8.1 The Ionic Theory.

Transcription

1 Chemistry Form 4 Page 21 Ms. R. Buttigieg 8.1 The Ionic Theory. What you should Learn: Writing formulae for cations and anions Writing ionic formulae for ionic substances Representing the change from atoms/molecules to ions, and vice/versa by means of ionic half equations. Ionic compounds are compounds containing metal atoms and non-metal atoms. Sodium chloride and copper(ii) sulphate are both ionic compounds. Working out the formula Although ions are charged particles, ionic compounds are neutral overall. You need to have equal numbers of positive charges and negative charges in an ionic compound. The diagrams show how you can work out the formulae for sodium chloride and iron(iii) oxide from the charges carried by their constituent ions Always write the symbol for the positive ion first and do not write any charges in the final formula. So, NaCl is correct but ClNa and NaCl - are wrong. Compound ions, such as OH -, contain more than one element. If you need two or more compound ions in your formulae, put brackets around them. Ionic Equations If we consider the reaction of sodium hydroxide and hydrochloric acid: NaOH (aq) + HCl (aq) NaCl (aq) + H 2 O (l) These are actually made of ions (except water), so this equation can be written: Na + OH - (aq) + H + Cl - (aq) Na + Cl - (aq) + H 2 O (l) The simplest equation you can have is: OH - + H + H 2 O An equation shows change, so anything which appears on both sides can be removed.

2 Chemistry Form 4 Page 22 Ms. R. Buttigieg Consider the reaction between silver nitrate and sodium chloride. Normally, this is written as, AgNO 3(aq) + NaCl (aq) NaNO 3(aq) + AgCl (s) Now let s separate the equation out and into its ions. Ag + (aq) + NO 3 - (aq) + Na + (aq) + C l - (aq) Na + (aq) + NO 3 - (aq) + AgCl (s) You can see that the ions NO 3 - (aq) and Na + appear on both sides of the equation. They have not changed and so have not reacted at all. Sometimes these are called spectator ions. So, the ionic equation reduces the chemical equation to just showing the ions that are actually participating in the reaction. In this case the ionic equation is Ag + (aq) + C l - (aq) AgCl (s) The positive ions come from atoms that have lost electrons and the negative ions form atoms that have gained electrons. Make sure there is equal charge on each side. In terms of ionic equations, you need to make sure that the charges on the ions balance. There are easy rules to follow: All group 1 ions = + All group 2 ions = 2+ All group 3 ions = 3+ All group 5 ions = 3- All group 6 ions = 2- All group 7 ions = - Don t forget that group 4 and 0 don t form ions and that the transition metals may form more than one type of ion. Once you know what charge the ion carries, the rest is easy. For example to make calcium chloride you need calcium ions and chlorine ions. Calcium is in group 2 so is Ca 2+ Chlorine is in group 7 so is Cl - The number of charges need to balance (cancel each other out) so one Ca 2+ needs 2 Cl -. Therefore calcium chloride would be CaCl 2.

3 Chemistry Form 4 Page 23 See Ms. C4U R. pg. Buttigieg 151, 305 GCSE Chem pg. 99 Writing ionic Equations An ionic equation tells you what is really happening. To write an ionic equation: 1. First write the equation 2. Then write the equation again: the ionic compounds write them in ion form, except when the ionic compound is a precipitate. 3. Cross out the spectator ions : i.e. the ions which are common on both sides of the equation. These ions, didn t take part in the reaction since they remained the same. 4. Write the ions, molecules and atoms which remain. This is the ionic equation. 2 FeCl 2 (aq) + Cl 2 (aq) 2 FeCl 3 (aq) Some Ionic Equations you should know Metal Hydroxide/Alkali with an Acid (also known as neutralization reaction) Metal Oxide with an Acid Metal with an Acid H + (aq) + OH (aq) H 2 O (l) 2H + (aq) + O - - (aq) H 2 O (l) Zn (s) + 2H + (aq) Zn ++ (aq) + H 2 (g) Displacement Reactions Metal displacing less reactive metal Zn (s) + Cu ++ (aq) Zn ++ (aq) + Cu (s) Metal displacing hydrogen from an acid Mg (s) + 2H + (aq) Mg ++ (aq) + H 2 (g) Halogen displacing less reactive halogen Cl 2 (g) + 2Br - (aq) 2Cl - (aq) + Br 2 (g) Carbonate with an Acid 2H + (aq) + CO (s) H 2 O (l) + CO 2 (g) Precipitation Reaction Ag + (aq) + Cl - (aq) AgCl (s) Write ionic equations for the following reactions (from JL 2003) (6 marks) a) NaCl (aq) + AgNO 3 (aq) NaNO 3 (aq) + AgCl (s) b) MgO (s) + 2 HNO 3 (aq) Mg(NO 3 ) 2 (aq) + H 2 O (l)

