributes both electrons involved in forming the H H covalent bond (see sub-topic 4.2). + TOK H The process of enquiry involves a variety of perspectives to deepen + N H your understanding and your ability H H to draw conclusions and develop theories. Competing acid–base H H N H theories represent dierent levels H of understanding and dierent perspectives. How do we judge these G.N. Lewis dened a Lewis acid as “an electron pair acceptor” and a theories? What criteria do we apply? Lewis base as “an electron pair donor”. Lewis focused on a more general Do we examine their universality, their denition of acids and bases than Arrhenius and Brønsted–Lowry, enabling simplicity, or the elegance of their a wider range of substances to be included. Ammonia and the hydroxide argument? ion are acting as Lewis bases, donating a pair of electrons to the hydrogen ion. The hydrogen ion is a Lewis acid, as it accepts the electron pair. Forming coordinate bonds In the reaction between boron triuoride and ammonia, no proton is involved. Neither compound in this reaction acts as a Brønsted–Lowry acid or base: F N H F B N H B H F H H F H Ammonia donates a lone pair of electrons to form a coordinate bond. 2 2 1 2 Boron, 1s 2s 2p , forms three sp hybrid orbitals, resulting in a vacant unhybridized 2p orbital (gure 1). The lone pair on the nitrogen atom z forms a coordinate bond with the empty 2p orbital of the boron atom. z 1s 2s 2p 1s 2s 2p 1s 2 2p 2sp z ground-state excited-state Figure 1 Hybridization of boron in boron triuoride 396
18 . 2 C a l C U l a T i O n S i n v O lv i n g a C i d S a n d b a S E S Transition elements have a partially occupied d subshell (sub-topic 13.2) so they can form complex ions with ligands that possess a lone pair of electrons. The metal atom or ion is acting as a Lewis acid and the ligand as a Lewis base: 2+ 2+ Co (aq) + 6H O(l) → [Co(H O) ] (aq) 2 2 6 2+ 2+ Ni (aq) + 6NH (aq) → [Ni(NH ) ] (aq) 3 3 6 ammonia 2+ 2+ water cobalt molecule nickel O N 2+ 2+ [Co(H O) ] [Ni(NH ) ] 36 2 6 hexaaquacobalt(II) hexaamminenickel(II) Figure 2 The ligands in these complex ions are acting as Lewis bases Water, H O and ammonia, NH act as Lewis bases in forming complexes 2 3 (gure 2). The cyanide ion, CN , chloride ion, Cl , and hydroxide ion, OH can also act as Lewis bases. As well as acting as ligands these species can also act as nucleophiles in nucleophilic substitution reactions (sub- topic 20.1). They are electron rich with at least one lone pair of electrons. An electrophile is an electron-decient species that can accept a lone A uceophe is a Lewis pair from a nucleophile, in the same way that the nickel(II) ion accepts base and an eectophe is a electron pairs from ammonia. Lewis acid. 18.2 Ccutos o cs ses Understandings Applications and skills ➔ The expression for the dissociation constant of + Solution of problems involving [H (aq)], ➔ a weak acid (K ) and a weak base (K ). a b [OH (aq)], pH, pOH, K , pK , K , and pK . a a b b ➔ For a conjugate acid base pair, K × K = K a b w ➔ Discussion of the relative strengths of acids ➔ The relationship between K and pK is a a and bases using values of K , pK , K , and pK a a b b (pK = -log K ), and between K and pK is a a b b (pK = -log K ). b b Nature of science ➔ Obtaining evidence for scientic theories strengths of acids and bases to be determined and related to their molecular structure. – application of the equilibrium law allows 397
18 ACIDS AND BA SE S (AHL) The strengths of acids and the acid dissociation Calculations involving acids constant and bases are communicated through the universal language Strong acids are a s s um e d to be c om p le t e ly i on i z e d i n wa t e r, t h e of mathematics. Some reaction effectively going to completion. The conjugate base of a scientic explanations exist strong acid has almost no afnity for a proton. The consequence only in mathematical form and is that to determine the concentrationof the hydrogen ion and their meaning is unbiased by subsequently the pH is a simple calculation. For a monoprotic acid, opinion or judgment. the concentrations of each of the two ions produced is the same as the initial concentration of the strong acid. States of matter are not required in the equilibrium expression. HCl(aq) + H O(l) → + + Cl (aq) 2 H O (aq) 3 3 concentration/mol dm 0.5 0.5 0.5 Weak acids, such as ethanoic acid, only partially ionize in water. At equilibrium a majority of the ethanoic acid molecules remains unreacted. CH COOH(aq) + H O(l) ⇋ CH COO (aq) + + H O (aq) 3 2 3 3 We can determine the concentration of the dissociated weak acid using the relationship between concentrations of reactants and products and the equilibrium position. The following general equilibrium constant expression, K , can be c written for the reaction of a weak acid HA with water: HA(aq) + H O(l) ⇋ A (aq) + + 2 H O (aq) 3 [A ][H + O] __3 K = c [HA][H O] 2 In this reaction [H O] is considered a constant, and can be removed 2 from the expression. The resulting expression represents the acid dissociation constant K : a [A + ][H O ] __3 K = a [HA] A weak base B will also ionize in water. The following expression represents the base dissociation constant K : b B(aq) + H O(l) ⇋ + + OH (aq) 2 BH (aq) Figure 1 Arboreal ants spray intruders K = + of their nests with methanoic acid b [BH ][OH ] (commonly called formic acid) __ [B] Calculating K and K a b There are many weak acids and bases that undergo partial ionization in water. The strength of a weak acid or weak base can be expressed Stuy tp quantitatively by determining the dissociation constant at a given Organic acids and bases and their dissociation constants are temperature. listed in section 21 of the Data booklet The stronger the acid, the greater the concentration of hydrogen ions in solution at equilibrium. This corresponds to a larger K value. Similarly, a the stronger the base, the larger the value of K . b 398
18 . 2 C a l C U l a T i O n S i n v O lv i n g a C i d S a n d b a S E S Worked examples: dissociation constant Stuy tp When performing equilibrium Example 1 calculations, always state any approximations and then Calculate the acid dissociation constant K at 298 K for a explain why they are valid. In Example 1, the a approximation is valid since the expression: 0.010moldm 3 solution of propanoic acid, CH CH COOH. 4 _3.7 _× 1_0 3 2 × 100% = 3.7%, The pH of this solution is 3.43. 0.010 which is less than 5% Solution Use of ppoxmtos + In Example 2, the Use the pH to calculate [H ] at equilibrium (sub-topics 8.3 and 17.1): approximation is invalid since the expression: + pH 3.43 4 [H ] = 10 = 10 4 = 3.7 × 10 _1.5 _× 1_0 CH CH COOH(aq) ⇋ CH CH COO + × 100% = 15%, (aq) + H (aq) 0.00100 which is greater than 5% 3 2 3 2 3 I/mol dm 0.010 0.000 0.000 3 4 4 4 +3.7 × 10 C/mol dm 3.7 × 10 +3.7 × 10 3 4 4 4 3.7 × 10 E/mol dm 0.010 3.7 × 10 3.7 × 10 [CH CH COO + ][H ] _3 2 _ K = a [CH CH COOH] 3 2 4 4 (3.7 × 10 )(3.7 × 10 ) ___ 5 3 mol dm K = = 1.4 × 10 a 0.010 In this calculation, because 3.7 × 10 4 is a very small value, the term 0.01 (3.7 × 10 4 ) is rounded to 0.01, which is a valid approximation within the boundaries of experimental error. Example 2 Calculate the base dissociation constant K at 298 K for a b 0.001 00 mol dm 3 solution of the base 1-phenylmethanamine, C H CH NH . The pH of this solution is 10.17. 6 5 2 2 Solution pH + pOH = 14 pOH = 14 - pH = 14 10.17 = 3.83 3.83 4 = 1.5 × [OH ] = 10 10 + C H CH NH (aq) + H O(l) ⇋ C H CH NH (aq) + OH (aq) 2 6 5 2 2 6 5 2 3 3 I/mol dm 0.001 00 0.00 0.00 3 4 4 4 C/mol dm 1.5 × 10 +1.5 × 10 +1.5 × 10 3 4 4 4 E/mol dm 0.001 00 1.5 × 10 1.5 × 10 1.5 × 10 4 2 (1.5 × 10 ) __ 5 K = = 2.6 × 10 b 4 0.001 00 1.5 × 10 399
18 ACIDS AND BA SE S (AHL) K and K for a conjugate acid–base pair a b The relationship between the acid dissociation constant for a weak acid and the base dissociation constant of its conjugate base can be useful in calculations. For example, ethanoic acid partially dissociates in water: CH COOH(aq) ⇋ CH COO + (aq) + H (aq) 3 3 [CH COO + ][H ] _3 _ K = a [CH COOH] 3 The conjugate base of ethanoic acid is the ethanoate ion, CH COO . It 3 reacts with water according to the following equation. CH COO (aq) + H O(l) ⇋ CH COOH(aq) + OH (aq) 3 2 3 [CH COOH][OH ] _3 _ K = b [CH COO ] 3 Combining these expressions: [CH COO + [CH COOH][OH ] ][H ] _3 _ _3 _ + [H ][OH KK = × = ]=K w a b [CH COOH] [CH COO ] 3 3 In summary: KK = K w a b Other forms of this equation are useful in applying the relationship: K K _w K = and K _w a K = b b K a Analysing these relationships reinforces the following conclusions: The stronger the acid: ● the larger the K a ● the weaker the conjugate base ● the smaller the K of the conjugate base. b The stronger the base: ● the larger the K b ● the weaker the conjugate acid The value of K at dierent w ● the smaller the K of the conjugate acid. temperatures can be found in a section 23 of the Data booklet. The temperature dependence of K w In sub-topic 8.3 the ionic product constant K was dened: w Tempetue/°C K pH w 7.17 7.08 7.00 H O(l) ⇋ + + OH (aq) 6.92 2 H (aq) 6.84 15 14 0.453 × 10 + 14 [H ][OH K = ] = 1.0 × 10 at 298 K w 20 14 0.684 × 10 25 14 The ionization of water is an endothermic process. In accordance with 1.00 × 10 Le Châtelier’s principle, a change in the temperature of the system will result in a change in the position of equilibrium. 30 14 1.47 × 10 35 14 2.09 × 10 A rise in temperature will result in the forward reaction being favoured, Table 1 The temperature dependence of K increasing the concentration of the hydrogen and hydroxide ions. This w represents an increase in the magnitude of K and a decrease in the and pH w pH (table 1). Here we make the distinction between the neutrality and 400 the pH of the solution. The pH of the solution decreases with an increase
18 . 2 C a l C U l a T i O n S i n v O lv i n g a C i d S a n d b a S E S in the concentration of hydrogen ions. However, as the concentration of hydroxide ions increases by an equal amount, the solution remains neutral. pK and pK a b The pH scale is a model that represents very small concentrations of hydrogen ions in a way that is easy to interpret, eliminating the use of negative exponents. In a similar way, while acid and base dissociation constants are good descriptors of the strengths of weak acids and bases their values can be very small and so difcult to compare; for example, the K of ethanoic a acid is 1.74 × 10 5 Therefore K and K values are represented as pK . a b a and pK respectively: b pKa log K = pK K = 10 a 10 a a pKb log K = pK K = 10 b 10 b b pK increases pK increases a b strong weak strong weak acid acid base base K increases K increases a b Figure 2 pK and pK values give a model of strength of acids and bases that is easy a b to interpret Worked example Calculate the pH of a solution of 0.080 mol dm 3 methanoic acid, HCOOH, for which pK = 3.75 at 298 K. a Solution pKa 3.75 4 K = 10 = 10 = 1.8 × 10 a HCOOH (aq) ⇋ HCOO + (aq) + H (aq) 3 I/mol dm 0.080 0.000 0.000 3 C/mol dm α +α +α 3 E/mol dm 0.080 - α 0.000 + α 0.000 + α Finding α would require solving a quadratic expression. K = + The small K value for this a [H ][HCOO ] __ a weak acid means that very [HCOOH] little dissociation occurs so the value of α is very small. K = 1.8 × 10 2 It is acceptable to use the 4 _α approximation that the initial a concentration of the weak = acid or base is equal to its equilibrium concentration. 0.080 - α 401 4 2 _α K = 1.8 × 10 = a 0.080 _______________ 4 α = √ 1.8 × 10 × 0.080 3 α = 3.8 × 10 + 3 [H ] = 3.8 × 10 pH = 2.42
18 ACIDS AND BA SE S (AHL) Sttees whe so Worked example c–se equum poems Calculate the pH of a 0.25 mol dm 3 All chemical equations written solution of triethylamine, should be fully balanced, paying attention to state (C H ) N. The pK of triethylamine at 298 K is 3.25. symbols. 2 5 3 b Solution (C H ) N (aq) + H O(l) ⇋ + + OH (aq) 2 (C H ) NH (aq) 2 5 3 2 5 3 3 I/mol dm 0.25 0.00 0.00 ● Use all the data provided 3 C/mol dm α +α +α in an IB question: no extraneous information will 3 E/mol dm 0.25 - α 0.00 + α 0.00 + α be included. + ] [(C H ) NH ][OH 2 5 3 __ K = b ● Show all your working clearly [(C H ) N] 2 5 3 to enable the examiner to pK 3.25 4 b= K = 10 10 = 5.6 × 10 follow your logic. b ● State all the equilibrium 4 2 _α = 5.6 × 10 = 0.25 - α constant expressions ______________ you use. α = √ 5.6 × 10 4 0.25 × ● Pay attention to signicant 3 = 1.2 × 10 gures. 3 [OH ] = 1.2 × 10 pOH = -log (1.2 × 10 3 ) = 1.9 10 pH = 14 - pOH = 14 - 1.9 = 12.1 402
18 . 3 P H C U r v E S 18.3 pH cu es Understandings Applications and skills ➔ The characteristics of the pH curves produced ➔ The general shapes of graphs of pH against by the dierent combinations of strong and volume for titrations involving strong and weak weak acid and bases. acids and bases with an explanation of their ➔ An acid–base indicator is a weak acid or a weak impor tant features. base where the components of the conjugate ➔ Selection of an appropriate indicator for a acid–base pair have dierent colours. titration, given the equivalence point of the ➔ The relationship between the pH range of an titration and the end point of the indicator. acid–base indicator, which is a weak acid, and ➔ While the nature of the acid–base buer its pK value. always remains the same, buer solutions a ➔ The buer region on the pH curve represents can be prepared either by mixing a weak acid/ the region where small additions of acid or base base with a solution of a salt containing its result in little or no change in pH. conjugate, or by par tial neutralization of a weak ➔ The composition and action of a buer solution. acid/base with a strong acid/base. ➔ Prediction of the relative pH of aqueous salt solutions formed by the dierent combinations of strong and weak acid and base. Nature of science ➔ Increased power of instrumentation and in pH meter technology has allowed for more reliable and ready measurement of pH. advances in available techniques – development Titration years. Quantitative data such as that resulting from titrations is subject to mathematical analysis As described in sub-topic 1.3, titration is a and can help chemists identify patterns and quantitative analytical technique used to formulate relationships. The many features determine the concentration of a reactant from shown in titration or pH curves unlock a wealth a reaction of known stoichiometry. Different of chemical knowledge. types of titration have been utilized in a range of industry and research settings for over 150 Buer solutions The addition of a single drop of a strong acid or base to water can result in a signicant change in pH. A buffer is a solution that resists a change in pH upon the addition of small amounts of a strong base or strong acid, or upon the dilution of the buffer through the addition of water. A buffer may be composed of a weak acid and its conjugate base, or a weak base and its conjugate acid. 403
