chem ch5

Worked solutions to student book questions Chapter 5 Analysing oxidants and reductants

Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

Q1. Name the chemicals that undergo oxidation in the following reactions. a 2Zn(s) + O2(g) 2ZnO(s) b Ca(s) + Cl2(g) CaCl2(s) c 2AgBr(s) 2Ag(s) + Br2(g)

A1. a Zn b Ca c Br ions in AgBr

Q2. Identify the oxidants and reductants in each of the reactions in Question 1.

A2.

(Oxidant is listed first.) a O2; Zn b Cl2; Ca c Ag+ in AgBr; Br in AgBr

Q3. State the oxidation number of carbon in: a CO b CO2 c CH4 d C (graphite) e HCO3

A3. a +2 b +4 c 4 d 0 e +4



Q4. Which one or more of the following substances contain manganese in the +6 oxidation state: MnCl2, MnCl3, MnO2, K2MnO4, KMnO4?

A4. K2MnO4 the oxidation state of K is +1, O is 2 2 +1 + ? + 4 2 = 0 hence ? = +6

Q5. Find the oxidation numbers of the elements in the following compounds or ions. Hint: For ionic compounds, use the charge on each ion to help you. a CaO b CaCl2 c HSO4 d MnO4 e F2 f SO32 g NaNO3 h K2Cr2O7

A5. a Ca: +2; O: 2 b Ca: +2; Cl: 1 c H: +1; S: +6; O: 2 d Mn: +7; O: 2 e F: 0 f S: +4; O: 2 g Na: +1; N: +5; O: 2 h K: +1; Cr: +6; O: 2

Q6. 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)

A6.

(Oxidant is listed first.) a Cl2; Mg b O2; SO2 c Fe2O3; CO d H2O2; Fe2+



Q7. 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.

A7.

a MnO2(s) + 4H+(aq) + 2e Mn2+(aq) + 2H2O(l) b MnO4(aq) + 4H+(aq) + 3e MnO2(s) + 2H2O(l) c SO42(aq) + 10H+(aq) + 8e H2S(g) + 4H2O(l) d SO2(g) + 2H2O(l) SO42(aq) + 4H+(aq) + 2e e H2S(g) S(s) + 2H+(aq) + 2e f SO32(aq) + H2O(l) SO42(aq) + 2H+(aq) + 2e

Q8. 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.

A8.

Cu2+(aq) + 2e Cu(s) Fe(s) Fe2+(aq) + 2e Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq)

Q9. When zinc powder is sprinkled into an acidified solution of potassium dichromate, a reaction occurs which 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.

A9.

a Zn(s) Zn2+(aq) + 2e b Cr2O72(aq) + 14H+(aq) + 6e 2Cr3+(aq) + 7H2O(l) c 3Zn(s) + Cr2O72(aq) + 14H+(aq) 3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l)



Q10. 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. 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.

A10.

a Fe2+(aq) Fe3+(aq) + e Cr2O72(aq) + 14H+(aq) + 6e 2Cr3+(aq) + 7H2O(l) Cr2O72(aq) + 14H+(aq) + 6Fe2+(aq) 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

b SO32(aq) + H2O(l) SO42(aq) + 2H+(aq) + 2e

MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) 5SO32(aq) + 2MnO4(aq) + 6H+(aq) 5SO42(aq) + 2Mn2+(aq) + 3H2O(l)

c MnO2(s) + 4H+(aq) +2e Mn2+(aq) + 2H2O(l) 2Cl(aq) Cl2(g) + 2e

MnO2(s) + 4H+(aq) +2Cl(aq) Mn2+(aq) + 2H2O(l) + Cl2(g)

Q11. The following equations are not balanced. i Identify the species that has been reduced and the species that has been oxidised. ii Write balanced half equations for the oxidation and reduction reactions. iii Combine the half equations to write a balanced equation.

