Chapter 8 Redox Reactions Solutions
Question - 1 : - Assign oxidation numbers to the underlined elements ineach of the following species ;
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12H2O
Answer - 1 : - 
Question - 2 : - What are the oxidation numbers of the underlinedelements in each of the following and how do you rationalise your results ?|
(a) KI3 (b) H2S4Oe (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
Answer - 2 : - (a) O.N. of iodine in KI3 : By conventional method, the oxidation number of iodine in the I3_ ionmay be expressed as :
Explanation : The oxidationnumber of iodine (I) comes out to be fractional which does not seem to bepossible. Let us consider the structure of I3~ ion. In it, two atomsof iodine are linked to each other by covalent bond (I—I). The iodide ion (I~)is linked to the molecule by co-ordinate bond [I—I <—I]“. The molecule maybe represented as K+[I—I <—1| . Now, in the anion, the oxidationnumber of the two I atoms is zero while I– ion has – 1oxidation number. The overall oxidation number of L“ ion is :(b) O.N. of S in H2S4O6 :By conventional method, the oxidation number of sulphur may be calculated as :
H+12 Sx4 O-26
2 + 4x + 6 (- 2) =0 or 4x = 12 – 2 = 10 or x = 5/2
Explanation : In order toaccount for the fractional value, let us assign oxidation numbers to differentsulphur atoms in the structural formula of the acid. The oxidation numbers ofthe two middle sulphur atoms is zero while the two atoms at the terminalpositions have + 5 oxidation number.
Average O.N. of S =—1/4 [5+ 0 + 0 +5] = 5/2
(c) O. N. of Fe in Fe3O4 : Byconventional method, oxidation number of Fe may be calculated as :
Fe3x 3 034
3x + 4(- 2) =0 or x = 8/3
Explanation : Fe3O4 isa mixed oxide and is an equimolar mixture of Fe+2 O-2 andFe+32 O-23
Average O.N. of Fe = 1/3(2 + 2 x 3) =8/3
(d) O.N. of C in CH3CH2OH : Byconventional method, the oxidation number of carbon in ethanol molecule may becalculated as :
CH3cH2OH or Cx2 H+16 O-2
2x + 6(+ 1) + (- 2) = 0 or 2x + 6 – 2 = 0 or 2x = – 4 or x = – 2
Explanation : Let uscalculate the oxidation number of both C1 and C2 atoms.
C2 : Thecarbon atom is attached to three H atoms (less electronegative than carbon) andone CH2OH group (more electronegative than carbon).
∴ O.N. of C2 = 3(+ 1) +x + (- 1) = O or x =-2
C1 : The carbon atom isattached to one OH group (O.N. = – 1), two H atoms (O.N. = + 1) and one CH3 group(O.N. = + 1)
O.N. of C1 = ( + 1) + x + 2 (+ 1) + 1(- 1) =0 or x = – 2 Average O.N. of C =l/2[ – 2 + (- 2)] = – 2
(e) O.N. of C in CH3COOH : By conventionalmethod the O.N. of carbon may be calculated as:
CH3COOH or c2 H4 62
2x + 4(+ 1) + 2(— 2) = 0 or x — 0
Explanation : Let uscalculate the oxidation number of both C2 and C1 atoms.
C1 : The carbon atom isattached to three H atoms (less electronegative than carbon) and one COOH group(more electronegative than carbon).
∴ O.N. of C2 =3 (+ 1) + x + 1(- 1) = 0 or x = – 2
C2 : Thecarbon atom is attached to one OH group (O.N. = – 1) one oxygen atom by doublebond
(O.N. = – 2) and one CH3 group (O.N. = + 1)
O.N. of C1 = 1( + 1) + Jt + 1 (-2) + 1(- 1) = O or x = + 2
∴ Average O.N. of C =1/2 [+ 2 + (- 2)] = 0
Question - 3 : - How will you justify that the following reactionsare redox reactions in nature ?
(a) CuO(s) + H2(g) → Cu(s) + H2O(g)
(b) Fe2O3(s) + 3CO(g) →2Fe(s) + 3CO2(g)
(c) 2K(s) + F2(g)→2K+F–(s)
(d) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Answer - 3 : - A chemical reaction may be regarded as redox reactionif one of the reacting species undergoes increase in O.N. (oxidation) and theother decrease in O.N. (reduction). Based upon this, let us try to justify thatthe reactions under study are redox reactions in nature.
Question - 4 : - Fluorine reacts with ice as follows :
H2O(s)+ F2(g) — HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Answer - 4 : - 
In this reaction, the F2 molecule has undergoneincrease in O.N. in changing to HOF and decrease in O.N. in change to HF. It istherefore, a redox reaction. Since the same reacting species has undergoneincrease as well as decrease in O.N., it is also called disproportionation reaction.