18 ACIDS AND BA SE S (AHL) How buer solutions work Consider a buffer made from ethanoic acid and its In this case the forward reaction is favoured, conjugate base, the ethanoate ion: replenishing hydronium ions and again maintaining the pH. CH COOH(aq) + H O(l) ⇋ CH COO (aq) 3 2 3 + + The buffer capacity or the effectiveness of H O (aq) thebuffer to resist changes in pH depends on the molar concentration of the acid and 3 conjugate base. The higher the concentration, the more effective the buffer at resisting For effective control of changes in pH the changes in pH. weak acid and its conjugate base are mixed in equimolar concentrations. The equilibrium position changes in accordance with Le Châtelier’s principle: the addition of small amounts of a strong acid will increase the A buffer can also be prepared from the partial neutralization of a weak acid with a strong concentration of hydronium ions. The reverse base.The weak acid is present in excess, producing a conjugate acid–base pair with the reaction is favoured and the additional hydrogen salt formed. ions are mopped up, maintaining the pH. In contrast, small amounts of added strong base react with the hydronium ions: CH COOH(aq) + NaOH(aq) → CH COONa(aq) 3 3 + + OH (aq) → 2H O(l) + H O(l) H O (aq) 2 2 3 Salt hydrolysis When an acid and a base react in a neutralization the ionization of water that results from reaction reaction, an ionic salt is formed by the combination with an ionic salt. The salts of strong and weak of the cation from the base and the anion from the acids and bases react in different ways with water. acid. Ionic salts completely dissociate in water. The relative strengths of the acid and base dictate the type and degree of the salt hydrolysis that The pH of the resulting solution is dependent on results and its effect on the solution pH. the reaction of the salt with water. Hydrolysis is pH curves 3 3 A typical acid–base titration might start with 25 cm of 0.1 mol dm acid in a conical (also called Erlenmeyer) ask, together with several drops of an appropriate indicator. 3 Into a burette is placed 0.1 mol dm alkali solution. As the titration progresses, quantitative and qualitative data are recorded. The pH of the neutralization reaction can be measured using data-logging equipment and the data recorded in a spreadsheet. This data can then be used to create a pH curve. We shall look at four examples of pH curve: ● strong acid and strong base ● weak acid and strong base ● weak base and strong acid ▲ Figure 1 A coloured indicator shows ● weak base and weak acid. the end point of a titration 404
18 . 3 P H C U r v E S The titration of a strong acid with a strong base Indicators and Consider the reaction of a strong acid with a strong base: end point HCl(aq) + NaOH(aq) → NaCl(aq) + H O(l) Advances in pH probe 2 instrumentation have enabled the pH change during acid–base or, in ionic form: titrations to be accurately tracked. However, it is often convenient to + (aq) → H O(l) visually determine the end point H (aq) + OH 2 of the titration using an acid-base indicator – an organic dye that + (aq) changes its colour in a narrow pH NaCl(aq) → Na (aq) + Cl range. Ideally, this colour change must occur when the equivalence Salts derived from a strong acid and a strong base do not react with point is reached. hydrogen ions or hydroxide ions and so will not undergo hydrolysis. The salt sodium chloride forms a neutral aqueous solution. E pot equece pot As described in the iUPaC go book Figure 2 shows the pH curve for the titration of hydrochloric acid with (http://goldbook.iupac.org), a ttto sodium hydroxide carried out as described above. determines the quantity of a substance A by adding measured increments of 14 substance B with which it reacts, with provision for some means of recognizing 12 high pH (indicating) the e pot at which 10 (strong base) essentially all of A has reacted. If the end point coincides with the addition 8 equivalence point of the exact chemical equivalence, pH pH = 7 it is called the equece pot (or theoretical endpoint), thus allowing the 6 amount of A to be found from known amounts of B added up to this point. 4 pH 1 intercept with axis 2 0 10 20 25 30 40 50 0 3 volume of NaOH added/cm ▲ Figure 2 pH cur ve for the titration of a strong acid with a strong base ● The starting point on the pH axis is an important feature of a pH curve as it is an indication of the relative strength of the acid. The strong acid gives an initial pH reading ≈ 1.0. ● There is a gradual rise in the pH as the titration approaches the equivalence point. ● The sharp rise in pH at the equivalence point (pH = 7.0) is described as the point of inection of the curve. ● Once there is no remaining acid to be neutralized, the curve attens and nishes at a high pH reecting the strong base. The titration of a weak acid with a strong base The reaction of the weak ethanoic acid with the strong base sodium hydroxide is shown in the following equation: CH COOH(aq) + NaOH(aq) → CH COONa(aq) + H O(l) 3 3 2 or, in ionic form: CH COOH(aq) + OH (aq) → CH COO (aq) + H O(l) 3 3 2 CH COONa(aq) → CH COO + (aq) + Na (aq) 3 3 405
18 ACIDS AND BA SE S (AHL) The aqueous solution resulting The sodium ion will not undergo hydrolysis but the ethanoate ion from the reaction between is the conjugate base of a weak acid and so has a strong afnity equal amounts of a weak acid for hydrogen ions. The ethanoate ions are hydrolysed with water, and a strong base is alkaline. producing hydroxide ions: CH COO (aq) + H O(l) ⇋ CH COOH(aq) + OH (aq) 3 2 3 Figure 3 shows the pH curve for the titration of ethanoic acid with sodium hydroxide. high pH (strong base) 14 12 10 8 equivalence point pH > 7 pH = pK pH a half-equivalence point 6 4 buer zone 2 pH 3 intercept with axis (weak acid) 0 0 10 20 30 40 50 3 volume of NaOH added/cm ▲ Figure 3 pH cur ve for the titration of a weak acid with a strong base ● The weak acid gives an initial pH reading ≈ 3.0. ● The initial rise is steep, as a strong base is being added to a weak acid and neutralization is rapid. ● As the weak acid begins to be neutralized the strong conjugate base sodium ethanoate is formed, creating a buffer that resists change in pH. Ethanoic acid is in equilibrium with the ethanoateion: CH COOH(aq) + H O(l) ⇋ CH COO (aq) + + H O (aq) 3 2 3 3 ● The continued addition of base to the solution uses up hydrogen ions, hence the forward reaction is favoured. This results in a very gradual change in pH in this region of the curve. ● The half-equivalence point is the stage of the titration at which half of the amount of weak acid has been neutralized: [CH COOH(aq)] = [CH COO (aq)] 3 3 [CH COO ][H + O] _3 _3 K = a [CH COOH] 3 K = [H + a O] 3 pK = pH a ● There is a sharp rise in pH at the equivalence point (pH > 7). The equivalence point is the result of salt hydrolysis. ● With no remaining acid to be neutralized, the curve attens and nishes at a high pH due to the presence of excess strong base. This section of the curve is identical to that in gure 2. 406
18 . 3 P H C U r v E S The titration of a weak base with a strong acid The reaction between hydrochloric acid and ammonia is shown in the following equation: HCl(aq) + NH (aq) → NH Cl(aq) 3 4 or, in ionic form: + + H (aq) + NH (aq) → NH (aq) 3 4 + NH Cl(aq) → NH (aq) + Cl (aq) 4 4 The chloride ion, Cl is the conjugate base of the strong acid hydrochloric acid, HCl and has almost no afnity for hydrogen ions. The aqueous solution resulting from the reaction between + equal amounts of a strong acid and a weak base is acidic. The ammonium ion, NH is the conjugate acid of the weak base 4 ammonia, NH . It will donate a proton in the reaction with water, 3 forming the hydronium ion: + + H O (aq) NH (aq) + H O(l) ⇋ NH (aq) + 2 3 3 4 Figure 4 shows the pH curve for the titration of ammonia with hydrochloric acid. In this titration ammonia is put into the conical ask and the burette is lled with hydrochloric acid. 14 pH 11 intercept with axis (weak base) 12 pOH = pK 10 b half-equivalence point 8 buffer zone pH 6 equivalence point pH < 7 4 2 low pH (strong acid) 0 10 20 30 40 50 0 3 volume of HCl added/cm ▲ Figure 4 pH cur ve for the titration of a weak base with a strong acid ● The weak base gives an initial pH reading ≈ 11.0. ● As the weak base begins to be neutralized, the ammonium ion NH + the conjugate acid, is created resulting in a buffer that resists , 4 change in pH. Ammonia is in equilibrium with the ammonium ion: + NH (aq) + H O(l) ⇋ NH (aq) + OH (aq) 3 2 4 ● At the half-equivalence point half of the amount of weak base has been neutralized. At this point, pOH = pK b ● There is a gradual fall in the pH due to the buffering effect as the titration approaches the equivalence point. ● The pH falls sharply at the equivalence point (pH < 7). The equivalence point is the result of salt hydrolysis. ● With no remaining base to be neutralized, the curve attens and ends at a low pH due to the presence of excess strong acid. 407
18 ACIDS AND BA SE S (AHL) The titration of a weak base with a weak acid Salts derived from a weak acid and a weak base will undergo hydrolysis in water and the resulting pH of the aqueous solution depends on the relative strengths of the acid ( K ) and base (K ). Ammonium ethanoate, a b CH COONH forms a neutral aqueous solution: 3 4 NH (aq) + CH COOH(aq) → CH COONH (aq) 3 3 3 4 or, in ionic form: + NH (aq) + CH COOH(aq) → CH COO (aq) + NH (aq) 3 3 3 4 Figure 5 shows the pH curve for the titration of ammonia with ethanoicacid. ● The weak base gives an initial pH reading ≈ 11.0. ● The change in pH throughout the titration is very gradual. ● The point of inection in the pH curve is not as steep as in the previous pH curves. The point of equivalence is difcult to determine, so this kind of titration has little or no practical use. ● With no remaining base to be neutralized, the curve attens and ends at a pH that indicates the presence of a weak acid. 14 pH 11 intercept with axis 12 (weak base) 10 8 equivalence point pH pH ≈ 7 6 4 2 0 10 20 30 40 50 0 3 volume of CH COOH(aq) added/cm 3 ▲ Figure 5 pH cur ve for the titration of a weak base with a weak acid Indicators An indicator is typically a weak acid or a weak base that displays a different colour in acidic or alkaline environments. Many indicators in aqueous solutions behave as weak acids: + (aq) HIn(aq) ⇋ H (aq) + In colour A colour B K = + a [H ][In ] _ [HIn] The above formula can be rearranged as follows: K = [In ] _a _ + [HIn] [H ] The midpoint of the colour change is observed when [HIn] = [In ]. 408
18 . 3 P H C U r v E S At this point: + = K [H ] a pH = pK a The colour change for most indicators takes place over a range of pH = pK ± 1. a The colour of a given indicator depends on the pH of the solution. In acidic solutions the indicator exists in protonated form HIn(aq), so colour A is observed. In basic solutions hydrogen ions in the equilibrium are consumed and the forward reaction is favoured. In this case, the indicator exists as In (aq) and colour B becomes dominant. An indicator can also be a weak base. + (aq) BOH(aq) ⇋ B (aq) + OH colour A colour B For such indicators, colour A is observed in alkaline solutions while ▲ Figure 6 Methyl orange indicator is red in acidic colour B appears in the presence of acids. solutions and yellow in alkaline solutions. Most indicators are weak acids but methyl orange is Selection of an indicator in fact a weak base The choice of indicator for an acid–base titration depends on the 14 relative strengths of the acid and base and therefore on the pH of the 13 equivalence point. The midpoint of an indicator’s colour change must 12 correspond to the equivalence point of the titration. 11 10 The titration of a strong acid with a strong base such as hydrochloric acid with sodium hydroxide has an equivalence point of pH 7.0. Phenol phenolphthalein pH range pH at equivalence 9 red has a pK of 7.9 and a pH range of 6.8–8.4. However, the titration a 8 curve of a strong acid with a strong base shows a very steep rise near pH 7 the equivalence point. This rise covers the pH range of most acid–base 6 indicators, so all common indicators, such as phenolphthalein or methyl 5 orange, can be used in such titrations. The titration of hydrochloric acid with the weak base ammonia has an 4 3 equivalence point at pH < 7.0. Methyl orange (pK = 3.7) is an effective 2 1 a indicator for this titration. The titration of the weak acid ethanoic acid with sodium hydroxide has 0 10 20 30 40 50 0 3 volume of NaOH/cm an equivalence point at pH > 7.0. Phenolphthalein (p K = 9.6) is an a effective indicator for this titration (gure 7): ▲ Figure 7 Phenolphthalein indicator is suitable for the titration of ethanoic acid with sodium Ttto icto pK pH e acc ake hydroxide 6.8–8.4 coou coou 3.1–4.4 yellow red strong acid– phenol red 7.9 strong base 8.3–10.0 red yellow Stuy tp strong acid– methyl orange 3.7 colourless pink weak base Examples of acid–base indicators, their pK values, and their colour weak acid– phenolphthalein 9.6 a strong base changes are listed in section 22 of the Data booklet, which will be available during the examination. ▲ Table 1 Some acid–base indicators commonly used in titrations 409