4 Chemistry Form 4 Page 24 Ms. R. Buttigieg c) CaCO 3 (s) + 2 HCl (aq) CaCl 2 (aq) + CO 2 (g) + H 2 O (l) d) (i) K2CO3 + 2HCl 2KCl + H2O + CO2 (i.e. an acid on a carbonate) (ii) NaOH + HNO3 NaNO3 + H2O (i.e. neutralisation) (4 marks) Electron-half-equations What is an electron-half-equation? When magnesium reduces hot copper(ii) oxide to copper, the ionic equation for the reaction is: You can split the ionic equation into two parts, and look at it from the point of view of the magnesium and of the copper(ii) ions separately. This shows clearly that the magnesium has lost two electrons, and the copper(ii) ions have gained them. These two equations are described as "electron-half-equations" or "half-equations" or "ionichalf-equations" or "half-reactions" - lots of variations all meaning exactly the same thing! The reaction between chlorine and iron(ii) ions (a metal and non-metal) Chlorine gas oxidises iron(ii) ions to iron(iii) ions. In the process, the chlorine is reduced to chloride ions. You start by writing down what you know for each of the half-reactions. In the chlorine case, you know that chlorine (as molecules) turns into chloride ions: The first thing to do is to balance the atoms that you have got as far as you possibly can: Now you have to add things to the half-equation in order to make it balance completely.

5 Chemistry Form 4 Page 25 Ms. R. Buttigieg All you are allowed to add are: electrons water hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead) In the chlorine case, all that is wrong with the existing equation that we've produced so far is that the charges don't balance. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. That's easily put right by adding two electrons to the left-hand side. The final version of the halfreaction is: Now you repeat this for the iron(ii) ions. You know (or are told) that they are oxidised to iron(iii) ions. Write this down: The atoms balance, but the charges don't. There are 3 positive charges on the right-hand side, but only 2 on the left. You need to reduce the number of positive charges on the right-hand side. That's easily done by adding an electron to that side: Combining the half-reactions to make the ionic equation for the reaction What we've got at the moment is this: It is obvious that the iron reaction will have to happen twice for every chlorine molecule that reacts. Allow for that, and then add the two half-equations together. But don't stop there!! Check that everything balances - atoms and charges.

