Heinemann Chemistry 2 Enhanced

0

chapter 05

Analysing oxidants and reductants

key knowledge

Sa

m

pl e

pa ge s

• volumetric analysis: redox titrations

chapter ch haptter out outcomes tcomes After completing this chapter, you should be able to: • recognise a redox reaction • identify chemicals undergoing oxidation and reduction in a redox reaction • write half equations for redox reactions • combine half equations to produce overall equations • assign oxidation numbers to elements in redox reactions and hence identify oxidants and reductants • describe appropriate procedures for volumetric analyses that involve redox titrations of some common consumer products • perform stoichiometric calculations for volumetric analysis involving redox reactions • perform calculations for volumetric analysis that involve back titrations.

chemistryy in action Alcohol and the road toll

pa ge s

m

Ethanol content (%v/v) 2.5 5.0 10 12 18 37

Sa

Drink ‘Light’ beer Beer Alcoholic cider Wines (including sparkling wines) Port, sherry Spirits (brandy, bourbon, gin, rum, vodka, whisky etc.) Pre-mixed vodka-based drinks

pl e

Figure 5.1 The blood alcohol content of motorists can be analysed using a breathalyser.

TABLE 5.1 Typical ethanol contents of selected alcoholic beverages

Drinks such as wine, beer, vodka and bourbon contain ethanol (CH3CH2OH). Ethanol acts as a depressant, slowing the functioning of the brain. When consumed in excess, the intoxicating qualities of alcoholic drinks can lead to antisocial behaviour and can damage a person’s health. The ethanol content of alcoholic drinks varies, as shown in Table 5.1. Regulations require that the ethanol content of alcoholic drinks be specified on their labels, since the content determines the quantity of drink that can be consumed without adverse effects. Because ethanol slows down a person’s reaction time, it seriously affects driving skills. It is estimated that alcohol has contributed to nearly 40% of all road accidents. Governments have responded by introducing penalties, such as fines and licence disqualification, for drivers who register a concentration of alcohol in the blood exceed-ing a certain level—typically 0.05% (w/v). Probationary or P-plate drivers are required to have a zero blood alcohol level. The introduction of penalties, a policy of randomly testing motorists for blood alcohol levels, and various advertising campaigns have all helped to increase public awareness of the link between alcohol consumption and road accidents. Figure 5.2 shows how the road toll in Victoria changed during the period when these various measures designed to discourage drink-driving were introduced. Chemists were involved in the invention of the ‘breathalyser’, an instrument designed for use by police to estimate blood alcohol content. Rather than analysing samples of blood, this instrument measures the concentration of alcohol in a person’s breath, which is closely related to the concentration of alcohol in their blood. If this screening test indicates a driver is over 0.05, more accurate measurements are taken either in a ‘booze bus’ or at a police station.The blood alcohol content may then be confirmed using instrumental techniques such as infrared spectroscopy (Chapter 7) or gas–liquid chromatography (Chapter 6). An alcohol fuel cell sensor (Chapter 27) is also used by some police departments. The first breathalysers operated by detecting the colour change that occurs when ethanol reacts with an acidified solution of potassium dichromate (K2Cr2O7), forming Cr3+ ions and ethanoic acid. The reaction can be represented by the equation:

9

2Cr2O72−(aq) + 3CH3CH2OH(g) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l) orange green This reaction can also be used to determine the ethanol content of drinks by volumetric analysis and is an example of a particular class of reactions known as oxidation–reduction (or redox) reactions. We will now look at redox reactions in more detail and examine how they are used in chemical analysis.

48 Chemical analysis C

chemistryy in action Alcohol and the road toll (continued) 10 9

0.05% law introduced

Fatality rate per 10 000 vehicles

8 50 km/h limit in residential areas introduced

7 6 5

Zero blood alcohol for probationary drivers

Increased blood alcohol penalties

4 3

40 km/h around schools

Random breath-testing introduced

2

3 km/h tolerance on speeding replaces 10% tolerance

Increased penalties

pa ge s

1 0

62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 00 02 04 06 08 10 Year

5.1

Figure 5.2 Deaths on Victorian roads, 1960–2009

Mg

_

2e

pl e

What is a redox reaction?

Sa

m

You may remember from your study of Unit 2 (Heinemann Chemistry 1, Chapter 16) that redox reactions involve the transfer of electrons from one chemical to another. They can often be considered as occurring in two parts. In these reactions: One of the reactants loses electrons—this process is called oxidation. One of the reactants gains these electrons—this is called reduction. Oxidation and reduction occur simultaneously. Redox reactions occur when, for example, fireworks are triggered. These substances usually contain magnesium metal, which burns with an intense light to produce a white, ionic solid—magnesium oxide. During the reaction, magnesium atoms lose two electrons to become Mg2+ ions (Figure 5.3):

Mg

O

Mg2+

O

O2

_

_

O2

Mg2+

_

2e

Electron transfer 2Mg(s) O2(g) + Reductant Oxidant (is reduced) (is oxidised)

2MgO(s)

Figure 5.3 A representation of the electron transfer reaction between magnesium and oxygen.