a Ce4+(aq) + H2S(g) Ce3+(aq) + S(s) + H+(aq) b NO3(aq) + H+(aq) + Cu(s) NO(g) + H2O(l) + Cu2+(aq) c H2O2(aq) + Br(aq) + H+(aq) Br2(l) + H2O(l) d MnO2(s) + H+(aq) + S(s) Mn2+(aq) + H2O(l) + SO2(g)

A11. a i Ce4+ is reduced to Ce3+ and H2S is oxidised to S

ii Ce4+(aq) + e Ce3+ (aq) H2S(aq) S(s) + 2H+(aq) + 2e

iii 2Ce4+(aq) + H2S(aq) 2Ce3+(aq) + S(s) + 2H+(aq) b i NO3 is reduced to NO and Cu is oxidised to Cu2+

ii NO3 (aq) + 4H+(aq) + 3e NO(g) + 2H2O(l) Cu(s) Cu2+(aq) + 2e

iii 2NO3 (aq) + 8H+(aq) + 3Cu 2NO(g) + 4H2O(l) + 3Cu2+(aq) c i H2O2 is reduced to H2O and Br is oxidised to Br2

ii H2O2(aq) + 2H+(aq) + 2e 2H2O(l) 2Br(aq) Br2(l) + 2e

iii H2O2(aq) + 2H+(aq) + 2Br(aq) 2H2O(l) + Br2(l)



d i MnO2 is reduced to Mn2+ and S is oxidised to SO2 ii 2MnO2(s) + 4H+(aq) + 2e Mn2+ + 2H2O(l)

S (s) + 2H2O(l) SO2(g) + 4H+(aq) + 4e iii MnO2(s) + 4H+(aq) + S(s) 2Mn2+ + 2H2O(l) + SO2(g)

Q12. 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?

A12.

From equation: n(H2O2) = 25 n(MnO4)

= 25

100025020 . mol

= 0.00125 mol

c(H2O2) = Vn

= 02.0

00125.0 M

= 0.0625 M

Q13. An artist uses 10.0 mL of 15.0 M HNO3 to etch a design into a sheet of copper. What mass of copper will have reacted with the acid?

Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 2H2O(l) + 2NO2(g)

A13.

From equation: n(Cu) = 41 n(HNO3)

= 41

10000.100.15 mol

= 0.0375 mol m(Cu) = mMr = 0.0375 63.54 g = 2.38 g



Chapter review

Q14. 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

A14. a +4 b 2 c +6 d +6 e +4 f +2

Q15.

Which of the following may be regarded as redox reactions? Give reasons for each of your answers. a BaCl2(aq) + H2SO4(aq) BaSO4(s) + 2HCl(aq) b 2Ag(s) + Cl2(g) 2AgCl(s) c 2FeCl3(aq) + SnCl2(aq) 2FeCl2(aq) + SnCl4(aq) d ZnCO3(s) ZnO(s) + CO2(g) e HPO32(aq) + I2(aq) + OH(aq) H2PO4(aq) + 2I(aq) f 2Cu+(aq) Cu2+(aq) + Cu(s) g CaF2(s) Ca2+(aq) + 2F(aq) h P4(s) + 6H2(g) 4PH3(g)

A15. b, c, e, f and h are redox reactions as the elements in the reactions undergo changes in oxidation number during the course of the reaction. The changes in oxidation number which occur are: b Ag from 0 to +1; Cl from 0 to 1 c Fe from +3 to +2; Sn from +2 to +4 e P from +3 to +5; I from 0 to 1 f Cu from +1 to +2; Cu from +1 to 0 h P from 0 to 3; H from 0 to +1



Q16. 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.

A16. a i 0

ii +2 b i +5

ii +4 c oxidant HNO3; reductant Cu

Q17. 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.

A17.

a Al(s) Al3+(aq) + 3e 2H+(aq) + 2e H2(g)

b Since equal numbers of electrons are produced and consumed during a redox reaction, two aluminium atoms must lose six electrons in order to reduce six hydrogen ions.

Q18.

As a result of a road traffic accident, residents in the Melbourne suburb of Kensington had to be evacuated when toxic fumes leaked from a container of sodium dithionite (Na2S2O4). The dithionite ion reacts with water according to the equation:

2S2O42(aq) + H2O(l) S2O32(aq) + 2HSO3(aq) 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.