6 Chemistry Form 4 Page 26 Ms. R. Buttigieg Consider the following list of metals. Zn, Mg, Cu, Ag, Fe, Al. (a) (i) Rewrite this list of metals, starting with the most reactive. (2 marks) (ii) Suggest the two metals between which hydrogen could be placed. and (1 mark) (b) Complete the following statements by inserting the name of one of the metals given above. (i) cannot displace magnesium but it displaces all the other metals form solutions of their salts. (ii) can displace silver but cannot displace the other metals. (2) (c) Some metals, for example calcium, can also displace hydrogen from dilute hydrochloric acid. The equation for the reaction is. Ca + 2HCl CaCl 2 + H2 (i) Write an ionic equation for this reaction (omitting spectator ions). (2 marks) (ii) Write the ionic half equations for this reaction. (2 marks) (iii) Give a reason why this displacement reaction is also a redox reaction. (1 mark) 2 Mg (s) + O 2 (g) > 2 MgO 1) What is happening to magnesium in this reaction? Answer: Magnesium is losing electrons to become a positive ion. This is the process of oxidation. The reaction we can write for this process is called the oxidation half - reaction. Mg (s) > Mg e - ( electrons ) Since two magnesium atoms are reacting a total of 4 electrons are lost by the magnesium 2) What is happening to the Oxygen in this reaction? Answer: Oxygen is gaining electrons to become a negative ion. This is the process of reduction. The reaction we write for this process is called the reduction half- reaction. O (g) > O e Since oxygen is a diatomic molecule two oxygen atoms are undergoing reduction and a total of 4 electrons are gained 1 1 Note: The total number of electrons lost in oxidation = the total number gained in reduction. This is true of all redox reactions

7 Chemistry Form 4 Page 27 Ms. R. Buttigieg This question is about oxidation and reduction. Say which substances are oxidized and which are reduced in each equation. 1. C(s) + H 2 O (g) CO(g) + H 2 (g) 2. C(s) + PbO (s) CO(g) + Pb(s) 3. Ca(s) + Cl 2 (g) CaCl 2 (s) 4. 2Ag + (aq) + Mg (s) 2Ag + (s) + Mg 2+ (aq) 5. Cl 2 (g) + 2KBr (aq) 2KCl(aq) + Br 2 (g) For the same equations state the oxidizing agent and the reducing agent. 1. C(s) + H 2 O (g) CO(g) + H 2 (g) 2. C(s) + PbO (s) CO(g) + Pb(s) 3. Ca(s) + Cl 2 (g) CaCl 2 (s) 4. 2Ag + (aq) + Mg (s) 2Ag + (s) + Mg 2+ (aq) 5. Cl 2 (g) + 2KBr (aq) 2KCl(aq) + Br 2 (g) Oxidation Number: The charge an atom has, or appears to have, when the electrons of the compound are counted in accordance with a set of rules. 1. Free elements are assigned an oxidation state of zero. 2. The sum of the oxidation states of all that atoms in a species must be equal to the net charge on the species. 3. The alkali metals (Li, Na, K, Rb, and Cs) in compounds are always assigned an oxidation state of Fluorine in compounds is always assigned an oxidation state of The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) and also Zn and Cd in compounds are always assigned an oxidation state of Hydrogen in compounds is assigned an oxidation state of Oxygen in compounds is assigned an oxidation state of Halogen in compounds is assigned an oxidation state of -1. Application of these rules to a compound must occur in the order given.

8 Chemistry Form 4 Page 28 Ms. R. Buttigieg What is the oxidation number of... 1) N in NO 3 2) C in CO 32 3) Cr in CrO 42 4) Cr in Cr 2 O 72 5) Fe in Fe 2 O 3 6) Pb in PbOH + 7) V in VO + 2 8) V in VO 2+ 9) Mn in MnO 4 10) Mn in MnO 42 Reactivity Series and Reactions - SUMMARY Metals Potassium Sodium Lithium Calcium Reaction with cold water These react Metal (s) + Water (l) Metal Hydroxide (aq) + Hydrogen (g) Reaction with steam React with steam Metal (s) + Steam (g) Metal hydroxide (aq) + Reaction with acid React. Metal (s) + Acid (aq) Salt (aq) + Hydrogen (g) Reduction of Heated Metal oxide by hydrogen Are not reduced Magnesium Aluminium Zinc Iron Lead Hydrogen Copper Silver Reacts very slowly. Form Hydroxide and Hydrogen Do not react Hydrogen (g) Do not react Aluminium is slow to react as it is protected by an aluminium oxide layer. These all displace hydrogen from acid as above it. Do not react Reduced. Hydrogen is passed over the heated oxide. Metal oxide + Hydrogen Metal (s) + Steam The reaction with iron is reversible Silver oxide is reduced to silver even by heat alone