Mg –—h Mg2+ + 2e−

This part of the overall reaction is oxidation. At the same time, oxygen molecules in the air (O2) each gain four electrons that were lost by the magnesium atoms and form O2− ions: O2 + 4e− –—h 2O2−

This part of the overall reaction involves reduction.

review The words ‘OIL RIG’ can be used to remember that ‘Oxidation Is Loss of electrons’ and ‘Reduction Is Gain of electrons’. Redox reactions are described in further detail in Heinemann Chemistry 1, Chapter 16.

49 Analysing oxidants and reductants

The number of electrons produced during the oxidation is the same as the number of electrons consumed in the reduction. To write an overall equation for the reaction we add the half equations, making sure that the number of electrons lost is equal to the number of electrons gained during the reaction. Overall equation: 2(Mg –—➤ Mg2+ + 2e−) O2 + 4e− –—➤ 2O2− 2Mg(s) + O2(g) –—➤ 2MgO(s)

pa ge s

Substances that cause oxidation to occur are called oxidising agents, or oxidants. In the reaction between magnesium and oxygen above, oxygen acts as the oxidant. Note that the oxidant causes oxidation but itself will always be reduced. Substances that cause reduction to occur are called reducing agents, or reductants. Magnesium is the reductant in our example. Reductants cause reduction but are always themselves oxidised. Redox reactions occur as part of many everyday processes, such as the bleaching of hair, the corrosion of metals, extraction of metals from their ores, respiration, photosynthesis and film-based photography.

chemistryy in action Photochromic sunglasses

Sa

m

pl e

Some people wear glasses fitted with photochromic lenses, which darken in bright sunlight and become more transparent when the light intensity drops (Figure 5.4). Use of these glasses means that a pair of sunglasses is not required, as the photochromic lenses can decrease the amount of transmitted light by up to 80%. Many silver compounds, including silver chloride, are sensitive to light. Tiny crystals of silver chloride are incorporated into the complex silicate-based structure of the glass used to make the photochromic lenses. On exposure to ultraviolet light, which is present in sunlight, the chloride ions are oxidised to chlorine atoms: Cl− → Cl + e− Electrons produced from this reaction are transferred to silver ions, causing the silver ions to be reduced to metallic silver: Ag+ + e− → Ag The silver metal formed causes light to be reflected, the lenses to darken, and the intensity of light reaching the eyes of a person wearing the glasses to be reduced. To prevent the silver metal and chlorine atoms forming silver chloride again immediately, copper(I) chloride is also added to the glass in tiny amounts. It reacts with the chlorine atoms, reducing the rate at which silver chloride can be re-formed: Cl + Cu+ → Cl− + Cu2+ The darkening process must be reversible for the glasses to be effective. In the absence of strong sunlight, silver ions are re-formed by a redox reaction involving the silver metal and Cu2+ ions: Cu2+ + Ag → Cu+ + Ag+ As a consequence, the lenses of the glasses recover their transparency.

Figure 5.4 The lenses of these photochromic glasses darken in sunlight as a consequence of redox reactions involving silver chloride.


summary

5.1

O Oxidation id ti involves i l a loss l off electrons. l t Reduction involves a gain of electrons. Oxidation and reduction reactions occur simultaneously.

An A oxidant id t causes oxidation id ti andd is i itself it lf reduced. d d A reductant causes reduction and is itself oxidised.

key questions c 2AgBr(s) → 2Ag(s) + Br2(g) 2 Identify the oxidants and reductants in each of the reactions in Question 1.

pa ge s

1 Name the chemicals that undergo oxidation in the following reactions. a 2Zn(s) + O2(g) → 2ZnO(s) b Ca(s) + Cl2(g) → CaCl2(s)

5.2

Oxidation numbers: classifying redox reactions C(s) + O2(g) –—➤ CO2(g)

pl e

We have seen that the reaction of magnesium with oxygen is a redox reaction. Similar reactions occur when other substances burn in oxygen. During a traditional barbecue, carbon in charcoal burns to form carbon dioxide:

m

If the air supply is limited, such as where charcoal is burning strongly in the centre of the fire, carbon monoxide may be formed:

Sa

2C(s) + O2(g) –—➤ 2CO(g)

Carbon monoxide is flammable and burns with a clean, blue flame in the outer regions of the fire: 2CO(g) + O2(g) –—➤ 2CO2(g)

These reactions are so similar to the combustion reaction of magnesium that chemists have agreed they should also be classified as redox reactions. However, since the products in the reactions are molecular rather than ionic, they cannot be classified as redox reactions according to our earlier definitions of oxidation and reduction as involving the transfer of electrons. This difficulty is overcome in a somewhat artificial manner. To determine whether a reaction is a redox reaction, oxidation numbers are assigned to elements involved in the reaction. These numbers are determined by applying a set of rules that treat molecular substances as if they were composed of ions. Oxidation numbers have no physical meaning. Oxidation numbers are also known as oxidation states.

