Chem ch28

Worked solutions to student book questions Chapter 28 Electrolysis: driving chemical reactions by electricity

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

E1. Use the Internet or another reference to find out why phenol (and, if possible, phenol sulfonic acid) is regarded as a hazardous material.

AE1. High levels of exposure of humans to phenol is reported to cause effects such as headaches, respiratory irritation, fatigue, liver damage, diarrhoea and haemolytic anaemia. It causes irritation to the lungs of animals. Phenol remains in the soil for about 25 days and in water for longer than 9 days. Hazards of phenol sulfonic acid include being corrosive, causing severe burns to the eyes and skin, damaging the liver and other organs, and causing pulmonary oedema (accumulation of fluids in the lungs).

E2. Before the steel plate enters the tin plating cell it is cleaned thoroughly. Why is this cleaning process essential?

AE2. In order for the tin plating to adhere to the metal, iron oxide and grease present on the steel surface must first be removed.

E3. Why does the presence of a strong acid in the electrolyte allow higher rates of tin plating than a weak acid?

AE3. There are more ions present in a solution of a strong acid compared with a solution of a weak acid of the same concentration. As a consequence, a strong acid solution will conduct electricity better and allow higher rates of plating.

Q1. A student wishes to copper-plate her iron locker key. a Design a suitable electroplating cell. Identify the anode and cathode, electrode

polarities and the electrolyte to be used. b If the student connects the cell to the wrong terminals of the power supply, what

happens to the locker key?



A1.

a

b If the student were to connect the iron key to the positive terminal of the power supply, it would be oxidised. Copper metal would plate the negative electrode.

Q2. Draw a diagram to show the migration of ions in a nickel-plating cell containing nickel sulfate solution.

A2.

Q3. Calculate the quantity of electric charge, in coulomb, represented by: a a current of 6.00 A flowing for 25.0 minutes b a current of 25 mA flowing for 2.0 days

A3. a Q = It Q = 6.00 A (25.0 60) s = 9.00 103 C b Q = It Q = (25 103) A (2 24 60 60) s = 4.3 103 C



Q4. The electrolyte in an electrolytic cell used to chromium-plate bicycle frames contains Cr3+ ions. If the cell operates at 30.0 A for 25.0 minutes, calculate: a the quantity of charge that passes through the cell b the amount, in mol, of electrons that pass through the cell, in mol c the mass of metal deposited on the frames

A4. a Q = It Q = 30.0 A (25 60) s = 4.50 104 C

b n(e) = FQ

n(e) = 14

mol C 50096C1050.4

= 0.466 mol c Step 1 Write the equation. Cr3+(aq) + 3e Cr(s) Step 2 Use stoichiometry to calculate the amount of Cr.

)e()Cr(

nn =

31

n(Cr) = 3

4663.0

= 0.1554 mol Step 3 Calculate the mass of Cr. m(Cr) = 0.1554 mol 52.00 g mol1 = 8.08 g



Q5. A 0.300 g mass of silver is to be used to plate a sporting trophy. How long should the trophy be left in a silver-plating cell using a current of 6.00 A?

A5. Step 1 Write the equation. Ag+(aq) + e Ag(s) Step 2 Calculate the amount of Ag.

n(Ag) = 1mol g 87.107g300.0

= 0.002 781 mol Step 3 Use stoichiometry to calculate the amount in mol of electrons.

)Ag()e(

nn =

11

So n(e) = 0.002781 mol Step 4 Calculate the quantity of electricity required. Q = n(e) F = 0.002 781 mol 96 500 C mol1 = 268 C

Step 5 Calculate the time required, using I = tQ .

t = IQ

= A 00.6C268

= 44.7 s



Q6. A solution used to gold-plate bathroom fittings contains Au3+ ions. What current would be required to plate 5.00 g of gold onto a tap in 30.0 minutes?

A6. Step 1 Write the equation. Au3+(aq) + 3e Au(s) Step 2 Calculate the amount of Au.

n(Au) = 1mol g 967.196g00.5

= 0.0254 mol Step 3 Use stoichiometry to calculate the amount of electrons.

)Au()e(

nn =

13

n(e) = 3 0.0254 mol = 0.076 mol Step 4 Calculate the quantity of electricity required. Q = n(e) F = 0.076 96 500 = 7348.9 C

Step 5 Calculate the current, using I = tQ .

I = s 600.30

C 9.7348

= 4.08 A

Q7.

Predict the products at each electrode during electrolysis (using unreactive electrodes) of the following 1 M solutions. The nitrate ion is not involved in any of the reactions. a copper(II) bromide b sodium iodide c lead(II) nitrate d zinc chloride e aluminium nitrate

A7. a Br2(aq); Cu(s) b I2(aq); H2(g) and OH(aq) c O2(g) and H+(aq); Pb(s) d O2(g) and H+(aq) (Cl2(g) is formed in practice); Zn(s) e O2(g) and H+(aq); H2(g) and OH(aq)



Q8. From the industrial electrolytic cells described in the text, select one that uses a molten electrolyte and another that uses an aqueous electrolyte. a Write the half reactions that occur at each electrode. b What materials are used for the anode and cathode in each cell? c Explain the reasons for the choice of the electrode materials. d Why is an aqueous electrolyte more energy efficient than a molten electrolyte? e Explain why aqueous solutions are not employed in the cell using a molten

electrolyte. f What factors may be important in deciding where to site an industrial cell of each

type?

A8. Refer to the appropriate descriptions of each cell in the text to answer this question.

Chapter review

Q9. a Explain what is meant by a faraday of charge. b State the number of faradays of charge needed to produce 1 mole of:

i silver atoms from silver nitrate solution ii zinc atoms from zinc nitrate solution iii chlorine molecules from molten potassium chloride iv hydrogen molecules from water

A9. a A faraday is the charge on 1 mole of electrons, 96 500 C. b i 1 F ii 2 F iii 2 F iv 2 F

Q10.

A car battery is charged at a current of 2.0 A for 9.00 hours. Calculate the mass of lead deposited at the cathode if the electrode reaction is represented by the equation:

PbSO4(s) + 2e Pb(s) + SO42(aq)

A10.

Step 1 Calculate the number of coulomb, using Q = It. Q = 2.0 A (9.00 60 60) s = 64 800 C

Step 2 Calculate the amount in mol of electrons, using n = FQ .

n(e) = 1mol C 500 96C 800 64

= 0.6715 mol



Step 3 Use stoichiometry to calculate the amount of Pb.

)e()Pb(

nn =

21

n(Pb) = 2

6715.0

= 0.3358 mol

Step 4 Calculate the mass of Pb, using n = Mm .

m(Pb) = 0.3358 mol 207.2 g mol1 = 69.57 g = 70 g

Q11. Sodium chloride solution is electrolysed in a diaphragm cell to manufacture chlorine. The cell operates at 100 000 A. For one days operation, calculate: a the volume of chlorine produced, measured at SLC b the mass of sodium chloride consumed

A11. a Step 1 Write a balanced equation. 2Cl(aq) Cl2(g) + 2e Step 2 Calculate the number of coulomb, using Q = It. Q = 100 000 A (24 60 60) s = 8.64 109 C


n(e) = 19

molC 500 96C 108.64

= 8.953 104 mol Step 4 Use stoichiometry to calculate the amount of Cl2.

)e()Cl( 2

nn =

21

n(Cl2) = 210953.8 4

= 4.477 104 mol

Step 5 Calculate the volume of Cl2, using n = mV

V .

V(Cl2) = (4.477 104) mol 24.5 mol L1 = 1.10 106 L b Step 1 Calculate the number of coulomb, using Q = It. Q = 100 000 A (24 60 60) s = 8.64 109 C




n(e) = 19

mol C 500 96C 108.64

= 8.953 104 mol Step 3 Use stoichiometry to calculate the amount of NaCl.

)e(

)NaCl(n

n = 11

n(NaCl) = 8.953 104 mol

Step 4 Calculate the mass of NaCl, using n = Mm .

m(NaCl) = 8.953 104 mol 58.5 g mol1 = 5.237 106 g = 5.24 tonnes

Q12.

