OXIDATION AND REDUCTION REACTIONS AS SOURCES OF ENERGY

Transcription

1 Section 4 OXIDATION AND REDUCTION REACTIONS AS SOURCES OF ENERGY 1. What is meant by electrochemistry? The branch of science which associates electric current and changes in matter. 2. Give an example of the production of electricity from chemical reactions. Any galvanic cell, e.g. a dry cell battery. 3. What is the activity series of metals? The series of metals listed in order of their reactivity. e.g. potassium, sodium, magnesium, zinc, lead, copper, silver are listed in decreasing order of activity. 4. How is it determined? It is determined by reacting the metals with reagents such as water, dilute acid and oxygen and comparing the rates of reaction. It can also be determined using displacement reactions, using the rule that an active metal will displace a less active metal from a solution containing the ions of the less active metal. 5. What are displacement reactions? Reactions of active metals with the ions of a less active metal. Zinc will displace copper from a solution of copper sulfate. Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) 6. What happens in every oxidation-reduction reaction? A transfer of electrons. 7. What is oxidation? Loss of electrons 8. What is reduction? Gain of electrons 9. Why do these reactions always occur together? Electrons will only move from the reducing agent if there is an oxidising agent available to pick them up. 10. What is a reducing agent (also reductant)? A reducing agent causes something else to be reduced and is itself oxidised (loses electrons). 11. What is an oxidising agent (also oxidant)? An oxidising agent causes something else to be oxidised and is itself reduced (gains electrons).

2 Refer to the Redox Table / Table of Standard Electrode Potentials provided at the back of this book. The format of this table is the same as that on recent HSC papers. Note that the format in text books varies from book to book. 12. Where, in the table, are the best reducing agents? Top of table, on the RHS. 13. Where, in the table, are the best oxidising agents? Bottom of the table, on LHS. 14. Which combination of chemical reagents listed on the table will have the greatest tendency to react together? Potassium metal, K, (top, right) with fluorine, F 2, (bottom left). 15. Which of the following gives up electrons most readily? Zn, Na, Cu, Ag, Mg, Fe. Why? How do you know? Na is closest to the top, on RHS Which of the above metals is the most active? Why? Na; closest to the top, on RHS. Activity of metals is determined by the tendency to lose electrons. Which is the least active? Why? Ag; lowest, on RHS. Ag has the smallest tendency (of those listed) to lose electrons. E oxidation = V 16. Which of the following gains electrons most readily? Cl 2, F 2, Cu 2+, Na +, Zn 2+, K + Fluorine (F 2 ) has the greatest tendency to gain electrons of all elements E reduction = V Which of these species gains electrons least readily? K + How do you know? The potassium ion is located at the top left of the redox table. It has the smallest tendency of all oxidising agents to gain electrons. 17. When iron is placed in a copper II sulfate solution, solid copper forms on the iron. Explain what is occurring in terms of the transfer of electrons. Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) Zinc metal is losing electrons and these are being gained by the copper ions. 18. In general terms, which metal ions will be displaced by solid metals when they come into contact. Metals ions will be displaced by metals which are more active. Cu 2+ will be displaced when they react with magnesium, and Cu (s), the metal, will form. Cu 2+ will not be displaced when with silver, as silver is a less active metal than copper.

3 19. Write balanced half-equations and full equations for the following oxidation-reduction reactions (using the redox table). If no reaction occurs, indicate this by writing no reaction. Magnesium + copper nitrate solution Mg (s) Mg 2+ (aq) + 2e - Cu 2+ (aq) + 2e - Cu (s) Overall: Mg (s) + Cu 2+ (aq) Mg 2+ (aq) + Cu (s) Copper + magnesium nitrate solution No reaction Zinc + silver nitrate solution Zn (s) Zn 2+ (aq) + 2e - Ag + (aq) + e - Ag (s) Overall: Zn (s) + 2Ag + (aq) Zn 2+ (aq) + 2Ag (s) Zinc + dilute hydrochloric acid solution Zn (s) Zn 2+ (aq) + 2e - 2H + (aq) + 2e - H 2 (g) Overall: Zn (s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Sodium + chlorine Na (s) Na + + e - Cl 2 (g) + 2e - 2Cl - Overall: 2Na (s) + Cl 2 (g) 2NaCl (s) Solid would form if the elements were heated. 20. Write half-equations for the oxidation and reduction reactions which combine to give the following overall equations. 5Fe 2+ (aq) + MnO 4 - (aq) + 8H + (aq) 5Fe 3+ (aq) + Mn 2+ (aq) + 4H 2 O (l) 5Fe 2+ (aq) 5Fe 3+ (aq) + 5e - MnO 4 - (aq) + 8H + (aq) + 5e - Mn 2+ (aq) + 4H 2 O (l) Cr 2 O 7 2- (aq) + 14H + (aq) + 6I - (aq) 2Cr 3+ (aq) + 3I 2 (aq) + 7H 2 O (l) Cr 2 O 2-7 (aq) + 14H + (aq) + 6e - 6I - (aq) 3I 2 (aq) + 6e - 2Cr 3+ (aq) + 7H 2 O (l)