! The oxidation number of an element is written above its symbol. The plus or minus sign precedes the number and so distinguishes the oxidation number from the charge on an −2

ion. The O 2− ion, for example, has a charge of 2− and an oxidation number of −2.


Oxidation number rules

review Electronegativity is a measure of the electronattracting ability of an element. The order of electron-attracting power is: F > O > Cl > N > other elements Metals generally have low electronegativities. Electronegativities of the elements are given in the periodic table at the end of this book.

Oxidation numbers are determined using the following rules: Free elements have an oxidation number of zero. 0

0

0

e.g. Na(s), C(s), Cl2(g) In ionic compounds composed of simple ions, the oxidation number is equal to the charge on the ion. +1 −1

+2

−1

+3

−2

e.g. NaCl, CaCl2, Al2O3 Some elements have oxidation numbers that are regarded as fixed, except in a few exceptional circumstances. − Oxygen usually takes −2 in compounds. In peroxides such as H2O2 and BaO2 it has −1. − Hydrogen takes +1 in compounds, except in metal hydrides such as NaH and CaH2 where it has −1. − Electronegative elements such as F, Cl and O take numbers equal to the charges on their simple ions (−1, −1 and −2, respectively) when part of a compound, provided that they are the most electronegative element present in the compound. The sum of the oxidation numbers in a neutral compound is zero. In a polyatomic ion the sum of the oxidation numbers is equal to the charge on the ion. The most electronegative element in a compound has the negative oxidation number.

!

pa ge s

Remember that oxidation numbers have no physical meaning. In the case of water, the hydrogen is assigned a value of +1 and the oxygen a value of −2, as if the compound were made of H+ and O2− ions. (Of course, water really consists of H2O molecules.) This is done so that reactions involving covalent substances, as well as those involving ionic substances, can be classified as redox reactions.

Worked example 5.2a

pl e

The sum of the oxidation numbers in the neutral compound HNO3 equals zero. H is fixed as +1 and O is fixed as −2. Find the oxidation number of nitrogen.

Solution

m

(oxidation number of H) + (oxidation number of N) + 3 × (oxidation number of O) = 0 So, (+1) + x + 3(−2) = 0 x = +5 +1 +5 −2

Sa

The oxidation numbers are H N O3.

Worked example 5.2b For CO32−, the sum of the oxidation numbers equals the ionic charge of 2−. The oxidation number of O is fixed as −2. Find the oxidation number of carbon.

Solution (oxidation number of C) + 3 × (oxidation number of O) = −2 So, x + 3(−2) = −2 x = +4 +4 −2 The oxidation numbers are C O32−.

Worked example 5.2c The sum of the oxidation numbers of F2O is zero. Find the oxidation number for each atom.

Solution F is always −1 and O is normally −2 in compounds. In this case F is more electronegative, so O becomes +2. As a result, the sum of the oxidation numbers is zero.


Variable oxidation number

chemfact Oxygen is transported through the bodies of crustaceans and molluscs by haemocyanin instead of the haemoglobin used in mammals. Haemocyanin is a blue compound containing copper ions. The transfer of oxygen from a crustacean’s blood to its cells employs an equilibrium reaction involving copper(I ) and copper(II ) ions: Cu2+ + e− Cu+

pa ge s

In all their compounds the group 1 metals only have an oxidation number of +1 and the group 2 metals only have an oxidation number of +2. The oxidation numbers of the transition metals vary depending upon the compound. The different oxidation states of the transition metals often have characteristic colours (Figure 5.5). The oxidation numbers in non-metallic elements may also vary from compound to compound. The oxidation number of nitrogen in NH3 is −3, in NO it is +2, in NO2 and N2O4 it is +4, while in HNO3 and N2O5 it is +5.

pl e

Figure 5.5 The test tubes contain solutions of vanadium compounds that have oxidation numbers from left to right of +5, +4, +3 and +2.

Using oxidation numbers to name chemicals

Sa

m

There are two compounds that can be called iron chloride: FeCl2 and FeCl3. To distinguish between the two iron chlorides, roman numerals representing the appropriate oxidation number are inserted in the name: FeCl2 is named iron(II) chloride. FeCl3 is named iron(III) chloride.