Iron is plated with tin in an electrolytic cell containing K2Sn(OH)6 as the electrolyte. A cell operates for 5.00 hours at a current of 25.0 A. a What is the charge on the tin ions in the electrolyte? b How many faradays of charge are required to produce 1.00 mole of tin? c Calculate the mass of tin deposited during this period.

A12. a Step 1 Write a balanced equation. Remember that K2SnF6 contains Sn4+ ions. Sn4+(aq) + 4e Sn(s) b Step 2 A faraday is the charge on 1 mol electrons. So the number of faraday is

the same as the mole of electrons. Use stoichiometry to calculate the amount of electrons.

)Sn()e(

nn =

14

n(e) = 4 1.00 mol = 4.00 mol So number of faraday = 4.00 F c Step 1 Calculate the number of coulomb, using Q = It. Q = 25.0 A (5.00 60 60) s = 450 000 C


n(e) = 1mol C 500 96C 000 450

= 4.663 mol Step 3 Use stoichiometry to calculate the amount of Sn.

)e()Sn(

nn =

41



n(Sn) = 4663.4

= 1.166 mol

Step 4 Calculate the mass of tin, using n = Mm .

m(Sn) = 1.166 mol 118.69 g mol1 = 139 g

Q13. A charge of 0.400 faraday was passed through 1.00 L of 1.00 M copper(II) sulfate solution using carbon electrodes. a Write equations for the reactions at each electrode. b Calculate the concentration of the copper ions in solution after electrolysis.

A13.

2H2O(l) 4H+(aq) + O2(g) + 4e

Cu2+(aq) + 2e Cu(s) b Step 1 Calculate the amount in mol of electrons, remembering that 1 F is the

charge on to 1 mol electrons. n(e) = 0.400 mol Step 2 Use stoichiometry to calculate the amount of Cu2+ deposited as Cu.

)e(

)Cu( 2

+

nn =

21

n(Cu2+) = 2400.0

= 0.200 mol Step 3 Calculate the amount of Cu2+ remaining. n(Cu2+)remaining = n(Cu2+)initial n(Cu2+)discharged = (1.00 L 1.00 M) mol 0.200 mol = 0.800 mol Step 4 Calculate the concentration of Cu2+ remaining.

c(Cu2+) = L 1.00mol 0.800

= 0.800 M



Q14. Three electrolytic cells containing silver nitrate solution, copper(II) sulfate solution and chromium(III) sulfate solution, respectively, were connected in series so that the same amount of electric charge passed through each cell. Metal was deposited at the cathode of each cell. If 10.0 g of silver was obtained from one cell, what mass of metal would be obtained from each of the other cells?

A14. Step 1 Write a balanced equation for each reaction. Ag+(aq) + e Ag(s) Cu2+(aq) + 2e Cu(s) Cr3+(aq) + 3e Cr(s)

Step 2 Calculate the amount of Ag deposited, using n = Mm .

n(Ag) = 1mol g 107.87g 10.0

= 0.09270 mol Step 3 Use stoichiometry to calculate the amount in mol of electrons that passed

through all cells.

)Ag()e(

nn =

11

n(e) = 0.09270 mol Step 4 Use stoichiometry to calculate the amount of Cu and Cr deposited.

)e()Cu(

nn =

21

n(Cu) = 2

0.09270

= 0.04635 mol

)e()Cr(

nn =

31

n(Cr) = 3

0.09270

= 0.0309 mol Step 5 Calculate the masses of Cu and Cr deposited. m(Cu) = 0.04635 mol 63.55 g mol1 = 2.946 g = 2.95 g m(Cr) = 0.0309 mol 52.00 g mol1 = 1.607 g = 1.61 g



Q15. Electrolysis of concentrated sodium chloride solution produced 15.0 L of chlorine gas, measured at SLC. What mass of silver would be produced if the same amount of electricity were passed through a silver-plating cell?

A15.

Step 1 Write a balanced equation. 2Cl(aq) Cl2(g) + 2e

Step 2 Calculate the amount of Cl2, using n = mV

V .

n = 1L mol 24.5L 15.0

= 0.6122 mol Step 3 Use stoichiometry to calculate the amount in mol of electrons released.

)Cl()e(

2nn =

12

n(e) = 2 0.6122 = 1.224 mol Step 4 Write a balanced equation. Ag+(aq) + e Ag(s) Step 5 Use stoichiometry to calculate the amount of Ag deposited.

)e()Ag(

nn =

11

n(Ag) = 1.224 mol

Step 6 Calculate the mass of Ag, using n = Mm .

m(Ag) = 1.224 g 107.87 g mol1 = 132 g



Q16. A nickel teapot, with a surface area of 900 cm2, is to be silver plated. a Which electrode should the teapot be connected to? b What is the polarity of this electrode? c If the plating is to be 0.005 00 cm thick and the density of silver is 10.5 g cm3,

how long should the pot be put in a silver-plating cell with a current of 0.500 A?

A16. a cathode b negative c Step 1 Calculate the volume of Ag required, using volume = area thickness. V(Ag) = 900 cm2 0.00500 cm = 4.5 cm3

Step 2 Calculate the mass of Ag required, using density = Vm .

m(Ag) = 10.5 g cm3 4.5 cm3 = 47.25 g Step 3 Calculate the amount of Ag.

n(Ag) = 1mol g 107.87g 47.25

= 0.4380 mol Step 4 Use stoichiometry to calculate the amount in mol of electrons required.

)Ag()e(

nn =

11

= 0.4380 mol

Step 5 Calculate the amount of charge required, using n(e) = FQ .

Q = 0.4380 mol 96 500 C mol1 = 42 269 C Step 6 Calculate the time required, using Q = It.

t = A 0.500C 269 42

= 84 539 s = 23.5 h



Q17. Electrolysis of a molten ionic compound with a current of 0.50 A for 30.0 minutes yielded 0.700 g of a metallic element at the cathode. If the element has a relative atomic mass of 150, calculate the charge on the metal ions.

A17.

Step 1 Calculate the charge which passed through circuit, using Q = It. Q = 0.50 A (30.0 60) s = 900 C

Step 2 Calculate the amount in mol of electrons, using n(e) = FQ .

n(e) = 1mol C 500 96C 900

= 0.009 326 4 mol

Step 3 Calculate the amount of metal, using n = Mm .

n = 1mol g 150g 0.700

= 0.004 667 mol Step 4 Calculate the amount of electrons per mole of metal.

)metal(

)e(n

n = mol004667.0mol 0093264.0

= 1.998 = 2 Step 5 Calculate the charge on metal ion. Charge must be a whole number. So it is 2+.



Q18. A student wishes to copper-plate a nickel medallion and sets up an electrolytic cell using 250 mL of 1.00 M copper(II) sulfate solution and a large copper anode. The student runs a current of 10.0 A through the cell for 20 minutes. a What mass of copper will be plated onto the medallion? b What will be the concentration of copper(II) sulfate remaining in solution?

The student replaces the copper anode with an inert graphite electrode and runs the same current through the cell for a further 20 minutes.

c What gas will be produced at the anode? d What will be the concentration of copper(II) sulfate remaining in solution?

A18. a Step 1 Write balanced half equations. Cathode: Cu2+(aq) + 2e Cu(s) Anode: Cu(s) Cu2+(aq) + 2e Step 2 Calculate the quantity of charge, using Q = It. Q = 10.0 A (20 60) s = 12 000 C


n(e) = 1mol C 500 96C 000 12

= 0.1244 mol Step 4 Use stoichiometry to calculate the amount of Cu2+ deposited as Cu.

)e(

)Cu( 2

+

nn =

21

n(Cu2+) = 2

0.1244

= 0.0622 mol Step 5 Calculate the mass of Cu deposited. m(Cu) = 0.0622 mol 63.5 g mol1 = 3.95 g b Cathode: Cu2+(aq) + 2e Cu(s) Anode: Cu(s) Cu2+(aq) + 2e Because the reverse reaction is occurring at each electrode [Cu] will remain

constant, 1.00 M. c Write balanced half equations after the change of electrode. Cathode: Cu2+(aq) + 2e Cu(s) Anode: 2H2O(l) O2(g) + 4H+(aq) + 4e O2(g) is produced



d Step 1 Calculate the quantity of charge, using Q = It. Q = 10.0 A (20 60) s = 12 000 C


n(e) = 1mol C 500 96C 000 12

= 0.1244 mol Step 3 Use stoichiometry to calculate the amount of Cu2+ deposited as Cu.

