Redox Reactions

Chapter 8

Redox Reactions Solutions

SECTION - A

Objective Type Questions (One option is correct) 1.

Which Wh ich of th the e fol follow lowing ing is a Red Redox ox rea react ction ion? ? (1) CaCO3  CaO + CO2

(2)

AgNO3 + NaCl  AgCl + NaNO3

(3) Na NaOH OH + H HCl Cl  NaCl + H2O

(4)

KClO3  KCl +

3 2

O2

Sol. Answer (4) KClO3  KCl +

3 2

O2

Here Cl is reduced from +5 to –1 oxidation state and oxygen is oxidized from –2 to zero oxidation state. 2.

In the reaction 2Ag + 2H2SO4  Ag2SO4 + 2H2O + SO2 Sulphuric acid acts as (1) Oxidising agent

(2)

Reducing agent

(3) Oxidising as well as reducing agent

(4)

Catalyst

Sol. Answer (1) In H2SO4, O.S. (S) = + 6 In SO2,

O.S. (S) = + 4

 Sulphur is getting reduced, hence, H 2SO4 acts as an oxidizing agent.

3.

Which Wh ich ca can n act act as an an oxidi oxidisin sing g as well well as as a reduc reducing ing age agent? nt? (1) HClO4

(2)

HNO3

(3)

H2SO4

(4)

H2O2

Sol. Answer (4) H2O2  H 2O +

1 2

O 2

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180 Redox Reactions

4.

Solution of Assignment (Set-2)

The num number ber of of electr electrons ons requ require ired d to balan balance ce the the follow following ing equa equatio tion n are NO3– + 4H+ + e–  2H 2O + NO is (1) 5

(2)

4

(3)

3

(4)

2

Sol. Answer (3) 5  N O3

5.

2

 4H  3e   2H2O  N O

Which Whic h of the the follow following ing state statemen mentt is corre correct ct about about oxid oxidatio ation n numbe number? r? (1)) Oxidation number of all atoms (1 atoms in elementa elementary ry state is zero (2)) The sum of Oxidation (2 Oxidation number number of all the atoms atoms in the formula of of a compound compound is always zero zero (3)) Alkali and alkaline (3 alkaline earth metals metals have have +1 and +2 oxidation oxidation states states respectively respectively (4) Al Alll of of the these se

Sol. Answer (4) O.S. (O2) = 0 Elementary state In the molecular formula, sum of the oxidation number of the atoms is zero. 6.

Whic Wh ich h can can ac actt as as a re redu duci cing ng ag agen ent? t? (1) HNO3

(2)

H2SO4

(3)

H2S

(4)

KMnO4 KM

Sol. Answer (3) 2 0 H2S  S

(reducing agent), as sulphur is oxidized from –2 to zero oxidation state. 7.

Oxid Ox idat atio ion n num numbe berr of of iron iron in Fe0.94O is (1) +2

(2)

+3

(3)



200 94

(4)) (4

8 3

Sol. Answer (3) In Fe0.94O  0.94 x – 2 = 0 2

x = 

8.

0.94

x (Oxidation number)



200 94

In thi this s re reactio ion n 4Al 4Al + 3O 3O2  4Al 3+ + 6O2–. Which of the following statement is incorrect? (1) It is a redox reaction

(2)

Metallic aluminium is a reducing agent

(3)) Met (3 Metallic allic alumi aluminium nium is oxidi oxidised sed to Al3+

(4)

Meta Me talli llic c alum alumini inium um is red reduc uced ed to Al3+

Sol. Answer (4) Metallic Al is oxidized to Al3+ not reduced. Al  Al3  (Oxidation)

 3e 



Redox Reactions 181


9.