Changing oxidation numbers Once oxidation numbers are assigned to the elements involved in a chemical reaction, we can look for a change in these numbers during a reaction. An increased oxidation number means the element has been oxidised. A decreased oxidation number means the element has been reduced. Both processes always occur in a redox reaction.

Worked example 5.2d Let us apply the rules for assigning oxidation numbers to the reactions that occur in a barbecue. 0

0

+4 −2

i C(s) + O2(g) → C O2(g) Since the oxidation number of carbon increases from 0 in elemental carbon to +4 in carbon dioxide, the carbon is oxidised. The oxidation number of oxygen decreases from 0 to −2, so oxygen is reduced. Carbon is the reductant and oxygen is the oxidant in this reaction.


0

0

+2 −2

+2 −2

0

+4 −2

ii 2C(s) + O2(g) → 2C O(g) Carbon is oxidised; oxygen is reduced. Carbon is the reductant and oxygen is the oxidant. iii 2 C O + O2(g) → 2C O2(g) Carbon in carbon monoxide is oxidised. (We often say the carbon monoxide is oxidised.) Molecular oxygen is reduced. Carbon is the reductant and oxygen is the oxidant.

summary

5.2

O Oxidation idation numbers n mbers can be used sed to identify identif species that ha havee been oxidised or reduced. Oxidation numbers have no physical meaning.

An element has been oxidised o idised if its oxidation o idation number n mber has increased. An element has been reduced if its oxidation number has decreased.

pa ge s

key questions a CaO b CaCl2 c HSO4− − d MnO4 e F2 f SO32− g NaNO3 h K2Cr2O7 6 Assign oxidation numbers to each element in these equations, and hence identify the oxidant and reductant: a Mg(s) + Cl2(g) → MgCl2(s) b 2SO2(g) + O2(g) → 2SO3(g) c Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) d 2Fe2+(aq) + H2O2(aq) + 2H+(aq) → 2Fe3+(aq) + 2H2O(l)

pl e

3 State the oxidation number of carbon in: a CO b CO2 c CH4 d C (graphite) e HCO3− 4 Which one or more of the following substances contain manganese in the +6 oxidation state: MnCl2, MnCl3, MnO2, K2MnO4, KMnO4? 5 Find the oxidation numbers of each element in the following compounds or ions. Hint: For ionic compounds, use the charge on each ion to help you.

m

5.3

Sa

Writing half equations Half equations that involve atoms or simple ions can be written with little difficulty. For instance, knowing that magnesium metal is oxidised to form Mg2+ ions, you can quickly write the half equation as:

Reactants

Products

Mg –—➤ Mg2+ + 2e−

Half equations involving polyatomic ions may be less obvious. The anaesthetic nitrogen(I) oxide or laughing gas (N2O), can be prepared by the reduction of nitric acid in an acid solution: 2NO3−(aq) + 10H+(aq) + 8e− –—➤ N2O(g) + 5H2O(l)

Figure 5.6 Remember, in balanced half and overall equations: the number of atoms of each element is equal on both sides the total charge on each side is equal.


Such equations can be deduced using the following procedure. Let us take the reduction of the nitrate ion as an example. 1 Balance all elements except hydrogen and oxygen in the half equation. 2NO3− –—➤ N2O

2 Balance the oxygen atoms by adding water. (Oxygen atoms react to form water in acidic solution.) 2NO3− –—➤ N2O + 5H2O

3 Balance the hydrogen atoms by adding H+ ions (which are present in acidic solution). 2NO3− + 10H+ –—➤ N2O + 5H2O

4 Note that the total charge on the left-hand side of this incomplete equation is (2−) + (10+), which equals 8+. The total charge on the righthand side is 0. Balanced equations should have the same total charge on each side. Balance the charge on both sides of the equation by adding electrons. Add states. 2NO3−(aq) + 10H+(aq) + 8e− –—➤ N2O(g) + 5H2O(l)

Worked example 5.3 When a pale green solution containing Fe2+ ions is mixed with a purple-coloured solution of MnO4− ions, the purple colour disappears. Fe3+ and Mn2+ ions are formed. Write a balanced equation for this reaction.