Answer - 5 : -
(1) Oxidationnumber of S in H2SO5
By conventional method, O.N. of sulphur may be calculated as :
2 x 1 + x + 5 X (-2)= 0 or 2 + x – 10 = 0 or x = + 8 (wrong)
But this cannot be true as maximum oxidation number for sulphur cannot exceed+6 as it has only six valence electrons. The oxidation number of the elementcan be calculated from the structure of the persulphuric acid.
The sulphur atom is linked one OH group (O.N. = – 1) two oxygen atoms (O.N. = –2) and one peroxide (- O – O -) linkage (O.N. = – 1)
O.N. of S = 1 (- 1) + x + 2(- 2) + 1( – 1) = 0
or
x = + 6
(2) Oxidation number of Cr in CrO5
Byconventional method, O.N. of chromium may be calculated as :
x + 5 (-2) = 0 or x = + 10 (wrong)
This is wrong because maximum oxidation number of Crcannot be more than +6, since it has only six electrons (3d54.s1)to take part in bond formation. The oxidation number of the metal atom can becalculated by taking into account its structure.
The Cr atom is linked with one oxygen atom by double bond (O. N. = – 2) andfour oxygen atoms by peroxide linkages (O.N. = -1)
.’. O.N. of Cr = 4(- 1) \x + 1(- 2) = 0 or x = + 6
(3) Oxidation number N in NO–3 ion
By conventional method, O.N. of N atom in the ion may be calculated as : x + 3(- 2) = – 1 or x = +5.This can be further made clear from the structure of theion. In it, the nitrogen atom is linked to one single bonded oxygen atom (O.N.= – 2), one double bonded oxygen atom
(O.N.= – 2) and one oxygen atom by dative bond (O.N. = – 2).
O.N. of N = 1 (- 2) + x + 1( – 2) + 1 (- 2) = – 1 or x = + 5
Question - 6 : - Write the formulas of 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 - 6 : -
The formulas have been shown by stock notations. The Romannumerals given in parenthesis represent the oxidation state of the metal atom.Taking into account the oxidation state of the anions, the chemical formulas ofthe compounds may be written as :
(a) HgCl2 (b) NiSO4 (c) SnO2 (d)T12SO4 (e) Fe2(SO4)3 (f)Cr2O3
Question - 7 : - While sulphur dioxide and hydrogen peroxide can act asoxidising as well as reducing agents in their reactions, ozone and nitric acidcan act only as oxidising agents. Why ?
Answer - 7 : - In sulphur dioxide (SO2) and hydrogenperoxide (H2O2), the oxidation states of sulphur andoxygen are +4 and -1 respectively. Since they can increase as well as decreasewhen these compounds take part in chemical reactions, they can act as oxidisingas well as reducing agents. For example
In ozone (O3), the oxidation state of oxygen is zerowhile in nitric acid (HNO3), the oxidation state is nitrogen is +5.Since both them can undergo decrease in oxidation state and not an increase inits value, they can act only as oxidising agents and not as reducing agents.
Question - 8 : - Consider the reactions :
(a) 6CO 2(g)+ 6H2O (l) → C6H12O6(s) + 6O2(g)
(b) O3(g) + H2O2(l) → H2O (l) + 2O2(g)
Why it is more appropriate to writethese reactions as :
(a) 6CO2(g) + 12H2O(l)→ C6H12O6(s)+ 6H2O(l) + 6O2(g)
(b) 63(g) + H2O2(l)→ H2O(l) + O2(g)+ O2(g)
Also suggest a technique to investigatethe path of the above (a) and (b) redox reactions.
Answer - 8 : - (a) 6CO2(g) + 6H2O(l) → C6H6O6(s)+ 6O2(g)
Let us try to balance the equation by ion-electron method.
Note:
(1) In this reaction O3 acts as an O.A. and H2O2 actsas a R.A.
(2) If a co-ordinate bond exist between two similar atoms, the donor atomacquires + 2 O.N. and the acceptor gets -2 O.N.
This explains the path of the reaction i.e., how electrons are lost and gainedas well as and why it is more appropriate to write the equation as above.
Answer - 9 : -
In AgFL the oxidationstate of silver is + 2 i.e., it exists as Ag2+ ion. This isquite unstable since the normal stable oxidation state of the metal is + 1 orit exists as Ag+ ion. This means that if the compound AgF2 beformed, it will undergo reduction by the gain of electrons. It is therefore, avery strong oxidising agent
Ag2+ + e– — Ag+
Question - 10 : - Whenever a reaction between an oxidising agent and a reducing agent iscarried out, a compound of lower oxidation state is formed if the reducingagent is in excess and a compound of higher oxidation state is formed if theoxidising agent is in excess. Justify this statement giving three illustrations
Answer - 10 : - 