2. What are the oxidation numbers of the underlined elements in each of the following

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1 1. Assign oxidation number to the underlined elements in each of the following species (a) NaH 2 PO 4 (b) NaHSO 4 (c) H 4_ P 2 O 7 (d) K 2 MnO 4 (e) CaO 2 (f) NaBH 4 (g) H 2 S 2 O 7 (h) KAl(SO 4 ) 2.12H 2 O (a) NaH 2 PO 4 = 1(+1) + 2(+1) + P + 4(-2) = P - 8 = 0 P - 5 = 0 O. N. of P = +5 (b) NaHSO 4 = 1(1) + 1(1) + S + 4(-2) = S - 8 = 0 O. N. of S = +6 (c) H 4 P 2 O 7 = 4(H) + 2(P) + 7(O) = 0 4(1) + 2P + 7(-2) = P - 14 = 0 2P - 10 = 0 O. N. of P = +5 (d) K 2 Mn O 4 = 2(k) + Mn + 4(0) = 0 2(1) + Mn + 4(-2) = Mn - 8 = 0 O. N. of Mn = +6 (e) Ca0 2 = 1(Ca) + 2(0) = 0 1(2)+2(0) = 0 2(0) = -2 O. N. of O = -1 (f) Na = 1 ; H = -1 ( it s a hydride) B = B - 4 = 0 B = 3 (g) H 2 S 2 O 7 = 2(H) + 2(S) + 7(0) = 0 2(1) + 2S + 7(-2) = S - 14 = 0 2S = +12 O. N. of S = +6 (h) K Al(SO 4 ) H 2 O 1(K) + 1(Al) + 2(S) + 8(O) = 0 1(1) + 1(3) + 2(S) + 8(-2) = S - 16 = 0 2S = +12 O. N. of S = What are the oxidation numbers of the underlined elements in each of the following

2 and how do you rationalize your results? (a) KI 3 (b) H 2_ S 4 O 6 (c) Fe 3 O 4 (d) CH 3 CH 2 OH (e) CH 3 COOH (a) KI 3 = 1(K) + 3(I) = 0 1(1) + 3I = 0 3I = -1 O. N. of I = -1/3 (b) H 2 S 4 O 6 = 2(H) + 4(S) + 6(O) = 0 2(1) + 4S + 6(-2) = 0 4S - 10 = 0 S = = O. N. of S = +5/2 (c) Fe 3 O 4 = 3Fe + 4(0) = 0 3Fe + 4(-2) = 0 3Fe = +8 Fe=+8/3 O. N. of Fe=+8/3 (d) CH 3 CH 2 OH = C 2 H 6 O 2C + 6(H) + 1(O) = 0 2C + 6(1) + 1(-2) = 0 2C + 4 = 0 C = -2 (e) CH 3 COOH = C 2 H 4 O 2 2C + 4(H) + 2(0) = 0 2C + 4(1) + 2(-2) = 0 2C = 0 2C = 0 C = 0 O. N. of C = 0 The above values are calculated on the basis of conventional methods. 3. Justify that the following reactions are redox reactions (a) CuO(s) + H 2 (g) Cu(s) + H 2 O(g) (b) Fe 2 O 3 (s) + 3CO(g) 2Fe(s) + 3CO 2 (g) (c) 4BCl 3 (g) + 3LiAlH 4 (s) 2B 2 H 6 (g) +3LiCl(s) + 3 AlCl 3 (s) (d) 2K(s) + F 2 (g) 2K + F - (s) (e) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) (a) and (b) on the basis of classical idea. (c ) on the basis of new concept of electrodensity. (d ) on the basis of change in O.N.