Solution

5.3

Sa

summary

m

pl e

pa ge s

The half equation for the reaction involving Fe2+ ions is written easily: Fe2+(aq) → Fe3+(aq) + e− We can work out the equation involving MnO4− ions by following the steps described above. 1 MnO4− → Mn2+ 2 MnO4− → Mn2+ + 4H2O 3 MnO4− + 8H+ → Mn2+ + 4H2O 4 MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) To write an overall equation, the half equations are multiplied so that the number of electrons in each is the same. They are then added together and simplified if required: 5 × (Fe2+(aq) → Fe3+(aq) + e−) MnO4−(aq) + 8H+(aq) + 5e− → Mn2+(aq) + 4H2O(l) MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

AAn oxidation id ti or a reduction d ti reaction ti can be b represented t d by b a half equation. An oxidation half equation and a reduction half equation can be combined by multiplying the half equations so that there is

an equall number b off electrons l t in i eachh half h lf equation. ti Th The hhalflf equations can then be added.

key questions 7 Write ionic half equations for the: a reduction of MnO2 to Mn2+ b reduction of MnO4− to MnO2 c reduction of SO42− to H2S d oxidation of SO2 to SO42− e oxidation of H2S to S f oxidation of SO32− to SO42− 8 When copper(II ) sulfate solution is stored in a steel (iron) container, the container gradually corrodes. Write ionic half equations and a balanced ionic equation to represent the reaction if the products of the reaction are copper and iron(II ) sulfate solution.

9 When zinc powder is sprinkled into an acidified solution of potassium dichromate, a reaction occurs that leaves zinc ions and chromium(III ) ions in solution. a Write the oxidation half equation for the reaction. b Write the reduction half equation for the reaction. c Use your answers to parts a and b to write a balanced ionic equation for the overall reaction. 10 Write the ionic half equations and the balanced overall ionic equation for the reaction in which: a a solution containing iron(II ) ions is oxidised by an acidified solution containing dichromate ions (Cr2O72−). The products include iron(III ) and chromium(III ) ions.

55 5 Analysing oxidants and reductants

key questions (cont.) b a solution containing sulfite ions (SO32−) reacts with an acidified solution of permanganate ions (MnO4−) to produce a colourless solution containing sulfate ions and manganese(II ) ions. c manganese dioxide (MnO2) reacts with concentrated hydrochloric acid to form chlorine gas and a solution containing manganese(II ) ions. 11 The following equations are not balanced. i Identify the species that has been reduced and the species that has been oxidised.

a b c d

ii Write balanced half equations for the oxidation and reduction reactions. iii Combine the half equations to write a balanced equation. Ce4+(aq) + H2S(g) → Ce3+(aq) + S(s) + H+(aq) NO3−(aq) + H+(aq) + Cu(s) → NO(g) + H2O(l) + Cu2+(aq) H2O2(aq) + Br−(aq) + H+(aq) → Br2(l) + H2O(l) MnO2(s) + H+(aq) + S(s) → Mn2+(aq) + H2O(l) + SO2(g)

Volumetric analysis

pl e

The concentration of solutions of substances that can undergo redox reactions can be found by volumetric analysis, the technique described in Chapters 3 and 4. Instead of reacting an acid with a base, a redox titration involves reaction of an oxidant with a reductant. As for acid–base titrations, one solution is usually pipetted into a conical flask and the other is dispensed into the flask from a burette. For some reactions, for example those involving the permanganate ion (MnO4−), the equivalence point will be indicated by a colour change in the reacting solutions, whereas for others an indicator must be added in order to detect the equivalence point. Volumetric analysis involving redox reactions can be used to determine the composition of many substances, including fruit juice, wine and bleach. Table 5.2 gives details of how the oxidants and reductants in some everyday substances are analysed. In some cases back titrations are used (p. 43).

Sa

m

Selecting a suitable indicator can be more difficult for redox titrations than for acid–base titrations. Redox indicators must behave as oxidants or reductants after the equivalence point has been reached and a small excess of the solution from the burette is present. They must also be highly coloured in either oxidised or reduced form. For example, the permanganate ion often used in redox titrations has a purple colour while the manganese(II ) ion is colourless: MnO4−(aq) + 8H+(aq) + 5e− purple → Mn2+(aq)+ 4H2O(l) colourless

pa ge s

5.4

chemfact

TABLE 5.2 Analysis of oxidants and reductants in common products Item Wine

Ingredient for analysis Ethanol

Wine Fruit juice Household bleach

Sulfur dioxide Vitamin C (ascorbic acid) Hypochlorite ion

Hair bleach

Hydrogen peroxide

Titrate with Iron(II) solution, after reaction with an excess of potassium dichromate solution Iodine solution Iodine solution Sodium thiosulfate solution, after reaction with an excess of acidified potassium iodide solution Potassium permanganate solution

Worked example 5.4 A 10.0 mL sample of white wine was placed in a volumetric flask and water was added to make 100.0 mL of solution. Then 20.0 mL aliquots of the diluted wine were titrated against 0.100 M acidified potassium dichromate (K2Cr2O7) solution. The mean titre was 24.61 mL. Calculate the concentration of ethanol in the sample of white wine. 2Cr2O72−(aq) + 3CH3CH2OH(aq) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)


Solution

Ethanol reacts only slowly with dichromate ions at room temperature. To overcome this problem, the ethanol content of wine is usually found by performing a back titration, rather than by using the method described here. Excess dichromate ions are first added to the wine and the mixture is heated so that reaction occurs. The amount of dichromate ions in excess is then found by titration with a solution of Fe2+ ions.