)e(

)Cu( 2

+

nn =

21

n(Cu2+) = 2

0.1244

= 0.0622 mol Step 4 Calculate the amount of Cu2+ ions remaining in solution. n(Cu2+)remaining = n(Cu)initial n(Cu)discharged = (0.250 1.00) mol 0.0622 mol = 0.1878 mol Step 5 Calculate the concentration of Cu2+ ions remaining in solution.

[Cu2+remaining] = L0.250mol 0.1878

= 0.7512 M = 0.75 M



Q19. A number of important metals are produced by electrolysis of molten salts. In one such process a current of 25 000 A produces 272 kg of metal in 24 hours. If the cation in the salt has a 2+ charge, what is the metal?

A19.

Step 1 Calculate the quantity of charge. Q = It = 25 000 24 60 60 = 2.16 109 C Step 2 Calculate the amount in mol of electrons.

n(e) = FQ

= 50096

1016.2 9

= 2.24 104 mol Step 3 Write a balanced equation. M2+ + 2e M i.e. mole ratio M2+ : e = 1 : 2 Step 4 Calculate the number of moles of M.

n(M) = 21 n(e)

= 21 2.24 104 mol

= 1.12 104 mol Step 5 Mass M produced = 2.72 105 g

Step 6 Using n = Mm

M = nm

= 45

1012.11072.2

= 24.3 g mol1 Step 7 Inspection of a periodic table shows that this divalent metal must be

magnesium.



Q20. Two electrolytic cells are connected in series as shown in Figure 28.25. A current of 5.00 A flows for 15.0 minutes.

Figure 28.25 Electrolytic cells in series.

a Calculate the charge flowing through each cell. b Write half equations for the reactions occurring at each electrode (AD) when the

current commences to flow. c Calculate the change in mass of electrode C after 15 minutes. d Calculate the volume of gas, measured at SLC, that is formed at electrode B after

15.0 minutes. e As the current flows through the cells, how does the reaction at electrode D

change?

A20. a Q = It Q = 5.00 A (15 60) s = 4500 C b A: 2H2O(l) O2(g) + 4H+(aq) + 4e

B: 2H2O(l) + 2e H2(g) + 2OH(aq) C: Cu(s) Cu2+(aq) + 2e D: 2H+(aq) + 2e H2(g)



c Step 1 Write the half equation. Cu(s) Cu2+(aq) + 2e


n(e) = 1molC 500 96C 4500

= 0.0466 mol Step 3 Use stoichiometry to calculate the amount of Cu consumed.

)e()Cu(

nn =

21

n(Cu) = 2

0.0466

= 0.0233 mol Step 4 Calculate the mass of Cu consumed. m(Cu) = 0.0233 mol 63.5 g mol1 = 1.48 g d Step 1 Write the half equation. 2H2O(l) + 2e H2(g) + 2OH(aq)


n(e) = 1molC50096C 4500

= 0.0466 mol Step 3 Use stoichiometry to calculate the amount of H2 produced.

)e()H( 2

nn =

21

n(H2) = 2

0466.0

= 0.0233 mol Step 4 Calculate the volume of H2 gas produced at SLC. V(H2) = 0.0233 mol 24.5 L mol1 = 0.570997 L = 0.571 L e Cu2+(aq) + 2e Cu(s)

Q21. For electrolytic and galvanic cells, compare: a the polarity of the anode and cathode b the direction of electron flow c the energy transformation occurring in the cells d the tendency of the cell reaction to occur spontaneously



A21. a In galvanic cells the anode is negative and the cathode is positive; in electrolytic

cells the anode is positive and the cathode is negative. b In galvanic cells the direction of electron flow is determined by the cell reaction;

in electrolytic cells the direction of electron flow is determined by the external power supply.

c In galvanic cells chemical energy is converted into electrical energy; in electrolytic cells electrical energy is converted into chemical energy.

d Galvanic cell reactions occur spontaneously; electrolytic cell reactions are non-spontaneous.

Q22. Sodium and chlorine are manufactured by passing direct current through molten sodium chloride. a Why would alternating current be unsuitable? b Why is it necessary for the electrolyte to be molten?

A22. a If electrodes were connected to a source of alternating current, the polarity of the

electrodes would change (typically 50 times per second), with the result that little reaction was observed at either electrode.

b The electrolyte needs to be molten, rather than solid, because ions must be able to move to create a current and allow reactions to occur at the electrodes.

Q23. Using the electrochemical series, complete the table by predicting the initial products of each of the following electrolysis experiments.

Experiment Electrolyte Electrodes Cathode reaction

Anode reaction

a Molten potassium iodide Platinum b Copper(II) sulfate

solution Platinum

c Potassium bromide solution

Copper

d A mixture of copper(II) nitrate and nickel(II) nitrate solutions

Carbon

e Lithium fluoride Platinum

A23. Cathode reaction Anode reaction a K+(l) + e K(l) 2I(l) I2(g) + 2e b Cu2+(aq) + 2e Cu(s) 2H2O(l) O2(g) + 4H+(aq) + 4e c 2H2O(l) + 2e H2(g) + 2OH(aq) Cu(s) Cu2+(aq) + 2e d Cu2+(aq) + 2e Cu(s) 2H2O(l) O2(g) + 4H+(aq) + 4e e 2H2O(l) + 2e H2(g) + 2OH(aq) 2H2O(l) O2(g) + 4H+(aq) + 4e



Q24. Early attempts to produce aluminium by electrolysis of aqueous solutions of aluminium compounds were unsuccessful. Use the electrochemical series to explain why.

A24.

The electrochemical series shows that the oxidising strength of H2O is greater than that of Al3+. If water is present in an electrolysis cell, it therefore reacts preferentially at the cathode and electrolysis of aqueous aluminium salts does not yield aluminium metal.

Q25.

A student attempts to electroplate magnesium using a cell made of a magnesium anode, iron cathode and magnesium nitrate solution as electrolyte. The student is dismayed to discover that no magnesium has plated onto the cathode, despite careful checks of all the electrical connections in the electrolytic cell. a Why didnt magnesium plate onto the cathode? b What reaction occurred at the iron cathode?

A25. a The electrolyte solution contains water. As water is a stronger oxidant than Mg2+,

water will be reduced in preference to the Mg2+. b 2H2O(l) + 2e H2(g) + 2OH(aq)

Q26.

Sodium and chlorine are produced by electrolysis of molten sodium chloride in the Downs cell. A typical Downs cell runs at 600C, 7 V and with a current of 30 kA. In a 24-hour period: a what mass of sodium metal will be produced? b what volume of chlorine gas (at SLC) will be produced?



A26. a Step 1 Write half equations. Na+(l) + e Na(l) 2Cl(l) Cl2(g) + 2e Step 2 Calculate the quantity of charge, using Q = It. Q = (30 103) A (24 60 60) s = 2.593 109 C


n(e) = 19

molC 500 96C 102.593

= 26 860.1 mol Step 4 Use stoichiometry to calculate the amount of Na.

)e()Na(

nn =

11

n(Na) = 26 860.1 mol Step 5 Calculate the mass of Na. m(Na) = 26 860.1 mol 23 g mol1 = 618 kg = 6.2 102 g b Step 1 Use stoichiometry to calculate the amount of Cl2.