18 ACIDS AND BA SE S (AHL) Questions 1 Which of the following could be added to a 5 Equal volumes and concentrations of solution of ethanoic acid to prepare a buffer? hydrochloric acid and ethanoic acid are titrated with sodium hydroxide solutions of the same A. Sodium hydroxide concentration. Which statement is correct? B. Hydrochloric acid a) The initial pH values of both acids are equal. C. Sodium chloride b) At the equivalence points, the solutions of D. More ethanoic acid [1] both titrations have pH values of 7. IB, May 2010 c) The same volume of sodium hydroxide is needed to reach the equivalence point. 2 Which mixture of acid and alkali would d) The pH values of both acids increase produce a buffer solution? equally until the equivalence points are reached. [1] Acid Alkali 3 3 IB, November 2010 A. 40 cm 0.1 60 cm 0.1 B. C. mol dm 3 mol dm 3 D. HCl NaOH 3 3 3 ammonia solution is placed 60 cm 0.1 40 cm 0.1 6 A 0.10 mol dm 3 3 3 HCl NaOH mol dm mol dm in a ask and titrated with a 0.10 mol dm 3 3 hydrochloric acid solution. 40 cm 0.1 60 cm 0.1 3 3 HCl mol dm mol dm NH 3 a) Explain why the pH of the ammonia 3 3 60 cm 0.1 40 cm 0.1 solution is less than 13. [2] 3 3 HCl mol dm mol dm NH 3 [1] b) Estimate the pH at the equivalence point for the titration of hydrochloric acid with IB, November 2009 ammonia and explain your reasoning. [2] c) State the equation for the reaction of 3 When the following 1.0 mol dm 3 aqueous solutions are arranged in order of increasing ammonia with water and write the K b [2] expression for NH (aq). pH, which is the correct order? 3 d) When half the ammonia has been neutralized I Ammonium chloride (the half-equivalence point), the pH of the II Ammonium ethanoate solution is 9.25. Deduce the relationship + III Sodium ethanoate between [NH ] and [NH ] at the half- 3 4 equivalence point. [1] A. I, II, III e) Determine pK and K for ammonia based B. II, I, III b b on the pH at the half-equivalence point. [3] C. III, I, II f) Describe the signicance of the half- D. III, II, I [1] equivalence point in terms of its IB, November 2003 effectiveness as a buffer. [1] IB, May 2010 4 Predict and explain, using equations where appropriate, whether the following solutions are 7 Which species can act as a Lewis acid? acidic, alkaline, or neutral. A. BF C. HO 3 2 3 a) 0.1 mol dm FeCl (aq) [1] 3 [1] [1] B. OH D. NH [1] 3 3 b) 0.1 mol dm NaNO (aq) 3 IB, Specimen paper c) 0.1 mol dm 3 Na CO (aq) 2 3 IB, Specimen paper 410
QUE STiOnS 8 Which statement explains why ammonia can 13 When these 1.0 mol dm 3 acidic solutions act as a Lewis base? are arranged in order of increasing strength (weakest rst), what is the correct order? A. Ammonia can donate a lone pair of electrons. B. Ammonia can accept a lone pair of electrons. acid in solution X K = 1.74 × 10 5 3 mol dm at 298 K C. Ammonia can donate a proton. a acid in solution Y D. Ammonia can accept a proton. [1] K = 1.38 × 10 3 3 mol dm at 298 K a IB, May 2011 acid in solution Z K = 1.78 × 10 5 3 mol dm at 298 K a 9 Which equation represents an acid–base A. X<Z<Y C. Z <X<Y reaction according to the Lewis theory but not B. X<Y<Z D. Y <X<Z [1] the Brønsted–Lowry theory? IB, May 2010 A. NH + HCl ⇋ NH Cl 3 4 B. 2H O ⇋ H + + OH 2 O 3 14 pK for water at 10 °C = 14.54. What is the pH w C. NaOH + HCl ⇋ NaCl + H O 2 of pure water at this temperature? D. CrCl + 6NH ⇋ [Cr(NH ) 3+ + 3Cl [1] 3 3 ] 3 6 A. 6.73 C. 7.27 IB, November 2003 B. 7.00 D. 7.54 [1] IB, May 2010 10 The equilibrium reached when ethanoic acid 15 What is K for the aqueous uoride ion is added to water can be represented by the following equation: b 14 + giventhat K is 1.0 × 10 and K for HF is H O (aq) a CH COOH(l) + H O(l) ⇋ CH COO (aq) + w 3 3 2 3 6.8× 10 4 at 298 K? Dene the terms Brønsted–Lowry acid and _1 A. Lewis base, and identify two examples of each 4 6.8 × 10 of these species in the equation. [4] 4 -14 B. (6.8 × 10 )(1.0 × 10 ) IB, November 2005 -14 _1.0 ×_10 C. 4 6.8 × 10 11 a) Dene a Brønsted–Lowry acid. [1] [1] D. 4 6.8 × 10 b) Deduce the two acids and their conjugate IB, May 2010 bases in the following reaction: [2] + H O(l) + NH (aq) ⇋ OH (aq) + NH (aq) 2 3 4 16 Ammonia acts as a weak base when it reacts with c) Explain why the following reaction can also water. What is the K expression for this reaction? b be described as an acid–base reaction. [2] + [NH ][OH ] [NH ] __3 F (g) + BF (g) ⇋ BF (s) _4 _ 3 4 A. C. + [NH ][H O] [NH ][OH ] 3 2 4 IB, May 2009 + [NH ][H O] [NH ][OH ] _3 _2 _4 _ B. D. [1] + [NH ][OH ] [NH ] 3 4 12 a) Dene a Lewis acid and state an example IB, May 2009 that is not a Brønsted–Lowry acid. [2] b) Draw structural formulas to represent the 17 Ammonia, NH , is a weak base. It has a 3 reaction between the Lewis acid named in pK value of 4.75. (a) and a Lewis base and identify the nature b of the bond formed in the product. [4] 2 3 mol dm Calculate the pH of a 1.00 × 10 IB, November 2009 aqueous solution of ammonia at 298 K. [4] IB, May 2011 411
18 ACIDS AND BA SE S (AHL) 18 Ammonia can be converted into nitric acid, 21 Which indicator would be the most appropriate HNO (aq), and hydrocyanic acid, HCN(aq). for titrating aqueous ethylamine, CH CH NH , 3 3 2 2 ThepK of hydrocyanic acid is 9.21. with nitric acid, HNO ? 3 a a) Distinguish between the terms strong and A. Bromophenol blue (p K = 4.1) a weak acid and state the equations used B. Bromothymol blue (p K = 7.3) a to show the dissociation of each acid in C. Phenol red (pK = 8.0) a aqueous solution. [3] D. Thymolphthalein (p K = 10.0) [1] a b) Deduce the expression for the ionization IB, November 2009 constant, K , of hydrocyanic acid and calculate a its value from the pK value given. [2] a c) Use your answer from part (b) to calculate 22 The graph below (gure 8) shows the titration + 3 3 the [H ] and the pH of an aqueous solution of hydrochloric curve of 25 cm of 0.100 mol dm of hydrocyanic acid of concentration 3 acid with sodium hydroxide, of 0.100 mol dm 0.108 mol dm 3 . State one assumption concentration. The indicator methyl orange was made in arriving at your answer. [4] used to determine the equivalence point. Methyl IB, November 2010 orange has a pH range of 3.1–4.4. 14 19 0.100 mol of ammonia, NH , was dissolved 3 3 in water to make 1.00 dm of solution. This 12 solution has a hydroxide ion concentration of 10 3 3 mol dm 1.28 × 10 8 a) Determine the pH of the solution. [2] pH b) Calculate the base dissociation constant, 6 K , for ammonia. [3] yellow b methyl orange indicator 4 IB, November 2009 red 2 20 Consider an acid–base indicator solution. 0 0 5 10 15 20 25 30 35 40 45 50 3 3 + volume of 0.100 mol dm NaOH(aq) added/cm HIn(aq) ⇋ H (aq) + In (aq) colour A colour B ▲ Figure 8 What is the effect on this acid–base indicator If the hydrochloric acid was replaced by when sodium hydroxide solution is added to it? ethanoic acid of the same volume and concentration, which property of the titration A. Equilibrium shifts to the right and more of would remain the same? colour B is seen. B. Equilibrium shifts to the left and more of A. The initial pH. colour B is seen. B. The pH at the equivalence point. C. Equilibrium shifts to the right and more of C. The volume of strong base, NaOH, needed colour A is seen. to reach the equivalence point. D. Equilibrium shifts to the left and more of D. The colour of the titration mixture just colour A is seen. [1] before the equivalence point is reached. [1] IB, May 2010 IB, May 2011 412
19 R E D O X P R O C E S S E S ( A H L ) Introduction types of electrochemical cell. The electrolysis of aqueous solutions will be discussed in In this topic we will explore further voltaic this topic and the quantitative aspects of and electrolytic cells which we rst met in electrolysis. topic 9. In particular we will see the role that standard electrode potentials play in both 19.1 Eecroceca c es Understandings Applications and skills ➔ A voltaic cell generates a voltage resulting in ➔ Calculation of cell potentials using standard the movement of electrons from the anode electrode potentials. (negative electrode) to the cathode (positive ➔ Prediction of whether a reaction is spontaneous electrode) via the external circuit. The voltage is or not using E values. termed the cell potential (E). ➔ Determination of standard free-energy changes ➔ The standard hydrogen electrode (SHE) consists (ΔG ) using standard electrode potentials. of an iner t platinum electrode in contact with 3 ➔ Explanation of the products formed during the 1 mol dm hydrogen ions and hydrogen gas electrolysis of aqueous solutions. at 100 kPa and 298 K . The standard electrode potential (E ) is the potential (voltage) of ➔ Perform laboratory experiments that could the reduction half-equation under standard include single replacement reactions in conditions measured relative to the SHE. Solute aqueous solutions. 3 concentration is 1 mol dm or 100 kPa for ➔ Determination of the relative amounts of gases. E of the SHE is 0 V. products formed during electrolytic processes. ➔ When aqueous solutions are electrolysed, ➔ Explanation of the process of electroplating. water can be oxidized to oxygen at the anode and reduced to hydrogen at the cathode. ➔ ΔG = –nFE . When E is positive, ΔG is Nature of science negative, indicative of a spontaneous process. ➔ Employing quantitative reasoning – electrode When E is negative, ΔG is positive, indicative potentials and the standard hydrogen of a non-spontaneous process. When E is 0, electrode. then ΔG is 0. ➔ Collaboration and ethical implications – ➔ Current, duration of electrolysis, and charge on scientists have collaborated to work on the ion aect the amount of product formed at electrochemical cell technologies and have the electrodes during electrolysis. to consider the environmental and ethical ➔ Electroplating involves the electrolytic coating implications of using fuel cells and microbial of an object with a metallic thin layer. fuel cells. 413
19 REDOX PROCE SSE S (Ahl) Voac (gavanc) ces EMF and the standard cell potential In topic 9 we examined voltaic (galvanic) cells, which convert chemical to electrical energy. As described by IUPAC an electromotive force (EMF) is the energy supplied by a source divided by the electric charge transported through the source. In a voltaic cell the EMF is equal to the electric potential difference for zero current through the cell. The EMF is the maximum voltage that can be delivered by the cell. A helpful analogy The idea of potential difference can often be difcult to understand. However, here is one useful analogy. Imagine you have two water barrels, A and B, with different volumes of water at two distinct levels in each barrel and with the two barrels connected by a pipe. When the connecting pipe is opened the water will ow from the barrel where the water is at a higher level (that is barrel A) through the open pipe to barrel B, where the water level is less. barrel A barrel B open pipe high PE low PE Figure 1 (a) Analogy for the idea of potential dierence, of water spontaneously flowing from one barrel, barrel A, where the water is at a higher level, through the pipe when open to barrel B, where the water level is lower. Barrel A could be described as having a greater potential energy (PE) than barrel B high PE e low PE Figure 1(b) The photograph to the left shows the Rio Latus waterfall in Ecuador. The same principle applies at the top of a waterfall with water flowing down into a lake when a suitable pathway is available for this to happen. At the top of the waterfall, the PE is greatest, whereas at the bottom of the waterfall the PE is lowest. In a similar way electrons flow downhill from high PE to low PE, once a suitable pathway is present, eg a conducting wire. The dierence in PE between the two electrodes is the EMF In a voltaic cell a cell potential is generated, resulting in the movement of electrons from the anode (negative electrode) to the cathode (positive electrode) via the external circuit. The cell potential is then dened 414
19 . 1 E l E C t R O C h E m i C A l C E l l S as the potential difference between the cathode and the anode when the cell is operating and is always less than the maximum voltage that Sudy p Remember reduction always takes can be delivered by the cell. The cell potential also depends on the place at the cathode and oxidation always takes place at the anode in any concentrations of the species involved (that is reactants and products) electrochemical cell (both voltaic cells and electrolytic cells). A useful way of and the operating temperature (taken in general as 298 K or 25 °C). remembering this is the mnemonic, CROA (as mentioned in topic 9). 3 CROA Under standard conditions (1 mol dm concentration for reactants in the standard cell potential (E ) cell In order to calculate the overall standard cell potential, we use the expression: E = E - E cell rhe lhe where: E represents the standard electrode potential at the cathode, which by rhe convention is taken as the right-hand side electrode in a voltaic cell; cathode reduction oxidation anode E represents the standard electrode potential at the anode, which by lhe convention is taken as the left-hand side electrode in a voltaic cell. Or another useful mnemonic involves the two animals: Section 24 of the Data booklet contains a number of standard electrode potentials, given in units of volts. By international agreement with the “a fat Red Ca” and “an Anorexic O x” scientic community standard electrode potentials are always expressed as reductive processes. In order to calculate E for a spontaneous cell, cell reduction cathode anode oxidation the cathode (and hence E ) is taken as the more positive value chosen rhe from two standard electrode potentials and the anode (and hence E ) is lhe taken as the least positive value. Let us return to the Daniell cell, rst introduced in topic 9 for redox 2+ 2+ processes. The two electrodes are Zn(s)|Zn (aq) and Cu(s)|Cu (aq). From section 24 of the Data booklet, the two standard electrode potentials are as follows: 2+ Zn (aq) + 2e ⇋ Zn(s) E = -0.76 V 2+ Cu (aq) + 2e ⇋ Cu(s) E = +0.34 V Notice that both are written as reduction half-equations with an equilibrium sign, and signs are always included ( + or ) in this tabular format. Based on the two E values, we see that, since +0.34 V is more 2+ positive, the Cu (aq)|Cu(s) electrode is the cathode (positive electrode), and therefore reduction will take place at this electrode. The half- equations corresponding to the processes occurring at the cathode and anode can then be written as follows: Cathode (positive electrode): Reduction takes place here. 2+ Cu (aq) + → Cu(s) E = +0.34 V rhe Anode (negative electrode): Oxidation takes place here. 2+ Zn(s) → Zn (aq) + E = -0.76 V lhe Overall cell reaction: 2+ 2+ Cu (aq) + Zn(s) → Zn (aq) + Cu(s) The overall standard cell potential for the Daniell cell, E , can be cell calculated as follows: E =E E = (+0.34 V) (–0.76 V) = +1.10 V cell lhe rhe 415