3 (e) Both classical as well as change of electron density. 4. Fluorine reacts with ice and results in the change H 2 O(s) + F 2 (g) HF(g) + HOF(g) Justify that this reaction is a redox reaction. We can explain this on the basis of change in oxidation no. F in HF and HOF. The oxidation no. of fluorine changes from 0 to -1 in HF (reduced) whereas it increases from 0 to +1 in HOF(oxidized).Hence this reaction is redox as well as a disproportionation reaction. 5. Calculate the oxidation number of sulphur, chromium and nitrogen in H 2 SO 5, Cr 2 O 7 2- and NO 3 -. Suggest structure of these compounds. Count for the fallacy. H 2 SO 5 = 2(H) + 1(S) + 5(O) = 0 2(1) + S + 5(-2) = S - 10 = 0 S = +8 O. N. of S = +8 Cr 2 O 7 2- = 2(Cr) + 7(0) = -2 2Cr + 7(-2) = -2 2Cr = = 12 O. N. of Cr = +6 NO - 3 = 1(N) + 3(0) = -1 N + 3(-2) = -1 N - 6 = -1 N = O. N. of N = Write formulas for the following compounds (a) Mercury (II) Chloride (b) Nickel (II) sulphate (c) Tin (IV) oxide (d) Thallium (I) sulphate (e) Iron (III) sulphate (f) Chromium (III) oxide. (a) HgCl 2 (b) NiSO 4 (c) SnO 2

4 (d) Tl 2 SO 4 (e) Fe 2 (SO 4 ) 3 (f) Cr 2 O 3 7. Suggest a list of the substances where carbon can exhibit oxidation states from 4 to +4 and nitrogen from 3 to +5. Carbon can exhibit oxidation states from -4 to +4 in the following compounds. The compounds of nitrogen with oxidation state from 3 to 5 are NH 3, NH 2 -NH 2, NH=NH, N N, N 2 O, NO, N 2 O 3, N 2 O 4, N 2 O 5 are the compounds of nitrogen where the oxidation state of nitrogen is -3,-2,-1,0,1,2,3,4,5 respectively. (i) sulphur in SO 2 has an oxidation state of +4. The maximum oxidation state of S is +6 (e.q., SO 3 ) and minimum is 2(H 2 S). Similarly, H 2 O 2 has oxygen atom in oxidation state of oxygen is +2 and minimum oxidation state is 2 (in oxides, H 2 O, etc.) (ii) In HNO 3 nitrogen is in its maximum oxidation state of +5 and hence can only act as oxidant. Similarly oxidation state of oxygen in ozone is zero and hence can change to its stable oxidation state of 2 i.e., act as an oxidant. 8. While sulphur dioxide and hydrogen per oxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why? Any substance can act as an oxidising and reducing agent if one of the element in the compound is present in an intermediate oxidation state so that it can either increase its oxidation number (reducing agent) or decrease its oxidation number (oxidising agent). sulphur in SO 2 has an oxidation state of +4. The maximum oxidation state of S is +6 (e.q., SO 3 ) and minimum is 2(H 2 S). Similarly, H 2 O 2 has oxygen atom in oxidation state of oxygen is +2 and minimum oxidation state is 2 (in oxides, H 2 O, etc.) In HNO 3 nitrogen is in its maximum oxidation state of +5 and hence can only act as oxidant. Similarly oxidation state of oxygen in ozone is zero and hence can change to its stable oxidation state of 2 i.e., acts as an oxidant. 9. Consider the reactions a. 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (aq)+ 6O 2 (g) b. O 3 (g) + H 2 O 2 (l) H 2 O(l) + 2O 2 (g) Why it more appropriate to write these reactions as: a)6co 2 (g) + 12H 2 O(l) C 6 H 12 O 6 (aq)+ 6H 2 O(l) )+ 6O 2 (g) b)o 3 (g) + H 2 O 2 (l) H 2 O(l) + O 2 (g) + O 2 (g) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions. 4