)e()Cl( 2

nn =

21

n(Cl2) = 2mol 860.1 26

= 13 430.05 mol Step 2 Calculate the volume of Cl2 at SLC. V(Cl2) = 13 430.05 mol 24.5 L mol1 = 329 036.225 L = 3.3 105 L

Q27. Account for the fact that: a most chlorine is produced by electrolysis of sodium chloride solution, rather than

by electrolysis of the molten salt b fluorine was first isolated from fluorine compounds by electrolysis c the concentration of Sn2+ ions in a tin-plating cell (Figure 28.3) remains constant

as the cell operates d although calcium chloride is present in the electrolyte in the Downs cell, calcium

metal is not formed at the cathode



A27. a It is cheaper to obtain chlorine by electrolysis of concentrated sodium chloride

solution than from molten sodium chloride because electrical energy is required in order to melt sodium chloride.

b Fluorine is the strongest oxidant known. Since a stronger oxidant than fluorine would be required to convert F ions into fluorine, the element cannot be made by direct reaction. However, it is generated at the anode during electrolysis of molten metal fluorides.

c As Sn2+ ions are reduced to tin metal at the cathode, the tin anode of the cell is oxidised to form Sn2+ ions. Since one electrode reaction is the opposite of the other, there is no overall change in the concentration of the electrolyte.

d Na+ ions are stronger oxidants than Ca2+ ions, so sodium metal is formed at the cathode in preference to calcium.

Q28. Magnesium and zinc are both extracted from their compounds by electrolysis. Sketch and label cells that you consider would be suitable for commercial extraction of these metals.

A28.



Q29. An electrolytic cell for the extraction of aluminium operates at a current of 150 000 A. In order to produce 1.00 tonne (106 g), calculate: a how long the cell must operate b the mass of carbon consumed at the anodes c the volume of carbon dioxide gas produced, measured at STP

A29. a Step 1 Write balanced half equations. Cathode: Al3+(in cryolite) + 3e Al(l) Anode: C(s) + 2O2(in cryolite) CO2(g) + 4e

Step 2 Calculate the amount of Al, using n = Mm .

n(Al) = 16

mol g 26.98g 101.00

= 3.706 104 mol Step 3 Use stoichiometry to calculate the amount in mol of electrons.

)Al()e(

nn =

13

n(e) = 3 3.706 104 mol = 1.112 105 mol

Step 4 Calculate the number of coulombs, using n = FQ .

Q = 1.112 105 mol 96 500 C mol1 = 1.073 1010 C Step 5 Calculate the time for which current flowed, using Q = It.

t = A 000 150

C 101073 10

= 71 530 s = 19.85 hours = 19 h 51 min



b Step 1 Use stoichiometry to calculate the amount in mol of electrons.

)Al()e(

nn =

13

n(e) = 3 3.706 104 mol = 1.112 105 mol Step 2 Use stoichiometry to calculate the amount of C.

)e()C(n

n = 41

n(C) = 4

101.112 5

= 2.78 104 mol

Step 3 Calculate the mass of C, using n = Mm .

m(C) = (2.78 104 ) mol 12 g mol1 = 3.334 105 g = 333 kg c Step 1 Use stoichiometry to calculate the amount of CO2.

)C(

)CO( 2n

n =

11

n(CO2) = 2.78 104 mol

Step 2 Calculate the volume of CO2, using n = mV

V .

V(CO2) = 2.78 104 mol 22.4 mol L1 = 6.23 105 L

Q30. Commercial operation of an aluminium smelter depends upon the availability of cheap electric power. a Calculate the electric charge required to produce 1.00 tonne of aluminium from

alumina (Al2O3) by electrolysis. b If the cell operates at 5.00 V, calculate the energy needed to make 1.00 tonne of

aluminium. (1 volt = 1 joule per coulomb) c Calculate the cost of the electricity required to make 1.00 tonne of aluminium if

the smelter purchased electricity at the domestic rate of 12 cents per kilowatt hour. (1 kilowatt hour = 3 600 000 J)

d Apart from the cost of electricity, name three other important costs involved in the production of aluminium.



A30. a Step 1 Write a balanced equation. Al3+(in cryolite) + 3e Al(l)


n(Al) = 16

mol g 26.98g 101.00

= 37 064 mol Step 3 Use stoichiometry to calculate the amount in mol of electrons.

)Al()e(

nn =

13

n(e) = 3 37 064 mol = 111 193 mol

Step 4 Calculate the electric charge required, using n(e) = FQ .

Q = 111 193 mol 96 500 C mol1 = 1.073 1010 C = 1.07 1010 C b Energy (J) = V C = 5.00 V (1.07 1010) C = 5.36 1010 J c Step 1 Calculate the number of kilowatt hours required.

Number of kWh = J0006003J 105.36 10

= 1.489 104 kWh Step 2 Calculate the cost of electricity. Cost = 1.489 104 $0.12 = $1787 = $1780 d Three other important costs in the production of aluminium are the capital cost of

the plant, the cost of producing the alumina and the cost of manufacturing the carbon anodes.

Q31. Chemists are investigating the possibility of replacing the carbon anodes in the electrolytic cell for aluminium extraction with anodes made of an unreactive material. a Write an equation for the anode reaction using unreactive electrodes. b What would be the advantages of using unreactive anodes?



A31.

a 2O2 (in cryolite) O2(g) + 4e b Unreactive electrodes would not be consumed in the anode reaction; frequent

replacement of the anodes would thus be avoided, reducing labour and materials. The manufacture of carbon anodes is a major activity in an aluminium smelter and its elimination could substantially reduce the cost of aluminium production. Emissions of carbon dioxide gas from the smelter would also be reduced since oxygen would be produced at the anode rather than carbon dioxide. However, higher potentials would be involved in such cells and the energy consumption would be greater.

Q32. The Portland aluminium smelter in Victoria produces about 1000 tonnes of aluminium a day. What volume of carbon dioxide gas (measured at SLC) is produced by the smelter each day?

A32. Step 1 Write a balanced equation. 2Al2O3(in cryolite) + 3C(s) 4Al(l) + 3CO2(g)


n(Al) = 16

mol g 26.98g 101.00

= 37 064 mol Step 3 Use stoichiometry to calculate the amount of CO2.

)Al()CO( 2

nn

= 43

n(CO2) = 4

064 373 mol

= 27 798 mol Step 4 Calculate the volume of CO2 at SLC. V(CO2) = 27 798 mol 24.5 L mol1 = 681 057.125 L = 6.811 105 L

Q33. Impure copper, called blister copper is produced by the reduction of copper ore in a blast furnace. Blister copper can be purified by the electrolytic process called electrorefining. Use the description of this process in the text to help you answer the following questions. a To which electrode is the blister copper attached? b What is the nature of the other electrode? c What is the composition of the electrolyte? d Write the half equations for the reactions involving copper that occur at the two

electrodes.



e What happens to impurities in the blister copper that are: i more reactive than copper, such as zinc? ii less reactive than copper, such as gold?

A33. a anode b a thin sheet of pure copper c sulfuric acid d anode Cu(s) Cu2+(aq) + 2e; cathode Cu2+(aq) + 2e Cu(s) e i More reactive impurities are oxidised and remain in solution as ions. ii Less reactive impurities fall to the bottom of the cell and are collected.

Q34.

Construct a concept map that includes the terms: electrolysis, electrolytic cell, chemical energy, electrical energy, anode, cathode, reductant, oxidant, reduction, oxidation, electrolyte and non-spontaneous reactions.

A34.

Q35. Under the title Using electrical energy to produce a useful chemical, write one or two paragraphs that explain to a Year 11 Chemistry student how a chemical of your choice can be produced industrially by electrolysis.

A35. You might choose to select one of the following chemicals: sodium, chlorine, sodium hydroxide, aluminium or copper. Refer to the appropriate descriptions of each electrolysis cell in the text to answer this question.

Q36. When constructing a galvanic cell in the laboratory, why are two half cells usually used whereas the reactants of an electrolysis cell are often placed in a single container?



A36. The reaction in a galvanic cell is spontaneous and if the reactants in the cell were in the one container and in contact with each other, the reaction between them could occur directly, releasing energy as heat rather than as electricity. In electrolysis cells, the reaction is non-spontaneous so that both electrode reactions can occur within the same container. The products of the electrolysis reaction should not be allowed to come into contact with each other, however, or a reaction may occur.