9 Chemistry Form 4 Page 29 Ms. R. Buttigieg 8.2 Electrolysis Electricity is a flow of charge. In fact current is the amount of charge flowing per second. In metals it is the electrons which flow. In solutions/melts of ionic compounds, it is the ions which carry the current. In electrolysis we use a direct current (d.c.) supply. Because if the polarity changes many times a seconds, the ions are neither attracted towards on electrode or the other. In a d.c. supply, electron flow is from the negative to the positive. Conventional current flow is from the positive to the negative. Metals and graphite are good conductors, because they have free electrons. Ionic compounds when molten or dissolved in water also conduct electricity because they have free ions. Moreover they are decomposed in the process. So they are called electrolytes. The chemical change brought about by electricity is called electrolysis. The positive electrode is called anode. The negative electrode is the cathode. Positive ions like Al 3+ and Cu 2+ move to the negative electrode (cathode), and negative ions like Cl - and O 2- move towards the positive electrode (anode). At the cathode the half equation: Cu 2+ (aq) + 2e Cu (s) At the anode the half equation is: 2Cl - (aq) Cl 2 (g) + 2e The overall equation is: HW: Chemistry for You pg. 116 no 1, pg. 117 no. 5, 6 CuCl 2 (aq) Cu (s) + Cl 2 (g)

10 Chemistry Form 4 Page 30 Ms. R. Buttigieg Complete the following: The positive electrode is called the and the negative electrode is called the. The chemical change brought about by electricity is called. are substances which conduct electricity and are decomposed in the process. Only compounds when or conduct electricity and are decomposed in the process. The ion which is attracted to the cathode is called the and the ion which is attracted to the anode is called the. Electrolysis of Molten Lead Bromide The Lead (II) ion, Pb ++ is attracted towards the cathode (-). There it is changed to a lead atom. Pb ++ (l) + 2e Pb (l) Half equation of what happens at the cathode The Bromide ion, Br - is attracted towards the anode (+). 2Br - (l) - 2e Br 2 (g) Half equation of what happens at the anode The overall equation is: PbBr 2 (l) Pb (l) + Br 2 (g)

11 Chemistry Form 4 Page 31 Ms. R. Buttigieg Electrolysis of Potassium Chloride. Potassium chloride must be heated until it is molten before it will conduct electricity. Electrolysis separates the molten ionic compound into its elements. The reactions at each electrode are called half equations. 2K + + 2e - 2K (potassium metal at the (-)cathode). 2Cl - - 2e - Cl 2 (chlorine gas at the (+)anode). Potassium ions gain electrons (reduction) to form potassium atoms. Chloride ions lose electrons (oxidation) to form chlorine atoms. The chlorine atoms combine to form molecules of chlorine gas. The overall reaction is 2K + Cl - (l) 2K (s) + Cl 2(g) Label the diagram: Write half equations for what is happening At the cathode and anode: At anode Molten zinc chloride At cathode Overall equation