4 Oxidation Number is a measure of the number of electrons which have been gained or lost by an atom in a species (by comparison with the number of electrons possessed by that atom in its elemental form). The O.N. of an atom in the element is therefore always zero. The charge on a simple ion is equal to its oxidation number. The total of all the oxidation numbers of the atoms making up a species equals the charge on that species, e.g. the oxidation number for: Na + +1 Ba Copper in Cu 2 O +1 Copper in CuO +2 - Manganese in MnO 4 +7 Chlorine in Cl 2 0 Chlorine in Cl - -1 Oxygen in O The oxidation number changes when redox reactions (transfer of electrons) occur. Oxidation is therefore increase in oxidation number. Reduction is decrease in oxidation number. 21. Give the oxidation state of iron in FeO and in Fe 2 O 3. Explain the oxidation states in terms of loss of electrons. Oxidation state (number) of Fe in FeO = +2 (the charge on the ion = the number of electrons it has lost when it was formed from the element, Fe) Oxidation state (number) of Fe in Fe 2 O 3 = +3 (the charge on the ion = the number of electrons it has lost when it was formed from the element, Fe) 22. Give the oxidation state of the metal atoms in each of the following: MnO 4 -, CuSO 4, MnCl 2 and Cr 2 O MnO CuSO 4 +2 MnCl 2 +2 Cr 2 O

5 23. Give the oxidation state (number) for each of the atoms involved in the following equations 2Na + Cl 2 2Na + + 2Cl K (s) + H 2 O (l) 2K + (aq) + 2OH - (aq) + H 2 (g) 0 H = +1 O = O= -2 H = Draw a diagram for the galvanic cell (known as the Daniell Cell) formed using Zn and Cu electrodes. Label the anode, cathode, direction of electron flow, direction of ion movement in the salt bridge. Write equations for the half-cell reactions and for the overall reaction. Predict the voltage of this cell under standard conditions. What changes would you expect to see after a galvanic cell had operated for a few minutes? The Daniell Cell (named after J. Daniell, 1836) is set up as shown in the diagram. The anode half-reaction is: Zn (s) Zn 2+ (aq) + 2e V The cathode half-reaction is: Cu 2+ (aq) + 2e - Cu (s) V The overall equation is: Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) E = V The solution in the cathode half-cell would become lighter blue. The zinc electrode would get smaller. The copper electrode would get bigger.

6 25. Draw a diagram for the galvanic cell formed using Ag and Cu electrodes. Label the anode, cathode, direction of electron flow, direction of ion movement in the salt bridge. Write equations for the half cell reactions and for the overall reaction. Predict the voltage of this cell under standard conditions. What changes would you expect to see after a galvanic cell had operated for a few minutes? electrons copper anode NO K + silver cathode KNO 3 (aq) Cu(NO 3) 2 (aq) AgNO 3(aq) electrolytes The anode half-reaction is: Cu (s) Cu 2+ (aq) + 2e V The cathode half-reaction is: Ag + (aq) + e - Ag (s) V The overall equation is: Cu (s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag (s) E = V The solution in the anode half-cell would become darker blue. The copper electrode would get smaller. The silver electrode would get bigger. 26. What is the purpose of the salt bridge in a galvanic cell? To complete the circuit, allowing the flow of ions between the half-cells, so that there is no build-up of charge in either half-cell. A build-up of charge would impede or stop the flow of electrons. 27. In which direction would electrons and ions flow in all galvanic cells? Electrons flow through the external wire from the anode to the cathode. Negative ions flow through the salt bridge from the cathode half-cell towards the anode halfcell. Positive ions flow through the salt bridge from the anode half-cell towards the cathode half-cell.

7 28. A galvanic cell is constructed as follows: Br 2 (l)/br - //Cl 2 (g)/cl -. Draw a diagram of this setup and label the anode the cathode direction of electron flow direction of ion flow in the salt bridge suitable electrodes Use the table of standard potentials to determine the voltage of this cell under standard conditions. inert (platinum) anode bromine Br 2 (l) NO 3 electrons K + KNO 3 (aq) KBr (aq) KCl (aq) inert (platinum) cathode chlorine Cl 2 (g) electrolytes The anode half-reaction is: 2Br - (aq) Br 2 (l) + 2e V The cathode half-reaction is: Cl 2 (g) + 2e - 2Cl - (aq) V The overall voltage of the cell is E overall = V 29. Use the table of standard potentials to determine the standard cell potential for each of the following galvanic cells: (Necessary to use the SI Chemical Data Book) Co Co 2+ Ag + Ag E overall = 0.28 V V = 1.08 V How will a decrease in Co 2+ concentration affect the cell potential? A decrease in the concentration of Co 2+, according to Le Chatelier s Principle, favours the oxidation of cobalt to cobalt ions. As a result, the forward reaction is favoured and the voltage will increase. Pt, H 2 OH - Cu 2+ Cu E overall = V V = 1.17 V How will a decrease in Cu 2+ affect the cell potential? A decrease in the concentration of Cu 2+, according to Le Chatelier s Principle, favours the reverse reaction. As a result, the voltage will decrease.