19 REDOX PROCE SSE S (Ahl) Sudy p Consider the Daniell cell: 2+ 2+ Zn(s)|Zn (aq) || Cu (aq)|Cu(s) switch e e V + NO + 3 Na Zn anode Cu cathode () (+) NO NO 3 3 2+ NO Zn 3 NO 2+ 3 Cu 2+ 2+ Zn(s) → Zn (aq) + 2e Cu (aq) + 2e → Cu(s) movement of cations movement of anions Note: In the representation of the half-equations taking place at both the cathode and anode electrodes, NO EQUiliBRiUm SiGNS are included, allthough equilibrium signs are given for the standard electrode potentials in section 24 of the Data booklet Caode (posve eecrode): 2+ Cu (aq) + → Cu(s) E = +0.34 V rhe Anode (negave eecrode): 2+ Zn(s) → Zn (aq) + E = –0.76 V lhe te sandard ydrogen eecrode It is not possible to measure the electrode potential of a single half- cell, as in order to measure the potential we require a potential energy difference for the electrons, which must be in two chemically different set-ups. For this reason, electrode potentials are measured relative to an internationally agreed standard, which has been chosen as the standard hydrogen electrode (SHE). The SHE is the universal reference electrode and is a gas electrode. The standard hydrogen electrode 3 (SHE) consists of an inert platinum electrode in contact with 1 mol dm hydrogen ions and hydrogen gas at 100 kPa. Therefore, the standard electrode potential (E ) is the potential (voltage) of the reduction half-equation under standard conditions measured relative to the SHE. The standard concentration for a solute is 1 mol dm 3 and under standard conditions the pressure is 100 kPa for gases. E of the SHE is taken as 0 V at all temperatures. The potentials of other electrodes are then compared to the SHE reference at the same temperature. The reduction half-equation corresponding to the SHE half-cell is: + ⇋ H (g) 2H (aq) + 2e 2 416
H (g) (at 100 kPa) 19 . 1 E l E C t R O C h E m i C A l C E l l S 2 Usefu resource Pt electrode The Chemical Education Research Group at the Depar tment of Chemistry at Iowa State University, USA have developed an excellent series of chemistry animations and simulations. The electrochemical simulations and animations are par ticularly relevant to both topics 9 and 19 and should be used in combination with the laboratory aspects related to electrochemical experiments. http://group.chem.iastate.edu/ Greenbowe/tg-research.html + 3 H (aq), 1 mol dm Figure 2 The standard hydrogen electrode (SHE) The standard electrode potential of another half-cell is then determined simply by connecting the half-cell, under standard conditions to the SHE, using a connecting wire with a voltmeter attached and a salt bridge. The cell potential can then be determined. 2+ Let us consider the following cell consisting of the Cu (aq)|Cu(s) half- cell connected to the SHE. Cathode (positive electrode): 2+ Cu (aq) + → Cu(s) E = +0.34 V tOK rhe Anode (negative electrode): The absolute electrode potential of the + standard hydrogen electrode under 2H (aq) H (g) → + E = 0.00 V 2 lhe standard conditions has the estimated This cell can be represented by the following cell diagram: value: E = (+4.44 ± 0.02) V abs + 2+ at 298 K . However, for comparison Pt(s)|H (g)|H (aq) || Cu (aq)|Cu(s) 2 purposes with all other electrode E =E E = (+0.34 V) (0.00 V) = +0.34 V cell lhe rhe reactions, hydrogen’s standard electrode potential is assigned as 0 V V a a eperaures. The SHE is a e universal reference electrode and is an example of an arbitrary reference. e KNO (aq) salt bridge Since the SHE is an example of an LHE 3 arbitrary reference, consider whether or not our scientic knowledge would Anode + Cu(s) Cathode be the same if we chose dierent () K RHE (+) references? NO 3 2+ Cu (aq) H (g) (100 kPa) Pt 3 + Can you think of other examples of 2 arbitrary references in chemistry, which we met already in an earlier 1 mol dm H (aq) topic? See Example 2 on page 419. + 3 ) H (1 mol dm 2+ Figure 3 A voltaic cell with a Cu (aq)|Cu(s) half-cell connected to the SHE 417
19 REDOX PROCE SSE S (AHL) Usefu anaogy The SHE has been described as “ an electrochemical sea-level”. some caution in science when using dierent terminology. You may like to think of the arbitrary reference of 0 V for the For example, when we use the term mean sea level we are not SHE as being analogous to arbitrarily assigning zero elevation suggesting that such a value is accurate for the entire planet, for sea level and then referring all other elevations as either x as seas are constantly moving and tidal movements, pressure, metres above or below sea level. For example, it is often stated etc. can have an impact. For this reason, geographers choose that 50% of the land mass in the Netherlands lies just 1 m a particular location and use this as a point where specic above sea level, or that 50% of the land mass of Bangladesh measurements can be made. Satellite technology has also lies 12.5 m below sea level. Of course, we need to exercise contributed to improvements in the availability of reliable data. Ce poena and Gbbs free energy Sudy p If the standard cell potential, , is positive, a redox reaction will be From physics, energy = cell potential × charge. The SI unit of potential (voltage) is the volt spontaneous. If the standard cell potential, E , is negative, the redox and the SI unit of charge is the cell coulomb. Since the joule is the SI unit of energy: reaction is non-spontaneous. In topic 15 we saw that Δ G , the standard 1J=1V×1C change in Gibbs free energy, is negative for a spontaneous reaction and positive for a non-spontaneous reaction. Δ G is related to E by the cell following expression: ΔG = –nFE cell where: n = amount, in mol, of electrons transferred in the balanced equation; 1 F = Faraday’s constant = 96500 C mol (given in section 2 of the Data booklet); E = standard cell potential (calculated from E =E E ); cell cell lhe rhe Faraday’s constant (F) is the charge in coulombs of 1 mol of electrons. Hence in the Daniell cell: 2+ 2+ Cu (aq) + Zn(s) → Zn (aq) + Cu(s) 1 ΔG = -nFE = -(2 mol e )(96500 C mol e)(+1.10 V) cell 5 = -212300 VC = -2.12 × 10 J. Since ΔG is negative, this reaction is spontaneous under standard conditions. Worked examples a) Deduce the species which is the strongest Example 1 oxidizing agent. Consider the following table of standard electrode 3+ potentials. b) Deduce the species which can reduce Cr (aq) to Cr(s) under standard conditions. E /V 4+ 1.66 0.74 c) Deduce the species which can reduce Sn (aq) 0.28 2+ 3+ +0.15 3+ +1.36 to Sn (aq) but not Cr (aq) to Cr(s) under Al (aq) + 3e ⇋ Al(s) standard conditions. 3+ Cr (aq) + 3e ⇋ Cr(s) d) The standard electrode potential for the 2+ Co (aq) + 2e ⇋ Co(s) half-cell made from cobalt metal, Co(s), in a 2+ solution of cobalt(II) ions, Co (aq) has the 4+ 2+ Sn (aq) + 2e ⇋ Sn (aq) value of 0.28 V. Explain the signicance of 1 Cl (g) + e ⇋ Cl (aq) the negative sign in 0.28 V. 2 2 418
19 . 1 E l E C t R O C h E m i C A l C E l l S Solution a) The higher the standard electrode potential, the greater the ability of the species to gain H (g) (at 100 kPa) 2 electrons, so the strongest oxidizing agent is Cl (g) with E = +1.36 V 2 3+ b) To reduce Cr (aq) to Cr(s), we need a species with an E of less than 0.74 V. The only species with such potential (E = 1.66 V) involves Al(s) as the reducing species. 4+ Pt electrode c) The species that can reduce Sn (aq) to 2+ 3+ + 3 Sn (aq) but not Cr (aq) to Cr(s) must have H (aq), 1 mol dm E lower than +0.15 V but greater than 0.74 V. The only such species in the table is Co(s) with E = -0.28 V. The temperature is often quoted as 298 K d) When a cell with a negative E is connected to o (that is 25 C) possibly due to the fact that the the SHE (E = 0.00 V), the SHE will act as the absolute electrode potential of the hydrogen cathode (positive electrode), and reduction will electrode under standard conditions has the 2+ take place here, whereas the Co(s)|Co (aq) estimated value: E = (4.44 ± 0.02) V at abs half-cell will act as the anode (negative 298 K. However, under such standard-state electrode), where oxidation takes place. This conditions the potential for the reduction of means that, at the anode, there will be a loss of + H (aq) to H (g), E , is taken to be 0 V at all 2 electrons and hence electrons will ow from the temperatures. 2+ Co(s)|Co (aq) half-cell, the anode, to the SHE, b) the cathode. Example 2 Note: the platinum electrode, Pt(s), a is actually platinized platinum, that is the platinum The standard hydrogen electrode (SHE) is an metallic surface is coated with nely divided example of an arbitrary reference. platinum, thereby increasing its surface area. The functions of the Pt(s) electrode are: a) Describe the SHE, using an annotated diagram. ● Platinum is an inert metal and does not corrode b) Describe the functions of the platinum or ionize. It will not act as an electrode in the electrode in the SHE. system. c) State one other example of an arbitrary ● Platinum can act as a heterogeneous catalyst. It reference in chemistry. provides a surface to allow the dissociation of Solution the molecules of hydrogen. Hydrogen absorbs a) on its surface (if the SHE acts as the anode): H (g) → + + 2e 2 2H (aq) ● The SHE consists of a platinum electrode, Pt(s), with hydrogen gas, H (g), at a pressure The platinum provides the surface where 2 transfer of electrons can occur. 3 + of 100 kPa bubbled into a 1 mol dm H solution (eg HCl). The conditions involved are standard-state conditions. Hydrogen Note that the reverse reaction would be the case if the SHE was acting as the cathode. is bubbled through the tube and into the ● An equilibrium is established between the solution, where the following reaction takes place: adsorbed molecules of hydrogen, H (g), on the 2 + Pt surface and the hydrogen ions, H (aq): + ⇋ H (g) 2H (aq) + 2e 2 + ⇋ H (g) 2H (aq) + 2e 2 419
19 REDOX PROCE SSE S (AHL) ● Platinum acts as an electrical conductor to the d) Calculate the standard potential, in V, for this cell. external circuit. e) (i) Determine ΔG , the standard change in c) Gibbs free energy at 298 K, in kJ, for the electrochemical reaction. ● Another example of an arbitrary reference in chemistry that we have already encountered (ii) Comment on the spontaneity of the reaction. in thermodynamics, covered in topic 5 on energetics and thermchemistry, is the standard Solution enthalpy change of formation of a substance, a) o ● Using the information given for the ΔH (substance). This is described relative to 2+ 4+ f Sn (aq)| Sn (aq) electrode and section 24 o the arbitrary reference of ΔH (element) of the f component elements of the substance, each of the Data booklet: 1 of which is taken as 0 kJ mol . Recall that the 4+ 2+ standard enthalpy change of formation of a Sn (aq) + 2e ⇋ Sn (aq) E = +0.15 V substance is dened as the enthalpy change when 3+ 2+ Fe (aq) + e ⇋ Fe (aq) E = +0.77 V 1 mol of a compound is formed from its elements 2+ 3+ ● Since E for the Fe (aq)|Fe (aq) half-cell is in their standard states (that is at 100 kPa). The o more positive, this is deemed the RHE, that is standard enthalpy of formation, ΔH could f the cathode where reduction takes place. Since be described as a “thermodynamic sea-level” 2+ 4+ E for the Sn (aq)| Sn (aq) half-cell is less reference just like the SHE could be described as positive, this is the LHE, that is the anode where an “electrochemical sea-level” reference. oxidation takes place. Hence the two half- Note: standard state itself is also an example of equations taking place at the two electrodes are an arbitrary reference. Recall that standard state is as follows: the most stable state of a substance under standard conditions. It is the state of a system chosen as Cathode (positive electrode): Reduction takes place here. standard for reference by convention. As dened by IUPAC, for a gas phase it is the hypothetical 3+ 2+ Fe (aq) + e → Fe (aq) E = +0.77 V rhe state of the pure substance in that state at 100 kPa. For a pure phase, or a mixture, or a solvent in the Anode (negative electrode): Oxidation takes place here. liquid or solid state, it is the state of the pure substance in the liquid or solid phase at 100 kPa. 2+ 4+ Sn (aq) → Sn (aq) + 2e E = +0.15 V lhe For a solute in solution, it is the hypothetical state 3 The electrons are not balanced, so the cathode half-equation must be multiplied by two to of the solute at 1 mol dm concentration at generate the overall reaction: 100kPa and showing innitely dilute solution behaviour. Cathode (positive electrode): Carbon 12, as discussed in topic 1, was chosen as 3+ 2+ the arbitrary reference standard for the SI unit of amount of substance (the mole). 2Fe (aq) + 2e → 2Fe (aq) E = +0.77 V rhe b) Oxidizing agent: Example 3 3+ Fe (aq). Consider the following electrochemical reaction, Reducing agent: which takes place in a voltaic cell at 298 K: 2+ Sn (aq). 2+ 3+ 4+ 2+ Sn (aq) + 2Fe (aq) → Sn (aq) + 2Fe (aq) 4+ 2+ c) Given: Sn (aq) + 2e ⇋ Sn (aq) E = +0.15 V Since all reacting species are present in the aqueous a) Identify the half-equations occurring at the phases, inert electrodes, Pt(s), must be used: cathode and anode electrodes. 2+ 4+ 3+ 2+ Pt(s)|Sn (aq), Sn (aq) || Fe (aq), Fe (aq)|Pt(s) b) Identify the oxidizing and reducing agents. LHE RHE anode cathode c) State the cell diagram convention for the cell. 420
