Redox Reactions, Chemical Cells and Electrolysis

Transcription

1 Topic 5 Redox Reactions, Chemical Cells and Electrolysis Part A Unit-based exercise Unit 18 Chemical cells in daily life Fill in the blanks 1 chemical; electrical 2 electrolyte 3 voltmeter; multimeter 4 a) negative; positive b) zinc; copper 5 a) anode b) cathode c) electrolyte 6 a) zinc b) carbon c) manganese(iv) oxide d) ammonium chloride e) a) lithium; graphite b) lithium cobalt oxide; lithium manganese oxide c) lithium d) a) hydrogen absorbing b) nickel(ii) hydroxide c) potassium hydroxide d) 1.2 True or false 9 F In a chemical cell, electrons flow from the negative electrode to the positive electrode in the external circuit. 10 T 125

2 11 F In a zinc-carbon cell, the carbon rod is the positive electrode (cathode). The cylinder casing is made of zinc, which serves as the negative electrode (anode). 12 F The electrolyte in an alkaline manganese cell is potassium hydroxide. 13 F Compared with zinc-carbon cells, alkaline manganese cells give a more steady voltage during discharge. 14 T 15 F Silver oxide cell is a primary cell. It is NOT rechargeable. 16 T 17 F The cathode of a nickel metal hydride cell is made of nickel(ii) hydroxide. 18 T 19 T 20 T Multiple choice questions 21 C Option A Electrons flow from the zinc strip to the copper strip in the external circuit. Option B Copper(II) ions in the solution near to the copper strip gain electrons and form copper atoms. As a result, a deposit of copper forms on the copper strip. The mass of the copper strip increases. Cu 2+ (aq) + 2e Cu(s) Option C Zinc atoms lose electrons and form zinc ions. These ions then go into the copper(ii) sulphate solution. As a result, the mass of the zinc strip decreases. Zn(s) Zn 2+ (aq) + 2e Option D Ions flow through the copper(ii) sulphate solution. 22 C Option A The copper strip is the positive electrode. Option B Electrons flow from the magnesium strip to the copper strip in the external circuit. Option D Chemical energy is converted into electrical energy. 23 C Options A & B Pure water and ethanol are non-conductors. 24 A Option C A chemical cell consists of two different metals and an electrolyte. The chemical cell makes the light bulb shine. 126

3 25 A Option C The maximum voltage of a zinc-carbon cell is 1.5 V. 26 D Option D The service life of a zinc-carbon cell is relatively short. 27 D Option D Lithium ion cells are commonly used in mobile phones. 28 A 29 B 30 B Option A Silver oxide cells are NOT rechargeable. 31 B Option A A lithium ion secondary cell has a high energy density. Option C The maximum voltage of a lithium ion secondary cell is 3.7 V. Option D The electrolyte of a lithium ion secondary cell is composed of a lithium salt dissolved in an organic solvent. The lithium salt is NOT dissolved in water because lithium reacts violently with water. 32 D Option A Potassium hydroxide acts as the electrolyte of a nickel metal hydride cell. 33 D Option B A lead-acid accumulator is mainly used for automotive starting, lighting and ignition applications. Option C The maximum voltage of a nickel metal hydride cells is 1.2 V. 34 B Option B The voltage of a zinc-carbon cell drops rapidly during discharge. 35 D 36 A (1) (3) Ions flow through the electrolyte. 127

4 37 D (1) The magnesium strip is the negative electrode. Electrons flow from the magnesium strip to the copper strip in the external circuit. (3) The lemon juice acts as the electrolyte. Potato juice can be used also. 38 B When the multimeter gives a positive voltage, the metal connected to the positive terminal of the multimeter is the positive electrode while the metal connected to the negative terminal of the multimeter is the negative electrode. (2) Electrons flow from electrode Y to electrode X in the external circuit. 39 D 40 B (2) Potassium hydroxide acts as the electrolyte. 41 B (1) A silver oxide cell is NOT rechargeable. (3) The voltage of a silver oxide cell remains steady during discharge. 42 A (2) In a lithium ion secondary cell, lithium atoms lying between graphite sheets act as the negative electrode. (3) A lithium ion secondary cell is not very robust and CANNOT take high charging currents. Using a high charging current may overheat the pack and lead to explosion. 43 A (3) Potassium hydroxide acts as the electroyte of a nickel metal hydride cell. 128

5 44 A (3) A 12 V accumulator consists of six cells joined in series. 45 A (3) A nickel metal hydride cell uses hydrogen absorbing alloys to make the negative electrode. 46 C (1) Lithium ion cell uses lithium salt dissolved in an organic solvent as the electrolyte. 47 C (1) The maximum voltage of alkaline manganese cells and zinc-carbon cells are both 1.5 V. 48 B (1) Both nickel metal hydride cells and nickel-cadmium cells are rechargeable. (3) Both nickel metal hydride cells and nickel-cadmium cells can maintain a constant voltage during discharge. 49 B (1) Nickel metal hydride cells can deliver high discharge currents while lithium ion cells cannot. (2) Lithium ion cells have a higher energy density than nickel metal hydride cells. (3) A lithium ion cell is not very robust and CANNOT take high charging currents. Using a high charging current may overheat the pack and lead to explosion. 50 C (1) Using new and used zinc-carbon cells at the same time in an electrical appliance is NOT dangerous, but this gives poorer results. (2) Charging a lithium ion secondary cell with a high current may overheat the pack and lead to explosion. 51 C In a magnesium-copper chemical cell, electrons flow from the magnesium electrode to the copper electrode in the external circuit. 52 D Zinc-carbon cells are NOT rechargeable. 53 A Ammonium chloride acts as the electrolyte in zinc-carbon cells. 54 C The electrolyte of a zinc-carbon cell is a moist paste of ammonium chloride. 55 D A lithium ion cell is not very robust and cannot deliver high discharge currents. Loading the cell with excess discharge current may overheat the pack and lead to explosion. 129

6 Unit 19 Simple chemical cells Fill in the blanks 1 a) electrons; magnesium ions b) electrons; copper c) i) Mg(s) Mg 2+ (aq) + 2e ii) Cu 2+ (aq) + 2e Cu(s) d) magnesium; copper e) decreases; increases 2 electrochemical 3 a) higher b) lower 4 a) zinc b) copper c) i) Zn(s) Zn 2+ (aq) + 2e ii) Cu 2+ (aq) + 2e Cu(s) d) zinc strip; copper container e) ions True or false 5 T 6 F In the electrochemical series, the position of calcium is higher than that of sodium. The order is different from that in the reactivity series. This is because, when compared with sodium, a calcium atom loses electrons more readily in cell reactions whereas in reactions with air, water and dilute acids it loses electrons less readily. 7 T A copper-silver chemical cell is shown below: In the copper-silver chemical cell, copper atoms lose electrons and form copper(ii) ions. These ions then go into the copper(ii) sulphate solution. 130