8 30. The following chemicals are mixed. Use the table of standard potentials to determine whether a spontaneous reaction will occur. If a spontaneous reaction occurs, determine the overall cell potential. Silver metal is added to a cobalt nitrate solution. No spontaneous reaction. (E overall is negative) Iron (II) sulphate solution is added to bromine water. Fe 2+ (aq) Fe 3+ (aq) + e V Br 2 (aq) + 2e - 2Br - (aq) V E overall = V Fluorine is bubbled through water. F 2 (aq) + 2e - 2F - (aq) V 2H 2 O (l) O 2 (g) + 4H + (aq) + 4e V E overall = V 31. Research the properties of the following cells and complete the table below. (Note that the HSC syllabus requires students to study 1 of dry cell and lead acid and 1 only of the other cells. Button cells fulfil the syllabus requirements and are easy to understand.) CELL TYPES CHEMISTRY COST AND PRACTICALITY DRY CELL LEAD-ACID BATTERY Anode: Zn (s) Zn e - Cathode: NH MnO 2 + H 2 O Mn(OH) 2 + NH 3 Electrolyte: NH 4 Cl, ZnCl 2 paste Anode: Pb (s) + SO 4 2- (aq) PbSO 4 (s) + 2e - Cathode: PbO 2 (s) + 4H + + SO e - PbSO 4 (s) + 2H 2 O Electrolyte: Conc H 2 SO 4 Inexpensive Best for infrequent use, as voltage drops as used Maximum voltage = 1.5 V Expensive but long lasting, as rechargeable. Heavy, so adds to weight of vehicle and running costs. 6 cells in series Max each cell = 2 V IMPACT ON SOCIETY Widely used for low energy, low cost, portable applications Used as car battery Extensive application ENVIRONMENT IMPACT No negative impact Sulfuric acid is corrosive, so care must be taken with disposal. Lead compounds have environmental and health risks.

9 CELL TYPES CHEMISTRY COST AND PRACTICALITY BUTTON CELL Anode: Zn (s) + 2OH - ZnO (s) + H 2 O (l) + 2e - Cathode: Ag 2 O (s) + H 2 O + 2e - 2Ag (s) + 2OH - Electrolyte: KOH paste Small. Stable voltage. More expensive than dry cell, but much longer lasting. Delivers 1.6 V IMPACT ON SOCIETY Useful when small batteries needed. Watches, camera, hearing aids. ENVIRONMENT IMPACT No environmental or disposal problems 32. Draw labelled diagrams of: (a) a dry cell OR a lead-acid battery (b) a button cell

10 Other Questions relating to Section 4 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1. Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) Metal Dark blue solution Light blue solution In the above displacement reaction, electrons are transferred (A) from Cu 2+ to Zn (B) from Zn to Cu 2+ (C) from Zn to Zn 2+ (D) to the water molecules 2. The light blue solution in Question 1 above will contain the following dissolved species (A) Zn (s), Cu 2+ and Zn 2+ (B) Zn (s), Cu 2+, Zn 2+ and Cu (s) (C) Zn (s) and Cu (s) (D) Cu 2+ and Zn Metal Activity Series K Na Li Ba Ca Mg Al Zn Fe Sn Pb X Y Pt Au Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) This displacement equation could be used to identify substances X and Y in the activity series shown above. Which of the following would be the most correct? X Y (A) Cl Ag (B) Ag Cu (C) Cu Cl (D) Cu Ag

11 4. Sn e - Sn 2+ In this equation, the tin IV ion (A) has been reduced (B) has been oxidized (C) changed to an oxidation number of 4 (D) was oxidized by the electrons For questions 5 to 8: Zn Salt bridge Cu Zn 2+ Cu 2+ Beaker A Beaker B 5. The most active metal in the above galvanic cell is (A) Zn (B) Zn 2+ (C) Cu (D) Cu In the above cell, the oxidant or oxidising agent is (A) Zn (B) Zn 2+ (C) Cu (D) Cu When the above cell is operating electrons will move through the (A) solution from Zn to Cu (B) solution from Cu to Zn (C) wire from Zn to Cu (D) wire from Cu to Zn 8. As the above cell continues to operate the (A) concentration of positive ions in beaker A will decrease (B) concentration of positive ions in beaker B will decrease and negative ions will also decrease (C) salt bridge will transfer Cu 2+ ions from beaker B to beaker A (D) overall concentration of positive and negative ions remains unchanged