12 Chemistry Form 4 Page 32 Ms. R. Buttigieg Electrolysis of Ionic Compounds Dissolved in Water Aqueous solutions of ionic compounds also conduct electricity because they have free ions. Some molecules of water break up into hydrogen ions, H + and hydroxide ions, OH -. So sodium chloride solution contains these four ions: Na +, Cl -, H + and OH -. The sodium ion and hydrogen ion migrate to the cathode but only one ion is discharged. The hydroxide ion and chloride ion migrated to the anode but only one is discharged. a) If we are electrolyzing concentrated sodium chloride solution, and the chloride ion and hydroxide ion are attracted to the anode, it is the chloride ion, which is discharged, because it is more concentrated. b) The hydrogen ion is preferred over the sodium ion at the cathode, because the ion of the less reactive element is discharged. An atom of a reactive element easily becomes an ion. So an ion of a reactive element is very difficult for it to become an atom. An atom of an unreactive element becomes an ion with difficulty. So an ion of an unreactive element becomes an atom with ease. So the copper(ii) ion and silver ion are discharged instead of the hydrogen ion. The less stable ion is discharged. The more stable ion remains an ion. So the hydroxide ion is discharged instead of a nitrate or a sulphate ion. c) Sometimes the electrode dictates what happens. So a copper anode in copper (II) sulphate solution is eroded and neither the sulphate nor the hydroxide ion are discharged. A mercury cathode prefers the sodium ion over the hydrogen ion. Electrolysing Acidified Water We acidify water with sulphuric acid, so that the products are the elements of water: hydrogen and oxygen in the proportion of 2:l. We do not use hydrochloric acid, because the product would be chlorine instead of oxygen. We call it the electrolysis of water because the products are the elements of water: hydrogen and oxygen in the proportion of 2:l. We use a Hoffman voltmeter to collect the gases given off. At the cathode (-): 4H'(aq) + 4e 2H 2 (g) At the anode (+): 4 OH - (aq) - 4e 2H 2 O (l) + O 2 (g) The acid becomes more concentrated.

13 Chemistry Form 4 Page 33 Ms. R. Buttigieg Electrolysing Copper (II) Sulphate Solution Using Carbon Electrodes. Ions present: Cu ++, SO 4 2-, H + and OH - At the cathode, the copper(ii)ion is discharged since it is the ion of the less reactive element and orange brown copper is formed. At the anode, the hydroxide ion is discharged because it is less stable and oxygen gas is given off. At the cathode (-): Cu ++ (aq) + 2e Cu (s) At the anode (+): 4 OH - (aq) - 4e 2H 2 O (l) + O 2 (g) The blue solution fades in colour. The solution becomes acidic since the hydroxide ions are decreasing and so there are more hydrogen ions present. Electrolysing Copper (II) Sulphate Solution Using a Copper Anode. Ions present: Cu ++, SO 4 2-, H + and OH - At the cathode, the copper(ii)ion is discharged since it is the ion of the less reactive element. At the anode, neither the sulphate nor the hydroxide ion is discharged. But the copper atoms become copper ions and the anode is eroded. At the cathode (-): Cu ++ (aq) + 2e Cu (s) At the anode (+): Cu (s) - 2e Cu ++ (aq) The solution remains blue because the number of copper ions is decreasing at the same rate as it is increasing. So the amount of copper ions remains the same. This set-up, i.e. having a copper anode in copper (II) sulphate solution is used in industry in electroplating and in the purification of copper. Electroplating This means covering a metal object with another metal using electricity To electroplate with copper, the metal object is the cathode; we use copper(ii)sulphate solution and a copper anode. Purification of Copper Copper, which is extracted, is not pure enough to be used for electrical wiring. Very pure copper is needed to make electrical wires so that it has very low resistance. To purify copper, copper(ii)sulphate solution is used, the impure copper is the anode and a thin strip of pure copper is placed as the cathode. The anode is eroded and becomes copper(ii)ions. The impurities fall to the bottom. Pure copper collects at the cathode.

14 Chemistry Form 4 Page 34 Ms. R. Buttigieg Purification of copper When copper is made from sulphide ores by the first method above, it is impure. The blister copper is first treated to remove any remaining sulphur (trapped as bubbles of sulphur dioxide in the copper - hence "blister copper") and then cast into anodes for refining using electrolysis. Electrolytic refining The purification uses an electrolyte of copper(ii) sulphate solution, impure copper anodes, and strips of high purity copper for the cathodes. The diagram shows a very simplified view of a cell. At the cathode, copper(ii) ions are deposited as copper. At the anode, copper goes into solution as copper(ii) ions. For every copper ion that is deposited at the cathode, in principle another one goes into solution at the anode. The concentration of the solution should stay the same. All that happens is that there is a transfer of copper from the anode to the cathode. The cathode gets bigger as more and more pure copper is deposited; the anode gradually disappears. In practice, it isn't quite as simple as that because of the impurities involved. What happens to the impurities? Any metal in the impure anode which is below copper in the electrochemical series (reactivity series) doesn't go into solution as ions. It stays as a metal and falls to the bottom of the cell as an "anode sludge" together with any unreactive material left over from the ore. The anode sludge will contain valuable metals such as silver and gold. Metals above copper in the electrochemical series (like zinc) will form ions at the anode and go into solution. However, they won't get discharged at the cathode provided their concentration doesn't get too high. The concentration of ions like zinc will increase with time, and the concentration of the copper(ii) ions in the solution will fall. For every zinc ion going into solution there will obviously be one fewer copper ion formed. (See the next note if you aren't sure about this.) The copper(ii) sulphate solution has to be continuously purified to make up for this.