7 Cu(s) Cu 2+ (aq) + 2e Electrons given up by copper atoms flow along the conducting wires to the silver strip. Copper(II) ions in the copper(ii) sulphate solution near to the silver strip gain these electrons and form copper atoms. As a result, a deposit of copper forms on the silver strip. Cu 2+ (aq) + 2e Cu(s) Copper is transferred from the copper strip to the silver strip. The concentration of copper(ii) ions in the electrolyte remains the same. The blue colour of the solution remains unchanged. 8 F A salt bridge can be prepared by soaking a piece of filter paper in an ionic salt solution. 9 T Sugar solution does NOT conduct electricity. Hence sugar solution CANNOT be used to prepare salt bridges. 10 F In a Daniell cell, electrons flow from the zinc strip to the copper container in the external circuit, i.e. a current flows from the copper container to the zinc strip. Multiple choice questions 11 A Zinc forms ions more readily than copper does. Hence zinc atoms would lose electrons and form zinc ions. Zn(s) Zn 2+ (aq) + 2e The zinc strip is the negative electrode. Copper(II) ions in the electrolyte near to the copper strip gain electrons and form copper atoms. Cu 2+ (aq) + 2e Cu(s) The copper strip is the positive electrode. 12 B Options A & C Pure water and ethanol are non-conductors. Option B Zinc and copper are farther apart in the electrochemical series than lead and copper. Hence the zinc-copper couple gives the highest voltage, making the light bulb the brightest. 13 C Option C The position of magnesium in the electrochemical series is the highest. Hence magnesium loses electrons to form ions most readily. 14 D Option D Electrons move from the zinc strip to the copper strip in the external circuit. 131

8 15 A In the above chemical cell, electrons flow from the strip of metal X to the copper strip in the external circuit. It can be deduced that metal X forms ions more readily than copper does. Option A The copper strip is the positive electrode and the strip made of metal X is the negative electrode. Option B At the copper strip, hydrogen ions in the acid gain electrons and form hydrogen gas. 2H + (aq) + 2e H 2 (g) Option C Atoms of metal X lose electrons and form ions. As a result, the mass of the strip of metal X decreases. Option D Metal X is at a higher position in the electrochemical series than copper. 16 D Magnesium forms ions more readily than zinc. Magnesium and copper are farther apart in the electrochemical series than zinc and copper. Therefore the voltage of the cell would increase if the zinc strip is replaced by a magnesium strip. 17 C Option A In the above chemical cell, electrons flow from the metal X rod to the iron rod in the external circuit. It can be deduced that metal X forms ions more readily than iron does. Option B Atoms of metal X lose electrons and form ions. As a result, the metal X rod dissolves gradually. Option C The iron rod is the positive electrode, i.e. the cathode. 18 D For the first two chemical cells, metal W is the positive electrode while metals X and Y are the negative electrodes. Therefore metals X and Y form ions more readily than metal W. 132 The Y / W couple gives a higher voltage than the X / W couple. Therefore the difference in the tendency to form ions between metal Y and metal W is greater than that between metal X and metal W. Hence metal Y forms ions most readily.

9 For the third chemical cell, the voltmeter gives a negative voltage. Therefore metal W is the negative electrode while metal Z is the positive electrode. Metal W forms ions more readily than metal Z. This gives the descending order of reactivity of the four metals: Y, X, W, Z. 19 D For a simple chemical cell, the farther apart the two metals are in the electrochemical series (or the reactivity series), the higher is the voltage of the cell. 20 C From the cell of the Mn / Fe couple, it can be deduced that Mn forms ions more readily than Fe. From the cells of the Fe / Ag couple and Fe / Cu couple, it can be deduced that Fe forms ions more readily than Ag and Cu. The Fe / Ag couple gives a higher voltage than the Fe / Cu couple. Therefore the difference in the tendency to form ions between Fe and Ag is greater than that between Fe and Cu. Hence Ag forms ions least readily. This gives the descending order of the tendency to form ions for the four metals: Mn, Fe, Cu, Ag. 21 D Option Metal X Direction of electron flow in the external circuit A copper from zinc to copper B iron from zinc to iron C magnesium from magnesium to zinc D silver from zinc to silver 22 A X displaces Z from Z(NO 3 ) 2 solution. Hence X is more reactive than Z and forms ions more readily. The following chemical cell uses X and Z as electrodes. Option A Atoms of metal X lose electrons and form ions. Electrons flow from X to Z in the external circuit. Options C and D Atoms of metal X lose electrons and form ions. As a result, the strip made of X dissolves gradually and its mass decreases. 133

10 23 A Nickel is more reactive than silver. Hence nickel forms ions more readily than silver does. In the above chemical cell, nickel atoms lose electrons and form nickel(ii) ions. Ni(s) Ni 2+ (aq) + 2e Electrons given up by nickel atoms flow along the conducting wires to the silver rod. Nickel(II) ions in the solution near to the silver rod gain these electrons and form nickel atoms. As a result, a deposit of nickel forms on the silver rod. Ni 2+ (aq) + 2e Ni(s) Hence the mass of the silver rod (i.e. the cathode) increases. Nickel was transferred from the nickel rod to the silver rod. The concentration of nickel(ii) ions in the electrolyte remains the same. The colour intensity of the solution would not change. 24 B The order of reactivity of the metals is: Q > T > R > S Q and S are the furthest apart in the electrochemical series. Hence the Q / S couple would give the highest voltage. 25 C Option C The zinc plate reacts with the dilute sulphuric acid. Hence colourless gas bubbles appear at the surface of the zinc plate. 26 B When the circuit is closed, electrons flow from the zinc plate to the copper plate in the external circuit. Zinc atoms lose electrons and form ions. As a result, the mass of the zinc plate (i.e. the anode) decreases. Zinc ions are colourless. Hence these is NO change in the colour of the dilute sulphuric acid. 134

11 27 A Nickel is more reactive than silver. Hence nickel forms ions more readily than silver does. In the above chemical cell, electrons flow from the nickel rod to the silver rod in the external circuit. Nickel is the negative electrode (i.e. the anode) of the chemical cell. Nickel atoms lose electrons and form nickel(ii) ions. Ni(s) Ni 2+ (aq) + 2e 28 C Option C Magnesium forms ions more readily than copper does. Magnesium is the negative electrode (i.e. anode) of the chemical cell. Magnesium atoms lose electrons and form magnesium ions. Mg(s) Mg 2+ (aq) + 2e 29 D Option D Magnesium atoms lose electrons and form magnesium ions. As a result, the mass of the magnesium electrode decreases. Electrons given up by magnesium atoms flow along the conducting wires to the copper electrode. Copper(II) ions in the copper(ii) sulphate solution near to the copper electrode gain these electrons and form copper atoms. As a result, a deposit of copper forms on the copper electrode. Cu 2+ (aq) + 2e Cu(s) Hence the mass of the copper electrode increases. 30 B 135