19 . 1 E l E C t R O C h E m i C A l C E l l S Question Sudy p Construct and annotate an example of a voltaic Note that the standard cell potential E values are no 2+ cell consisting of a Ni(s)|Ni (aq) half-cell and a aected by coecients and you should never multiply 2+ Cu(s)|Cu (aq) half-cell. the E values by the integer mole ratios. The reason for a) Identify the half-equations occurring at this is that E is an example of an nensve proper y, the cathode and anode electrodes. that is it is independent of quantity of sample (other examples of intensive proper ties include density, b) Deduce the equation for the spontaneous temperature, and melting point). This diers from reaction occurring in this cell. ex ensve proper es (such as volume and mass) c) State the cell diagram convention for the cell. which depend on amount of substance. In contrast, the standard Gibbs free energy change, ΔG , is an d) Identify the direction of the movement of example of an extensive proper ty and if an equation electrons and ion ow, both in solution and in is multiplied by a factor, n will change and hence also the salt bridge. ΔG , from the expression ΔG = -nFE , as seen in cell e) Calculate the standard potential, in V, for this cell. (e) in Example 3. f) Determine ΔG , the standard change in Gibbs If we return to the analogy of the waterfall conveying free energy, in J, for the electrochemical reaction. the idea of PE dierence, it does not matter whether 3 3 15000 dm of water or 30000 dm of water falls from the top to the bottom of the waterfall – as long as Note that the spontaneity of the reaction can also be there exists a pathway for the water to gush down the deduced from the sign of E . A positive E is indicative cell cell waterfall, the dierence in height between the top and of a spontaneous redox reaction whereas a negative E cell bottom will stay constant. Hence, if an equation is is indicative of a non-spontaneous redox reaction. multiplied by a factor, the stoichiometry coecients will change and hence the number of electrons will change but no the potential dierence through which electron transfer occurs. Sudy p Relationships between ΔG and E (ΔG = –nFE ) cell cell d) ΔG E Reacon under sandard- sae condons cell E =E E = (+0.77 V) – (+0.15 V) cell lhe rhe = +0.62 V negative positive spontaneous, so will favour formation of products e) 1 (i) ΔG = –nFE = –(2 mol e )(96500 C mol e) cell positive negative non-spontaneous, so will (+0.62 V) = –119660 V C favour formation of reactants 5 2 = –1.2 × 10 J = –1.2 × 10 kJ. zero zero both products and reactants will be favoured equally (ii) Since ΔG is negative this reaction is spontaneous under standard conditions. Sudy p Note that temperature is no a formal requirement in You should remember sandard-sae condons which o can be summarized simply as follows: the description of standard state, but 298 K (25 C) is (s), (l), (g) as pure substances at a pressure of 100 kPa often quoted in thermodynamic tables as the specied temperature! Solutes at 1 mol dm 3 concentration 421
19 REDOX PROCE SSE S (AHL) Eecroyc ces In topic 9 we also examined a second type of electrochemical cell, the electrolytic cell, which converts electrical to chemical energy. In topic 9, we looked at one type of electrolytic cell, the electrolysis of a molten salt. We shall now examine another type of electrolysis, namely the electrolysis of aqueous solutions. We will consider the following examples of electrolysis of aqueous solutions: a) Electrolysis of aqueous sodium chloride Sudy p (i) Concentrated solution Remember the mnemonic CNAP for electrolytic cells: (ii) Dilute solution Cathode – Negative b) Electrolysis of aqueous copper(II) sulfate Anode – Positive (i) Using inert graphite (carbon) electrodes This diers from voltaic cells, where the cathode is (ii) Using active copper electrodes the positive electrode and the anode is the negative c) Electrolysis of water electrode. However, for both electrochemical cells, (a) Electrolysis of concentrated aqueous sodium chloride reduction takes place at the cathode and oxidation takes Unlike the electrolysis of molten sodium chloride discussed in topic 9, in place always at the anode. the electrolysis of concentrated aqueous sodium chloride there is an additional species to be considered, namely water! Let us consider the species present at each electrode: Cathode (negative electrode): + H O(l) Na (aq), 2 Anode (positive electrode): Cl (aq), H O(l) 2 ● In order to determine the most relevant half-equation corresponding to each electrode process, you should rst write down the two half- equations taking place at each electrode and the corresponding values using section 24 of the Data booklet. (Remember that in the Data booklet the standard electrode potential values relate to reductive processes. Hence, since reduction takes place at the cathode, the sign of E taken from section 24 in the Data booklet will be correct for the reductive process. However, when you write any half-equation for an oxidation reaction taking place at the anode, the sign of E will have to be switched if using the section 24 Data booklet values .) ● In addition, if you examine section 24, you will pick out two different equations involving water as a single species on either side of the equilibrium sign (you can ignore any other equations involving water where it is not written on its own on either side of the equilibrium sign ): H O(l) + e ⇋ _1 (g) + OH (aq) E = –0.83 V 2 H 2 2 Water on its own as a single species on one side of the equilibrium sign _1 (g) + + + 2e ⇋ H O(l) E = +1.23 V O 2H (aq) 2 2 2 422
19 . 1 E l E C t R O C h E m i C A l C E l l S Let us look rst at the possible half-equations that take place at the cathode, as there will be no change in the sign of E here, since it will be a reductive process. Cathode (negative electrode): + → Na(s) E = –2.71 V Sudy tp Na (aq) + e Always try to use your H O(l) + – _1 (g) + OH (aq) E = –0.83 V chemical intuition in working 2 e→ H out chemical reactions, products and processes. It 2 would be very unlikely here to have sodium metal forming, 2 since sodium reacts vigorously with water generating The half-equation for water is chosen as written, since H O(l) will be hydrogen gas and sodium 2 hydroxide solution! Thinking as a real chemist makes the species present at the cathode (not O (g), as seen in the other half- IB chemistry much more 2 accessible and fun as you discover the power of chemical equation showing the reduction process). prediction! Anode (positive electrode): 2Cl (aq) → Cl (g) + 2e E = –1.36 V 2 H O(l) → _1 (g) + + + 2e E = –1.23 V 2 O 2H (aq) 2 2 At the anode, oxidation takes place, so any sign of E for a reductive process taken from section 24 of the Data booklet will have to be inverted (as will the half-equation) to indicate the oxidative process. This is the reason why the signs have been changed above. In addition, since we have used the rst of the two half-equations for water given in the table to describe the cathode half-equation, we now use the second one here to describe a possible anode half-equation. We now have to decide which of these two half-equations is preferred . If you look at the two E values, In general, the more positive E value would indicate the preferred reaction, the one with the more positive suggesting that oxidation of water would be preferential. HOWEVER, in the case E will correspond to the of concentrated aqueous sodium chloride this is not as simple as the rule favoured reduction. Hence, the suggests. The reason for this is the phenomenon termed overvoltage. cathode half-equation will be: In an electrolytic experiment the applied potential needed to carry out – 1 H O(l) + e → H (g) + OH (aq) the electrolysis is always greater than the potential calculated from the 2 2 2 standard redox potentials. This extra difference in potential or voltage is E = -0.83 V the overvoltage. Many reactions taking place at electrodes are extremely slow. Therefore, this additional voltage effectively is the extra voltage required for a reaction with a slow rate to proceed at a reasonable rate in an electrochemical cell. As a result the oxidation reaction taking place at the anode would actually require a potential greater than 1.23 V to occur (the overvoltage for oxygen gas formation is quite high compared to chlorine gas formation), suggesting that, at the anode in a concentrated solution of sodium chloride, the chloride ions, Cl (aq), are actually reduced to chlorine gas, Cl (g), which is what is observed experimentally. 2 Note that this can only be conrmed from experimental evidence. In the case of a dilute aqueous solution of sodium chloride, overvoltage does not play such a role and the half-equation taking place at the anode can simply be worked out using E values. The half-equation taking place at the anode in the electrolysis of concentrated aqueous sodium chloride will be: 2Cl (aq) → Cl (g) + 2e E = –1.36 V 2 Hence, let’s combine the two electrode half-equations to generate the overall cell reaction: 2H O(l) + 2Cl (aq) → Cl (g) + H (g) + 2OH (aq) E = -2.19 V 2 2 2 cell 423
19 REDOX PROCE SSE S (AHL) Sodium is not involved in the electrode reactions and simply acts as a spectator cation. In essence, as the electrolytic process progresses, the solution of sodium chloride is converted to a solution of sodium hydroxide. The electrolysis of concentrated sodium chloride solution (brine) is a very important industrial process, the basis of the chlor-alkali industry. Three very important industrial products are produced by this process: chlorine gas, hydrogen gas and sodium hydroxide. Uses: ● Uses of corne: Chlorine can be used to make the polymer polyvinylchloride (PVC), which is used in pipes, oor tiles, transparent lm for the packaging of meats and rain jackets (sub-topic A .5). Chlorine can also be used as a bleaching agent (used in the textile and paper industries) and as a disinfectant. It can be used in the purication of water. ● Uses of ydrogen: Hydrogen is a valuable fuel and can be used in the Haber process to produce ammonia: N (g) + 3H (g) ⇋ 2NH (g). The ammonia 2 2 3 produced is impor tant in the manufacturing of fer tilizers such as ammonium nitrate, NH NO (see topic 7). 4 3 ● Uses of sodu ydroxde: Sodium hydroxide is used in the manufacturing of soap and paper. Observations at each electrode Caode (negave eecrode): Bubbles of colourless hydrogen gas are obser ved. You could test the gas by taking a sample in a closed test tube and lighting a match in the gas. The 2H O(l) + 2e → H (g) + 2OH (aq) gas will ignite with a small popping sound heard. The sample will be mainly 2 2 pure hydrogen, whereas typically a much louder pop is obtained if a mixture of hydrogen and air is present, which you might have carried out in the laboratory, in a separate experiment. Anode (posve eecrode): Bubbles of chlorine gas are observed (pale yellow colour may be seen perhaps). 2Cl (aq) → Cl (g) + 2e 2 Pungent odour of chlorine gas can be experienced (similar to odour found in bleach). Note that the chlorine formed at the anode can combine with sodium chloride to form bleach which is sodium hypochlorite, NaOCl. This can be tested by using some moist blue litmus paper which can be eectively bleached. At higher temperatures, sodium chlorate, NaClO , may form instead 3 of NaOCl. Safey noe: Chlorine gas is a toxic gas so, when working with it even in small amounts, a fumehood should be used in the chemical laboratory. Eecroye: pH of the electrolyte will increase due to the formation of OH (aq), producing a more basic solution. This can be observed by testing the solution with 2H O(l) + 2Cl (aq) → indicator paper. 2 Cl (g) + H (g) + 2OH (aq) 2 2 424
19 . 1 E l E C t R O C h E m i C A l C E l l S + iner t Pt electrodes + e anode cathode Cl (aq) H O(1) 2 bubbles of Cl (g) bubbles of H (g) 2 2 NaCl(aq) Figure 4 Electrolysis of concenraed aqueous sodium chloride, NaCl(aq). (b) Electrolysis of aqueous copper(II) sulfate ()Usng ner grape (carbon) eecrodes The formula of aqueous copper(II) sulfate is CuSO (aq). 4 Let us consider the species present at each electrode. Cathode (negative electrode): 2+ Cu (aq), H O(l) 2 Anode (positive electrode): 2 SO (aq), H O(l) 2 4 Let us look next at the possible half-equations that may take place at the cathode, Cathode (negative electrode): Reduction takes place here. 2+ Cu (aq) + 2e → Cu(s) E = +0.34 V H O(l) + e → _1 (g) + OH (aq) E = -0.83 V 2 H 2 2 The reduction with more positive E value will be favoured. Hence, the cathode half-equation will be: 2+ Cu (aq) + 2e– → Cu(s) E = +0.34 V Anode (positive electrode): Oxidation takes place here. Sulfates do not tend to oxidize. In sulfate, the oxidation state of sulfur is +6, corresponding to the stable noble gas core of [Ne] which it will not want to give up. At the anode the following reaction therefore takes place: H O(l) → _1 (g) + + + 2e E = -1.23 V 2 O 2H (aq) 2 2 Hence, let’s combine the two electrode half-equations to generate the overall cell reaction: 425
19 REDOX PROCE SSE S (Ahl) Cathode (negative electrode): 2+ Cu (aq) + → Cu(s) E = +0.34 V Anode (positive electrode): H O(l) → _1 (g) + + + 2e E = 1.23 V 2 O 2H (aq) 2 2 Overall cell reaction: 2+ _1 + O 2H (aq) Cu (aq) + H O(l) → Cu(s) + (g) + E = -0.89 V 2 cell 2 2 Observations at each electrode Caode (negave eecrode): Layer of pink–brown colour of solid copper seen deposited on cathode. 2+ Cu (aq) + 2e → Cu(s) Anode (posve eecrode): Bubbles of colourless oxygen gas observed at anode. _1 + (g) + 2H (aq) + 2e H O(l) → O 2 2 2 Eecroye: pH of the electrolyte will decrease due to an increase in the concentration 2+ _1 + + (g) + 2H (aq) of H (aq). This can be observed by testing with an indicator. Cu (aq) + H O(l) → Cu(s) + O 2 2 2 2+ Mediterranean blue colour of Cu (aq) ions fades in colour due to the 2+ discharge of Cu (aq). + iner t graphite electrodes + e anode cathode 2+ Cu (aq) bubbles of O (g) pink–brown layer 2 of Cu(s) CuSO (aq) 4 Figure 5 Electrolysis of aqueous copper(II) sulfate, CuSO (aq) using ner grape 4 (carbon) electrodes ()Usng acve copper eecrodes Using copper instead of graphite electrodes means that the copper electrodes now participate in the electrolysis process (they are termed active electrodes as opposed to inert electrodes) and the following will be the corresponding half-equations taking place at each electrode: Cathode (negative electrode): 2+ Cu (aq) + 2e → Cu(s) Anode (positive electrode): 2+ Cu(s) → Cu (aq) + 2e 426
19 . 1 E l E C t R O C h E m i C A l C E l l S Observations at each electrode Caode (negave eecrode): Layer of pink–brown colour of solid copper deposited on cathode (this 2+ copper will be pure). Mass of cathode increases. Cu (aq) + 2e → Cu(s) Copper anode seen to disintegrate Anode (posve eecrode): since the mass of the anode decreases. 2+ At the bottom a sludge of impurities is Cu(s) → Cu (aq) + 2e seen to form. Eecroye: Mediterranean blue colour of solution does not change, since the 2+ concentration of Cu (aq) ions remains constant . One use of this type of electrolysis is in the electrorening of copper, as the purication of copper takes place. In electrical wires the purity of copper needs to be very high. If impure copper wiring is used, the electrical resistance increases. In this electrolysis the anode consists of impure copper metal and the cathode consists of pure copper metal. The impure copper at the anode is converted into pure copper at the cathode and the residue of impurities (typically platinum, gold and silver) forms a sludge below the anode as seen in Figure 6 below. More easily oxidized impurities, such as iron and 2+ 2+ zinc, remain in solution as Fe and Zn species. + e active copper electrodes cathode ( ) anode (+) 2+ Quck queson Suggest why the sludge might Cu (aq) be reprocessed. impure copper pink–brown layer 427 of pure Cu(s) sludge of impurities CuSO (aq) at anode 4 Figure 6 Electrolysis of aqueous copper(II) sulfate, CuSO (aq) using acve 4 copper electrodes