5 a) 6CO 2 (g) + 12H 2 O(l) C 6 H 12 O 6 (aq)+ 6H 2 O(l) )+ 6O 2 (g) The above reaction is an overall photosynthetic reaction. It is appropriate to write it this way because 12 H 2 O are used per molecule of carbohydrate formed and 6 H 2 O are produced as the product of photosynthesis. b) O3(g) O 2 (g)+[o] Hence it is more appropriate to write the reaction as 2 O 2 molecules are produced by two different reactants. O 3 (g) + H 2 O 2 (l) H 2 O(l) + O 2 (g) + O 2 (g) 10. The compound AgF 2 is an unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why? AgF 2 can be synthesized by fluorinating Ag 2 O with elemental Flourine. AgF 2 is light sensitive and reacts vigorously with water and hence it is considered as an unstable compound. It should be stored in Teflon, a passivated metal container, or a quartz tube. AgF 2 acts as a very strong oxidising agent because Ag is in +2 oxidation state in AgF 2 which is very unusual. Ag in AgF 2 is a powerful electron (1e - ) acceptor to get converted to Ag 1+ state hence making AgF 2 a powerful oxidising agent. 11. How do you account for the following observations? (a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction. (b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapours of bromine. Why? (a) Alcoholic KMnO 4 is used as an oxidant in the manufacture of benzoic acid because organic reactions must be performed in a medium of organic solvent (like alcohol) in which the oxidizing agent (inorganic salt) is able to dissolve. (b) When conc.h 2 SO 4 is added to an inorganic mixture containing chloride, a red vapour of bromine and SO 2 (colourless) is evolved. HBr can be formed only if dil.h 2 SO 4 is used. 12. Identify the substances oxidized and reduced, the oxidising agent and reducing agent for each of the following reactions: (a) 2AgBr (s) + C 6 H 6 O 2 (aq) 2Ag(s) + 2HBr (aq)+ C 6 H 4 O 2 (aq)

6 (b) HCHO(l) + 2[Ag (NH 3 ) 2 ] + (aq) + 3OH - (aq) 2Ag(s) + HCOO - (aq) + 4NH 3 (aq) + 2H 2 O(l) (c) HCHO (l) + 2Cu 2+ (aq) + 5 OH - (aq) Cu 2 O(s) + HCOO - (aq) + 3H 2 O(l) (d) N 2 H 4 (l) + 2H 2 O 2 (l) N 2 (g) + 4H 2 O(l) (e) Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) 2PbSO 4 (s) + 2H 2 O(l) Substances which undergo oxidation act as a good reducing agent while the ones which undergoes reduction act as good oxidising agent. (a) Silver is reduced while C of compound C 6 H 6 O 2 is oxidised. Therefore, oxidising agent is Ag and reducing agent is C 6 H 6 O 2. (b) HCHO undergoes oxidation while [Ag (NH 3 ) 2 ]+(Tollens reagent) undergoes reduction. Oxidising agent is Tollen s reagent; reducing agent is HCHO. (c) HCHO undergoes oxidation while Cu 2+ (from Fehling s solution) undergoes reduction. oxidizing agent is Fehling s solution and reducing agent is HCHO. (d) Nitrogen in N 2 H 4 undergoes oxidation (reducing agent), H 2 O 2.undergoes reduction (oxidising agent). (e) Pb undergoes oxidation (red. Agent) ; PbO 2 undergoes reduction (oxidising agent). 13. Consider the reactions 2S 2 O 2-3 (aq) + I 2 (s) S 4 O 2-6 (aq) + 2I - (aq) S 2 O 2-3 (aq) + 2Br 2(l) + 5H 2 O(l) SO 2-4 (aq) + 4Br - (aq) + 10H + (aq) Why does the same reductant, thiosulphate react differently with iodine and bromine? Bromine acts as an oxidizing agent. It oxidizes sodium thio sulphate to sodium sulphate. S 2 O 3 2- (aq) + 2Br 2(l) + 5H 2 O(l) SO 4 2- (aq) + 4Br - (aq) + 10H + (aq) Iodine being a weaker oxidizing agent reacts with thio sulphate and forms sodium tetrathionate. 2Na 2 S 2 O 3 (aq) + I 2 (s) Na 2 S 4 O 6 (aq) + 2NaI (aq) 14. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant. Fluorine is the best oxidant among all halogens due to its smaller atomic size and greater nuclear attraction it attracts electrons more readily than other halogens.