A18. a i +3

ii +2 iii +4

b S2O42(aq) + 2H2O(l) 2HSO3(aq) + 2H+(aq) + 2e

S2O42(aq) + 2H+(aq) + 2e S2O32(aq) + H2O(l)

Q19.

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.

A19.

a Zn(s) Zn2+(aq) + 2e

Pb2+(aq) + 2e Pb(s) Zn(s) + Pb2+(aq) Zn2+(aq) + Pb(s)

b Fe2+(aq) Fe3+(aq) + e

MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) 5Fe2+(aq) + MnO4(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

c SO2(aq) + 2H2O(l) SO42(aq) + 4H+(aq) + 2e

I2(aq) + 2e 2I(aq) SO2(aq) + 2H2O(l) + I2(aq) SO42(aq) + 4H+(aq) + 2I(aq)

d OCl(aq) + 2H+(aq) + 2e Cl(aq) + H2O(l) 2I(aq) I2(aq) + 2e

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

Q20. 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.

A20. a Half equations:

Zn(s) Zn2+(aq) + 2e

2MnO2(s) + 2H+(aq) + 2e Mn2O3(s) + H2O(l)



Overall equation: Zn(s) + 2MnO2(s) + 2H+(aq) Zn2+(aq) + Mn2O3(s) + H2O(l)

b Step 1 Calculate the amount of MnO2 in the dry cell.

n(O2) = 1mol g 86.94g 8.0

= 0.09202 mol Step 2 From the equation, 1 mol Zn reacts with 2 mol MnO2.

)(MnO

(Zn)

2nn =

21

n(Zn) = 2

mol 0.09202

= 0.04601 mol Step 3 Calculate the mass of Zn. m(Zn) = 0.04601 mol 65.38 g mol1 = 3.0 g (two significant figures)

Q21.

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)

A21. Step 1 Write a balanced equation. Ag(s) + 2HNO3(aq) AgNO3(aq) + H2O(l) + NO2(g) Step 2 Calculate the amount of the given reactant, HNO3. n(HNO3) = 2.000 L 10.00 M = 20.00 mol Step 3 From the equation, 1 mol Ag reacts with 2 mol HNO3.

)(HNO

(Ag)

3nn =

21

n(Ag) = 2

mol 20.00

= 10.00 mol Step 4 Calculate the mass of Ag. m(Ag) = 10.00 mol 107.87 g mol1 = 1079 g (four significant figures)

Q22. The thermite process can be used to weld lengths of railway track together. A mould is placed over the two ends of rails to be joined and it 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. b Is the half equation you wrote for part a an oxidation or 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?

A22.

a Fe2O3(s) + 6e 2Fe(l) + 3O2(s) b reduction c Fe2O3(s) + 2Al(s) 2Fe(l) + Al2O3(s) d Step 1 Write a balanced equation. Fe2O3(s) + 6e 2Fe(l) + 3O2(s) Step 2 Calculate the amount of the given reactant, Fe.

n(Fe) = 1mol g 55.847g 3.70

= 0.066252 mol Step 3 From the equation, 1 mol of Fe2O3 produces 2 mol Fe.

)(Fe

)O(Fe 32n

n =

21

n(Fe2O3) = 2mol 0.066252

= 0.033126 mol Step 4 Calculate the mass of Fe2O3. m(Fe2O3) = 0.033126 mol 159.694 g mol1 = 5.29 g (three significant figures)

Q23. 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.

A23. a Step 1 Write a balanced equation. C6H8O6(aq) + I2(aq) C6H6O6(aq) + 2H+(aq) + 2I(aq) Step 2 Calculate the amount of the given reactant, I2. n(I2) = 0.0500 M 0.02840 L = 0.00142 mol Step 3 From the equation, 1 mol vitamin C reacts with 1 mol I2.