12 For questions 9 to 11: Cu Salt bridge Ag Beaker A Beaker B 9. An appropriate electrolyte for beaker A would be (A) Cu(NO 3 ) 2 (aq) (B) AgNO 3 (aq) (C) Cu (s) (D) CuO (s) 10. Within the above cell, reduction is occurring (A) in the salt bridge (B) in the voltmeter (C) at the Cu electrode (D) at the Ag electrode 11. Which of the following most accurately describes the situation at each of the electrodes in the above cell? Cu Ag (A) oxidant cathode (B) anode cathode (C) anode reductant (D) cathode anode 12. Which is the correct statement relating to the following reaction? Mg (s) + CuSO 4 (aq) MgSO 4 (aq) + Cu (s) (A) (B) (C) (D) Magnesium is reduced and copper ions are the oxidant. Copper ions are reduced and magnesium is the reductant. Sulfate ions are oxidised and magnesium is the oxidant. Magnesium ions are the reduction product and copper is the reductant.

13 Short Response Questions relating to Sections (similar to past HSC questions) 1. Four metals (Pb, x, y and z) were connected in pairs in a galvanic cell and the voltage recorded. The following results were obtained: List the 4 metals in decreasing order of activity. (1 mark) z, y, Pb, x 2. A student investigated the following electrochemical cell. List THREE correct observations the student could make as the cell operated. The aluminium electrode would get smaller. The copper electrode would get bigger. The solution in Beaker 2 would become a lighter blue.

14 3. A galvanic cell was made by connecting two half-cells. One half-cell was made by putting a copper electrode in a copper(ii) nitrate solution. The other half-cell was made by putting a silver electrode in a silver nitrate solution. The electrodes were connected to the voltmeter as shown in the diagram. - ions + ions (a) Complete the above diagram by drawing a salt bridge and show the direction of movement of positive and negative ions in the salt bridge. (2 marks) See above (b) Using the standard potentials table (at the back of book), calculate the theoretical voltage (at standard conditions) of this galvanic cell. (2 marks) The anode half-reaction is: Cu (s) Cu 2+ (aq) + 2e V The cathode half-reaction is: Ag + (aq) + e - Ag (s) V The overall equation is: Cu (s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag (s) E = V (c) Explain what is meant by standard conditions with respect to measurement of reduction potentials. (3 marks) Standard conditions are at 25 C and at 1 atm pressure Electrolyte concentrations must be 1 mol/l

15 4. (a) Identify the anode in this diagram. (1 mark) Lead (b) Write the equation for the overall cell reaction and calculate the overall cell voltage (assume standard conditions). (2 marks) The anode half-reaction is: Pb (s) Pb 2+ (aq) + 2e V The cathode half-reaction is: Ag + (aq) + e - Ag (s) V The overall equation is: Pb (s) + 2Ag + (aq) Pb 2+ (aq) + 2Ag (s) E = V 5. Cell 1 Cell 2 Choose ONE of the cells shown above and for that cell (a) Identify the cell. Cell 1 is a dry cell (1 mark) (b) State ONE problem associated with using the cell. (1 mark) The voltage fluctuates because the ions in the electrolyte do not easily migrate between the half-cells; the battery becomes polarised (charge build-up) and the current drops. (c) Describe, using half-equations, the chemistry of the chosen cell. (3 marks) Anode: Zn (s) Zn 2+ (aq) + 2e - Cathode: NH 4 + (aq) + MnO 2 (s) + H 2 O (l) + e - Mn(OH) 3 (s) + NH 3 (aq) The cathode is a carbon rod in contact with manganese dioxide and the electrolytic paste containing ammonium ions and zinc chloride. The anode is zinc.

16 6. Students performed the following first-hand investigation. The beaker initially contained ml of mol L -1 copper sulfate solution. After several hours the colour of the solution had become lighter and a reddish deposit had formed on the surface of the zinc. The reddish deposit was removed from the zinc and dried. The deposit was found to weigh g. (a) Explain the observations and write a balanced overall equation for the reaction which occurred. (3 marks) The solution became lighter as the concentration of the blue copper sulfate decreased. Cu 2+ ions were reduced to form copper (a reddish solid), which deposited on the surface of the zinc. The overall equation is: Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s) E = V (b) Calculate the concentration of the copper sulfate solution remaining after the reaction. (3 marks) Initial moles of copper sulfate in solution = c x V = x mol = mol Moles of copper formed = g/63.55 g = moles Since 1 mole of copper is formed from 1 mole of copper ions, Moles of copper ions remaining in solution = = mol Volume of solution = ml Hence concentration of final solution = / = mol/l (2 s. f.)

