Chapter – 7
Definitions If S.I Unit
1. Assign oxidation number to the underlined elements in each of the following species:
(a)\(\begin{array}{l}NaH_2\underline{P}O_4\end{array} \)
(b)\(\begin{array}{l}NaH\underline{S}O_{4}\end{array} \)
(c)\(\begin{array}{l}H_{4}\underline{P}_{2}O_{7}\end{array} \)
(d)\(\begin{array}{l}K_{2}\underline{Mn}O_{4}\end{array} \)
(e)\(\begin{array}{l}Ca\underline{O}_{2}\end{array} \)
(f)\(\begin{array}{l}Na\underline{B}H_{4}\end{array} \)
(g)\(\begin{array}{l}H_{2}\underline{S}_{2}O_{7}\end{array} \)
(h)\(\begin{array}{l}KAl(\underline{S}O_{4})_{2}.12 H_{2}O\end{array} \)
Answer:
(a) \(\begin{array}{l}NaH_2\underline{P}O_4\end{array} \)Let x be the oxidation no. of P.Oxidation no. of Na = +1O xidation no. of H = +1O xidation no. of O = -2 Q.1(a)Then,1(+1) + 2(+1) + 1(x) + 4(-2) = 01 + 2 + x -8 = 0 x = +5 Therefore, oxidation no. of P is +5.(b) \(\begin{array}{l}NaH\underline{S}O_{4}\end{array} \)
(b)Let x be the oxidation no. of S.Oxidation no. of Na = +1Oxidation no. of H = +1Oxidation no. of O = -2Then,1(+1) + 1(+1) + 1(x) + 4(-2) = 01 + 1 + x -8 = 0x = +6Therefore, oxidation no. of S is +6.
(c)\(\begin{array}{l}H_{4}\underline{P}_{2}O_{7}\end{array} \)NCERT Solutions for Class 11 Chemistry- Chapter 8 Redox Reactions Q.1(c)Let x be the oxidation no. of P.Oxidation no. of H = +1Oxidation no. of O = -2Then,4(+1) + 2(x) + 7(-2) = 04 + 2x – 14 = 02x = +10x = +5Therefore, oxidation no. of P is +5.
(d)\(\begin{array}{l}K_{2}\underline{Mn}O_{4}\end{array} \)Therefore, oxidation no. of S is +6.ORIgnore the water molecules because it is neutral. Then, the summation of the oxidation no. of all atoms of water molecules can be taken as 0. Hence, ignore the water molecule.1(+1) + 1(+3) + 2(x) + 8(-2) = 01 + 3 + 2x -16 = 02x = 12x = +6Therefore, oxidation no. of S is +6.
(e)Let x be the oxidation no. of O.Oxidation no. of Ca = +2Then,(+2) + 2(x) = 02 + 2x = 02x = -2x = -1Therefore, oxidation no. of O is -1.
(f)Let x be the oxidation no. of B.Oxidation no. of Na = +1Oxidation no. of H = -1Then,1(+1) + 1(x) + 4(-1) = 01 + x -4 = 0x = +3Therefore, oxidation no. of B is +3.
(g)Let x be the oxidation no. of S.Oxidation no. of H = +1Oxidation no. of O = -2Then,2(+1) + 2(x) + 7(-2) = 02 + 2x – 14 = 02x = +12x = +6Therefore, oxidation no. of S is +6.
(h)Let x be the oxidation no. of S.Oxidation no. of K = +1Oxidation no. of Al = +3Oxidation no. of O= -2 Oxidation no. of H = +1 Then,1(+1) + 1(+3) + 2(x) + 8(-2) + 24(+1) + 12(-2) = 01 + 3 + 2x – 16 + 24 – 24 = 02 x = + 12 x = + 6 Therefore, oxidation no. of S is +6.ORIgnore the water molecules because it is neutral. Then, the summation of the oxidation no. of all atoms of water molecules can be taken as 0. Hence, ignore the water molecule.1(+1) + 1(+3) + 2(x) + 8(-2) = 01 + 3 + 2x -16 = 02x = 12x = +6Therefore, oxidation no. of S is +6.
Question 2. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
Answer:
In Kl3, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K+[I —I <— I]–. Here, a coordinate bond is formed between I2 molecule and I– ion. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine forming the coordinate bond is -1. Thus, the O.N. of three I atoms, atoms in Kl3 are 0, 0 and -1 respectively.