15 Chemistry Form 4 Page 35 Ms. R. Buttigieg Industrial Production of Sodium Hydroxide Electrolysis of Brine (concentrated sodium chloride solution) is done. Hydrogen is produced at the cathode and chlorine at the anode. Sodium hydroxide solution is left. The cell (the place where the electrolysis takes place) is constructed so that the chlorine and sodium hydroxide solution do not mix. Otherwise they would react. Calculating the Amount of Substance Produced. 1 mole of electrons carry 96, 500 coulombs of charge and is equal to 1 faraday (F). Current, I = Charge, Q Time,t e.g. What is the mass of aluminium produced when 20,000 amperes pass for 10 hours? Quantity Symbol SI unit Current I Ampere (A) Charge Q Coulomb (C) Time t second Q = I x t = 20,000 x (10 x 60 x 60) = 7.2 x 10 8 C Al e Al 3 moles of electrons produce 1 mole of aluminium Convert moles of electrons to coulombs, and mole of aluminium to grams, because you know the coulombs that passed and you need to know the mass of aluminium 3 x C produce 27g C produce 27g 7.2 X 10 8 C produce? 7.2 X 10 8 x 27 = 67150g = kg So much electricity for so little aluminium produced! That's why aluminium is expensive.

16 Chemistry Form 4 Page 36 Ms. R. Buttigieg

17 Chemistry Form 4 Page 37 Ms. R. Buttigieg 3. The diagram below shows the electrolysis of concentrated sodium chloride solution in the laboratory. carbon electrodes bubbles of gas Y bubbles of gas Z sodium chloride solution If a few drops of universal indicator are added to the solution before electrolysis starts, the indicator is green. As electrolysis happens, the indicator turns blue around the cathode while around the anode first it turns red, then colourless. (a) (i) Name the gases Y and Z. (2 marks) (ii) Explain, in terms of the ions present in solution, what causes the indicator to go blue at the cathode. (2 marks) (iii) Suggest a reason for the colour changes at the anode. (1 mark) (b) Name a compound produced on an industrial scale by the electrolysis of sodium chloride solution. (1 mark)

18 Chemistry Form 4 Page 38 Ms. R. Buttigieg Cobalt has been used for electroplating in a similar way to copper. Complete the labelling of the diagram below, which could be used to electroplate a spoon with cobalt. The object to be plated is made the _ The metal being used for plating is made the _. containing ions of the metal being used for plating. Rewrite the following equations as an ionic equation, (omitting spectator ions). (a) (i) Ca + 2HCl CaCl 2 + H 2 (ii) Cl 2 + 2NaBr 2NaCl + Br 2 (4 marks) (b) Explain, in terms of electrons, why (i) calcium is oxidised in equation (a)(i) above. (ii) chlorine is reduced in equation (a)(ii) above (2 marks) Complete this table: (4 marks) chemical copper nitric acid large-scale use manufacture of bleaches construction of aircraft