17 Section 5 NUCLEAR ENERGY AS A SOURCE OF MATERIALS 1. What is radioactivity? The spontaneous emission of particles and/or energy from the nucleus of an atom. 2. What is the difference between a stable and a radioactive isotope? Do radioactive isotopes have the same chemical and physical properties as stable isotopes of the same element? Explain your answer using examples of stable and radioactive isotopes of the same element. A stable isotope does not emit particles from its nucleus, whereas a radioactive nucleus is unstable and, emitting particles or energy, moves to a more stable arrangement of particles within the nucleus. Unstable nuclei are either too big (like uranium, with 92 protons and a mass number 235) or have a neutron/proton ratio which is too large. Isotopes with a high n/p ratio tend to emit beta particles (which reduces the ratio by a neutron changing into a proton plus a high energy electron, the beta particle). Atoms which have a big nucleus tend to emit alpha particles, which reduces the total number of protons and neutrons. Further changes then emit beta particles and these reduce the neutron/proton ratio. Radioisotopes and stable isotopes have identical chemical and physical properties (apart from a slight difference in atomic mass). Carbon-12 is stable, whereas carbon-14 is radioactive and emits beta and gamma particles. 3. When is a nucleus unstable? Describe conditions for the formation of alpha and for beta particles. Unstable nuclei are either too big (like uranium, with 92 protons and a mass number 235) or have a neutron/proton ratio which is too large. Isotopes with a high n/p ratio tend to emit beta particles (which reduces the ratio by a neutron changing into a proton plus a high energy electron, the beta particle). Atoms which have a big nucleus tend to emit alpha particles, which reduces the total number of protons and neutrons. Further changes then emit beta particles and these reduce the neutron/proton ratio.

18 4. Draw up a table to compare alpha, beta and gamma radiation in terms of particles, mass, charge, ionising ability and penetrating ability. Write a suitable nuclear equation to show the formation of each type of radiation. Type of radiation Particle or not Alpha Beta Gamma Particle, 2 protons, 2 neutrons (equivalent to a helium nucleus) Particle, high energy electron formed from the break-up of a neutron into a proton plus electron (the beta particle) Energy only, no particle Relative Mass 4 (mass of 2 protons plus 2 neutrons) 1/2000 (as an electron has only 1/2000 mass of a proton or neutron) 0 (no mass) Charge (no charge) Ionising ability High (high charge and mass) High (charged and high speed) Low (no mass or charge so no momentum) Penetrating ability Very low; stopped by tissue paper and 1 cm of air Does not penetrate skin Moderate; 10 cm of air and will pass through Al foil but stopped by Pb Penetrates skin High; stopped by 5 cm of lead or 15 cm of concrete Penetrates body tissues Equation for formation Rn 84 Po He Ra Ac + 0 e 89 1 No equation as no change in particles

19 5. Write an equation for the alpha decay of radon Rn Po + 4 He Write an equation for the beta decay of radium Ra Ac + 0 e What is meant by the half-life of a radioactive isotope and what is its relevance? The half-life is a measure of the rate of disintegration of radioisotopes. Each isotope has its own half-life. The half-life of an isotope is the time required for half the atoms in a given sample to undergo radioactive decay. For any particular radioisotope the half-life is independent of the initial amount of the isotope present. Radioisotopes which have a short half-life will disintegrate rapidly. The level of radiation from a sample drops rapidly. Isotopes, such as technetium-99m which has a half-life of 6 hours, lose their radioactivity very quickly, and are useful for medical diagnosis. Iodine-131 has a half-life of 8 days, long enough for treatment, but short enough that the patient does not need to be isolated for too long. Cobalt-60, used in gamma ray treatment of cancer cells from an instrument outside the body, has a half-life of 5.3 years. The isotope which releases the gamma rays does not need to be replaced in the instrument too often. 8. Consider the following graph representing the decay of strontium-90. (a) What is the approximate half-life of this isotope? 27 years (b) A sample which contained 4 g of this isotope was analysed. How much strontium-90 would remain in the sample after approximately 80 years had elapsed? 0.5 g (2 g left after 27 years, 1 g left after 54 years, 0.5 g after 81 years)