12 Electrons flow from the magnesium electrode towards the copper electrode in the external circuit. The nitrate ions in the salt bridge move towards the magnesium electrode. 31 C Option D Ammonium and nitrate ions are NOT involved in the reactions of the half-cells. 32 A Iron forms ions more readily than copper does. Iron is the negative electrode of the chemical cell. Iron atoms lose electrons and form iron ions. Fe(s) Fe 2+ (aq) + 2e 33 B Nickel is more reactive than silver. Hence nickel forms ions more readily than silver does. In the chemical cell, nickel atoms lose electrons and form nickel(ii) ions. Electrons given up by nickel atoms flow along the conducting wires to the silver electrode. Silver ions in the silver nitrate solution near to the silver electrode gain these electrons and form silver atoms. Ag + (aq) + e Ag(s) The silver electrode is the positive electrode, i.e. the cathode. 34 D Option A 136 Electrons flow from the nickel electrode to the silver electrode in the external circuit.

13 Options B and D Nickel atoms lose electrons and form nickel(ii) ions. Ni(s) Ni 2+ (aq) + 2e NO bubbles of gas are given off from the nickel electrode. The concentration of nickel(ii) ions in the nickel(ii) nitrate solution increases. Hence the green colour of the solution becomes more intense. Option C Silver ions in the silver nitrate solution near to the silver electrode gain electrons and form silver atoms. Hence the concentration of silver ions in the solution decreases. 35 B Option Combination X Y Z Direction of electron flow in the external circuit A magnesium copper magnesium sulphate solution from X to Y B carbon magnesium magnesium sulphate solution from Y to X, as shown in the diagram C copper silver silver nitrate solution from X to Y D magnesium carbon silver nitrate solution from X to Y 36 C Electrons flow from X to Y in the external circuit. It can be deduced that X forms ions more readily than Y does. Options A and C In the above chemical cell, atoms of X lose electrons and form ions. These ions then go into the X(NO 3 ) 2 solution. Hence the strip made of X gradually dissolves. The strip made of X is the negative electrode, i.e. the anode. Option B In the electrochemical series, the position of X is higher than that of Y. Options A and D Electrons given up by atoms of X flow along the conducting wires to the strip made of Y. Y 2+ ions in the Y(NO 3 ) 2 solution near to the strip made of Y gain these electrons and form atoms. As a result, a deposit of Y forms on the strip. Y 2+ (aq) + 2e Y(s) 137

14 37 C The above chemical cell is made up of copper electrodes and copper(ii) sulphate solution of different concentrations. Electrons flow in such a direction that the concentration of Cu 2+ (aq) ions in each half-cell becomes the same evantually, i.e. electrons flow from X to Y in the external circuit (i.e. Option C is correct). At electrode X Cu(s) Cu 2+ (aq) + 2e At electrode Y Cu 2+ (aq) + 2e Cu(s) Option A Electrode Y is the positive electrode. Option B The mass of electrode X decreases. Option D Electrons flow in the external circuit, NOT via the salt bridge. 38 C Option A In the above chemical cell, electrons flow from electrode A to electrode B in the external circuit. At electrode A Na(l) Na + (aq) + e At electrode B S(l) + 2e S 2 (l) Option D A current flows from electrode B to electrode A in the external circuit. 138

15 39 D The zinc strip gradually dissolves. It can be deduced that zinc atoms lose electrons and form zinc ions. Zn(s) Zn 2+ (aq) + 2e Option D Electrons flow from the zinc strip to the cobalt strip in the external circuit. Hence the cobalt strip is the positive electrode (i.e. the cathode) of the chemical cell. Electrons given up by zinc atoms flow along the conducting wires to the cobalt electrode. Cobalt(II) ions in the solution near to the cobalt electrode gain these electrons and form cobalt atoms. Co 2+ (aq) + 2e Co(s) 40 A (2) Electrons flow in the external circuit, NOT in the salt bridge. (3) The salt bridge provides ions that can move into the half-cells to prevent the build-up of excess positively or negatively charged ions in the solutions. 41 C Option C Cations (i.e. positive ions) move from the nickel half-cell to the palladium half-cell. It can be deduced that nickel atoms lose electrons and form nickel(ii) ions. Ni(s) Ni 2+ (aq) + 2e Hence the nickel electrode is the negative electrode, i.e. the anode. 139

16 42 A (1) The palladium electrode is the positive electrode, i.e. the cathode. (3) Electrons flow in the external circuit, NOT through the porous barrier. 43 D Option A Electrons flow from the zinc strip to the copper container in the external circuit, i.e. a current flows from the copper container to the zinc strip in the external circuit. Option B The zinc strip is the negative electrode, i.e. the anode. Option D The copper container is the positive electrode. The copper(ii) ions in the copper(ii) sulphate solution gain electrons from the external circuit to form copper metal. Cu 2+ (aq) + 2e Cu(s) The concentration of copper(ii) ions in the solution decreases. Hence the blue colour of the copper(ii) sulphate solution becomes less intense. 44 B (2) The porous pot is different from a salt bridge. It does NOT provide ions to balance excess charges in the solutions of the cell. 45 A (1) and (3) In the copper-silver chemical cell, copper atoms lose electrons and form copper(ii) ions. These ions then go into the silver nitrate solution. Cu(s) Cu 2+ (aq) + 2e Hence the mass of the copper electrode decreases. The colour of the silver nitrate solution would change as the blue copper(ii) ions go into the solution. (2) Electrons given up by copper atoms flow along the conducting wires to the silver electrode. Silver ions in the solution near to the silver electrode gain these electrons and form silver atoms. As a result, a deposit of silver forms on the silver electrode. Hence the mass of the silver electrode increases. 46 B Electrons flow from the electrode made of metal X to the copper electrode in the external circuit. It can be deduced that metal X forms ions more readily than copper does. (1) The electrode made of metal X is the negative electrode, i.e. the anode. (3) The dilute sulphuric acid would NOT turn blue as NO copper(ii) ions are formed. 47 D Electrons flow from Y to X in the external circuit. It can be deduced that Y forms ions more readily than X does. 140