Oxid Ox idat atio ion n num numb ber of Cr in K3CrO8 is (1) +6

(2)

+5

(3)

+3

(4)

+2

Sol. Answer (2) Oxidation state of Cr is +5. There are 4 peroxide ions, each one has a charge of –2 10. Peroxides are are basic in nature and and they form form hydrogen hydrogen peroxides peroxides on treatment treatment with acid. What volume of 0.5 M H 2SO4 solution is required to neutralise a solution containing 7.2 g of CaO 2? (1) 400 ml

(2)

300 ml

(3)

200 ml

(4)

100 ml

Sol. Answer (3) Equ. of H2SO4 = Equ. of CaO 2 

0.5  2  V



7.2 36

 V = 0.2 I or 200 ml

11.

What is the difference difference in oxidation state of nitrogen in between between hydroxyl hydroxyl amine amine (NH2OH) and hydrazine (N2H4)? (1) +5

(2)

+3

(3)

–3

(4)

1

(4)

Cr2 O72–  2Cr 3+

Sol. Answer (4) Hydrazine NH2 – NH 2 hydroxyl amine –NH 2 – OH O.S. of N in NH2OH is 1 and O.S. of N in N 2H4 is –2, hence difference in O.S is 1 12. In which which of the the following following changes, changes, there there is a transfer transfer of five five electrons? electrons? (1) MnO4–  Mn+2

(2)

CrO42–  Cr 3+

(3)

MnO4–  MnO2

Sol. Answer (1)  MnO4   Mn2

x + 4 × –2 = –1 x=7 Change in oxidation state = 7 – 2 = 5 2 CrO4   Cr  3

x + 4 × –2 = –2 x=6 Change in oxidation state = 6 – 3 = 3 7 4   Mn O 2 Mn O 4 

x + 2 × –2 = 0 x=4 Change in oxidation state = 7 – 4 = 3 6 3 2 Cr 2 O7   2 C r 3 

2x + 7 × –2 = –2 2x = 12 x = +6 Total change in oxidation state = 2 (6 – 3) = 2 × 3 = 6 Aakash Educational Services Pvt. Ltd.


182 Redox Reactions


13. Ho How w man many y gm of K 2Cr 2O7 is present in 1 L of its N/10 solution in acid medium? (1) 4.9

(2)

49 49

(3)

0.49 0.

(4)

3.9 3.

Sol. Answer (1) 6 K 2 Cr 2 O7   Cr  3

n factor = 6 Number of eq. = NV (litre) 

1

1

1 

10

10

= 0.1

w = no. of eq. × Eq. wt = 4.9 g 14. When Cu2S is converted into Cu 2+ & SO2 then equivalent weight of Cu 2S will be (M = mol.wt. of Cu 2S) (1) M

(2)

M 2

M

M

(3)) (3

(4)) (4

4

8

Sol. Answer (4) +1 –2

2+

Cu2S – –2 × 1 e

Cu

+4

+ SO2

–

–6e M

So, equivalent wt. of Cu 2S =

8

15. Which of the the followin following g change changes s involve involve reduc reduction? tion? (1)) The conversio (1 conversion n of ferrous ferrous sulphate sulphate to ferric ferric sulphate sulphate (2)) The con (2 conver versio sion n of of H2S to S (3)) The con (3 conver versio sion n of of Cl Cl2 to NaCl (4)) The conv (4 conversi ersion on of of Zn to ZnSO ZnSO4 Sol. Answer (3) 2

3

Fe SO4  Fe2 (SO4 )3 [Oxidation] 2 o H2S   S [Oxidation] 0 1  Na Cl [Reduction] Cl2 

2 Zn   Zn SO 4 [Oxidation] 0

16. FeS  Fe 3+ + SO 3 Eq.wt of FeS for this change is (mol.wt. of FeS = M) (1)) (1

M

M

(2)) (2

1

5

(3)) (3

M 7

M

(4)) (4

9

Sol. Answer (4) 2 2 Fe S   Fe 3



6 S O3



Redox Reactions 183


n-factor of FeS =

1 (3 (3  2)  1 (6 ( 6  ( 2))     

   