19 Chemistry Form 4 Page 39 Ms. R. Buttigieg It is important to keep in mind that when metals are extracted from their ores, the process involved is a reduction process. Extraction of Aluminium Chemistry for You pg. 107 Aluminium ore is called bauxite. Bauxite contains aluminium oxide, water, iron oxide and other impurities. The purified dry ore, called alumina, is aluminium oxide - Al 2 O 3. Electrolysis of the alumina/cryolite solution gives aluminium at the cathode and oxygen at the anode. aluminium oxide The overall reaction is aluminium + oxygen. 2Al 2 O 3(l) 4Al (l) + 3O 2(g) This process is very expensive as a lot of electricity is needed. At the cathode Al +++ (l) + 3e Al (l) At the anode 2O -- (l) - 4e O 2 (g) Work out Chemistry for You pg. 138 number 18, pg. 140 number 23, 24 In the electrolysis of bauxite Al 2 O 3, aluminium is the main product. Calculate the mass of aluminium (R.A.M. = 27) produced if a current of 25A is passed for 30 minutes. Calculating the volume of gas liberated during electrolysis. (Chem 4U pg. 354,355) Moles of Gas = volume of gas (cm 3 ) At r.t. and pressure 1 mole of any gas takes a volume of 24l What volume does 8g of oxygen gas occupy at room temperature and pressure?

20 Chemistry Form 4 Page 40 Ms. R. Buttigieg Answer the following: A pure specimen of molten lead (II) bromide was electrolysed in a suitable apparatus using inert electrodes. a. Give the formulae of the ions present in the liquid. (i) (ii) b. What changes would you expect to SEE at each of the electrodes? Negative electrode (cathode) Positive electrode (anode) c. Write ionic equations to show the changes taking place at each of the electrodes: Negative electrode (cathode) Positive electrode (anode) d. In such an experiment, a current of 0.2 A was passed thourhg the molten lead (II) bromide for four minutes. i. What quantity of electricity is passed? ii. What would be the mass of the product at the negative electrode? iii. Assuming all the product liberated at the positive electrode to be in the form of a gas, what volume would it occupy at room temperature and pressure? (Don t forget to write the half equations for (ii) and (iii))

21 Chemistry Form 4 Page 41 Ms. R. Buttigieg 8.3 Reactivity Series This was already discussed in previous sections (make sure its clear) Producing electrical energy from a chemical reaction - limited to the simple cell. The Fuel Cell Chem 4U pg. 125 A scientist noted when electrolyzing sulphuric acid that when he switched off the current, the cell itself began producing electricity. So he concluded that if you have a cell, and you feed it oxygen and hydrogen, these are changed to water and produce electricity. What happens in the fuel-cell is the reverse of what happens when you electrolyze water (acidified with sulphuric acid). In electrolysis of water: electrolysis breaks up water into hydrogen and oxygen. In the fuel cell, hydrogen and oxygen are turned to water and produce electricity. At anode: 2H 2 (g) 4H + (aq) + 4e At cathode: 2H 2 O (l) + O 2 (g) + 4e 4OH - (aq) A simple cell - One of the first practical batteries is called the 'Daniel cell' - GCSE pg, 192 imp o It uses a half-cell of copper dipped in copper(ii) sulphate, o and in electrical contact with a 2nd half-cell of zinc dipped in zinc sulphate solution. o The zinc is the more reactive, and is the negative electrode, releasing electrons because on it zinc atoms lose electrons to form zinc ions, Zn (s) Zn 2+ (aq) + 2e - o The less reactive metal copper, is the positive electrode, and gains electrons from the negative electrode through the external wire connection and here.. the copper(ii) ions are reduced to copper atoms, Cu 2+ (aq) + 2e - Cu (s) o Overall the reactions is: Zn (s) + CuSO 4(aq) + ZnSO 4(aq) + Cu (s) or ionically: Zn (s) + Cu 2+ (aq) + Zn 2+ (aq) + Cu (s) o The overall reaction is therefore the same as displacement reaction, and it is a redox reaction involving electron transfer and the movement of the electrons through the external wire to the bulb or voltmeter etc. forms the working electric current. Work out GCSE Chemistry pg. 193 no. 1,2. With the help of diagrams, explain the basic steps in the purification of copper. (see GCSE chemistry pg. 88, 89) Work out GCSE Chemistry pg. 92 no. 1-4