Unit 4 Area of Study 2 review

Multiple-choice questions

Q1. Which one of the following lists includes only renewable energy sources? A ethanol, coal, natural gas B animal waste, the sun, crude oil C uranium, hot rocks, hydrogen D hot underground springs, biochemical fuels, wind

A1. D. Non-renewable energy sources are being used up at a faster rate than they can be replaced and include fossil fuels and uranium. Renewable energy sources are continually being replaced by natural processes, e.g. wind, hot underground springs, and photosynthesis in plants from which biochemical fuels such as ethanol (made from sugar cane) or bio diesel (from vegetable oils) are derived.

Q2.

Two gaseous fuels that have been investigated as possible alternative energy sources are hydrogen gas, H2, and biogas, which consists mostly of methane, CH4. The table shows the energy released by the combustion of 1 mol of hydrogen and methane.

Fuel Equation for combustion reaction Energy released per mole of fuel (kJ mol1)

Hydrogen H2(g) +

21 O2(g) H2O(l)

286

Methane CH4(g) + 2O2(g) CO2(g) + 2H2O(l) 890

Consider the following situations that may apply to the use of the fuels. I Limiting greenhouse gas emissions is very important. II Equal masses of the two gases are to be stored. III Equal volumes of the gases are to be stored (at the same temperature and

pressure). Hydrogen gas would be the preferred fuel in situations: A I, II and III B I and II



C II and III D I only.

A2.

B. Only methane releases the greenhouse gas CO2, increased levels of which in the atmosphere are thought to cause global warming. 2 g of hydrogen releases 286 kJ, therefore 1 g releases 143 kJ of energy. However, 16 g of methane releases 890 kJ, therefore 1 g releases 56 kJ. Therefore, hydrogen releases more energy per unit mass than methane; equal volumes of the two gases contain equal amounts in mol and, on this basis, methane releases more energy than hydrogen.

Q3.

Some eminent politicians and scientists have recently proposed that we should consider utilising our reserves of uranium to generate electricity for our cities, instead of the current situation where most electricity is supplied by coal-fired power stations. An argument in favour of greater reliance on nuclear power, rather than coal, could be based on: A whether the energy source is renewable or non-renewable B the relative amounts of greenhouse gases produced by the power stations C the safe disposal of waste materials from the power plant D the number of energy conversions involved between the energy source and the

electrical energy that is generated

A3. B. Burning coal releases carbon dioxide gas into the atmosphere, increasing levels of which are thought to be causing global warming. Sulfur dioxide can also be released when coal is burned. Nuclear power stations dont generate any air pollutants. However the storage or radioactive waste is a contentious issue. Both fuels are non-renewable and a similar number of energy conversions are involved in the two types of power station. The following information relates to Questions 4 and 5. As a response to rising petrol prices the Australian Government in 2006 offered car owners a $2000 incentive to convert their cars from petrol to LPG. Propane is one of the gases in the mixture sold as LPG. The thermochemical equation for the complete combustion of propane is:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l); H = 2220 kJ mol1



Q4. Which energy profile diagram best represents the energy changes that take place during this reaction? A B

C D

A4. C. As this is an exothermic reaction the energy of the reactants will be less than energy of the products. There is 2220 kJ mol1 difference between energy of the reactants and products. Bond breaking always requires an input of energy, i.e. energy is needed to break the bonds. Energy is always released when new bonds are formed. This exothermic reaction involves a net release of energy, so more energy is released during bond formation than was used up to break bonds of the reactants.

Q5. When 100 g of propane undergoes complete combustion: A 5.04 103 kJ of energy is absorbed B 5.04 103 kJ of energy is released C 3.92 101 kJ of energy is absorbed D 3.92 101 kJ of energy is released.

A5.

B. This exothermic reaction (with a negative H value) involves a net release of energy. Step 1 The amount of propane that reacted is calculated.

n(C3H8) = Mmass =

11.44100 = 2.26 mol

Step 2 The amount of energy released by 100 g of propane is calculated. = 2.26 2220 = 5.024 103 kJ



Q6. Which one of the following sources of power provides the highest quantity of energy per gram of fuel? A coal B natural gas C nuclear fission D biochemical fuels

A6.

C. 235U has a significantly energy per kilogram (energy density) than fossil or biochemical fuels. The energy density of 235U is 9 107 MJ kg1, compressed natural gas 54 MJ kg1, coal 24 MJ kg1 and a biochemical fuel such as ethanol 23 MJ kg1.

Q7. A temperature rise of 2.0C occurred when 1.50 103 mol of ethane gas burnt in a bomb calorimeter.

2C2H6(g) + 7O2(g) > 4CO2(g) + 6H2O(l); H = 3120 kJ mol1 The calibration constant of the calorimeter, in J oC1, is A 1170 B 3120 C 2340 D 4680

A7.

A. 2 mole C2H6 generates 3120 kJ of energy. 1 mole generates 1560 kJ. Energy produced by 1.50 103 mol = 1.50 103 1560 103 = 2340 J Calibration factor = energy/T = 2340/2 = 1170 J oC1



Q8. The following equations involve ions of the transition metal vanadium. The half cell potential E(V) is shown for each equation. E(V) VO2+(aq) + 2H+(aq) + e VO2+(aq) + H2O(l) +1.00 VO2+(aq) + 2H+(aq) + e V3+(aq) + H2O(l) +0.36

V3+(aq) + e V2+(aq) 0.25 Of the ions listed, the strongest oxidant and the strongest reductant are:

Strongest oxidant Strongest Reductant

A VO2+ V2+ B VO2+ VO2+ C V3+ VO2+ D V2+ VO2+

A8. A. An oxidant gains electrons and is shown on the left side of the half equation in the electrochemical series. The strongest oxidant is on the left in the half-cell reaction that has the highest E0. A reductant loses electrons and is shown on the right side of the half equation. The conjugate redox pair containing the strongest reductant has the lowest E0.

Q9.

Use the electrochemical series to determine which one of the following would not be expected to occur to an appreciable extent. A 2H+(aq) + Fe(s) Fe2+(aq) + H2(g) B 2Ag+(aq) + Ni(s) 2Ag(s) + Ni2+(aq) C Br2(aq) + 2Fe2+(aq) 2Br(aq) + 2Fe3+(aq) D 2I(aq) + Pb2+(aq) I2(s) + Pb(s)

A9.

D. A reaction will only occur if the oxidant of the conjugate redox pair with the higher E is added to the reductant of the conjugate redox pair having the lower E. In D, a reaction will not occur because the E of the half reaction involving the reductant I is higher than the E of the half reaction involving the oxidant, Pb2+.



Q10. Which of the following best describes the features of an anode in a galvanic cell? polarity electrode reaction A positive oxidation B positive reduction C negative oxidation D negative reduction

A10.

C. The oxidation reaction at the anode produces electron giving the electrode a negative polarity.

Q11. An electrochemical cell was made by dipping a copper rod into a solution of 1 M CuSO4 in one beaker and dipping a nickel rod into a solution of 1 M NiSO4 in another beaker. The metals were connected with wire and the two solutions were connected by a piece of paper towel that had been soaked in a potassium nitrate solution. The cell is shown in the diagram.

The solution in beaker I was initially coloured blue, owing to the presence of Cu2+ ions. The solution in beaker II was initially coloured green because of the presence of Ni2+ ions. After the galvanic cell has been discharging for a period of time, it might be possible to detect the following changes. A The green colour in beaker II has faded and the mass of the copper electrode has

increased. B The blue colour in beaker I has faded and the mass of the nickel electrode has

increased. C The green colour in beaker II has faded and the mass of the copper electrode has

decreased. D The blue colour in beaker I has faded and the mass of the nickel electrode has

decreased.



A11. D. The electrode half reactions are predicted as:

Cu2+(aq) + 2e Cu(s) thus causing this electrode to increase in mass (more Cu formed) and the solutions colour to fade (less Cu2+ is present); and

Ni(s) Ni2+(aq) + 2e this causing the electrode to decrease in mass (less Ni) and the solution colour to deepen (more Ni2+(aq)). The following information relates to Questions 12 and 13. A car manufacturer has developed a prototype of an electric car that uses hydrogen fuel cells to generate electricity.