17 (1) In the above chemical cell, atoms of Y lose electrons and form ions. These ions then go into Y(NO 3 ) 2 (aq). Hence the strip made of Y gradually dissolves. Electrons given up by atoms of Y flow along the conducting wires to the strip made of X. X 2+ (aq) ions in X(NO 3 ) 2 (aq) near to the strip made of X gain these electrons and form atoms. As a result, a deposit of X forms on the strip. Hence the strip made of X does NOT dissolve. (2) Negative ions, i.e. anions, in the salt bridge migrate towards the strip made of Y. (3) Y is more reactive than X. Hence Y can displace X from X(NO 3 ) 2 solution. 48 D When nickel is placed in a zinc sulphate solution, no reaction occurs. It can be deduced that zinc is more reactive than nickel, i.e. zinc forms ions more readily than nickel does. (1) and (3) Electrons flow from the zinc electrode to the nickel electrode in the external circuit. The zinc electrode is the negative electrode, i.e. the anode. (2) Positive ions, i.e. cations, in the salt bridge migrate towards the nickel half-cell. 141

18 49 B (1) X can displace Y from Y(NO 3 ) 2 solution. It can be deduced that X is more reactive than Y. (2) The position of X in the electrochemical series is higher than that of Y. (3) The following chemical cell uses X and Y as electrodes. Atoms of X lose electrons and form ions. Electrons flow from X to Y in the external circuit. Hence X is the negative electrode. 50 A (1) Copper forms ions more readily than silver does. Hence electrons flow from the copper electrode to the silver electrode in the external circuit. The copper electrode is the negative electrode, i.e. the anode. (2) A porous pot is different from a salt bridge. It does NOT provide ions to balance excess charges in the solutions of the cell. (3) The porous pot CANNOT be replaced by a glass cylinder. A glass cylinder does NOT allow ions to move between the two solutions. 51 D Iron forms ions more readily than copper does. In a chemical cell with a copper-iron couple, the copper is the positive electrode. 52 C The position of calcium is higher than that of sodium in the electrochemical series. Calcium atom loses electrons more readily in cell reactions than in reactions with air, water and dilute acids. 53 A The position of nickel in the electrochemical series is higher than that of silver. 142

19 54 C A copper-silver chemical cell is shown below: In the copper-silver chemical cell, copper atoms lose electrons and form copper(ii) ions. These ions then go into the copper(ii) sulphate solution. Cu(s) Cu 2+ (aq) + 2e Electrons given up by copper atoms flow along the conducting wires to the silver strip. Copper(II) ions in the copper(ii) sulphate solution near to the silver strip gain these electrons and form copper atoms. As a result, a deposit of copper forms on the silver strip. Cu 2+ (aq) + 2e Cu(s) Copper is transferred from the copper strip to the silver strip. The concentration of copper(ii) ions in the electrolyte remains the same. The blue colour of the solution remains unchanged. 55 D In a copper-zinc chemical cell using dilute sulphuric acid as the electrolyte, the zinc atoms lose electrons to form zinc ions. Zinc ions are colourless. Hence there is NO change in the colour of the dilute sulphuric acid. The copper electrode is the positive electrode. The copper atoms would NOT lose electrons to form copper(ii) ions. 56 A 57 D Electrons flow in the external circuit of the chemical cell, NOT through the salt bridge. 58 D In a Daniell cell, a glass container CANNOT be used to contain the zinc sulphate solution. The glass container does NOT allow ions to move between the two solutions. 143

20 Unit 20 Oxidation and reduction Fill in the blanks 1 reduction 2 oxidation 3 reducing 4 oxidizing 5 reducing 6 a) bromide b) Br 2 (aq) + 2e 2Br (aq) c) yellow-brown 7 a) iron(iii) b) Fe 2+ (aq) Fe 3+ (aq) + e c) yellow-brown 8 a) manganese(ii) b) MnO 4 (aq) + 8H + (aq) + 5e Mn 2+ (aq) + 4H 2 O(l) c) purple 9 a) chromium(iii) b) Cr 2 O 2 7 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) c) orange; green 10 a) yellow-brown; bromine b) Cl 2 (aq) + 2Br (aq) 2Cl (aq) + Br 2 (aq) 11 a) brown; iodine b) Cl 2 (aq) + 2I (aq) 2Cl (aq) + I 2 (aq) 12 a) chloride; hypochlorite b) Cl 2 (g) + 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H 2 O(l) 13 a) oxidizing agent b) acid 14 a) i) bromide ii) NaBr(s) + H 2 SO 4 (l) b) i) bromine ii) 2HBr(g) + H 2 SO 4 (l) NaHSO 4 (s) + HBr(g) Br 2 (g) + SO 2 (g) + 2H 2 O(l) 15 a) orange; green b) Cr 2 O 2 7 (aq) + 3SO 2 (aq) + 2H + (aq) 2Cr 3+ (aq) + 3SO 2 4 (aq) + H 2 O(l) 144

21 True or false 16 F Not all chemical reactions are redox reactions. For example, neutralization reactions are not redox reactions. 17 T 18 T The position of potassium in the electrochemical series is higher than that of sodium. 19 F Dilute nitric acid is an oxidizing agent. It will compete with the potassium permanganate solution for the reducing agent in reactions. Hence it is NOT used to acidify potassium permanganate solution. Usually dilute sulphuric acid is used to acidify potassium permanganate solution. 20 T The ionic half-equation for the reduction of acidified potassium dichromate solution is Cr 2 O 7 2 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) The oxidation of chromium in the Cr 2 O 7 2 ion is +6 while that in the Cr 3+ ion is +3. Hence the oxidation number of chromium changes from +6 to T Acidified potassium permanganate solution is a strong oxidizing agent. When it reacts with iron(ii) sulphate solution, the permanganate ions will be reduced to manganese(ii) ions. The permanganate solution turns from purple to colourless while the iron(ii) sulphate solution changes from pale green to yellow-brown. On the other hand, there is NO observable change when acidified potassium permanganate solution is mixed with iron(iii) sulphate solution. Hence acidified potassium permanganate solution can be used to distinguish between iron(ii) sulphate solution and iron(iii) sulphate solution. 22 F In the reaction between aqueous chlorine and potassium iodide solution, chlorine is reduced to chloride ions while iodide ions are oxidized to iodine. The resulting reaction mixture is brown in colour due to the presence of iodine. Cl 2 (aq) + 2e 2Cl (aq) 2l (aq) I 2 (aq) + 2e 23 F Both aqueous chlorine and acidified potassium permanganate solution are oxidizing agents. There is NO reaction between them. 24 T 25 T When aqueous bromine is mixed with sodium sulphite solution, the yellow-brown aqueous bromine becomes colourless. This is because the yellow-brown bromine is reduced to colourless bromide ions. On the other hand, there is NO observable change when aqueous bromine is mixed with sodium sulphate solution. Hence aqueous bromine can be used to distinguish between sodium sulphate solution and sodium sulphite solution. 145