20 9. What is a transuranic element? An element with atomic number >93 i.e. after uranium, in the Periodic Table. The majority of these elements are artificially made (by bombardment with neutrons in a nuclear reactor or positively charged particles in an accelerator). However, there are traces of plutonium and neptunium in natural samples of uranium. 10. Describe how transuranic elements are produced in a nuclear reactor. Give 2 examples. Transuranic elements (e.g. neptunium) are made in a reactor by bombardment of other nuclei by neutrons 238 U n U 0 e Np Plutonium is bombarded with neutrons to produce an isotope of americium, which is an alpha emitter and is used in domestic smoke detectors. 239 Pu + 2( n) Am + 0 e Describe how artificial isotopes are produced in a cyclotron. Give 2 examples. Artificial isotopes are produced by bombardment of nuclei by high speed positive particles such as helium (alpha particles), protons or nuclei of other atoms, such as carbon, in linear accelerators or in cyclotrons. Californium-245 is made by bombardment of curium-242 with alpha particles. 242 Cm He Cf + 1 n 98 0 Phosphorus-30 is formed by bombardment of aluminium with alpha particles. 27 Al He P + 1 n How is radiation detected? Describe 3 different methods of detection and the principle which allows this method of detection. Geiger counter. This method depends on the ionising ability of radiation, especially alpha and beta emission. The charged particles enter the detecting tube and the noble gas is ionised. An electric current flows and is detected. Photographic film. Gamma rays are a form of electromagnetic radiation and can change the silver compounds on a photographic film. Scintillation counter. This method involves the use of certain chemicals which give out a flash of light when hit by radiation such as alpha, beta or gamma rays.

21 13. Complete the table to summarise the uses of at least 2 radioactive isotopes in industry and 2 in medicine. Choose the 2 examples in the medicine category so that 1 isotope is used for diagnosis and the other for therapy. Use in : INDUSTRY Radioisotope used Co-60 Type of radiation Beta and gamma Properties of this type of radiation Gamma - Destroys large molecules Explanation of how it is used Sterilisation of surgical instruments Bacteria are killed by the gamma radiation Problems associated with its use Workers must be protected from gamma radiation MEDICINE Am-241 Alpha Ionises air Long halflife (>400 years) In smoke detectors. Current flows when air ionised. Smoke blocks flow of ions, no current. Alarm goes off. Special precautions for disposal of smoke detectors. Diagnosis I-123 Gamma Gamma rays penetrate out through skin to be detected outside body. Short halflife (13 hours) Ingested, goes to thyroid. Concentrates in thyroid, allows photo image of thyroid. Shielding of workers, others for short period. Treatment Co-60 Gamma (beta) Penetrating, kills living tissues, including cancer cells Moderate half-life (5 years) Irradiation of affected area over a period of weeks kills tumours. Isotope lasts in instrument. Shielding of workers.

22 14. The table below provides a summary of transuranic element synthesis. ELEMENT AND SYMBOL ATOMIC NUMBER MASS NUMBER YEAR DISCOVERED SYTHESISED FROM SYNTHESIS METHOD Neptunium (Np) U-238 Neutron in, electron out Plutonium (Pu) Np-239 Electron out Americium (Am) Pu neutrons in, electron out Curium (Cm) Pu-239 particle in, neutron out Berkelium (Bk) Am-241 particle in, 2 neutrons out Californium (Cf) Cm-242 particle in, neutron out Einsteinium (Es) U neutrons in, 7 electrons out Fermium (Fm) U neutrons in, 8 electrons out Mendelevium (Md) Es-253 particle in, neutron out Nobelium (No) Cm-246 C-12 in, 4 neutrons out Lawrencium (Lw) Cf-252 B-10 in, 5 neutrons out Choose 3 transuranic elements from the above list and write nuclear equations to show how they are synthesised. Ensure that the 3 examples involve bombardment by different particles. Americium: Formed by neutron bombardment of plutonium Pu + 2( n) Am + 0 e 95 1 Californium-245 is made by bombardment of curium-242 with alpha particles. 242 Cm He Cf + 1 n 98 0 Nobellium-254 is made by bombardment of curium-246 with a carbon nucleus. 246 Cm C No n 102 0