Q12. A reaction that might occur at the cathode of a fuel cell using hydrogen and oxygen as reactants is A 2H2(g) + O2(g) 2H2O(g) B O2(g) + 2H2O(g) + 4e 4OH(aq) C H2(g) 2H+(aq) + 2e D H2(g) + 2OH(aq) 2H2O(g) + 2e

A12. B. At the cathode, oxygen is reduced to hydroxide ions in the presence of water.

Q13. Consider the following features of the fuel cell powered car. I Two half cell reactions occur spontaneously at two separated locations (the

electrodes). II Oxygen behaves as an oxidant in the reaction that supplies energy. III Hydrogen gas stores 143 kJ g1 of fuel. The fuel cell powered car differs from a conventional petrol-driven car with respect to: A I, II and III B I and II C I and III D II and III

A13. C. A fuel cell uses an electrochemical process, whereas petrol reacts directly with oxygen in the internal combustion engine. Oxygen oxidises both fuels; petrol has a different energy density from that of hydrogen, approximately 47 kJ g1.



The following information relates to Questions 14 and 15. The diagram shows an electroplating cell that was devised by a student to place a silver coating onto a key. An electric current is passed through a solution of 1 M silver nitrate, AgNO3, using an inert carbon electrode as the anode.

Q14. The reaction that occurs at the electrode connected to the negative terminal of the external power supply is: A Ag+(aq) + e Ag(s) B 2H2O(l) + 2e H2(g) + 2OH(aq) C 2Cl(aq) Cl2(g) + 2e D 2H2O(l) O2(g) + 4H+(aq) + 4e

A14. A. Electrons are supplied to the negative electrode to allow reduction (electron gain) to occur; Ag+ is a stronger oxidant than H2O so it is reduced.

Q15. If 0.055 g silver is deposited on the surface of the key when a current is passed through the cell for 7.0 minutes, what current is required? A 0.12 A B 0.75 A C 7.0 A D 49 A



A15.

A. Cathode reaction: Ag+(aq) + e Ag(s)

n(Ag) = Mm

= 87.107

055.0 = 5.1 104 mol

Since n(e) = n(Ag+), 5.1 104 F of charge is required Q = 5.1 104 96 500 = 49 C

I = tQ

= s 60 7.0

C 49

= 0.12 A

Q16.

Three beakers contain solutions of 1.0 M chromium(III) nitrate, 1.0 M copper(II) nitrate and 1.0 M silver nitrate. Each solution has 0.60 F of electric charge passed through it, using carbon electrodes and a power pack. In each beaker a different metal is deposited onto the cathode. In decreasing order, the amount of metal (in moles) deposited onto the cathode in each beaker will be: A Cr > Cu > Ag B Ag > Cu > Cr C Cr > Ag > Cu D the same in all three beakers

A16. B. 0.60 F is carried by 0.60 mol of electrons. Cr3+ + 3e Cr; therefore, 0.20 mol Cr is produced Cu2+ + 2e Cu; therefore, 0.30 mol Cu is produced Ag+ + e Ag; therefore 0.60 mol Ag is produced

Q17.

A series of experiments is described below. I Zinc granules are added to a solution of tin(II) chloride II Copper turnings are added to tin(II) chloride solution III Electrolysis of molten tin(II) chloride IV Electrolysis of an aqueous solution of tin(II) chloride The experiments that would result in the production of metallic tin, Sn, are: A all of them B I, III and IV C I and II D only II



A17. B. Tin metal can be produced by a direct reaction with a stronger reductant than Sn, such as Zn (but not Cu), or by electrolysis of an aqueous solution of a tin(II) salt or the molten salt. In all cases the reduction involves Sn2+ + 2e Sn.



Short-answer questions

Q18. a During the 20th century and continuing to the present day, atmospheric carbon

dioxide concentrations have increased significantly due mainly to the burning of fossil fuels. i Describe two undesirable atmospheric pollutants, other than carbon dioxide

emissions, produced by the burning of fossil fuels. ii What is thought to be the environmental impact of increased levels of

atmospheric carbon dioxide? iii Identify one fossil fuel and briefly discuss the sustainability of its continued

use into the future. Your response should include a statement about the extent of the fossil fuels reserves and an outline one advantage and one disadvantage of its continued use.

b The main use of coal is in the generation of electricity. Electricity can also be generated using non-fossil fuel alternatives. Name one of theses alternatives and identify one advantage (other than less CO2 emission) and one disadvantage compared to burning fossil fuels for electricity generation.

c Some countries use nuclear fission instead of fossil fuels for electricity production. i Explain what the term nuclear fission means. ii What energy transformation occurs when 235U undergoes fission?

d There has been an increased awareness of the use of biochemical fuels as an alternative to fossil fuels. i How does a biochemical fuel differ from a fossil fuel? ii Give two examples of biochemical fuels and describe briefly how they are

produced. iii Why are biochemical fuels considered by some to be carbon dioxide neutral?

A18.

a i Carbon monoxide, carbon (soot), oxides of sulfur and nitrogen, sulfuric acid, ozone

ii Increased levels of atmospheric carbon dioxide have been linked to global warming

iii Estimated global fossil fuel reserves and their expected lifetimes are given in Table 23.3.

b Refer to Chapter 24 (Table 24.3 Some advantages and disadvantages of energy sources available in Australia).

c i Nuclear fission is the splitting of the nucleus of an atom ii Nuclear binding energy or nuclear mass heat and kinetic energy d Fossil fuels are derived from the remains of plants and animals that died millions

of years ago. Biochemical fuels are manufactured from living plant materials, e.g. ethanol from sugar cane, biodiesel from vegetable oils, biogas from the decay of plant and animal matter.



Q19. The heat of combustion of brown coal was measured in a bomb calorimeter that had been calibrated by passing a current of 3.00 amperes at 5.00 volts through the electrical heater for 2.00 minutes. The temperature rose by 1.1C. When a 0.50 g sample of freshly crushed brown coal was completed burned in the calorimeter, the temperature rose by 3.0C. a Calculate the value of the calorimeter constant. b Determine the heat of combustion of brown coal in kJ g1 c Another sample of brown coal was heated at 100C for 2 hours. A sample,

weighing 0.5 g, was then burned completely in the calorimeter. Would this sample release more or less energy per gram than the freshly crushed sample? Explain your answer.

d A brown coal fired power station produces 1.7 tonnes of carbon dioxide gas for ever megawatt of electricity generated. A black coal fired power station releases 0.9 tonnes of carbon dioxide into the atmosphere per megawatt of electricity. Explain the difference in the amount of carbon dioxide released to produce the same amount of electricity. Assume that both types of power stations operate at the same level of efficiency.

A19. a Energy supplied during calibration = VIt = 5.00 V 3.00 A (2.00 60) s = 1800 J

Since this caused the temperature to rise by 1.1C, the energy needed for a 1C

temperature change = 1.1

1800 = 1636 J C1

b When the coal burned, the energy released = calibration constant temperature change

= 1636J C1 3.0C = 4909.1 J = 4.91 kJ

0.500 g of coal supplies 4.91 kJ of energy.

Therefore, 1 g of coal will give 4.91 1.000.50

= 9.82 kJ

i.e. the heat of combustion is 9.82 kJ g1. c Brown coal contains between 30% and 70% water. Heating the coal prior to

combustion in the calorimeter drives off some of this moisture. Since the mass of the pre-dried and dried samples combusted are the same the dried sample will contain a greater proportion of carbon and will produce more energy per gram.

d Carbon dioxide is a combustion product of both black and brown coal. The energy content of brown coal is about a third of the energy content of the black coal used in power stations. Black coal has a much higher carbon content than brown coal and its water content is significantly less. To produce one megawatt of electricity less black coal needs to be burned and consequently less carbon dioxide will be released into the atmosphere.



Q20. One of the boilers at a power station burns 75.6 kg of brown coal every second. Each kilogram of brown coal produces 9.82 MJ of energy. The electrical output of the generator connected to this boiler is 300 MJ s1. a Calculate the amount of energy produced by burning 75.6 kg of coal each second. b Give two reasons why the answer to part a is much greater than the output of the

generator. c Over three-quarters of the electricity generated in Victoria comes from brown

coal, a non-renewable energy source. Explain what is meant by the term non-renewable energy source.

d Describe briefly how one renewable energy source can be used to generate electricity.