22 26 F Magnesium reacts with dilute nitric acid to give nitrogen monoxide. 3Mg(s) + 2NO 3 (aq) + 8H + (aq) 3Mg 2+ (aq) + 2NO(g) + 4H 2 O(l) 27 T The equation for the reaction between zinc and concentrated nitric acid is: Zn(s) + 2NO 3 (aq) + 4H + (aq) Zn 2+ (aq) + 2NO 2 (g) + 2H 2 O(l) The oxidation number of nitrogen in the NO 3 ion is +5 while that in NO 2 is +4. Hence the oxidation number of nitrogen changes from +5 to F Nitrogen monoxide gas is colourless. When this gas is exposed to air, it reacts with oxygen in the air to give brown nitrogen dioxide gas. 2NO(g) + O 2 (g) colourless 2NO 2 (g) brown 29 F Concentrated sulphuric acid liberates steamy fumes of hydrogen chloride when it reacts with sodium chloride. NaCl(s) + H 2 SO 4 (l) NaHSO 4 (s) + HCl(g) Hydrogen chloride is a weak reducing agent. It is NOT oxidized by concentrated sulphuric acid. Hence NO chlorine (halogen) is produced in the reaction. 30 T Aqueous sulphur dioxide is a good reducing agent. NO reaction occurs when sulphur dioxide gas is bubbled into potassium iodide solution. Multiple choice questions 31 D Suppose the oxidation number of Cl in HClO 4 is x. (+1) + x + ( 2) x 4 = 0 x = C Sum of oxidation number of all atoms in the OH ion = charge on the ion = 1 Suppose the oxidation number of Al in [Al(OH) 4 ] is x. x + ( 1) x 4 = 1 x = C Suppose the oxidation number of Co in Co(NH 3 ) 6 Cl 3 is x. x ( 1) x 3 = 0 x = D Option Compound Oxidation number of S in the compound A ZnS 2 B Na 2 S 2 O 3 +2 C NaHSO 3 +4 D H 2 S 2 O sulphur exhibits the highest oxidation number in H 2 S 2 O 7.

23 35 C Option Substances Oxidation number of underlined element A B C D PbO 2 +4 MnO 4 +7 Fe 2 O 3 +3 CO 2 +4 NO 2 +4 CO OH 2 2 Cr 2 O 7 +6 in NO 2 and CO 3 2, the oxidation number of the underlined element is the same. 36 C Option Compound Oxidation number of chromium in the compound Cr 2 O 3 +3 A K 2 CrO 7 +6 Na 2 CrO 4 +6 CrCl 2 +2 B Cr 2 O 3 +3 K 2 Cr 2 O 7 +6 Cr 2 O 3 +3 C CrCl 3 +3 Cr(NO 3 ) 3 +3 Na 2 CrO 4 +6 D CrO 3 +6 Cr(NO 3 ) 3 +3 in Cr 2 O 3, CrCl 3 and Cr(NO 3 ) 3, the oxidation number of chromium is the same. 37 A Option Stage Conversion Change in oxidation number of sulphur A 1 S 0 +4 S O 2 4 B 2 +4 S O 2 +6 S O 3 2 C 3 +6 S O 3 H +6 2 S O 4 0 D 4 H +6 2 S O 4 (NH 4 ) +6 2 S O 4 0 Stage 1 involves the largest change in oxidation number of sulphur. 147

24 38 C Option Conversion Change in oxidation number A C +3 r 3+ C +2 r 2+ 1 unit B C +1 lo C +3 lo 2 2 units C C +6 r 2 O 7 2 D M +7 no 4 C +3 r 3+ M +2 n 2+ 3 units 5 units the conversion Cr 2 O 7 2 Cr 3+ involves an oxidation number change of 3 units. 39 D The unbalanced ionic half-equation is: Br 2 2BrO 3 Balance the ionic half-equation with respect to the number of atoms. Balance the ionic half-equation with respect to the number of charges. To balance the 6 oxygen atoms in 2BrO 3, add 6H 2 O on the left-hand side. Br 2 + 6H 2 O 2BrO 3 To balance the 12 hydrogen atoms in 6H 2 O, add 12H + on the right-hand side. Br 2 + 6H 2 O 2BrO H + Charge on left-hand side = 0 Charge on right-hand side = 2 x ( 1) + 12 x (+1) = +10 add 10e on the right-hand side to balance the charge. The balanced ionic half-equation is: Br 2 + 6H 2 O 2BrO H e x is 6, y is 12 and z is B Writing the redox equation using ionic half-equations 1 Write down the oxidizing agent and the reducing agent involved. 2 a) Write an ionic halfequation for the reduction process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. NO 3 is the oxidizing agent and Fe 2+ is the reducing agent. The unbalanced ionic half-equation for NO 3 is: NO 3 NO To balance the 3 oxygen atoms in NO 3, add 2H 2 O on the right-hand side. NO 3 NO + 2H 2 O To balance the 4 hydrogen atoms in 2H 2 O, add 4H + on the left-hand side. NO 3 + 4H + NO + 2H 2 O Charge on left-hand side = ( 1) + 4 x (+1) = +3 Charge on right-hand side = 0 add 3e on the left-hand side to balance the charge. The balanced ionic half-equation for acidified NO 3 is: NO 3 + 4H + + 3e NO + 2H 2 O...(i) 148

25 2 b) Write an ionic halfequation for the oxidation process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 3 Make the number of electrons gained in one ionic half-equation equal to that lost in the other. 4 Combine the two ionic halfequations and eliminate the electrons. The unbalanced ionic half-equation for Fe 2+ is: Fe 2+ Fe 3+ The number of atoms is balanced. Charge on left-hand side = +2 Charge on right-hand side = +3 add 1e on the right-hand side to balance the charge. The balanced ionic half-equation for Fe 2+ is: Fe 2+ Fe 3+ + e...(ii) (ii) x 3 (i) NO 3 + 4H + + 3e NO + 2H 2 O (ii) x 3 3Fe 2+ 3Fe e 3Fe H + + NO 3 3Fe H 2 O + NO x is 3, y is 4 and z is 2. Writing the redox equation using oxidation number method 1 Write down the oxidizing agent and the reducing agent involved. Determine their products. 2 Assign oxidation numbers to all atoms. 3 Notice atoms which undergo a change in oxidation number. Determine the number of electrons lost or gained per formula unit. 4 Insert an appropriate coefficient before the formula of each reagent on the left-hand side of the equation to make the number of electrons gained equal to that lost. 5 Add appropriate coefficients on the right to balance the number of atoms which have gained or lost electrons. NO 3 is the oxidizing agent and Fe 2+ is the reducing agent. NO 3 NO Fe 2+ Fe 3+ N +5 O 3 + F +2 e 2+ N +2 O + F +3 e 3+ reduction: gain of 3e per NO 3 N +5 O 3 + F +2 e 2+ N +2 O + +3 F e 3+ oxidation: loss of 1e per Fe 2+ NO 3 + 3Fe 2+ NO + Fe 3+ NO 3 + 3Fe 2+ NO + 3Fe