Question 3. Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g) —–> Cu(s) + H20(g)
(b) Fe2O3(s) +3CO(g) —-> 2Fe(s) + 3CO2(g)
(c) 4 BCl 3(g) +3 LiAl H 4(s) ——> 2 B2H 6(g) + 3LiCl(s) + 3AlCl3(s)
(d) 2K(s) +F2(g)——> 2K+F–(s)
Answer:
(a) Here, O is removed from CuO, therefore, it is reduced to Cu while O is added to H2 to form H20, therefore, it is oxidised. Further, O.N. of Cu decreases from + 2 in CuO to 0 in Cu but that of H increases from 0 in H2 to +1 in H20. Therefore, CuO is reduced to Cu but H2 is oxidised to H20. Thus, this is a redox reaction
(b) Here O.N. of Fe decreases from +3 if Fe2O3 to 0 in Fe while that of C increases from +2 in CO to +4 in CO2. Further, oxygen is removed from Fe2O3 and added to CO, therefore, Fe2O3 is reduced while CO is oxidised. Thus, this is a redox reaction.
(c) Here, O.N. of B decreases from +3 in Br Cl 3 to -3 in B2H6 while that of H increases from -1 in LiAlH 4 to +1 in B2H6. Therefore, BCl3 is reduced while LiAlH4 is oxidised. Further, H is added to BCl3 but is removed from LiAlH4, therefore, BC13 is reduced while LiAlH4 is oxidised. Thus, it is a redox reaction.
(d) Here, each K atom as lost one electron to form K+ while F2 has gained two electrons to form two F– ions. Therefore, K is oxidised while F2 is reduced. Thus, it is a redox reaction.By chemical bonding, C2 is attached to three H-atoms (less electronegative than carbon) and one CH2OH group (more electronegative than carbon), therefore,O.N. of C2 = 3 (+1) + x + 1 (-1) = 0 or x = -2 C2 is, however, attached to one OH (O.N. = -1) and one CH3 (O.N. = +1) group, therefore, O.N. of C4 = + 1 + 2 (+1) + x + 1 (-1) = 0 or x = -2
Question 4. Fluorine reacts with ice and results in the change:H20(S) + F2 (g) ——-> HF(g) + HOF(g)Justify that this reaction is a redox reaction.
Answer:
Writing the O.N. of each atom above its symbol, we have, Here, the O.N. of F decreases from 0 in F2 to -1 in HF and increases from 0 in F2 to +1 in HOF. Therefore, F2 is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction
Question 5. Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2 and NOT. Suggest structure of these compounds. Count for the fallacy.
Answer:
O.N. of S in H2SO5. By conventional method, the O.N. of S in H2SO5 is 2 (+1) + x + 5 (-2) = 0 or x = +8 This is impossible because the maximum O.N. of S cannot be more than six since it has only six electrons in the valence shell. This fallacy is overcome if we calculate the O.N. of S by chemical bonding method. The structure of H2SO5 is Thus, there is no fallacy about the O.N. of N in N03–whether one calculates by conventional method or by chemical bonding method.
Question 6.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.
Answer:
(a) Hg(II)Cl2,
(b) Ni(II)SO4,
(c)Sn(IV)O2
(d) T12(I)SO4,
(e) Fe2(III)(S04)3,
(f) Cr2(III)O3.
Question 8. While sulphur dioxide and hydrogen peroxide can act as an oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:
(i) In S02 , O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in S02 can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent.
(ii) In H2O2, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O2F2and +2 in OF2). Therefore, O in H2O2 can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H2O2 acts both as an oxidising as well as a reducing agent.
(iii) In O3, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O3 acts only as an oxidant.
(iv) In HNO3, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only
Question 9. Consider the reactions:
(a) 6CO2(g) 6H2O(l) ———> C6H12O6(s) + 6O6(g)
(b) O3(g) + H2O2(l) H2O(l) + 2O2(g)Why it is more appropriate to write these reactions as:(a) 6CO2(g) + 12H2O(l) ————-> C6H12O6(s) + 6H2O(l) + 6O2(g)(b) O3(g) + H2O2 (l) ———–> H2O(l) + O2(g) + O2(g)Also suggest a technique to investigate the path of above (a) and (b) redox reactions.
Answer:
(a) Therefore, it is more appropriate to write the equation for photosynthesis as (iii) because it emphasises that 12H2O are used per molecule of carbohydrate formed and 6H2O are produced during the process.