)(I

)OH(C

2

686

nn

= 11

n(C6H8O6) = 0.00142 mol Step 4 Calculate the mass of vitamin C. m(C6H8O6) = 0.00142 mol 176.124 g mol1



= 0.2500 g = 0.250 g (three significant figures)

b % C6H8O6 = 1.3060.2500 100

= 19.1% (three significant figures) c Fillers, binders, sweeteners, flavours and colouring.

Q24.

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.

A24.

a 5Fe2+(aq) + MnO4(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(aq) b Step 1 Calculate the amount of the given reactant, MnO4.

n(MnO4) = 0.0300 M 0.02022 L = 0.006066 mol Step 2 From the equation, 5 mol of Fe2+ reacts with 1 mol of MnO4.

)(MnO

)(Fe

4

2

+

nn =

15

n(Fe2+) = 5 0.006066 mol = 0.003033 mol Step 3 The 0.200 g wire sample was dissolved, making up a 25 mL solution,

which was titrated. n(Fe2+)in 25 mL solution = 0.00303 mol (three significant figures)

c % Fe = g 0.200

mol g 55.847 mol 0.003033 1 100

= 84.7 % (three significant figures) d The end point is readily recognised by the first permanent tinge of pink caused by

excess permanganate ions. e Wear safety glasses and a laboratory coat; hold pipette close to the top when

fitting pipette filler.



Q25. 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:

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 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 mL1. Would this product conform with the regulations for low-alcohol beer?

A25. a Step 1 Write a balanced equation. 2Cr2O72(aq) + 3CH3CH2OH(aq) + 16H+(aq) 2Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l) Step 2 Calculate the average titre.

Average titre = 3

9.22 9.16 9.20 ++ mL

= 9.193 mL = 0.00919 L Step 3 Calculate the amount of the Cr2O72 present in the average titre.

n(Cr2O72) = 0.0500 M 0.00919 L = 0.0004595 mol = 0.000460 mol (three significant figures) b From the equation, 3 mol of CH3CH2OH reacts with 2 mol of Cr2O72.

)O(Cr

OH)CH(CH2

72

23n

n =

23

n(CH3CH2OH)in 20.00 mL = 20.0004595 3 mol

= 0.00068925 mol = 0.000690 mol c Calculate the amount of CH3CH2OH in the 250.0 mL flask, remembering that

only 20.00 mL was removed from the 250.0 mL flask for the titration. This is the amount of CH3CH2OH in the 10 mL beer sample.

n(CH3CH2OH)in 250.0 mL = 0.00068925 00.200.250 mol

= 0.008615 mol = 0.00862 mol (three significant figures)



d Calculate the mass of CH3CH2OH in the beer sample. m(CH3CH2OH) = 0.008615 mol 46 g mol1 = 0.3963 g = 0.396 g (three significant figures)

e Step 1 Calculate the mass of the 10.00 mL sample of beer, using the density

given and the formula, density = Vm , so m = density V.

m(CH3CH2OH) = 1.10 g mL1 10.00 mL = 11.00 g

Step 2 Calculate the percentage by mass of CH3CH2OH in the sample of beer.

% CH3CH2OH = 11.000.3963 100%

= 3.60% (three significant figures) No; this product would not conform with the regulations for low-alcohol beer.

Q26. 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. 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 manufacturers specification? List two possible sources of error that could account for any discrepancy.

A26.

a Fe2+(aq) Fe3+(aq) + e b MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) c 5Fe2+(aq) + MnO4(aq) + 8H+(aq) 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) d Step 1 Calculate the amount of MnO4.

n(MnO4) = 0.0200 M 0.01295 L = 0.000259 mol Step 2 From the equation, 5 mol of Fe2+ reacts with 1 mol of MnO4.