26 6 Balance the number of all other atoms except O and H. 7 Add H + to the side deficient in positive charges to make the number of charges on both sides equal. 8 Add H 2 O to the appropriate side to balance the number of O atoms. The number of all other atoms except O is balanced. Total charge on left-hand side = ( 1) + 3 x (+2) = +5 Total charge on right-hand side = 3 x (+3) = +9 Add 4H + to the left-hand side. NO 3 + 4H + + 3Fe 2+ NO + 3Fe 3+ 3Fe H + + NO 3 3Fe H 2 O + NO x is 3, y is 4 and z is B Number of H atoms on left-hand side = number of H atoms on right-hand side i.e. 2x = 2 x x = 5 Number of O atoms on left-hand side = number of O atoms on right-hand side i.e. 2 x 4 + x + 3y = 2 x y x 2 = y y = 3 42 C Writing the redox equation using ionic half-equations 1 Write down the oxidizing agent and the reducing agent involved. 2 a) Write an ionic halfequation for the reduction process. AuCl 4 is the oxidizing agent and Sn 2+ is the reducing agent. The unbalanced ionic half-equation for AuCl 4 is: 2AuCl 4 2Au i) Balance the ionic halfequation with respect to the number of atoms. To balance the 8 chlorine atoms in 2AuCl 4 8Cl on the right. 2AuCl 4 2Au + 8Cl on the left-hand side, add ii) Balance the ionic halfequation with respect to the number of charges. Charge on left-hand side = 2 x ( 1) = 2 Charge on right-hand side = 8 x ( 1) = 8 add 6e on the left-hand side to balance the charge. The balanced ionic half-equation for AuCl 4 is: 2AuCl 4 + 6e 2Au + 8Cl...(i) 150

27 2 b) Write an ionic halfequation for the oxidation process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 3 Make the number of electrons gained in one ionic half-equation equal to that lost in the other. 4 Combine the two ionic halfequations and eliminate the electrons. The unbalanced ionic half-equation for Sn 2+ is: Sn 2+ Sn 4+ The number of atoms is balanced. Charge on left-hand side = +2 Charge on right-hand side = +4 add 2e on the right-hand side to balance the charge. The balanced ionic half-equation for Sn 2+ is: Sn 2+ Sn e...(ii) (ii) x 3 (i) 2AuCl 4 + 6e 2Au + 8Cl (ii) x 3 3Sn 2+ 3Sn e 2AuCl 4 + 3Sn 2+ 3Sn Au + 8Cl the value of x is 3. Writing the redox equation using oxidation number method 1 Write down the oxidizing agent and the reducing agent involved. Determine their products. 2 Assign oxidation numbers to all atoms. 3 Notice atoms which undergo a change in oxidation number. Determine the number of electrons lost or gained per formula unit. 4 Insert an appropriate coefficient before the formula of each reagent on the left-hand side of the equation to make the number of electrons gained equal to that lost. 5 Add appropriate coefficients on the right to balance the number of atoms which have gained or lost electrons. AuCl 4 is the oxidizing agent and Sn 2+ is the reducing agent. AuCl 4 Au + Cl Sn 2+ Sn 4+ A +3 ucl 4 + S +2 n 2+ A 0 u + Cl + S +4 n 4+ reduction: gain of 3e per AuCl 4 A +3 ucl 4 + S +2 n 2+ A 0 u + Cl + S +4 n 4+ oxidation: loss of 2e per Sn 2+ 2AuCl 4 + 3Sn 2+ Au + Cl + Sn 4+ 2AuCl 4 + 3Sn 2+ 2Au + 8Cl + 3Sn 4+ the value of x is

28 43 D Writing the redox equation using ionic half-equations 1 Write down the oxidizing agent and the reducing agent involved. 2 a) Write an ionic halfequation for the reduction process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 2 b) Write an ionic halfequation for the oxidation process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 3 Make the number of electrons gained in one ionic half-equation equal to that lost in the other. 4 Combine the two ionic halfequations and eliminate the electrons. H 3 AsO 4 is the oxidizing agent and Zn is the reducing agent. The unbalanced ionic half-equation for H 3 AsO 4 is: H 3 AsO 4 AsH 3 To balance the 4 oxygen atoms in H 3 AsO 4, add 4H 2 O on the right-hand side. H 3 AsO 4 AsH 3 + 4H 2 O To balance the 8 hydrogen atoms in 4H 2 O, add 8H + on the left-hand side. H 3 AsO 4 + 8H + AsH 3 + 4H 2 O Charge on left-hand side = 8 x (+1) = +8 Charge on right-hand side = 0 add 8e on the left-hand side to balance the charge. The balanced ionic half-equation for H 3 AsO 4 is: H 3 AsO 4 + 8H + + 8e AsH 3 + 4H 2 O...(i) The unbalanced ionic half-equation for Zn is: Zn Zn 2+ The number of atoms is balanced. Charge on left-hand side = 0 Charge on right-hand side = +2 add 2e on the right-hand side to balance the charge. The balanced ionic half-equation for Zn is: Zn Zn e...(ii) (ii) x 4 (i) H 3 AsO 4 + 8H + + 8e AsH 3 + 4H 2 O (ii) x 4 4Zn 4Zn e H 3 AsO 4 + 4Zn + 8H + AsH 3 + 4Zn H 2 O x is 4, y is 8 and z is

29 Writing the redox equation using oxidation number method 1 Write down the oxidizing agent and the reducing agent involved. Determine their products. 2 Assign oxidation numbers to all atoms. 3 Notice atoms which undergo a change in oxidation number. Determine the number of electrons lost or gained per formula unit. 4 Insert an appropriate coefficient before the formula of each reagent on the left-hand side of the equation to make the number of electrons gained equal to that lost. 5 Add appropriate coefficients on the right to balance the number of atoms which have gained or lost electrons. 6 Balance the number of all other atoms except O and H. 7 Add H + to the side deficient in positive charges to make the number of charges on both sides equal. 8 Add H 2 O to the appropriate side to balance the number of O atoms. H 3 AsO 4 is the oxidizing agent and Zn is the reducing agent. H 3 AsO 4 AsH 3 Zn Zn 2+ H 3 A +5 so 4 + Z 0 n A 3 sh 3 + Z +2 n 2+ reduction: gain of 8e per H 3 AsO 4 H 3 A +5 so 4 + Z 0 n A 3 sh 3 + Z +2 n 2+ oxidation: loss of 2e per Zn H 3 AsO 4 + 4Zn AsH 3 + Zn 2+ H 3 AsO 4 + 4Zn AsH 3 + 4Zn 2+ The number of all other atoms except O is balanced. Total charge on left-hand side = 0 Total charge on right-hand side = 4 x (+2) = +8 Add 8H + to the left-hand side. H 3 AsO 4 + 4Zn + 8H + AsH 3 + 4Zn 2+ H 3 AsO 4 + 4Zn + 8H + AsH 3 + 4Zn H 2 O x is 4, y is 8 and z is