23 15. Research two recent (since 1969) discoveries of elements. Identify the: atomic number of the new element. year of discovery. process used to discover the new element. Give a brief description of the nuclear chemistry involved for each element researched. (Note that the discovery of element 118 has since been disregarded by the original discovery team on the basis that the experimental results have not been able to be repeated.) Information which might be useful for this research task (extracted from an article on transuranium elements at Transuranium elements are radioactive elements with atomic numbers greater than that of uranium (at. no. 92). All the transuranium elements of the actinide series were discovered as synthetic radioactive isotopes at the Univ. of California at Berkeley or at Argonne National Laboratory. In order of increasing atomic number they are neptunium, plutonium, americium, curium, berkelium, californium, einsteinium, fermium, mendelevium, nobelium, and lawrencium. Of these only neptunium and plutonium occur in nature; they are produced in minute amounts in the radioactive decay of uranium. Much of the study of the transuranium elements has taken place at the Lawrence Berkeley National Laboratory (at Berkeley, Calif.) and at the Joint Institute for Nuclear Research in Dubna, Russia; workers at both locations share credit for the independent discovery of rutherfordium, dubnium, and seaborgium (at. no. 104, 105, and 106 respectively), which are the first three transactinide elements. A German team at the Institute for Heavy Ion Research at Darmstadt discovered bohrium, hassium, meitnerium, darmstadtium, roentgenium, and ununbium (at. no. 107 through 112). The Dubna laboratory, with assistance from Berkeley, claims to have synthesized ununquadium (at. no. 114), and working jointly with the Lawrence Livermore National Laboratory (at Livermore, Calif.) claims to have produced ununtrium (at. no. 113) and ununpentium (at. no. 115). The Berkeley team claimed to have produced ununhexium (at. no. 116) and ununoctium (at. no. 118), but later retracted the claim for ununoctium after other laboratories failed to reproduce Berkeley's results and a reanalysis of their data did not show the production of the element. Other research teams have since synthesized ununhexium directly. Up to and including fermium (at. no. 100), the transuranium elements are produced by the capture of neutrons; the transfermium elements are synthesized by the bombardment of transuranium targets with light particles or, more recently, by projecting medium-weight elements at targets of other medium-weight elements. Isotopes of the transuranium elements are radioactive because their large nuclei are unstable, and the superheavy elements in particular have very short half-lives. In initial experiments, the formation of elements 118 and 116 was attempted by accelerating a beam of krypton-86 ( 86 36Kr) ions to an energy of 449 million electron volts and directing the beam onto targets of lead-208 ( Pb) Pb Kr Uuo + 1 n It was hypothesised that element 118 nucleus would decay less than a millisecond after its formation by emitting an -particle. This would result in an isotope of element 116 (mass number 289, containing 116 protons and 173 neutrons). This isotope of element 116, would also undergo further -decay processes to an isotope of element 114 and so on down to at least element 106.

24 Other Questions relating to Section 5 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1. A radioactive isotope is (A) only found after uranium on the periodic table (B) an atom that has an unstable nucleus (C) a version of an element that has different chemical properties (D) an element that has a stable nucleus and only emits radiation irregularly C is a stable isotope but 14 C is radioactive. The reason for this is (A) 14 C is exactly the same as 14 N but with fewer electrons (B) 12 C was discovered before 14 C (C) the ratio of neutrons to protons is too high in 14 C (D) there can only be one isotope of the element inside the zone of stability for protonneutron ratios 3. An example of a transuranic element is (A) 207 Pb (B) 235 U (C) 239 Np (D) 14 C 4. A transuranic element (A) generally has a stable nucleus (B) is made when a heavy nucleus is bombarded by neutrons or high speed particles (C) will have a molar mass less than 235 g/mole (D) is formed when a fissionable element like uranium 235 undergoes nuclear breakdown 5. An example of a commercially produced radioisotope used in medicine is (A) 235 U (B) 14 C (C) 206 Pb (D) 131 I 6. Technetium-99m is commonly used in medicine for the diagnosis of certain diseases. It is produced from the radioactive decay of molybdenum-99. Molybdenum-99 (rather than technetium-99m) is transported to the hospitals around Australia. A possible reason for this would be (A) technetium is too poisonous to be transported (B) 99 Mo is less dangerous (C) 99m Tc is too heavy to transport because of its density (D) 99 Mo has a longer half-life than 99m Tc

25 7. Most commercially produced isotopes have short half-lives. For instance, 131 I is used in the diagnosis and treatment of thyroid gland diseases. It has a half-life of 8 days. If 1 gram of 131 I was supplied to a hospital, how many grams of 131 I would be left after 24 days? (A) 0.5 g (B) 0.25 g (C) 1.0 g (D) g 8. The earliest means of detecting radiation was (A) AAS (B) the Geiger-Muller counter (C) the scintillation counter (D) a photographic film 9. A scintillation counter (A) measures the potential difference across a gas that has been ionized by radiation (B) displays a streak of water or alcohol droplets in the path of some radioactive particle (C) detects the flash given off when certain nuclei break down (D) detects the dark spots produced on photographic film that has been affected by radiation 10. Which of the following radioisotopes is not likely to be used in medicine for the purpose of diagnosis? (A) 238 U (B) 131 I (C) 18 F (D) 99m Tc 11. Identify the transuranic element. (A) calcium (B) cerium (C) chromium (D) curium 12. Which instrument is used to detect radiation from radioactive isotopes? (A) ph meter (B) Geiger counter (C) burette (D) atomic absorption spectrophotometer (AAS) 13. Consider the following nuclear reaction U 90 Identify X (A) -radiation (B) a beta particle (C) an alpha particle (D) a neutron