(b) The purpose of writing O2 two times suggests that O2 is being obtained from each of the two reactants.1 The path of reactions (a) and (b) can be determined by using H20218 or D20 in reaction(a) or by using H20218 or O318in reaction (b).
Question 10. The compound AgF2 is unstable. However, if formed, the compound acts as a very strong oxidising agent. Why?
Answer:
In AgF2 oxidation state of Ag is +2 which is very very unstable. Therefore, it quickly accepts an electron to form the more stable +1 oxidation state.Ag2+ + e– ————–> Ag+Therefore, AgF2, if formed, will act as a strong oxidising agent.
Question 11. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if oxidising agent is in excess. Justify this statement giving three illustrations.
Answer:
(i) C is a reducing agent while O2 is an oxidising agent. If excess of carbon is burnt in a limited supply of O2, CO is formed in which the oxidation state of C is +2. If, however, excess of O2 is used, the initially formed CO gets oxidised to CO2 in which oxidation state of C is + 4.
(ii) P4 is a reducing agent while Cl2 is an oxidising agent. When excess of P4 is used, PCl3 is formed in which the oxidation state of P is + 3. If, however, excess of Cl2 is used, the initially formed PCl3 reacts further to form PCl5 in which the oxidation state of P is +5.1
(iii) Na is a reducing agent while 02 is an oxidising agent. When excess of Na is used, sodium oxide is formed in which the oxidation state of O is -2. If, however, excess of 02 is used, Na2O2 is formed in which the oxidation state of O is -1 which is higher than -2.
Question 12. 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 vapour of bromine. Why?
Answer:
(a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral media according to the following redox equations: In the laboratory, benzoic acid is usually prepared by alkaline KMnO4 oxidation of toluene. However, in industry alcoholic KMnO4 is preferred over acidic or alkaline KMnO4 because of the following reasons:
(i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH- ions) are produced in the reaction itself.
(ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO4 (due to its polar nature) and toluene (because of its being an organic compound).
(b) When cone. H2S04 is added to an inorganic mixture containing chloride, a pungent smelling gas HCl is produced because a stronger acid displaces a weaker acid from its salt.1 Since HCl is a very weak reducing agent, it can not reduce H2S04 to S02 and hence HCl is not oxidised to Cl2.However, when the mixture contains bromide ion, the initially produced HBr being a strong reducing agent than HCl reduces H2S04to S02 and is itself oxidised to produce red vapour of Br2.
Question 14. Consider the reactions: Why does the same reductant, thiosulphate react difforerently with iodine and bromine?
Answer:
The average O.N. of S in S 2O32- is +2 while in S4O62- it is + 2.5. The O.N. of S in SO42- is +6. Since Br 2 is a stronger oxidising agent that I2, it oxidises S of S2O32- to a higher oxidation state of +6 and hence forms SO42- ion. I2, however, being weaker oxidising agent oxidises S of S2O32- ion to a lower oxidation of +2.5 in S4O62- ion. It is because of this reason that thiosulphate reacts differently with Br2 and I2.
Question 15. Justify-giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic add is the best reductant.
Answer:
Halogens have a strong tendency to accept electrons. Therefore, they are strong oxidising agents. Their relative oxidising power is, however, measured in terms of their electrode potentials. Since the electrode potentials of halogens decrease in the order: F2 (+2.87V) > Cl2 (+1.36V) > Br2 (+1.09V) > I2 (+0.54V), therefore, their oxidising power decreases in the same order.This is evident from the observation that F2 oxidises Cl– to Cl2, Br–to Br2, I – to I2 ; Cl2 oxidises Br–to Br2 and F to I2 but not F– to F2. Br2, however, oxidises F to I2 but not F– to F2 , and Cl– to Cl2.F2(g) + 2Cr(aq) ———–> 2F–(aq) + Cl2(g); F2(g) + 2Br–(aq) ———-> 2F–(aq) + Br2 (Z)F2(g) + 2I–(aq) ———-> 2F–(aq) + I2(s); Cl2 (g) + 2Br–(aq) ————> 2Cl–(aq) + Br2 (Z)Cl2(g) + 2I–(aq) ———–> 2Cl– (aq) + I2(s) and Br2 (Z) + 2F ———> 2Br– (aq) + I2(s)Thus, F2 is the best oxidant.Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. Since the electrode potentials of halide ions decreases in the order: I–(-0.54 V) > Br– (-1.09 V) > Cl–(-1.36 V) > I2 (-2.87 V), therefore, the reducing power of the halide ions or their corresponding hydrohalic acids decreases in the same order: HI > HBr > HCl > HF. Thus, hydroiodic acid is the best reductant. This is supported by the following reactions. For example, HI and HBr reduce H2S04 to S02 while HCl and HF do not.2HBr + H2S04 —–> Br2+ S02 + 2H2O; 2HI + H2S04 ——> I2 + S02 + 2H2OFurther F reduces Cu2+ to Cu+ but Br does not.2Cu2+(aq) + 4I–(aq) >Cu2I2(s) + I2(aq); Cu2+(aq) + 2Br–> No reaction.Thus, HI is a stronger reductant than HBr.Further among HCl and HF, HCl is a stronger reducing agent than HF because HCl reduces MnO2 to Mn2+ but HF does not.MnO2 (s) + 4HCl(aq) ——-> MnCl2(aq) + Cl2(aq) + 2H2OMnO2 (s) + 4HF(l) ———–> No reaction.Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF.