30 44 B Writing the redox equation using ionic half-equations 1 Write down the oxidizing agent and the reducing agent involved. 2 a) Write an ionic halfequation for the reduction process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 2 b) Write an ionic halfequation for the oxidation process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 3 Make the number of electrons gained in one ionic half-equation equal to that lost in the other. 4 Combine the two ionic halfequations and eliminate the electrons. MnO 4 is the oxidizing agent and H 2 O 2 is the reducing agent. The unbalanced ionic half-equation for MnO 4 is: MnO 4 Mn 2+ To balance the 4 oxygen atoms in MnO 4, add 4H 2 O on the right-hand side. MnO 4 Mn H 2 O To balance the 8 hydrogen atoms in 4H 2 O, add 8H + on the left-hand side. MnO 4 + 8H + Mn H 2 O Charge on left-hand side = ( 1) + 8 x (+1) = +7 Charge on right-hand side = +2 add 5e on the left-hand side to balance the charge. The balanced ionic half-equation for MnO 4 is: MnO 4 + 8H + + 5e Mn H 2 O...(i) The unbalanced ionic half-equation for H 2 O 2 is: H 2 O 2 O 2 To balance the hydrogen atoms, add 2H + on the right-hand side. H 2 O 2 O 2 + 2H + Charge on left-hand side = 0 Charge on right-hand side = 2 x (+1) = +2 add 2e on the right-hand side to balance the charge. The balanced ionic half-equation for H 2 O 2 is: H 2 O 2 O 2 + 2H + + 2e...(ii) (i) x 2 (ii) x 5 (ii) x 5 5H 2 O 2 5O H e (i) 2MnO H e 2Mn H 2 O 5H 2 O 2 + 2MnO H + 5O H + + 2Mn H 2 O Since H + appears on both sides of the equation, simplify the equation by collecting like terms. 5H 2 O 2 + 2MnO 4 + 6H + 5O 2 + 2Mn H 2 O x is 2, y is 6 and z is

31 Writing the redox equation using oxidation number method 1 Write down the oxidizing agent and the reducing agent involved. Determine their products. 2 Assign oxidation numbers to all atoms. 3 Notice atoms which undergo a change in oxidation number. Determine the number of electrons lost or gained per formula unit. 4 Insert an appropriate coefficient before the formula of each reagent on the left-hand side of the equation to make the number of electrons gained equal to that lost. 5 Add appropriate coefficients on the right to balance the number of atoms which have gained or lost electrons. 6 Balance the number of all other atoms except O and H. 7 Add H + to the side deficient in positive charges to make the number of charges on both sides equal. 8 Add H 2 O to the appropriate side to balance the number of O atoms. MnO 4 is the oxidizing agent and H 2 O 2 is the reducing agent. MnO 4 Mn 2+ H 2 O 2 O 2 M +7 no 4 + H 2 O 1 2 M +2 n 2+ + O 0 2 reduction: gain of 5e per MnO 4 M +7 no 4 + H 2 O 1 2 M +2 n 2+ + O 0 2 oxidation: loss of 2e per H 2 O 2 2MnO 4 + 5H 2 O 2 Mn 2+ + O 2 2MnO 4 + 5H 2 O 2 2Mn O 2 The number of all other atoms except O and H is balanced. Total charge on left-hand side = 2 x ( 1) = 2 Total charge on right-hand side = 2 x (+2) = +4 Add 6H + to the left-hand side. 5H 2 O 2 + 2MnO 4 + 6H + 5O 2 + 2Mn 2+ 5H 2 O 2 + 2MnO 4 + 6H + 5O 2 + 2Mn H 2 O x is 2, y is 6 and z is

32 45 D Writing the redox equation using ionic half-equations 1 Write down the oxidizing agent and the reducing agent involved. 2 a) Write an ionic halfequation for the reduction process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 2 b) Write an ionic halfequation for the oxidation process. i) Balance the ionic halfequation with respect to the number of atoms. ii) Balance the ionic halfequation with respect to the number of charges. 3 Make the number of electrons gained in one ionic half-equation equal to that lost in the other. 4 Combine the two ionic halfequations and eliminate the electrons. IO 3 is the oxidizing agent and SO 2 is the reducing agent. The unbalanced ionic half-equation for IO 3 is: 2IO 3 I 2 To balance the 6 oxygen atoms in 2IO 3, add 6H 2 O on the right-hand side. 2IO 3 I 2 + 6H 2 O To balance the 12 hydrogen atoms in 6H 2 O, add 12H + on the left-hand side. 2IO H + I 2 + 6H 2 O Charge on left-hand side = 2 x ( 1) + 12 x (+1) = +10 Charge on right-hand side = 0 add 10e on the left-hand side to balance the charge. The balanced ionic half-equation for IO 3 is: 2IO H e I 2 + 6H 2 O...(i) The unbalanced ionic half-equation for SO 2 is: SO 2 SO 4 2 To balance the oxygen atoms, add 2H 2 O on the left-hand side. 2 SO 2 + 2H 2 O SO 4 To balance the 4 hydrogen atoms in 2H 2 O, add 4H + on the right-hand side. SO 2 + 2H 2 O SO H + Charge on left-hand side = 0 Charge on right-hand side = ( 2) + 4 x (+1) = +2 add 2e on the right-hand side to balance the charge. The balanced ionic half-equation for SO 2 is: SO 2 + 2H 2 O SO H + + 2e...(ii) (ii) x 5 (i) 2IO H e I 2 + 6H 2 O (ii) x 5 5SO H 2 O 5SO H e 2IO H + + 5SO H 2 O I 2 + 6H 2 O + 5SO H + Since H + and H 2 O appear on both sides of the equation, simplify the equation by collecting like terms. 2IO 3 + 5SO 2 + 4H 2 O I 2 + 5SO H + x is 5, y is 4 and z is