19 REDOX PROCE SSE S (Ahl) This idea of using active electrodes is also the basis of the electrochemical process of electroplating, which involves using electrolysis to deposit a 3 4 thin layer (typically 10 to 10 mm thick) of one metal onto the cathode of another. This is usually done either to prevent corrosion or for decorative purposes, thereby enhancing the appearance of a particular object. Electroplating is widely used in jewellery and in the plating of steel bumpers in cars with chromium. Let’s consider the electroplating of an object with silver using a solution of Na[Ag(CN) ] as the electrolyte. Jewellery is commonly electroplated 2 with gold whereas cutlery is typically electroplated with silver (hence the name “silver service” in the restaurant business). You might be inclined to think that silver nitrate, AgNO , would be an appropriate 3 solution to use for this purpose, but it has been found that the rate at which the silver deposits is too quick and hence has been found not to adhere effectively to the object being plated. For this reason, a more appropriate solution involves the complex, sodium dicyanoargentate(I), Na[Ag(CN) ]. The anode consists of a bar of silver which disintegrates. 2 The cathode is the metal object to be plated (for example, a spoon). The following are the half-equations corresponding to the cathode and anode processes: Anode (positive electrode): Ag(s) + 2CN (aq) → [Ag(CN) ] (aq) + e 2 Cathode (negative electrode): [Ag(CN) ] (aq) + e → Ag(s) + 2CN (aq) 2 + e impure silver cathode ( ) anode (+) [Ag(CN) ] (aq) 2 spoon to be plated Figure 7 Electroplating of a spoon with silver (c) Electrolysis of water Pure water is a poor conductor of electricity, but when even a tiny amount of ions are present the electrical conductivity of water increases. The electrolysis of water can be carried out using a dilute solution of sulfuric acid, H SO (aq) or a dilute solution of sodium hydroxide, 2 4 NaOH(aq), using inert Pt electrodes. 428
19 . 1 E l E C t R O C h E m i C A l C E l l S Let us consider electrolysis of water using dilute sulfuric acid: Consider the species present at each electrode. Cathode (negative electrode): + H (aq) Anode (positive electrode): 2 Quck queson SO (aq), H O(l) 2 4 What is the purpose of the dilute Cathode (negative electrode): H SO (aq) (eg 0.1 mol dm 3 )? 2 4 + → _1 (g) E = 0.00 V H (aq) + e H 2 2 Anode (positive electrode): As stated previously sulfates do not tend to oxidize, so at the anode the following reaction therefore takes place: H O(l) → _1 (g) + + E = -1.23 V 2 O 2H (aq) + 2e 2 2 Hence, let’s combine the two electrode half-equations to generate the overall cell reaction: Cathode (negative electrode): + + e → _1 (g) E = 0.00 V H (aq) H 2 2 This needs to be multiplied by two to balance the number of electrons. + + 2e → H (g) E = 0.00 V 2H (aq) 2 Anode (positive electrode): H O(l) → _1 (g) + + + 2e E = -1.23 V 2 O 2H (aq) 2 2 Overall cell reaction: H O(l) → _1 (g) + H (g) E = -1.23 V 2 O 2 2 2 Observations at each electrode Caode (negave eecrode): Bubbles of colourless hydrogen gas observed. + → H (g) pH at cathode increases with the discharge 2H (aq) + 2e 2 + of H (aq). Figure 8 Electrolysis of water using a Homan apparatus. Reactions at the two electrodes Anode (posve eecrode): Bubbles of colourless oxygen gas (black hooks dipped in the beaker of water) obser ved. are powered by the electric current from the _1 + battery (lower left). Oxygen and hydrogen (g) + 2H (aq) + 2e + gas bubbles are evolved at the anode (left H O(l) → O pH at anode decreases because H (aq) electrode) and cathode (right electrode) respectively. As water molecules consist of 2 2 ions are produced two hydrogen atoms and one oxygen atom, twice as much hydrogen as oxygen is trapped 2 in the test tubes (upper right). Use of a burning splint will ignite the hydrogen gas, Eecroye: The ratio of the two gases by volume is: while the oxygen will relight a glowing splint. _1 _1 H O(l) → H (g) + O (g) 1H (g) : O (g) or 2 2 2 2 2 2 2 2H (g) : O (g) 2 2 429
19 REDOX PROCE SSE S (AHL) Hydrogen cells and research The electrolysis of water is an important source of dioxide, CO (g) are not produced,) there are other hydrogen gas (sub-topic C.6). The term “hydrogen 2 economy” was coined by Professor John Bokris, an electrochemist born in South Africa during a considerations which need to be considered such as presentation given to General Motors in 1970, the safe storage of the hydrogen fuel used, which is who proposed generating energy using hydrogen. a highly ammable substance. Other fuel cells such Many countries are trying to move away from as the DMFC, the direct methanol fuel cell, do the generation of energy using fossil fuels. There generate carbon dioxide. These fuel cells do have are extensive energy demands associated with some advantages over conventional batteries. They the production of hydrogen. The availability of are lighter and hence are often used in smart phone hydrogen is limited for this important use in fuel technologies. Market cost, environmental issues, cells. Remember that hydrogen is an energy carrier and ethical aspects play a role in the development and is not a resource per se. of any new technology. If liquid methanol is used in fuel cells it can be produced from biomass as a carbon neutral fuel (one which does not contribute to the greenhouse effect). The most common type of fuel cell, the Many electrochemical cells can act as energy hydrogen–oxygen fuel cell involves the sources alleviating the world’s energy problems reaction between hydrogen and oxygen to yield but some cells such as super-efcient microbial water as the product: fuel cells (MFCs also termed biological fuel cells sub-topic C.6) can also reduce the environmental H (g) + _1 (g) → H O(l) impact of human activities. 2 O 2 2 2 In order to produce electricity in the fuel cell, ● How do national governments and the hydrogen is required. Hydrogen can be produced international community decide on research by the electrolysis of water in a solar-powered priorities for funding purposes ? Although the electrolytic cell. In the development of this intended outcomes may be clear to a large technology extensive collaborative research extent in the development of strategic and between scientists and engineers from different applied research, do governments and funding elds (chemists, biologists, material scientists, etc.) agencies ignore the ongoing development of is necessary. In the design of such technologies basic research at their peril? Can you think of scientists often have to consider environmental, any examples from electrochemistry (or the socio-economic, safety and ethical aspects of broader elds of physical chemistry, inorganic energy production. Although vehicles powered by chemistry or materials science) where this hydrogen–oxygen fuel cells are environmentally might be the case? preferred (greenhouse gases such as carbon Worked example Construct and annotate the electrolytic cell for the Solution electrolysis of dilute sodium chloride. a) First, consider the ions present, noting that a) Identify the half-equations occurring at this is a dilute solution of sodium chloride the cathode and anode electrodes and the (not concentrated). The ions present are equation for the overall cell reaction. generated from: b) State a suitable material that can be used for + NaCl(aq) → Na (aq) + Cl (aq) and each electrode. H O(l) ⇋ + + OH (aq) 2 H (aq) c) Identify the direction of the movement of electrons and ion ow. Water slightly dissociates into hydrogen and hydroxide ions. d) State what would be observed at each electrode. 430
19 . 1 E l E C t R O C h E m i C A l C E l l S Cathode (negative electrode): + Ion ow: H (aq) ow from anode to cathode (in + H O(l), + solution) Na (aq), 2 H (aq) + Anode (positive electrode): Cl (aq), H O(l) 2 Possible processes at the cathode e (negative electrode): Reduction takes cathode place here iner t Pt electrodes + → Na(s) E = 2.71 V Na (aq) + e 1 anode + H O(l) + e → _1 (g) + OH (aq) E = -0.83 V 2 H 2 2 2 OH (aq) + → H (g) E = 0.00 V 2H (aq) + 2e 2 3 + H (aq) The third process takes place since E is the bubbles of O (g) bubbles of H (g) 3 2 2 most positive. Possible processes at the anode NaCl(aq) (positive electrode): Oxidation takes Figure 9 Electrolysis of due aqueous sodium chloride, place here NaCl(aq) 2Cl (aq) → Cl (g) + 2e E = -1.36 V 2 1 d) Observation at cathode: bubbles of H O(l) → _1 (g) + + + 2e E = -1.23 V colourless gas (hydrogen). 2 O 2H (aq) 2 2 2 Since E > E , the second process takes place. Observation at anode: bubbles of colourless 2 1 gas (oxygen), two times fewer bubbles than those of hydrogen. Note: There is no overvoltage as the solution is dilute which differs from the electrolysis of Queson concentrated sodium chloride where chlorine gas is evolved at the anode. Consider the electrolysis of water in dilute sodium hydroxide. Overall cell reaction: _1 a) Identify the half-equations occurring at O H O(l) → H (g) + (g) 2 2 2 2 the cathode and anode electrodes and the equation for the overall cell reaction. b) Material for electrodes: inert metal, such as Pt(s), or graphite, C(s). b) (i) State what would be observed at each electrode. c) Electrolytic cell: Electron ow: from anode to cathode (through (ii) Deduce the ratio by volumes of any external circuit). species produced. Quanave aspecs of eecroyss The following factors affect the amount of product formed at the electrodes during electrolysis: 1 Current (I) 2 Duration of electrolysis ( t) 3 Charge on the ion (z) 431
19 REDOX PROCE SSE S (AHL) Let us consider each of these separately: 1 mol of e carries an 1 Current (I) approximate charge of 96500 C. This known as Faraday’s From physics, the charge, Q (in C) is related to the current, I (in A), consan and is given in and the time, t (in s), as follows: section 2 of the Data booklet Q = It 1 1 F = 96500 C mol Since I α Q, this means that I will also be proportional to the number of electrons passing through the external circuit. Faraday’s rs aw of eecroyss states that the 3+ mass of an element deposited during electrolysis is directly eg Al (aq) + 3e → Al(s) propor tional to the quantity of electricity (that is the charge, 3 mol of e 1 mol of Al(s) Q) passing through during the electrolysis. To generate 1 mol of Al(s), 3 mol of e must pass through the external circuit. If I is doubled at time t, Q will be doubled and hence the number of electrons will be doubled. This then means that the amount, in mol, of Al(s) will also increase. 2 Duration of electrolysis ( t) From the equation above t α Q. This means that t will also be directly proportional to the number of electrons passing through the external circuit. 3 Charge on the ion ( z) The amount, in mol, of e needed to discharge 1 mol of an ion at an electrode is equal to the charge on the ion, z. This is termed Faraday’s second law Worked examples _1 Example 1 1 mol e ≡ mol Cu Calculate the mass, in g, of copper produced at 2 the cathode when a current of 1.50 A is passed through a solution of aqueous copper(II) sulfate 0.182 mol e ≡ (0.182 × 0.5) mol for 3.25 hours. Cu = 0.0910 mol Cu ● n = m/M, so m = n × M m = (0.0910 mol)(63.55 g mol 1 ) = 5.78 g Solution ● Q = It Example 2 Q = 1.50 × 3.25 × 60.0 × 60.00 = 17550 C Two electrolytic cells are connected in series, so that the same current passes through each 1 individual cell. The rst cell, cell A, contains silver electrodes in a silver nitrate solution. The ● 1 F = 96500 C mol second cell, cell B, contains copper electrodes in copper(II) sulfate solution. In an experiment, So 96500 C equates to 1 mol of e 0.658 g of silver was found to deposit in cell A. Calculate the mass, in g, of copper deposited in 1 C equates to (1/96500) mol of e cell B, showing your working. 17550 C equates to (17550/96500) mol of e = 0.182 mol of e (correct to three signicant gures) ● At the cathode the following reduction half- reaction takes place: Electrolytic cells connected n seres means that they are connected one after another in a circuit so that the 2+ same current passes through each one. Cu (aq) + 2e → Cu(s) 2 mol e ≡ 1 mol Cu 432
19 . 1 E l E C t R O C h E m i C A l C E l l S Solution Solution ● Q = It ● In cell A: + Q = (2.55) × (3.00 × 60.0 × 60.0) = 27540 C Ag (aq) + e → Ag(s) 1 ● 1 F = 96500 C mol 1 mol e ≡ 1 mol Ag(s) So 96 500 C equates to 1 mol of e 107.87 g Ag(s) ≡ 1 mol e 1 C equates to ___1 __ mol of e _0.658 () () 96500 0.658 g Ag(s) ≡ mol e 107.87 27540 C equates to _2_7_54_0_ mol of e = 0.285 () 96500 3 = 6.10 × 10 mol e mol of e (correct to three signicant gures) ● In cell B: ● At the anode the following oxidation half- 2+ Cu (aq) + 2e → Cu(s) reaction takes place: 6.10 × 10 3 ≡ (0.5 × 6.10 × 10 3 2Cl (aq) → Cl (g) + 2e mol e ) mol 2 Cu(s) 0.285 mol e ≡ (0.285 × 0.5) mol ≡ (0.5 × 6.10 × 10 3 = 0.143 mol Cl (g) mol) 2 (63.55 g mol 1 ) Cu(s) ● Using section 2 of the Data booklet: = 0.194 g Molar volume of an ideal gas at 100 kPa and Example 3 3 1 273 K = 22.7 dm mol A current of 2.55 A is passed through a 1 mol ≡ 22.7 3 Cl (g) dm 2 concentrated aqueous solution of sodium 3 chloride for 3.00 h. Calculate the volume, 3 0.143 mol ≡ 22.7 × 0.143 dm volume in dm , of chlorine gas produced at the of Cl (g) = 3.25 3 2 dm anode at a pressure of 100 kPa and a temperature of 273 K. 433