Question 17. Consider the reactions:
(a) H3P02(aq) + 4AgNO3(aq) + 2H2O(l) ————->H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
(b) H3P02(aq) + 2CuS04(aq) + 2H2O(l) ————->H3P04(aq) + 2Cu(s) + H2S04(aq)
(c) C6H5CHO(l) + 2[Ag(NH3)2]+(aq) + 30H–(aq)———–> C6H5COO–(aq) + 2Ag(s) + 4NH3(aq) + 2H20(l)
(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH–(aq) ———–> No change observedWhat inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions?
Answer:
Reactions (a) and (b) indicate that H3P02 (hypophosphorous acid) is a reducing agent and thus reduces both AgNO3 and CuS04 to Ag and Cu respectively. Conversely, both AgNO3 and CuS04 act as oxidising agent and thus oxidise H3P02to H3P04 (orthophosphoric acid) Reaction
(c) suggests that [Ag(NH3)2]+ oxidises C6H5CHO (benzaldehyde) to C6H5COO– (benzoate ion) but reaction
(d) indicates that Cu2+ ions cannot oxidise C6H5CHO to C6H5COO–. Therefore, from the above reactions, we conclude that Ag+ ion is a strong deoxidising agent than Cu2+ ion.
Question 18. Balance the following redox reactions by ion-electron method.
(a) MnO4–(aq) +I–(aq) ———>Mn02(s) + I2 (s) (in basic medium)
(b) MnO4–(aq) + S02(g) ——-> Mn2+(aq) +H2S04–(in acidic solution)
(c) H2O2(aq) + Fe2+(aq) ———-> Fe3+(aq) + H2O(l) (in acidic solution)
(d) Cr2O72- (aq) + S02 (g)——> Cr3+ (aq) + SO42-(aq) (in acidic solution)
Answer:
(a) Do it yourself.
(b) The balanced half reaction equations are:Oxidation half equation:S02(g) + 2H2O(l) ——–> HS04– (aq) + 3H+(aq) +2e– …(i)Reduction half equation:MnO4–(aq) + 8H+(aq) + 5e– ——–> Mn2+(aq) + 4H2O(l) ………..(ii)Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have,2MnO4–(aq) + 5S02(g) + 2H20(l) + H+(aq) ————> 2Mn2+(aq) + 5HSO4–(aq)
(c) Oxidation half equation: Fe2+(aq) ———> Fe3+(aq) + e– …(i)Reduction half equation: H2O2(aq) + 2H+(aq) + 2e– ———> 2H2O(l) …(ii)Multiply Eq. (i) by 2 and add it to Eq. (ii), we have,H2O2(aq) +2Fe2+(aq) +2H+(aq) ——-> 2Fe3+(aq) + 2H2O(l)
(d) Following the procedure detailed on page 8/23, the balanced half reaction equations are:Oxidation half equation:SO2(g) + 2H2O(l) ————> SO42-(aq) + 4H+(aq) + 2 e– …(i)Reduction half equation:Cr2O72–(aq) + 14H+(aq) + 6e– ————> 2Cr3+(aq) + 7H20(l) …(ii)Multiply Eq. (i) by 3 and add it to Eq. (ii), we have,Cr2O72–(aq) + 3SO2(q) + 2H+(aq) ————> 2Cr3+(aq) + 3SO42-(aq) + H20(l)