33 Writing the redox equation using oxidation number method 1 Write down the oxidizing agent and the reducing agent involved. Determine their products. 2 Assign oxidation numbers to all atoms. 3 Notice atoms which undergo a change in oxidation number. Determine the number of electrons lost or gained per formula unit. 4 Insert an appropriate coefficient before the formula of each reagent on the left-hand side of the equation to make the number of electrons gained equal to that lost. 5 Add appropriate coefficients on the right to balance the number of atoms which have gained or lost electrons. 6 Balance the number of all other atoms except O and H. 7 Add H + to the side deficient in positive charges to make the number of charges on both sides equal. 8 Add H 2 O to the appropriate side to balance the number of O atoms. IO 3 is the oxidizing agent and SO 2 is the reducing agent. 2IO 3 I 2 SO 2 2 SO 4 2I +5 O 3 + S +4 O 0 2 I2 + S +6 2 O 4 reduction: gain of 10e per 2IO 3 2I +5 O 3 + S +4 O 2 I S +6 O 4 2 oxidation: loss of 2e per SO 2 2IO 3 + 5SO 2 I 2 + SO 4 2 2IO 3 + 5SO 2 I 2 + 5SO 4 2 The number of all other atoms except O is balanced. Total charge on left-hand side = 2 x ( 1) = 2 Total charge on right-hand side = 5 x ( 2) = 10 Add 8H + to the right-hand side. 2IO 3 + 5SO 2 I 2 + 5SO H + 2IO 3 + 5SO 2 + 4H 2 O I 2 + 5SO H + x is 5, y is 4 and z is A Combine the two ionic half-equations to obtain a redox equation: (i) Cr 2 O 2 7 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) (ii) x 3 3C 2 O 2 4 (aq) 6CO 2 (g) + 6e Cr 2 O 7 2 (aq) + 14H + (aq) + 3C 2 O 4 2 (aq) 2Cr 3+ (aq) + 7H 2 O(l) + 6CO 2 (g) According to the equation, 1 3 ions. mole of Cr 2O 7 2 (aq) ions will react completely with one mole of C 2 O 4 2 (aq) 157

34 47 B Combine the two ionic half-equations to obtain a redox equation: (i) x 4 4MnO 4 (aq) + 32H + (aq) + 20e 4Mn 2+ (aq) + 16H 2 O(l) (ii) x 5 5As 2 O 3 (s) + 25H 2 O(l) 10AsO 4 3 (aq) + 50H + (aq) + 20e 4MnO 4 (aq) + 32H + (aq) + 5As 2 O 3 (s) + 25H 2 O(l) 4Mn 2+ (aq) + 16H 2 O(l) + 10AsO 4 3 (aq) + 50H + (aq) Since H + (aq) and H 2 O(l) appear on both sides of the equation, simplify the equation by collecting like terms. 4MnO 4 (aq) + 5As 2 O 3 (s) + 9H 2 O(l) 4Mn 2+ (aq) + 10AsO 4 3 (aq) + 18H + (aq) According to the equation, 4 moles of MnO 4 (aq) ions will react completely with 5 moles of As 2 O 3 (s), i.e mole of MnO 4 (aq) ions will react completely with 1 mole of As 2 O 3 (s). 48 B Option Reaction Oxidation number(s) exhibited by nitrogen A N 3 H 3 + HN +5 O 3 N 3 H 4 N +5 O 3 3, +5 B 3N +4 O 2 + H 2 O 2HN +5 O 3 + N +2 O +2, +4, +5 C 3Zn + 8HN +5 O 3 3Zn(N +5 O 3 ) 2 + 2N +2 O + 4H 2 O +2, +5 D (N 3 H 4 ) 2 SO 4 N 3 H 4 HSO 4 + N 3 H 3 3 in the reaction in Option B, nitrogen exhibits three different oxidation numbers. 49 B Option Reaction Oxidation number(s) exhibited by sulphur A CaS +4 O 3 + H +4 2 S O 3 Ca(HS +4 O 3 ) 2 +4 B 2H 2 2 S + S +4 O 2 2H 2 O + 3S 0 2, 0, +4 C Cu + 2H +6 2 S O 4 CuS +6 O 4 + 2H 2 O + S +4 O 2 +4, +6 D 2H +6 2 S O 4 + S +6 O 3 H +6 2 S 2O 7 +6 in the reaction in Option B, sulphur exhibits three different oxidation numbers. 50 C Option Conversion Change in oxidation number of underlined element A F +3 e 3+ (aq) F +2 e 2+ (aq) 1 unit B N +5 O 3 (aq) N +2 O(g) 3 units C M +7 no 4 (aq) M +2 n 2+ (aq) 5 units D H 2 S +6 O 4 (l) S +4 O 2 (g) 2 units the conversion in Option C involves the greatest change in oxidation number of the underlined element. 158

35 51 B Option Conversion Is the conversion an oxidation? A S 0 2 The oxidation number of S decreases from 0 to 2, H 2 S the conversion is a reduction. B V +4 O 2+ V +5 O 2 + C C +6 ro 4 2 C +6 r 2 O 7 2 D CH 3 COOH CH 3 CH 2 OH The oxidation number of V increases from +4 to +5, the conversion is an oxidation. The oxidation number of Cr remains unchanged, the conversion is not an oxidation. CH 3 COOH loses oxygen in the conversion, the conversion is a reduction. 52 B 2S +2 2O I 0 2 S O I 1 The oxidation number of S increases from +2 to +2.5 while that of I decreases from 0 to 1. Hence the equation represents a redox reaction. 53 B C +2 a(o 2 H +1 ) 2 + S +4 O 2 2 C +2 as +4 O H +1 2O 2 Oxidation numbers of all elements remain unchanged in the reaction. Hence this is NOT a redox reaction. 54 D Option Equation Does the underlined substance become oxidized? Oxidation numbers of all elements in KOH remain A K +1 O 2 H +1 + HNO 3 K +1 NO H +1 2O unchanged, KOH does NOT become oxidized. The oxidation number of Cl decreases from 0 to 1, B 8NH 3 + 3C 0 l 2 6NH 4 C 1 l + N 2 Cl 2 does NOT become oxidized. The oxidation number of Fe decreases from +2 to 0, C Zn + F +2 eso 4 ZnSO 4 + F 0 e FeSO 4 does NOT become oxidized. The oxidation number of Br increases from 1 to 0, D 2HB 1 r + H 2 SO 4 SO 2 + 2H 2 O + B 0 r 2 HBr becomes oxidized. 55 A The position of calcium in the electrochemical series is the highest among calcium, magnesium, sodium and zinc. Hence calcium is the strongest reducing agent. 56 D Element Atomic number Name of element W 10 neon X 12 magnesium Y 14 silicon Z 17 chlorine Chlorine is an oxidizing agent. 159