7 Hydroiodic acid is the best reductant. Due to the greater size of I 2 Hydrogen can be easily knocked out from HI hence it will be a supplier of H+ ion and therefore used as reducing agent. Because Cl 2 is more electronegative than iodine. 15. Consider the reactions (a) H 3 PO 2 (aq) + 4 AgNO 3 (aq) + 2H 2 O(l) H 3 PO 4 (aq) + 4Ag(s) +4HNO 3 (aq) (b) H 3 PO 2 (aq) + 4 CuSO 4 (aq) + 2H 2 O(l) H 3 PO 4 (aq) + 2Cu(s) +H 2 SO 4 (aq) (c) C 6 H 5 CHO(l) + 2[Ag(NH 3 ) 2 ] + (aq) + 3OH - (aq) C 6 H 5 COO - (aq) + 2Ag(s)+ 4NH 3 (aq) + 2H 2 O (d) C 6 H 5 CHO (l) + 2Cu 2+ (aq) + 5OH - (aq) No change observed What inference do you draw about the behaviour of Ag+ and Cu 2+ from these reactions? In reactions (a) and (c). Ag + is acting as electron acceptor ie, oxidising agent and get converted to Ag. In reaction (b) Cu 2+ is converted to Cu ie, 2e - s are accepted by Cu 2+ ion. Therefore it played the role of oxidising agent. In reaction c [Ag(NH 3 ) 2 ] + (aq) is reduced to metallic silver. In reaction (d) Cu 2+ remains unchanged i.e, neither reduced nor oxidized. 16. Balance the following redox reactions by ion- electron method (a) MnO - 4 (aq) + I - (aq) MnO 2 (s) + I 2 (s) (in basic medium) (b) MnO - 4 (aq) + SO 2 (g) Mn 2+ (aq) + HSO - 4 (aq) (in acidic solution) (c) H 2 O 2 (aq) + Fe 2+ (aq) Fe 3+ (aq) + H 2 O (l) (in acidic solution) (d) Cr 2 O SO 2 (g) Cr 3+ (aq) + SO 2-4 (aq) (in acidic solution)

8 (a) 2MnO I - + 4H 2 O 2MnO 2 + 3I 2 + 8OH - (b) 5SO 2 +2MnO H 2 O + H + 5HSO Mn +2 (c) 2Fe +2 + H 2 O 2 + 2H + 2Fe +3 +2H 2 O (d) 3SO 2 + Cr 2 O H + 2Cr SO 2-4 +H 2 O 17. Balance the following equations in basic medium by ion- electron method and oxidation number methods and identify the oxidising agent and the reducing agent. (a) P 4 (s) +OH - (aq) PH 3 (g) + HPO - 2 (aq) (b) N 2 H 4 (l) + ClO - 3 (aq) NO(g) + Cl - (g) (c) Cl 2 O 7 (g) + H 2 O 2 (aq) ClO - 2 (aq) + O 2 (g) + H + (a) 4P + 3OH - + 3H 2 O - PH 3 + 3H 2 PO 2 - (b) 3N 2 H 4 + 4ClO 3 4Cl - + 6H 2 O + 6NO (c) Cl 2 O 7 + 4H 2 O 2 + 2OH - 2ClO H 2 O + 4O What sorts of informations can you draw from the following reaction? (CN) 2 (g) + 2OH - (aq) CN - (aq) + CNO - (aq) + H 2 O(l) Informations 1. One cyanide molecule in (CN) 2 is undergoing reduction by accepting one electron forming CN - ion. 2. Another Cyanide molecule in (CN) 2 is undergoing oxidation by losing one electron forming CNO - ion. 3. The oxygen atom in CNO - ion is supplied by OH - ion. 4. The H 2 O molecules formed is due to the presence of OH - ions on the reactant side. 19. Consider the elements:cs, Ne, I and F (a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only positive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the negative nor does the positive oxidation