19 REDOX PROCE SSE S (AHL) Quesons 2+ 3+ 1 What conditions are necessary to D. 2Al(s) + 3Ni (aq) → 2Al (aq) + 3Ni(s) directly measure a standard electrode E = 1.43 V [1] potential (E )? IB, May 2011 I. A half-cell with an electrode in a 1.0 mol 3 dm solution of its ions. 4 How do the products compare at each electrode II. Connection to a standard hydrogen 3 when aqueous 1 mol dm magnesium electrode. bromide and molten magnesium bromide are III. A voltmeter between half-cells to measure electrolysed? potential difference. A. I and II only E /V –2.37 2+ +1.07 +1.23 B. I and III only Mg (aq) + 2e ⇋ Mg(s) C. II and III only _1 (l) + e ⇋ Br (aq) Br 2 2 D. I, II and III [1] _1 (g) + + + 2e ⇋ H O(l) O 2H (aq) 2 2 IB, May 2010 2 2 Consider the following standard electrode Posve eecrode Negave eecrode (anode) (caode) potentials. A. same same B. same dierent 2+ C. dierent same D. dierent dierent Zn (aq) + 2e ⇋ Zn(s) E = – 0.76 V Cl (g) + 2e ⇋ 2Cl (aq) E = +1.36 V 2 2+ Mg (aq) + 2e ⇋ Mg(s) E = – 2.37 V What will happen when zinc powder is IB, November 2009 [1] added to an aqueous solution of magnesium chloride? 5 What condition is necessary for the electroplating A. No reaction will take place. of silver, Ag, onto a steel spoon? B. Chlorine gas will be produced. A. The spoon must be the positive electrode. C. Magnesium metal will form. B. The silver electrode must be the negative D. Zinc chloride will form. [1] electrode. IB, May 2010 C. The spoon must be the negative electrode. D. The electrolyte must be acidied. [1] 3 The standard electrode potentials for two metals IB, May 2010 are given below. 3+ Al (aq) + 3e ⇋ Al(s) E = –1.66 V 6 The same quantity of electricity was passed 2+ Ni (aq) + 2e ⇋ Ni(s) E = –0.23 V through separate molten samples of sodium bromide, NaBr, and magnesium chloride, What is the equation and cell potential for the MgCl . Which statement is true about the spontaneous reaction that occurs? 2 amounts, in mol, that are formed? 3+ 2+ A. 2Al (aq) + 3Ni(s) → 2Al(s) + 3Ni (aq) A. The amount of Mg formed is equal to the E = 1.89 V amount of Na formed. 2+ 3+ B. 2Al(s) + 3Ni (aq) → 2Al (aq) + 3Ni(s) B. The amount of Mg formed is equal to the E = 1.89 V amount of Cl formed. 2 3+ 2+ C. 2Al (aq) + 3Ni(s) → 2Al(s) + 3Ni (aq) C. The amount of Mg formed is twice the E = 1.43 V amount of Cl formed. 2 434
QUE StiONS D. The amount of Mg formed is twice the (i) Identify the half-equations occurring at the cathode and anode electrodes and the amount of Na formed. [1] equation for the overall cell reaction. IB, May 2011 (ii) State a suitable material that can be used for each electrode. 7 What is the mass, in g, of copper produced at the cathode when a current of 1.00 A is passed (iii) Identify the direction of the movement of electrons and ion ow. through a solution of aqueous copper(II) sulfate for 60 minutes? (iv) State what would be observed at each 1 (1 F = 96500 C mol ) electrode. A. 63.55 × 3600 __ 2 × 96500 B. _63.55 ×_3600 10 A current of 2.35 A is passed through an 96500 electrolytic cell for the electrolysis of water, using a dilute sulfuric acid solution, for a duration of 5.00 h. C. _63.55_× 60 2 × 96500 a) Construct and annotate a suitable electrolytic cell for this experiment. D. _63.55 ×60 1 b) Identify the half-equations occurring at the cathode and anode electrodes and the 8 a) Construct and annotate an example of a equation for the overall cell reaction. + voltaic cell consisting of a Ag(s) |Ag (aq) 2+ c) dentify the direction of the movement of half-cell and a Co(s)|Co (aq) half-cell. electrons and ion ow. 2+ Given: Co (aq) + 2e ⇋ Co(s) E = -0.28 V d) 3 Determine the volume, in cm , of the two i) Identify the half-equations occurring at the cathode and anode electrodes. gases generated in the process at SATP conditions, using information from section 2 of the Data booklet ii) Deduce the equation for the spontaneous reaction occurring in this cell. iii) State the cell diagram convention for the cell. 11 Electroplating is an important application of electrolytic cells with commercial implications. iv) Identify the direction of the movement of Copper may be plated using an electrolytic cell with an aqueous acidied copper(II) sulfate electrons and ion ow, both in solution electrolyte. For the copper plating of tin to make jewellery, state the half-equation at each and in the salt bridge. electrode. v) Calculate the standard potential, in V, for this cell. Assume the tin electrode is inert. Suggest two vi) Determine ΔG , the standard change observations that you would be able to make as in Gibbs free energy, in J, for the the electroplating progresses. [4] electrochemical reaction. IB, May 2010 9 a) Construct and annotate the electrolytic cell for the electrolysis of concentrated potassium iodide. 435
19 REDOX PROCE SSE S (Ahl) 12 Two electrolytic ce l l s a r e co nne c te d i n s e r i e s + as shown in the d i a g r a m b e low. In o n e th e r e is molten magnesium chloride and in the other, dil ute sodium hy d r o x id e s o lut io n . Both cells have inert electrodes. If 12.16 g of magnesium is produced in the rst cell, deduce the identity and mass of products crucible produced at the positive and negative MgCl (l) 2 electrodes in the second cell. [4] NaOH(aq) IB, May 2010 heat Figure 9 436
20 O R G A N I C C H E M I S T R Y ( A H L ) Introduction Synthesis and associated reaction mechanisms lie are examined. Stereoisomerism concludes the at the core of organic chemistry. An understanding topic with a broad examination of a range of of the properties of organic compounds, their different types of isomers, their nature, properties, reactions and the mechanisms by which they react nomenclature and importance to the eld of is fundamental to the advancement of research chemistry. In both topics 10 and 20, students will in such important elds of science as medicine, learn how to represent organic compounds as both biotechnology, food production and the energy two- and three-dimensional structures, describe industry. In this topic, we examine nucleophilic key organic chemistry reactions using balanced substitution, electrophilic substitution, addition equations and draw reaction mechanisms using and reduction reactions. Synthetic routes and curly arrow notation to represent the synthetic the methodologies involved in retro-synthesis reactions they undergo. 20.1 Te et Understandings Nucleophilic substitution reactions: ➔ Markovnikov’s rule can be applied to predict the major product in electrophilic addition reactions ➔ S 1 represents a nucleophilic unimolecular N of unsymmetrical alkenes with hydrogen halides substitution reaction and S 2 represents a N and interhalogens. The formation of the major nucleophilic bimolecular substitution reaction. S 1 N product can be explained in terms of the relative involves a carbocation intermediate. S 2 involves a N stability of possible carbocations in the reaction concerted reaction with a transition state. mechanism. ➔ For ter tiary halogenoalkanes the predominant Electrophilic substitution reactions: mechanism is S 1 and for primary ➔ Benzene is the simplest aromatic hydrocarbon N compound (or arene) and has a delocalized halogenoalkanes it is S 2. Both mechanisms N occur for secondary halogenoalkanes. structure of π bonds around its ring. Each carbon ➔ The rate-determining step (slow step) in an S 1 N to carbon bond has a bond order of 1.5. Benzene reaction depends only on the concentration of is susceptible to attack by electrophiles. the halogenoalkane, rate = k[halogenoalkane]. For S 2, rate = k[halogenoalkane][nucleophile]. Reduction reactions: N ➔ Carboxylic acids can be reduced to primary S 2 is stereospecic with an inversion of N conguration at the carbon. alcohols (via the aldehyde). Ketones can ➔ S 2 reactions are best conducted using aprotic, be reduced to secondary alcohols. Typical N polar solvents and S 1 reactions are best reducing agents are lithium aluminium hydride N conducted using protic, polar solvents. (used to reduce carboxylic acids)and sodium borohydride. Electrophilic addition reactions: ➔ An electrophile is an electron-decient species that can accept electron pairs from a nucleophile. Electrophiles are Lewis acids. 437
Applications and skills Nucleophilic substitution reactions: Electrophilic substitution reactions: ➔ Explanation of why hydroxide is a better ➔ Deduction of the mechanism of the nitration nucleophile than water. (electrophilic substitution) reaction of benzene ➔ Deduction of the mechanism of the nucleophilic (using a mixture of concentrated nitric acid and sulfuric acid). substitution reactions of halogenoalkanes with aqueous sodium hydroxide in terms of S 1 and S 2 Reduction reactions: N N mechanisms. Explanation of how the rate depends ➔ Writing reduction reactions of carbonyl- on the identity of the halogen (i.e. the leaving group), containing compounds: aldehydes and whether the halogenoalkane is primary, secondary, ketones to primary and secondary alcohols and or tertiary and the choice of solvent. carboxylic acids to aldehydes, using suitable ➔ Outline of the dierence between protic and reducing agents. aprotic solvents. ➔ Conversion of nitrobenzene to aniline via a two- Electrophilic addition reactions: stage reaction. ➔ Deduction of the mechanism of the electrophilic addition reactions of alkenes with halogens/ interhalogens and hydrogen halides. Nature of science ➔ Looking for trends and discrepancies – by ➔ Collaboration and ethical implications – understanding dierent types of organic scientists have collaborated to work on reaction and their mechanisms, it is possible investigating the synthesis of new pathways and to synthesize new compounds with novel have considered the ethical and environmental proper ties which can then be used in several implications of adopting green chemistry. applications. Organic reaction types fall into a number of dierent categories. 438
20 .1 T y p E s of or g a ni c r E a c T ion Organic synthesis the responsibility to place rst the wellbeing of the public who put their trust in the scientic Organic synthesis reactions are of fundamental community and its methodologies. Many importance in the eld of drug design, leading new synthetic compounds are the product of to the production of new synthetic drugs for collaboration, ranging from small-scale projects the treatment of many different conditions. in the laboratories of research institutes to Knowledge and understanding of the properties of international ventures on a historic scale. organic compounds and the mechanisms of their reactions is central to the development of new As well as the development of new drugs, useful compounds. the elds of food science and nutrition, biotechnology, plastics and textiles, fuels and Associated with the development of new explosives, paints and dyes, and pesticides and synthetic compounds are ethical considerations fertilizers are among the applications of synthetic concerning the implications of their use. A risk– organic chemistry. benet analysis is stringently applied to all new drugs under consideration and scientists have Nucleophilic substitution reactions The substitution reactions of saturated alkanes (sub-topic 10.2) involve Elemet Eletetvt X homolytic ssion, creating free-radicals that possess unpaired electrons. C The formation of chloroethane from the reaction between ethane and F chlorine is an example: 2.6 Cl C H (g)+ Cl (g) UV light C H Cl(g)+ HCl(g) Br 4.0 _____ → I 3.2 2 6 2 2 5 3.0 The chloroalkane that is produced has very different properties from those 2.7 of the alkane, and can undergo substitution reactions, producing a wide variety of compounds with different functional groups. The reason for this ▲ Table 1 Halogen atoms have high increased reactivity is the highly electronegative chlorine atom and the electronegativity and form polar bonds polar halogen–carbon bond (table 1 and gure 1). with carbon The partial positive charge makes the carbon atom electron decient and therefore susceptible to attack by nucleophiles, electron-rich species that are capable of donating a pair of electrons to form a covalent bond (sub-topic 10.2). There are two types of nucleophilic substitution (S ). The mechanism N of an S reaction depends on whether the halogenoalkane is primary, N secondary, or tertiary. δ+ δ- S 2 reactions and primary halogenoalkanes C X N ▲ Figure 1 Representation of the par tial Nucleophilic substitution in primary halogenoalkanes proceed in one step. charges within the polar carbon– The rate-determining step (slow-step) (sub-topic 16.1) involves both the halogen bond halogenoalkane and the nucleophile so the rate of reaction is dependent on the concentrations of both reactants. It is described as a second-order reaction. rate = k[halogenoalkane][nucleophile] As there are two reactant molecular entities involved in the ‘microscopic chemical event’ termed the elementary reaction, the molecularity is described as bimolecular. Understand that molecularity is not the same as the order of the reaction (sub-topic 16.1). 439
20 ORGANIC CHEMISTRY (AHL) ce ted et For example, the reaction between bromoethane and aqueous hydroxide ion yields ethanol and the leaving group, the bromide ion (gure 2). The S 2 reaction is an example of a N - concer ted reaction. It is a single-step H reaction through which reactants are conver ted directly into products. The - - mechanism does not involve an Br intermediate. HO C Br HO HO + The Wlde ve H When a chemical species with a CH CH CH 3 3 3 3 sp carbon center and tetrahedral transition state geometry undergoes a backside attack by a nucleophile in a S 2 reaction, a ▲ Figure 2 S 2 mechanism for the reaction between the N N congurational change occurs. Imagine primary halogenoalkane bromoethane and hydroxide ion the shape of an umbrella. The handle of the umbrella points towards the attacking The hydroxide nucleophile attacks the partially positively charged carbon nucleophile and the ribs that suppor t atom, forming a transition state that involves both the halogenoalkane the fabric layer are representative of and the nucleophile. This has a partially formed covalent bond between the shape formed by the three atoms the nucleophile and the carbon atom, and a weakened carbon–bromine bonded to the carbon. The leaving group is bond that has not completely broken. represented by the tip of the umbrella. Can you picture this? Just as the strong wind In this “backside attack” the nucleophile attacks the electrophilic centre of a storm can blow an umbrella inside out, at 180° to the position of the bromine leaving group, the large halogen the approaching nucleophile results in an atom creating steric hindrance. Such hindrance by bulky substituents inversion of conguration. This inversion prevents “frontal attack” by a nucleophile. As the reaction proceeds, provides room for the entering nucleophile the entering nucleophile causes an inversion of conguration , in while the product has the same relative the same way that an umbrella blows inside out in a storm. This is conguration as the reactant. known as the Walden inversion. Hence, the S 2 reaction is said to be N stereospecic. This is a reaction where starting reagents differing only in their conguration are converted into stereoisomeric products. Drawing mechanisms for S 2 reactions N ● The curly arrow from the nucleophile originates from its lone pair or negative charge, terminating at the carbon atom. ● The curly arrow representing the bromine leaving group originates at the bond between the carbon and bromine atoms. This can be shown on bromoethane or on the transition state. ● Partial bonds, HO C Br, are represented by dotted lines. ● The transition state is enclosed in square brackets with a single negative charge. ● The formation of the product and the leaving group must be shown. ▲ Figure 3 The inversion of an umbrella is S 1 reactions and ter tiary halogenoalkanes visually analogous to the inversion of N Tertiary halogenoalkanes undergo nucleophilic substitution reactions conguration that occurs in S 2 reactions. that involve two steps. The rate-determining step involves only the N halogenoalkane: the bond to the leaving group breaks, forming a carbocation. The reaction is a rst-order reaction. rate = k[halogenoalkane] For example, the reaction between 2-chloro-2-methyl propane and aqueous hydroxide ion yields 2-methylpropan-2-ol and the chloride ion (gure 4). The reaction has a molecularity of one, as there is only 440