5.4

pl e

summary

chemfact

pa ge s

n(Cr2O72−) = c(Cr2O72−) × V(Cr2O72−) = 0.100 mol L−1 × 0.024 61 L = 0.002 461 mol 2− From the equation, 2 mol of Cr2O7 reacts with 3 mol of CH3CH2OH. n(CH3CH2OH) 3 So, the ratio = n(Cr2O72−) 2 3 n(CH3CH2OH) = × 0.002 461 = 0.003 692 mol 2 The amount of ethanol in the 20.0 mL volume of diluted wine is 0.003 692 mol. Since 100.0 this volume of wine was taken from a total volume of 100.0 mL, there would be 20.0 (or 5.000 times) this amount in the original 10.0 mL sample. 100 mol n(CH3CH2OH) in original 10.0 mL sample = 0.003 692 × 20 = 0.018 46 mol n(CH3CH2OH) c(CH3CH2OH) = V(CH3CH2OH) 0.018 46 mol = 0.100 L = 1.846 mol L−1 The concentration of alcohol in the wine is 1.85 M.

key questions

m

VVolumetric l t i analysis l i can be b usedd to t determine d t i the th concentration t ti or mass off substances bt that th t undergo d redox d reactions. ti

Sa

12 Potassium permanganate reacts with hydrogen peroxide: 2MnO4−(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g) 25.0 mL of 0.02 M KMnO4 solution is reduced by 20.0 mL of H2O2 solution. What is the concentration of the hydrogen peroxide solution? 13 An artist uses 10.0 mL of 15.0 M HNO3 to etch a design into a sheet of copper: Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g) What mass of copper will have reacted with the acid?


chapter review 05

key terms half equation oxidant oxidation

oxidation number oxidation state oxidising agent

redox reaction redox titration reducing agent

reductant reduction

Stoichiometry • Volumetric analysis: calculations • More stoichiometric calculations

a container of sodium dithionite (Na2S2O4). The dithionite ion reacts with water according to the equation: 2S2O42–(aq) + H2O(l) → S2O32–(aq) + 2HSO3–(aq)

pa ge s

Oxidation numbers

a State the oxidation number of sulfur in the following ions: i S2O42– ii S2O32– iii HSO3– b Write the ionic half equations for the oxidation and reduction reactions that occur when sodium dithionite is mixed with water. 19 During each of the following analyses of a substance, redox reactions occurred. Write half equations for the oxidation and reduction reactions. Use these half equations to write an overall equation for each reaction. a Zinc was analysed by reacting it with a solution of Pb2+ ions. Lead metal was precipitated and Zn2+ ions were formed. b Fe2+ ions in iron tablets were determined by oxidising them to Fe3+ ions, using an acid solution of MnO4– ions, which were themselves reduced to Mn2+ ions during the reaction. c Sulfur dioxide (SO2), a preservative in dried fruit, was determined by oxidation to SO42– using a solution of I2. Iodide (I–) ions were produced. d An acidic solution of bleach, which contains OCl– ions, was titrated against a solution of I– ions. The reaction products included Cl– and I2.

Sa

m

pl e

14 What is the oxidation number of sulfur in each of the following compounds? a SO2 b H2S c H2SO4 d SO3 e Na2SO3 f Na2S2O3 15 Which of the following may be regarded as redox reactions? Give reasons for each of your answers. a BaCl2 + H2SO4 → BaSO4 + 2HCl b 2Ag + Cl2 → 2AgCl c 2FeCl3 + SnCl2 → 2FeCl2 + SnCl4 d ZnCO3 → ZnO + CO2 d HPO32– + I2 + OH– → H2PO4– + 2I– f 2Cu+ → Cu2+ + Cu g CaF2 → Ca2+ + 2F– h P4 + 6H2 → 4PH3 16 Copper bowls and trays can be decorated by etching patterns on them using concentrated nitric acid. The overall reaction is: Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

a What is the oxidation number of copper: i before reaction? ii after reaction? b What is the oxidation number of nitrogen: i before reaction? ii after reaction? c Name the oxidant and the reductant in this process.