A20. a 1 kg coal supplies 9.82 MJ of energy. Therefore, 75.6 kg of coal will supply

9.82 75.6 = 742 MJ of energy, i.e. 742 MJ s1. b Coal-fired power stations are only about 40% efficient at converting coals

chemical energy into electrical energy. Heat is lost from the chimney stacks at power station with the waste gases and from the cooling towers in the power plant.

c A non-renewable energy source is one that is being used up at a faster rate than it can be replaced.

d Wind power: kinetic energy of moving air is converted to mechanical energy of spinning turbines that are connected to electricity generators.

Q21. When a NiCad cell is converting chemical energy to electrical energy (discharge), the electrode reactions are best described as follows:

positive electrode: NiOOH(s) + H2O(l) + e Ni(OH)2(s) + OH(aq) negative electrode: Cd(s) + 2OH(aq) Cd(OH)2(s) + 2e

a Which metal, nickel or cadmium, is at the anode of this cell as it is discharging? b Give the formula of a salt that might be expected to be found in the electrolyte

paste of a NiCad cell. NiCad cells are superior to alkaline cells as they can be recharged.

c Explain why cells such as the NiCad can be recharged. d Write equations for the half reactions and the overall cell reaction occurring when

a NiCad cell is recharged. e Describe the energy transformation that occurs when a NiCad cell is recharged.

A21. a Cadmium, Cd, since this is the electrode at which oxidation takes place. b KOH (or NaOH) c These cells can be recharged because the products of the discharge half reactions

remain in contact with the electrodes in a convertible form.



d Positive electrode: Ni(OH)2(s) + OH(aq) NiOOH(s) + H2O(l) + e Negative electrode: Cd(OH)2(s)+ 2e Cd(s) + 2OH(aq) Overall: 2 Ni(OH)2(s) + Cd(OH)2(s) 2NiOOH(s) + 2H2O(l) + Cd(s) e electrical energy chemical energy

Q22. Zincair cells have been developed for a range of different uses, including electric vehicles, heart pacemakers and laptop computers. They offer a favourable alternative to other batteries because of their non-toxic contents and high energy density. In the cells, zinc metal is oxidised at the negative electrode and oxygen gas is reduced to water at the positive electrode. A zincair cell used for a pacemaker operates with a steady current of 3.5 105 A. a Write a half equation for the half reaction at the negative electrode. b If the cell will continue operating until 1.5 g of zinc in the negative electrode has

been used up, determine how long the pacemaker will last before it needs replacing.

A22.

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

b n(Zn) = Mm =

38.655.1 = 0.023 mol

Therefore, n(e) released from anode = 2 0.023 = 0.046 mol Q = n(e) F = 0.046 96 500 = 4.4 103 C

t = IQ =

A 0.000035C 4400

= 1.3 108 s = 4.0 years

Q23. An alcometer is a small, hand-held device that can be used to give reliable readings of blood alcohol levels when a person blows air from their lungs into the instrument. It relies on the equilibrium established between ethanol dissolved in the blood and gaseous ethanol in the lungs:

C2H5OH(aq) (in blood) C2H5OH(g) (in lungs) which has an equilibrium constant, K = 4.35 104 at 37C. The alcometer is a fuel cell and air from the lungs is directed onto the anode of the cell where ethanol, C2H5OH, is oxidised to carbon dioxide, CO2. It consists of a porous glass plate containing a phosphoric acid electrolyte and coated with layers of platinum metal on the top and bottom. The fuel cells voltage depends on the concentration of ethanol in the air blown onto the anode, so the instrument can be calibrated to switch on a red light if blood alcohol concentration is greater than 0.05%, or a green light for zero blood alcohol concentration.



The net reaction that occurs in this fuel cell is given by the equation: C2H5OH(g) + 3O2(g) 2CO2(g) + 3H2O(l)

a By assigning oxidation numbers to carbon in both ethanol and carbon dioxide, verify that ethanol is oxidised in the fuel cell reaction.

b The half reaction that occurs at the cathode is: O2(g) + 4H+(aq) + 4e 2H2O(l) What is the equation for the half reaction that occurs at the anode in the alcometer?

c On the diagram, place an arrow to show the direction of electron movement through the external circuit.

d Give two different functions of the layers of platinum metal in the fuel cell.

A23. a Ethanol: the CH3 carbon is in the 3 oxidation state and the CH2OH carbon is

in 1 oxidation state (attributing zero effect to the CC bond) Carbon dioxide: carbon is in the +4 oxidation state. Carbon is oxidised.

b C2H5OH(g) + 3H2O(l) 2CO2(g) + 12H+(aq) + 12e



c

d They are the sites of the oxidation and reduction half reactions; they behave as catalysts to increase the rate of the electrode half reactions.

Q24. An experiment was carried out as follows to compare the effectiveness of two different methods of generating energy from liquid methanol (CH3OH) as a source of fuel. a The heat of combustion of methanol was determined in bomb calorimeter which

had a calibration factor of 8.26 kJ C1. When 2.0 g of methanol was completely burned in excess oxygen the temperature rose from 20.0C to 25.5C. i How much energy is released when 2.0 g of methanol is burned in excess

oxygen? ii How much energy is released when 1.0 mole of methanol is burned in excess

oxygen? iii What is the value of H for the reaction: 2CH3OH(l) + 3O2(g) CO2(g) + 4H2O(l)

b The methanol fuel cell generates electricity by consuming methanol and oxygen. The equations for the reactions occurring at the electrodes are: CH3OH (l) + H2O(l) CO2(g) + 6H+(aq) + 6e Equation 1 O2(g) + 4H+(aq) + 4e 2H2O(l) Equation 2 i Which equation represents the reaction occurring at the anode? ii Write an equation that represents the overall cell reaction. iii Calculate the electric current generated by the methanol fuel cell when 0.64 g

of methanol is consumed in 10 minutes.



iv How much energy in kJ is delivered per mole of methanol from a fuel cell that generates 1.26 V?

c Methanol can also be used as a fuel for an internal combustion engine driving an electrical generator. i Give one reason why the fuel cell would produce electricity more efficiently

than a generator driven by an internal combustion engine. ii Give one reason why fuel cells are not used to generate electricity on a large

scale.

A24. a i energy = calibration factor temperature change = 8.26 5.5 = 45.43 kJ ii 2 g produces 45.43kJ.

322 mole = 45.43

1 mole = 45.43 2

32 kJ = 726.88 kJ = 727 kJ (three significant figures)

iii The equation as written involves two moles. H = 2 727 = 1454 kJ mol1 b i Oxidation occurs at the anode. The oxidation state of carbon increases from

2 in CH3OH to + 4 in CO2, i.e. carbon has been oxidised. The anode reaction is:

CH3OH(l) + H2O(l) CO2(g) + 6H+(aq) + 6e ii The half equations can be combined when both have the same number of

electrons. There will be equal numbers of electrons when equation 1 is multiplied by 2 and equation 2 is multiplied by 3. 2CH3OH(l) + 2H2O(l) CO2(g) + 12H+(aq) + 12e 3O2(g) + 12H+(aq) + 12e 6H2O(l) 2CH3OH (l) + 2H2O(l) + 3O2(g) + 12H+(aq) + 12e CO2(g) + 12H+(aq) + 12e + 6H2O(l) This simplifies to: 2CH3OH (l) + 3O2(g) CO2(g) + 4H2O(l)

iii 1 mole CH3OH generates 6 mole of electrons = 6 96 500 C

0.64 g CH3OH = 3264.0 = 0.02 mol

0.02 mol CH3OH = 0.02 6 96 500 = 11580 C

current, I = tQ = 11580

10 60 = 19.3 A

iv Since 6 electrons are involved in the half equation, the charge of 6 mol of electrons is 6 96 500C. The energy produced when the cell generates 1.26 V is: 1.26 6 96 500 = 730 kJ mol1

c In a fuel cell, the energy conversions are chemical electrical energy. In a combustion engine, heat is lost at each step in the energy conversions: chemical kinetic mechanical electrical.