20 .1 T y p E s of or g a ni c r E a c T ion CH CH CH 3 3 3 C+ - OH CH C Cl CH CH C OH + - 3 3 3 Cl CH CH CH 3 3 3 ▲ Figure 4 S 1 mechanism for the reaction between a ter tiary halogenoalkane and aqueous N hydroxide ion one molecular entity involved in the elementary reaction and hence is termed unimolecular. The atoms present in a molecule inuence the stability of possible A special arrow is used in organic chemistry to represent the polarization intermediates (S 2) and carbocations (S 1) and help determine the likely of a bond and the movement of electron density within the σ bond (gure 5). N N reaction mechanism. Inductive effects in organic compounds have a signicant effect on which nucleophilic substitution mechanism occurs between a nucleophile + C CH and a halogenoalkane. The most important factor is differences in 3 electronegativity between atoms present in the molecule. In the C H bond the carbon atom has slightly greater electronegativity than hydrogen, ▲ Figure 5 Representing the movement of electron density in the inductive eect creating a weak dipole and a shift in position of the bonding electrons closer to the carbon atom. Other atoms, such as more electronegative halogens, have far greater polarizing effects on the sigma bond. Alkyl groups bonded to a carbocation have a positive inductive effect, 3° 2° 1° stabilizing the charged carbocation by donating electron density and R R R reducing the positive charge on the carbon atom. In a primary carbocation just one alkyl group contributes to the inductive effect so it receives the least + + + stabilization. A tertiary carbocation is bonded to three alkyl groups so will C C C be more stable (gure 6). This is one reason why tertiary halogenoalkanes have a tendency to undergo reactions via the S 1 mechanism. R' R'' R' H H H N Drawing mechanisms for S 1 reactions ▲ Figure 6 Decreasing stability due to diminishing inductive eects moving from N ter tiary to primary carbocations ● The curly arrow representing the halogen leaving group originates at the bond between the carbon and the halogen. ● The representation of the carbocation clearly shows a positive charge centred on the carbon atom. ● The curly arrow from the nucleophile originates from its lone pair or negative charge, terminating at the carbon atom. ● The formation of the product and the leaving group must be shown. Factors aecting the rate of nucleophilic substitution cb-hle Bd ethl/ bd The rate of a nucleophilic substitution reaction depends on three main factors. 1 kJ ml C F 492 (1) The identity of the halogen C Cl 324 The presence of a good leaving group in a reactant undergoing nucleophilic C Br 285 substitution increases the rate of reaction of both S 1 and S 2 mechanisms. In N N C I 228 both cases the rate-determining step involves the heterolytic ssion of the ▲ Table 2 Bond dissociation energies of carbon–halogen bond, in which the two bonding electrons move to the more carbon halogen bonds electronegative atom. The quicker this rate-determining step is completed, the higher the rate of reaction and a better leaving group will help achieve this. 441
20 ORGANIC CHEMISTRY (AHL) Bond strength as well as electronegativity is important in choosing a Hlelke Mehm leaving group (table 2). Fluoroalkanes are virtually inert due to the short primary S2 –1 secondary N length (138 pm) and the high strength of the C F bond (492 kJ mol ). ter tiary S 2/S 1 As you move down the halogen group the strength of the carbon–halogen N N bond decreases as the size of the halogen atom increases. A larger halogen S1 N results in longer, weaker bonds. Additionally, the stability of the halogen ▲ Table 3 Prevalence of the S 2/S 1 mechanisms anion formed during these reactions is directly related to its effectiveness as N N in dierent classes of halogenoalkanes the leaving group. The larger iodine atom can better dissipate the negative charge compared with the chlorine atom and so iodine is a better leaving group than uorine or chlorine. H H (2) The classes of halogenoalkane: Primary, secondary, H H or ter tiary Cl As we have seen, tertiary halogenoalkanes predominantly undergo nucleophilic substitution via the S 1 mechanism while primary N halogenoalkanes favour the S 2 mechanism. For secondary N halogenoalkanes, both mechanisms are possible (table 3). The class of halogenoalkane has a direct effect on the rate of formation of the carbocation (the rate-determining step) and hence the overall rate of reaction. A tertiary carbocation has greater stability than a primary carbocation as a consequence of the inductive effects of the alkyl groups bonded to the carbon atom. Formed following cleavage of the carbon- halogen bond, the more stable tertiary carbocation rapidly forms and immediately reacts with the nucleophile. O O (3) The choice of solvent H H S 2 reactions are best performed in aprotic, polar solvents while H H N S 1 reactions are carried out in protic, polar solvents. N Aprotic, polar solvents are suitable for S 2 reactions because they: N ● possess no O H or N H groups so they cannot form a hydrogen bond to the nucleophile ● cannot solvate the nucleophile so leaving it “naked” and maintaining its effectiveness as a nucleophile in forming the transition state. Examples of aprotic solvents include ethyl ethanoate, CH COOCH CH , 3 2 3 and propanone, CH C(O)CH 3 3 Protic, polar solvents are suitable for S 1 reactions because they: N H ● are polar in nature due to the presence of polar bonds H H H H H C ● possess either an O H or N H group so can form hydrogen bonds C with the nucleophile O O H + ● solvate the nucleophile, thus inhibiting its ability to attack Na H + electrophiles such as the δ carbon atom. H O C Solvation is the process by which solvent molecules surround the H dissolved ions. The smaller the nucleophile, the more effective the H solvation. Because the nucleophile is encapsulated by the solvation C shell, it is less effective as a nucleophile in forming an S 2 intermediate; H N H H therefore S 1 reactions are favoured. N H ▲ Figure 7 Solvation by the polar, protic solvent methanol 442
20 .1 T y p E s of or g a ni c r E a c T ion Examples of such solvents include methanol, CH OH (gure 7), water, H O, 3 2 ammonia, NH , methanoic acid, COOH, and hydrogen uoride, HF. gee hemt 3 Green chemistry is an approach that focuses on designing synthetic What makes a good nucleophile? processes so that they are sustainable, and do not have a negative impact on A nucleophile is an electron-rich species capable of donating a pair of electrons the environment and society through the to an electrophile to create a covalent bond. The strength of a nucleophile production of toxic substances. depends on the ease with which it can make these electrons available. It is understood that the hydroxide ion is a better nucleophile than the The 12 le ee hemt water molecule. These two nucleophiles can be used to demonstrate the factors that inuence the strength of a nucleophile. 1 Prevent waste 2 Use of renewable feedstock In summary: 3 Atom economy ● both nucleophiles possess at least one pair of electrons, so by denition, they can act as Lewis bases (sub-topic 18.1) 4 Reduce derivatives the negatively charged hydroxide ion is a stronger nucleophile than 5 Less hazardous waste the water molecule which is its conjugate acid (sub-topic 8.1). A ● negatively charged ion has a far greater attraction for an electrophile than does a neutral molecule. 6 Catalysts 7 Design benign chemicals 8 Design for degradation Since the early 1990s governments and scientic organizations have been 9 Benign solvents and auxiliaries suppor ting the development of green chemistry, recognizing the need to reduce the impact of rapid global development on the environment. The American 10 Real-time analysis for pollution Chemical Society (ACS) formulated the 12 principles of green chemistry. Examples of their incorporation into synthetic organic chemistry include the pharmaceutical prevention industry reducing the need for toxic organic solvents in manufacturing processes and the food industry’s development of biodegradable food packaging from corn 11 Design for energy eciency starch, as an alternative to plastics. Adhesives derived from soya proteins are being used in the building industry in place of adhesives and resins that contain 12 Inherently benign chemistry for the carcinogen methanal (formaldehyde), a product of the petrochemical industry. accident prevention Supercritical carbon dioxide is of increasing impor tance as an industrial and commercial solvent; see sub-topic D.6 for more information. With a supercritical minimal environmental impact, its list of applications continues to grow. The decaeination of coee has traditionally been achieved by solvent extraction critical point using dichloromethane, a known carcinogen. Increasingly, the coee industry is using supercritical carbon dioxide as a non-toxic, green alternative. P uid c 73 )mta( erusserp liquid solid 10 Electrophilic addition reactions 5.2 atm gas 5 In contrast to a nucleophile, an electrophile is an electron-decient species that will accept a pair of electrons, acting as a Lewis acid. 1 + + Electrophiles include the nitronium ion, NO and the methyl cation,CH . 2 3 78 57 0 31 Electrophiles have either a formal positive charge (a cation) or a temperature (°C) partial positive charge (δ+) generated by the presence of a highly electronegative species resulting in the polarization of the bond. ▲ Figure 8 Carbon dioxide phase diagram. In its supercritical state, carbon dioxide Alkenes are unsaturated compounds that contain electron-rich exhibits both gas and liquid proper ties. carbon–carbon double bonds. They undergo addition reactions in The temperature, pressure, and additives which the double bond breaks and two additional atoms bond with the that control the polarity of the liquid can be molecule, creating a saturated compound. An electrophile can act as the varied, resulting in an increasing range of its source of the new additional atoms. applications as a solvent 443
20 ORGANIC CHEMISTRY (AHL) 2 C sideways overlap C 2 A carbon–carbon double bond contains both a sigma ( σ) bond and a pi sp p p sp (π) bond (gure 9). A sigma bond is formed by the end-to-end overlap of head to head atomic orbitals and electron density is centred between the nuclei of the 2 overlap 2 bonding atoms, along the inter-nuclear axis (sub-topic 14.1). A pi bond sp sp is formed by the sideways overlap of atomic orbitals and electron density sideways overlap found above and below the plane of the nuclei of the bonding atoms. The VSEPR theory (sub-topic 14.1) regards a double bond as a region of 2 high electron density. With sp hybridization in the carbon atoms and pi bond a bond angle of approximately 120 °, the carbon–carbon double bond provides a reduced level of steric hindrance to the attacking electrophile. sigma bond Distinction must be made between the strength of a bond and its reactivity. In terms of bond dissociation energy, a double bond is stronger than a ▲ Figure 9 Carbon–carbon double bonds contain single bond. However, the high density of electrons in a double bond both sigma and pi bonds means that the bond is more susceptible to electrophilic attack. Markovnikov ’s rule The major products of the electrophilic addition of hydrogen halides to unsymmetrical alkenes (see below) can be predicted using Markovnikov’s rule. The hydrogen atom will preferentially bond to the carbon atom of the alkene that is already bonded to the largest number of hydrogen substituents. This comes about because the carbocation formed when the pi bond is broken has its positive charge centred on the most substituted carbon. A tertiary carbocation has greater stability than a primary carbocation due to the reduction in density of the positive charge through the inductive effects of the three alkyl substituents (table 4). Type of carbocation 3° 2° 1° Level of stability most stable least stable Structure CH H H 3 + + + C C C CH CH CH CH CH H 3 3 3 3 3 ▲ Table 4 The relative stabilities of primary, secondary and ter tiary carbocations form the basis of Markovnikov ’s rule CH CH CH CH H 2 3 H 2 3 C C C + Electrophilic addition of hydrogen halides H H C H to alkenes H I I In the electrophilic addition reaction between but-1-ene and hydrogen iodide, the major product is 2-iodobutane as the 2 ° carbocation is formed preferentially. Hydrogen iodide is split heterolytically, creating the H CH CH 2 3 + hydrogen cation, H and the iodide anion, I . The initial attack on the H C C H pi electrons of the C=C bond comes from the cation, followed by rapid reaction between the unstable carbocation and the halogen ion (gure 10). H I ▲ Figure 10 The mechanism of the electrophilic Drawing mechanisms for electrophilic addition reactions addition of hydrogen iodide to but-1-ene forming 2-iodobutane ● A curly arrow originates from the carbon–carbon double bond to the hydrogen atom of hydrogen iodide. 444
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