Writing and combining half equations 17 Aluminium metal can reduce the hydrogen ions in a solution of hydrochloric acid to hydrogen gas, according to the equation:

2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)

a Write half equations for the oxidation and reduction processes involved in this reaction. b Use these half equations to explain why six hydrogen ions are reduced for every two aluminium atoms oxidised. 18 As a result of a traffic accident, residents in a Melbourne suburb had to be evacuated when toxic fumes leaked from

58 Chemical analysis

Calculations involving redox reactions 20 In dry cells commonly used in torches, an electric current is produced from the reaction of zinc metal with MnO2. During this reaction, Zn2+ ions and Mn2O3 are formed. a Write half equations, and hence an overall equation, for the reaction. b Calculate the mass of zinc that would be needed to react completely with 8.0 g of MnO2 in a dry cell. 21 Find the mass of silver metal that will react with 2.000 L of 10.00 M nitric acid, according to the equation:

Ag(s) + 2HNO3(aq) → AgNO3(aq) + H2O(l) + NO2(g)

22 The thermite process can be used to weld lengths of railway track together. A mould placed over the ends of the two rails to be joined is filled with a charge of aluminium powder and iron(III) oxide. When the mixture is ignited, a redox reaction occurs to form molten iron, which joins the rails together. a Write a half equation for the conversion of iron(III) oxide to metallic iron and oxide ions. b Is the half equation you wrote for part a an oxidation or a reduction process? c Write the overall equation for the thermite process. d What mass of iron(III) oxide must be present in the charge if each joint requires 3.70 g of iron to weld it together? 23 A vitamin C tablet with a mass of 1.306 g was crushed and dissolved in de-ionised water. The solution was titrated against 0.0500 M iodine solution, using starch solution as indicator, to determine the vitamin C (C6H8O6) content of the tablet. The reaction can be represented by the equation: C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I–(aq)

The end point occurred when 28.40 mL of iodine solution had been added. a Find the mass of vitamin C in the tablet. b Calculate the percentage of vitamin C in the tablet by mass. c Suggest the function of the other substances that make up the remainder of the mass of the tablet. 24 The iron content in a 0.200 g sample of fencing wire was determined by dissolving the wire in dilute sulfuric acid and making up the resulting pale green solution of Fe2+ ions to 25 mL. The solution was titrated with 0.0300 M potassium permanganate (KMnO4) solution, which is purple in colour. A titre of 20.22 mL was obtained. The solution of Mn2+ and Fe3+ ions produced by the reaction was almost colourless. a Write an overall equation for the titration reaction. b Calculate the amount, in mol, of Fe2+ ions in the 25 mL volume of solution. c Calculate the percentage, by mass, of iron in the wire. d An indicator was not required for this titration. Why not? e Briefly describe two safety precautions that should be observed when carrying out this titration. 25 A food and drugs authority analysed a sample of light beer to see if it conformed with the regulation of a maximum of 2% by mass of alcohol (ethanol). The alcohol content was determined by volumetric analysis according to the reaction:

Sa

m

pl e

pa ge s

Volumetric analysis

dichromate (K2Cr2O7). Three separate titrations gave titres of 9.20 mL, 9.16 mL and 9.22 mL. Calculate: a the amount, in mol, of Cr2O72– present in the average of the titres b the amount, in mol, of ethanol present in each 20.00 mL aliquot c the amount, in mol, of ethanol in the original 10.00 mL sample of beer d the mass of ethanol in the original sample e the percentage by mass of alcohol in the beer, if the density of light beer is 1.10 g mL–1. Would this product conform with the regulations for low-alcohol beer? 26 Megavit multivitamin tablets contain iron(II) sulfate. The iron content of a sample of Megavit tablets was analysed. Ten tablets were crushed and ground into a paste with a little dilute sulfuric acid. All of the paste was carefully transferred into a 250.0 mL volumetric flask and the solution was made up to the mark with more dilute sulfuric acid. 20.00 mL aliquots of the solution were titrated with 0.0200 M potassium permanganate solution and a mean titre of 12.95 mL was obtained. a Write the half equation for the oxidation of Fe2+ to Fe3+. b Write the half equation for the reduction of purple MnO4– to colourless Mn2+ in acidic solution. c Write a balanced ionic equation to represent the overall reaction. d Calculate the amount, in mol, of Fe2+ present in each aliquot. e Calculate the amount of Fe2+ in the 250.0 mL volumetric flask. f Calculate the mass of iron(II) sulfate present in each tablet. g The label says that each tablet contains 250 mg FeSO4. How do these results compare with the manufacturer’s specification? List two possible sources of error that could account for any discrepancy. 27 The amount of vitamin C in fruit juice can be determined by titration with a standard 0.0100 M iodine solution:

2Cr2O72–(aq) + 3C2H5OH(aq) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)

The beer was tested by taking a 10.00 mL sample and making it up to 250 mL in a standard flask. 20.00 mL aliquots were titrated against a 0.0500 M solution of potassium

C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I–(aq)

If the maximum concentration of vitamin C is likely to be 0.000 50 g mL–1, describe how you would perform the analysis. You should mention: i the volume of fruit juice used ii the maximum titre of iodine you would expect to obtain. 28 The alcohol content of an imported brandy was found by taking 10.0 mL and diluting it to 500 mL. 20.00 mL aliquots of this solution were then titrated against 0.100 M acidified potassium dichromate (K2Cr2O7) solution. The mean titre was 17.98 mL.