26 Questions relating to Section (similar to past HSC questions) 1. A radioactive isotope is likely to demonstrate the following property: (A) too many protons for the number of electrons in the atom (B) too many protons and neutrons in the atom (C) too many electrons in the outer shell of the atom (D) too many electrons for the number of neutrons in the atom 2. Describe the conditions for a nucleus to be unstable. (2 marks) The nucleus is too big (too many protons and neutrons) or the balance of neutrons and protons is outside the zone of stability. 3. The following diagram shows the decay series for uranium-238. Use the diagram to identify and write the nuclear equation for: (a) an alpha decay process Radon-222 decays to polonium Rn Po + 4 He 84 2 (b) a beta decay process (2 marks) Thorium-234 decays to protactinium Th 91 0 e 1

27 4. Describe the different ways both commercial isotopes and transuranic elements are produced. (4 marks) Commercial isotopes and transuranic elements are produced by bombardment of nuclei by either neutrons (in a reactor) or charged particles (in an accelerator, such as a cyclotron). Artificial isotopes are produced by bombardment of nuclei by high speed positive particles such as helium (alpha particles), protons or nuclei of other atoms, such as carbon, in linear accelerators or in cyclotrons. The charged particles are accelerated to enormous speeds by moving through a magnetic field. When the optimum speed is reached, the charged particle crashes into the target nucleus. Phosphorus-30 is formed by bombardment of aluminium with alpha particles. 27 Al He P + 1 n 15 0 Transuranic elements (e.g. neptunium) are made in a reactor by bombardment of other nuclei by neutrons 238 U n U 0 e Np Plutonium is bombarded with neutrons to produce an isotope of americium, which is an alpha emitter and is used in domestic smoke detectors. 239 Pu + 2( n) Am + 0 e 95 1 Some transuranic elements are produced by bombardment with charged particles. Californium-245 is made by bombardment of curium-242 with alpha particles. 242 Cm He Cf + 1 n Discuss the benefits and problems associated with the use of ONE named radioactive isotope in industry. (4 marks) Cobalt-60. Benefits Used for sterilising medical supplies, as gamma rays are emitted. These destroy large molecules such as DNA and hence kill bacteria on instruments, bandages, etc. The isotope has a relatively long half-life (5 years), so the isotope does not need to be replaced for many years, yet the source is able to produce a reasonable level of intensity. Problems Gamma radiation is very penetrating (gamma penetrates 5 cm of lead and 15 cm of concrete) and causes cell damage and/or death of cells. Workers in industry (here the medical staff) need to be protected from the radiation source by thick lead shields.

28 6. Discuss the criteria considered when choosing radioisotopes for use in medicine. (4 marks) 3 criteria considered when choosing an appropriate isotope are: the type of radiation emitted the half-life of the isotope the chemical specificity of the isotope (which organs it targets). If the isotope is used for diagnosis, it must be a gamma emitter, as the radiation must be detected outside the body. Only gamma rays readily penetrate through tissues to reach the external detector (film). The isotope should have a short half-life (hours), so that the radiation level drops rapidly and the patient is only exposed to radiation for a short time. The isotope must target particular organs or tissue types in the body. Technetium-99m targets tissues where cells are dividing rapidly. As a result, tumours, infection or inflammation can be detected. If the isotope is used as treatment (to destroy cells in the body), the rays emitted are normally gamma, often in conjunction with beta. It is the gamma radiation that kills the target tissues. The half-life of the isotope should be longer than for diagnosis (8 days in the case of iodine- 123, used to treat thyroid problems) as days, rather than hours, are needed for successful treatment. As with diagnosis, the isotope should be chemically specific in that it targets the normal biochemistry of the organ to be treated (calcium or strontium isotopes are used to treat bone cancers). If a beam of gamma rays, from an instrument outside the body, is used to destroy tumour cells (as in the case of cobalt-60), then the isotope must be a gamma emitter as it must penetrate the body tissues to the target organ and destroy the target cells. The half-life is relatively long (5 years for Co-60), so that the isotope does not need to be replaced regularly but the intensity of the beam is still strong enough to destroy the cells targeted. Since the isotope used is external to the body, it does not need to be matched to the chemical metabolism of the target cells. 7. Assess the impact of the production of radioisotopes for society and the environment. (5 marks) The answer must include an overall assessment statement. (Deduct 1 mark if not included) Benefits of the use of isotopes for society should be identified, using specific examples. Generalisations should include medical diagnosis AND treatment AND use in industry and at least ONE specific example from medicine and ONE from industry should be used. (3 marks) Problems for society and the environment. Answer could include discussion of problems associated with the long half-life of some isotopes, the problems for storage of isotopes or radioactive wastes and the need for shielding of researchers and workers when using isotopes. Issues associated with the safety of nuclear reactors could also be included. (2 marks)