)(MnO

)(Fe

4

2

+

nn =

15

n(Fe2+) in each 20.00 mL aliquot = 5 0.000259 mol = 0.001295 mol = 0.00130 mol



e Calculate the amount of Fe2+ in the 250.0 mL flask, remembering that only 20.00 mL was removed from the 250.0 mL flask for the titration.

n(Fe2+) in 250.0 mL = 0.00130 20.00250.0 mol

= 0.0161875 mol = 0.0162 mol (three significant figures)

f From part e, this is the amount of Fe2+ in 10 tablets, which equals the amount of FeSO4 in 10 tablets. Calculate the mass of FeSO4 in the 10 tablets, and then the mass of Fe2+ in 1 tablet. m(FeSO4) in 10 tablets = 0.0161875 mol 151.847 g mol1 = 2.4580 g

m(FeSO4) in 1 tablet = 10g 2.4580

= 0.246 g (three significant figures) g Low results, due to loss of vitamin paste during transfer, or aliquots being less

than the correct volume.

Q27. The amount of vitamin C in fruit juice can be determined by titration with a standard 0.0100 M iodine solution:

C6H8O6(aq) + I2(aq) C6H6O6(aq) + 2H+(aq) + 2I(aq) If the maximum concentration of vitamin C is likely to be 0.00050 g mL1, describe how you would perform the analysis. You should mention:

the volume of fruit juice used the maximum titre of iodine you would expect to obtain.

A27.

If the maximum concentration of vitamin C in fruit juice were 0.00050 g mL1, titration of an aliquot of 50.00 mL of fruit juice with 0.0100 M iodine solution would give a maximum titre of 14.20 mL. The titration is a direct one using starch solution as the indicator. It involves the following steps. Step 1 Place the standard solution of iodine in a burette. Record the initial volume. Step 2 Place an aliquot of the juice in a conical flask. Step 3 Add 23 drops of indicator to the juice. Step 4 Titrate the juice with the iodine solution. Record the volume of solution used

to reach the end point. Step 5 Repeat steps 14 to obtain three concordant titres (titres within 0.1 mL).



Q28. 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.

A28.

Step 1 Write a balanced equation. 2Cr2O72(aq) + 3CH3CH2OH(aq) + 16H+(aq) 2Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l) Step 2 Calculate the amount of Cr2O72 used in the titration. n(Cr2O72) = 0.100 M 0.01798 L = 0.001798 mol Step 3 From the equation, 3 mol of CH3CH2OH reacts with 2 mol Cr2O72.

)O(Cr

OH)CH(CH2

72

23n

n =

23

n(CH3CH2OH) in 20.00 mL aliquots = 23 n(Cr2O72)

= 23 0.001798 mol

= 0.002697 mol Step 4 Calculate the amount of CH3CH2OH in the 500.0 mL flask, remembering

that only 20.00 mL was removed from the 500.0 mL flask for the titration.

n(CH3CH2OH) in 500.0 mL = 0.002697 00.200.500 mol

= 0.067425 mol Step 5 This is the same amount of CH3CH2OH as is present in the 10.00 mL sample.

Calculate the concentration of CH3CH2OH in the 10.00 mL sample.

c(CH3CH2OH) in the 10.00 mL sample = L0.0100mol 0.067425

= 6.74 M (three significant figures)



Q29. The active ingredient in bathroom mould-killers is the bleaching agent 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:

OCl(aq) + 2I(aq) + 2H+(aq) I2(aq) + Cl(aq) + H2O(l) The iodine formed by this reaction was titrated using 0.750 M sodium thiosulfate (Na2S2O3) solution.

I2(aq) + 2S2O32(aq) 2I(aq) + S4O62(aq) 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 mL1.)

A29. a Step 1 Write a balanced equation for the titration. I2(aq) + 2S2O32(aq) 2I(aq) + S4O62(aq) Step 2 Calculate the amount of S2O32 used in the titration. n(S2O32) = 0.750 M 0.02510 L = 0.018825 mol Step 3 From the equation, 1 mol I2 reacts with 2 mol S2O32.

)O(S

)(I2

32

2n

n = 21

n(I2) = 20.018825

= 0.0094125 mol = 0.00941 mol b Step 1 Write a balanced equation for the reaction where the bleach produces I2. OCl(aq) + 2I(aq) + 2H+(aq) I2(aq) + Cl(aq) + H2O(l) Step 2 From the equation, 1 mol of NaOCl produces 1 mol of I2.