2Cr2O72–(aq) + 3CH3CH2OH(aq) + 16H+(aq) → 4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)

Calculate the concentration of ethanol in the sample of brandy. Analysing oxidants and reductants

59

29 The active ingredient in bathroom mould-killers is the bleach sodium hypochlorite (NaOCl). The concentration of this chemical in a 20.00 mL sample was determined by adding an acidified solution containing an excess of I– ions to the sample. This reacted according to the equation:

The iodine formed by this reaction was titrated using 0.750 M sodium thiosulfate (Na2S2O3) solution. I2(aq) + 2S2O32–(aq) → 2I–(aq) + S4O62–(aq)

! a Write a balanced equation eq for the reaction between the permanganate and oxalate ions. b A solution containing 0.161 g of sodium oxalate, Na2(COO)2, reacted with 26.7 mL of a potassium permanganate solution. Determine the concentration of this solution. 34 Liquid household bleach contains sodium hypochlorite (NaOCl) as its active ingredient. The manufacturer maintains quality control by performing volumetric analysis on random samples taken from the end of the production line. In the analytical laboratory a 25.00 mL sample was diluted to 250.0 mL using distilled water. 20.00 mL aliquots were then pipetted into clean conical flasks and excess acidified potassium iodide solution was added to each flask to cause the reaction:

pa ge s

25.10 mL of the thiosulfate solution was required to reach the end point. Calculate: a the amount, in mol, of I2 reacting with the S2O32– ions b the amount of NaOCl in the sample c the percentage of NaOCl, by mass, in the mould-killer. (Assume the density of the solution is 1.0 g mL–1.)

Connecting the main ideas

Sa

m

pl e

30 What is the oxidation number of the underlined element in each of the following ions? a Fe2+ b Cr3+ c CrO42– d Cr2O72– 2– 3– – e SO3 f PO4 g NO3 h MnO4– 2+ – i VO j OCl 31 Use oxidation numbers to determine whether each of the following equations represents a redox reaction and, if it is a redox reaction, identify the oxidant and reductant. a 3Mg(s) + Fe2O3(s) → 3MgO(s) + 2Fe(s) b AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) c Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) d Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(aq) 32 For each of the following reactions write: i the oxidation half equation ii the reduction half equation iii the balanced ionic equation. a Chlorine gas is bubbled into sodium iodide solution to form a solution of sodium chloride and iodine. b A piece of copper wire is dipped into silver nitrate solution, producing silver crystals and copper(II) nitrate solution. 33 Potassium permanganate is used in many redox titrations. No indicator is required as the permanganate ion is purple while the manganese(II) ion is colourless. MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

Potassium permanganate cannot be used as a primary standard because it is slightly unstable. Prior to use for analytical purposes a potassium permanganate solution must be standardised. Sodium oxalate, Na2(COO)2, can be used as a

Chemical analysis C

(COO)22–(aq) → 2CO2(g) + 2e– Oxalic acid, (COOH)2, is a toxic diprotic acid found in rhubarb leaves.

OCl–(aq) + 2I–(aq) + 2H+(aq) → I2(aq) + Cl–(aq) + H2O(l)

60 06

primary standard for this purpose. During the analysis oxalate ions are oxidised to CO2:

ClO–(aq) + 2I–(aq) + 2H+(aq) → Cl–(aq) + I2(aq) + H2O(l)

The free iodine was then titrated with 0.1000 M sodium thiosulfate solution (Na2S2O3), using starch indicator solution. This indicator gives a blue-black colour in the presence of iodine. The titration reaction is represented by: 2S2O32–(aq) + I2(aq) → S4O62–(aq) + 2I–(aq) The average titre was 20.42 mL. a What is the oxidation state of chlorine in: i ClO–? ii Cl–? b Explain whether sodium hypochlorite acts as an oxidant or reductant when it is mixed with potassium iodide solution. c What colour change is observed at the end point of the titration? d What amount of sodium thiosulfate, in mol, was used in the titration with the free iodine? e What amount of iodine (I2), in mol, was released by the reaction of the sodium hypochlorite? f Deduce the amount, in mol, of sodium hypochlorite present in each 20.00 mL aliquot of diluted bleach solution. g What amount of sodium hypochlorite, in mol, was present in the original 25.00 mL sample of bleach? h What mass of sodium hypochlorite was present in the original 25.00 mL sample of the household bleach? i What is the concentration of sodium hypochlorite in g L–1?