9 state. (a) I can exhibit ve oxidation state as it is highly electronegative element. (b) Cs is highly electropositive element and exhibits +ve oxidation state. (c) I can show both -ve as well as +ve oxidation states. For example, in HI it is 1 and Icl 3 it is +3. (d) Ne is a noble gas with stable electronic configuration and hence neither exhibits Ve nor +ve oxidation state. 20. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water. The balanced equation is SO 2 + H 2 O H 2 SO 3 H 2 SO 3 + NaOH NaHSO 3 + H 2 O (or) HSO 3 + NaOH Na 2 SO 3 + H 2 O Na 2 SO 3 + Cl 2 +H 2 O Na 2 SO 4 + 2HCl. 21. Refer to the periodic table given in your book and now answer the following questions: (a) Select the possible non metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction. (a) 3NaOH + P 4 + H 2 O 3NaH 2 PO 2 + PH 3 2H 2 O 2 2H 2 O + O 2 Cl 2 + 2NaOH NaClO + NaCl +H 2 O Cu 2 O + H 2 SO 4 Cu + CuSO 4 + H 2 O 3S + NaOH + H 2 O Na 2 SO 3 + 2H 2 S 3NO 2 + H 2 O 2H + + 2NO 3 - +NO Disproportionation reaction is a redox reaction in which some atoms of a single element in a reactant

10 are oxidized and others are reduced. (b) The above mentioned reactions are a few examples of disproportionation reactions. Cu, Mn, Sn are the metals which can undergo dispoportionation reaction because they do exhibit variable oxidation states. Cl, O, I, S, N, P are some of the possible non-metals which can undergo disproportionation reaction. 22. In Ostwald s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00g. of ammonia and 20.00g of oxygen? 4NH 3 + 5O 2 4NO + 6H 2 O Molecular mass of NH 3 = 17g mol -1 Molecular mass of O 2 = 32g mol -1 Molecular mass of NO = 30g mol -1 (4 17 =) 68g of NH 3 reacted with (5 32 =) 160g of O 2 to form (4 30=)120g of NO. 68 g of NH 3 reacts with oxygen = 160 g 10 g of NH 3 react with O 2 = = 23.5 g The quantity of O 2 available is less than what is required for the reaction.. Hence O 2 is the limiting reagent and the calculations must be based on the quantity of O 2. According to the equation, 160 g of O 2 produce NO = 120 g 20 g of O 2 will produce No = = 15 g. 23. Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn. Zn + Cu 2+ Zn 2+ + Cu Reducing Agent Oxidising agent

11 2Al + 3Z n 2+ 2Al Zn Reducing Oxidising agent Agent Al > Zn> Cu decreasing order of reducing strength. 24. Given the standard electrode Potentials, K + /K = -2.93V, Ag + /Ag = 0.80V, Hg 2+ /Hg = 0.79V Mg 2+ /Mg = -2.37V. Cr 3+ /Cr = -0.74V Arrange these metals in their increasing order of reducing power. The increasing order of reducing power is given below Ag + Ag = 0.8 V < Hg 2+ /Hg = 0.79v < Cr 3+ /Cr = V < Mg 2+ /Mg = V<K + /K = Depict the galvanic cell in which the reaction Zn(s) + 2Ag + (aq) Zn 2+ (aq)+2ag(s) takes place, further show (i) Which of the electrode is negatively charged? (ii) The carriers of the current in the cell, and (iii) Individual reaction at each electrode. (i) Ag electrode is negatively charged (ii) Migration of Ag + ions towards cathode (iii) At anode, Zn Zn e - At cathode, Ag + Ag 2Ag + +2e - 2Ag

Class XI Chapter 8 Redox Reactions Chemistry

Class XI Chapter 8 Redox Reactions Chemistry Question 8.1: Assign oxidation numbers to the underlined elements in each of the following species: (a) NaH 2 PO 4 (b) NaHSO 4 (c) H 4 P 2 O 7 (d) K 2 MnO 4 (e) CaO 2 (f) NaBH 4 (g) H 2 S 2 O 7 (h) KAl(SO