Q25. a Complete the half equations for the reactions predicted when each of the

following aqueous solutions undergo electrolysis, using the electrodes described in the table. The process was carried out in a U-tube, as shown in the diagram.

Solution Electrode materials Half equation for the reaction at the

Positive electrode Negative electrode

Positive electrode

Negative electrode

KI(aq) carbon carbon PbCl2(aq) carbon carbon AlCl3(aq) copper copper

b For one of the solutions described in the table, explain how you might verify that the predicted products were being formed at each electrode.

A25.

a Positive Negative KI(aq) 2I(aq) I2(s) + 2e 2H2O(l) + 2e H2(g) + 2OH(aq)PbCl2(aq) 2H2O(l) O2(g) + 4H+(aq) + 4e Pb2+(aq) + 2e Pb(s) AlCl3(aq) Cu(s) Cu2+(aq) + 2e 2H2O(l) + 2e H2(g) + 2OH(aq)



b Positive Negative KI(aq) Look for the yellowbrown

colour of iodine in the solution. Use an acid-base indicator to test for OH, look for bubbles of H2 gas, collect and test gas with burning string pop confirms H2 gas.

PbCl2(aq) Use an acidbase indicator to test for presence of H+(aq), look for gas bubbles, collect gas and test with glowing splint, if this ignites confirms presence of O2.

A grey deposit of lead will form on the electrode.

AlCl3(aq) Blue colour will appear in solution as Cu electrode is consumed.

Use an acid-base indicator to test for OH, look for bubbles of H2 gas, collect and test gas with burning string pop confirms H2 gas.

Q26. A 100 mL solution containing a mixture of magnesium chloride, MgCl2, and tin(II) chloride, SnCl2, is prepared by dissolving 0.025 mol of each salt in water. Platinum electrodes are placed in the solution and a small current is passed through it, as shown in the diagram.

a Write half equations for the electrode reactions occurring just after the electrolysis is started i at the anode ii at the cathode

b After electrolysis has been occurring for a considerable period of time all of the metal that was first plated on the cathode will have been used up and a new electrode reaction will occur at the cathode. Write a half equation for the next electrode reaction that occurs at the cathode.

c What major difference would occur in the cell if the solution that was used had been a saturated solution of tin(II) chloride?



A26. a i In dilute solutions water, the stronger reductant will be oxidised at the anode

rather than Cl(aq). 2H2O(l) O2(g) + 4H+(aq) + 4e ii Sn2+(aq) is a stronger oxidant than either water or Mg2+(aq) and will be

reduced Sn2+(aq) + 2e Sn(s) b Once all the Sn2+(aq) has been reduced, water, being a stronger oxidant than

Mg2+(aq), will then be reduced. 2H2O(l) + 2e H2(g) + 2OH(aq) c Chlorine gas would be released at the positive electrode (anode) due to the very

high concentration of chloride ions in the solution. 2Cl(aq) Cl2(g) + 2e

Q27. In an electrolysis experiment, a student is provided with a solution of nickel(II) nitrate, Ni(NO3)2. The electrodes to be used were a carbon rod, as the positive electrode, and a metal spatula, as the negative electrode. a Will the nickel coating appear on the carbon rod or the metal spatula during the

experiment described? b Write equations for the half reactions that occur at each electrode. c A current of 2.5 A was passed through this nickel-plating cell for 15 minutes.

Calculate the mass of nickel that could be plated on the cathode.

A27. a On the metal spatula: electrons are supplied to the negative electrode, where

reduction of Ni2+ to Ni will occur. b Spatula (negative): Ni2+(aq) + 2e Ni(s) Carbon rod (positive): 2H2O(l) O2(g) + 4H+(aq) + 4e c Q = It = 2.5 A (15 60) s = 2250 C

n(e) delivered to negative electrode = FQ =

500962250 = 0.023 mol

n(Ni) formed = 21

n(e) = 0.012 mol

m(Ni) = n M = 0.012 58.69 = 0.68 g



Q28. Lithium metal is prepared by electrolysis of a molten mixture of lithium chloride and potassium chloride. a Write a half equation for the reaction occurring at the cathode in this cell. b Write a half equation for the reaction occurring at the anode in the cell. c Why is it not possible to produce lithium metal by the electrolysis of an aqueous

solution of lithium chloride? d Suggest why a mixture of lithium chloride and potassium chloride is used, rather

than pure lithium chloride, in this electrolytic process. e Assuming that lithium reacts in a similar way to sodium, describe two precautions

that would need to be taken in this preparation.

A28.

a Li+(l) + e Li(l) b 2Cl(l) Cl2(g) + 2e c Li+ is a weaker oxidant than H2O (Li+/Li has a lower E); therefore the negative

electrode reaction is: 2H2O(l) + 2e H2(g) + 2OH(aq)

rather than Li+(l) + e Li(l). d Addition of KCl lowers the melting temperature of the electrolyte, thus lowering

the energy used in the process and lowering costs. e The products of the half reactions would need to be kept separated from one

another (as in the Downs cell) and water would need to be excluded from the entire process as it is a more powerful oxidant than Li

Q29.

This question involves both electrochemical cells and electrolytic cells. a Compare the energy transformation occurring in an electrochemical cell with that

occurring in an electrolytic cell. b Explain why the negative electrode of an electrochemical cell is the site of

oxidation, whereas the negative electrode of an electrolytic cell is the site of reduction.

A29.

a Electrochemical cell: chemical energy electrical energy Electrolytic cell: electrical energy chemical energy

b In the electrochemical cell, electrons produced at the site of oxidation make that electrode develop a negative polarity, i.e. the half reaction causes the negative polarity of the electrode. However, in the electrolytic cell, the negative polarity of one electrode is imposed on it by the external power source, which forces (negatively charged) electrons towards this electrode and allows reduction to occur there.



Q30. Some industrially important chemicals are produced by the electrolysis of a molten electrolyte while others are produced by electrolysis of an aqueous solution. a Name one chemical produced by the electrolysis of:

i a molten electrolyte ii an aqueous solution.

b For each chemical that you have selected: i Write an equation for the reaction occurring at the anode ii Write an equation for the cathode reaction. iii Describe one feature of the electrolytic cell used to manufacture of each

selected chemical. c Explain why some chemicals can be produced by the electrolysis of a molten

electrolyte but not from the electrolysis of an aqueous solution.

A30. a i sodium, aluminium

ii chlorine, sodium hydroxide, copper b Equations depend on chemicals selected in part a (Refer to Chapter 28.) c In the electrolysis of aqueous solutions the possibility that water may be oxidised

at the anode or reduced at the cathode needs to be considered. Water is a stronger oxidant than either sodium or aluminium and will be reduced at the cathode in preference to the metal ions. These reactive metals are produced by the electrolysis of a molten salt.

Q31. Write equations to represent the anode and cathode reactions that occur when an electric current is passed through: a a dilute solution of sodium chloride b a concentrated solution of sodium chloride c molten sodium chloride

A31. a Anode. Both the reductants, Cl and H2O need to be considered. Water is the

stronger reductant and would be expected to oxidised in preference to Cl(aq). 2H2O(l) O2(g) + 4H+(aq) + 4e

Cathode. Water is the stronger oxidant and would be expected to be reduced in preference to Na+(aq).

2H2O(l) + 2e H2(g) + 2OH(aq) b Anode: The electrochemical series can be used as a guide to predict reactions at

concentrations of 1 M. In the case of a concentrated sodium chloride solution, Cl(aq) is oxidised rather than water.

2Cl(aq) Cl2(g) + 2e Cathode: Water is the stronger oxidant and would be expected to be reduced in preference to Na+(aq).

2H2O(l) + 2e H2(g) + 2OH(aq)



c Anode: Chloride ions are oxidised. 2Cl(aq) Cl2(g) + 2e Cathode: Sodium ions are reduced. Na+(l) + e Na(l)

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Chem ch28