22 Chemistry Form 4 Page 42 Ms. R. Buttigieg Limestone, Quicklime and slaked lime Limestone, chalk and marble are calcium carbonate (CaCO 3 ) Quicklime is calcium oxide. Slaked lime is calcium hydroxide Limewater is calcium hydroxide solution. Calcium hydroxide is slightly soluble in water. a) Limestone itself is used as a building material, to make iron and steel, to neutralize acid soil and in fertilizers. b) Limestone is used to make: i) Calcium oxide by heating calcium carbonate strongly in a lime kiln (pg. 129 C4U) CaCO 3 (s) CaO (s) + CO 2 (g) ii) Quicklime than absorbs water to form calcium hydroxide. CaO (s) + H 2 O (l) Ca(OH) 2 (s) Slaked lime is the cheapest industrial alkali. It is used to make sodium hydroxide, bleaching powder and mortar. It is also used to control soil acidity. Mortar is a mixture of calcium hydroxide, sand and water. Mortar sets by drying out, then hardens by reacting with carbon dioxide. Limewater is a dilute alkaline solution of calcium hydroxide. Carbon dioxide turns limewater milky: carbon dioxide, an acidic oxide, reacts with calcium hydroxide, an alkali to form a salt, calcium carbonate, which is a white solid, insoluble in water. Ca(OH) 2 (s) + CO 2 (g) CaCO 3 (s) + H 2 O (l) If we continue to pass carbon dioxide through the milky limewater, the milkiness disappears. This happens as calcium carbonate changes into soluble calcium hydrogencarbonate. CaCO 3 (s) + H 2 O (l) + CO 2 (g) Ca(HCO 3 ) 2 (aq) In the laboratory, limestone is changed to quicklime by heating strongly for a long time on a strong Bunsen flame. The appearance doesn t change. CaCO 3 (s) CaO (s) + CO 2 (g) When water is added drop by drop onto the quicklime this puffs up and then crumbles to calcium hydroxide powder. A hissing sound is heard, because some of the water is changed to steam. Because the reaction between quicklime and water is exothermic and changes some of the water into steam. CaO (s) + H 2 O (g) Ca(OH) 2 (s) See Chemistry for You pg Work out pg. 134 numbers 1, 3, 4.

23 Chemistry Form 4 Page 43 Ms. R. Buttigieg Work out the following for the Holidays: GCSE Chemistry pg 94 number 1, 2 (see GCSE chemistry pg. 88, 89), 5, 8a-d. Chemistry for You pg. 117 numbers 3, 4, 8. pg. 141 numbers 26, Underline the correct answer: 1. Aluminium is extracted using (electrolysis, carbon monoxide in the blast furnace) because it is a fairly (reactive/unreactive) metal. 2. Iron is extracted using (electrolysis, carbon monoxide in the blast furnace) because it is a fairly (reactive/unreactive) metal. 3. Aluminium is (more/less) abundant in the earth s crust than iron. 4. Aluminium is (more/less) expensive than iron due to the extraction process. 2. Complete the following table: Electrolyte Cathode type Anode type Product at cathode Molten lead (II) Graphite Graphite bromide Sodium chloride Graphite Graphite solution Sulfuric acid Platinum Platinum solution Copper (II) Platinum Platinum sulphate solution Copper (II) Copper Copper sulphate solution Product at anode 3. Work out the oxidation number of the underlined element, then state whether it has been oxidized or reduced. PbO 2 + 4HCl PbCl 2 + 2H 2 O + Cl 2 Oxidation number The element has been. ZnO + C Zn + CO Oxidation number The element has been. H 2 S + Br 2 2HBr + S Oxidation number The element has been. Give 2 other ways in which one can determine if an element has been oxidized or reduced.