)(I

)(OCI

2nn =

11

n(OCl) = 0.0094125 mol = 0.00941 mol (three significant figures) c Step 1 Calculate the mass of NaOCl in the 20.00 mL sample of mould-killer. m(NaOCl) = 0.0094125 mol 74.44 g mol1 = 0.7005 g Step 2 The mass of the 20.00 mL sample of mould-killer is 20.00 g, as the

density of mould-killer is 1.0 g mL1. Calculate the percentage of NaOCl in the mould-killer sample.

% NaOCl = 20.000.7005 100

= 3.5% (two significant figures)



Q30. What is the oxidation number of the underlined element in each of the following ions? a Fe2+ b Cr3+ c CrO42 d Cr2O72 e SO32 f PO43 g NO3 h MnO4 i VO2+ j OCl

A30. a +2 b +3 c +6 d +6 e +4 f +5 g +5 h +7 i +4 j +1

Q31. 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)

A31.

a redox reaction; oxidant Fe2O3; reductant Mg b not a redox reaction c redox reaction; oxidant HCl; reductant Zn d redox reaction; oxidant Cl2; reductant KBr

Q32.

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 solution and iodine.



b A piece of copper wire is dipped into silver nitrate solution, producing silver crystals and copper(II) nitrate solution.

A32.

a i 2I(aq) I2(aq) + 2e ii Cl2(g) + 2e 2Cl(aq) iii 2I(aq) + Cl2(g) I2(aq) + 2Cl(aq)

b i Cu(s) Cu2+(aq) + 2e ii Ag+(aq) + e Ag(s) iii Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

Q33.

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

(COO)22(aq) 2CO2(g) + 2e a Write a balanced equation 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.

Oxalic acid, (COOH)2, is a highly toxic diprotic acid found in rhubarb leaves.

A33.

a MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) multiply by 2 (COO)22(aq) 2CO2(g) + 2e multiply by 5 then add the equations 2MnO4(aq) + 16H+(aq) + 10e + 5(COO)22(aq)

2Mn2+(aq) + 8H2O(l) + 10CO2(g) + 10e simplify 2MnO4(aq) + 16H+(aq) + 5(COO)22(aq) 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

b n(Na2(COO)2) = 134161.0 mol

from the equation: n(MnO4) = 52 n((COO)22)

= 52

134161.0 mol

= 0.000481 mol

c(MnO4) = Vn

= 0267.0000481.0

M = 0.0180 M



Q34. 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:

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 L1?

A34. a i +1

ii 1 b oxidant c blue-black to colourless d n(S2O32) = 0.1000 M 0.02042 L = 0.002042 mol (four significant figures)



e From the equation for the titration reaction, 1 mol of I2 reacted with 2 mol of S2O32. Calculate the amount of free I2. This is I2 released from reaction of 20.00 mL aliquots of diluted bleach with I.

)O(S)(I2

32

2n

n = 21

n(I2) = 2

)O(S 232n

= 2

002042.0

= 0.001021 mol f Step 1 Write the balanced equation for the reaction of I with ClO. ClO(aq) + 2I(aq) + 2H+(aq) Cl(aq) + I2(aq) + H2O(l) Step 2 From the equation, 1 mol of ClO produces 1 mol of I2. Calculate the

amount of ClO in the 20.00 mL sample of diluted bleach.

)(I

)(ClO

2nn =

11

n(ClO) in 20.00 mL sample of diluted bleach = 0.001021 mol

g n(ClO) in 25 mL of bleach = n(ClO) in 20.00 mL of diluted bleach 00.20

0.250

= 0.001021 00.20

0.250 mol

= 0.01276 mol h m(NaOCl) = 0.01276 mol 74.5 g mol1 = 0.950806 g = 0.9506 g (four significant figures) i Convert mass from part h from g/25.00 mL to g/1000 mL.

c(NaOCl) = (0.9506 00.25

1000 ) g L1

= 38.02 g L1 (four significant figures)

chem ch5

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