Fe

s

=1+8=9 E

M





n factor

M 9

17. The re reac acti tio on Cl2 + S2O32– + OH–  SO 42 – + Cl – + H2O Starting with 0.15 mole Cl 2, 0.010 mole S 2O32– and 0.30 mole OH –, mole of Cl 2 left in solution will be (1) 0.11

(2)

0.01

(3)

0.04

(4)

0.09

Sol. Answer (1) 2 2 Cl2  S2O3 0

nf = 2

 OH

6  SO 42

1  Cl   H2O

nf = 8

Number of eq. of Cl 2 = moles × n factor = 0.15 × 2 = 0.30 Number of eq. of

2 S2O3 =

0.010 × 8

= 0.08 Number of eq. of OH – = 0.30 ×1 = 0.30 2 S2O3 has

∵

minimum equivalents hence

2 S 2 O3

is limiting reagent.

Number of eq. of Cl 2 remaining = 0.3 – 0.08 = 0.22 Number of moles of Cl 2 remaining =

0.22

 0.11

2

18.. The 18 The no. no. of mo mole les s of of KMn KMnO O4 that will be needed to react completely with one mole of ferrous oxalate in acidic solution is (1) 3/5

(2)

2/5 2/

(3)

4/5 4/

(4)

1/5 1/

Sol. Answer (1) 2 3 Fe C2 O4

7

 K Mn O4   Fe 3  acidic

4

C O2

 Mn2

nf 5

n factor of FeC 2O4 =

(3  2)  2  ( 4  3)  

   

Fe

C

=1+2=3 Number of eq. of of FeC2O4 = Number of eq. of KMnO 4 1×3=n×5 n



3 5



184 Redox Reactions


19.. 20 19 200 0 ml ml of of 0.0 0.01 1 M KM KMnO nO4 oxidise 20 ml of H 2O2 sample in acidic medium. The volume strength of H 2O2 is (1) 2.8 volume

(2)

5.6 volume

(3)

0.5 volume

(4)

0.25 volume

Sol. Answer (1) 7 K Mn O4 nf

2

 H 2O 2   M n2  H2O  O 2 acidic

medium

5

nf



2

Number of eq. of KMnO 4 = Number of eq. of H 2O2 0 .0 1  5 

200

20

 N

1 000

1000

N = 0.5 Volume strength = 5.6 × N = 5.6 × 0.5 = 2.8 20.. Equa 20 Equall volum volumes es of of 1 M eac each h of KMn KMnO O 4 and K2Cr 2O7 are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidised will be (1) Mo More re wit with h KMnO KMnO4

(2)

More with K2Cr 2O7

(3) Equal with both oxidising agent

(4)

Cannot be determined

Sol. Answer (2) 2

7

acidic

3 3

Fe (II)  KMnO 4   Fe Medium

nf  1

 Mn

nf  5

2

6

acidic

3 3

 Fe Fe (II)  K 2Cr2O7   Medium

nf  1

2 2

3 3

 Cr

nf  2  (6 (6  3) 6

let V litre, each of KMnO 4 and K2Cr 2O7 are taken number of equivalent of Fe +2 = Number of eq. of KMnO4 n1 × 1 = 1 × 5 × V n1 = 5 V Number of eq. of Fe+2 = Number of eq. of K 2Cr 2O7 n2 ×1 = 1 × 6 × V n2 = 6 V  more moles of Fe +2 will be oxidised with K 2Cr 2O7.

21.. 4 mol 21 mol of a mixt mixture ure of of Mohr’ Mohr’s s salt salt and and Fe2(SO4)3 required 400 ml of 0.5 M KMnO 4 for complete oxidation in acidic medium. The mole % of Mohr’s salt in the mixture is (1) 40%

(2)

5% 5%

(3)

50% 50

(4)

25% 25

Sol. Answer (4) Mohr’s salt is FeSO4.(NH4)2SO4.6H2O x m ole

( 4 – x ) m ole

FeSO4 .(NH4 )2 SO4 .6H2 O  Fe Fe2 (SO4 )3  KMnO4    Fe3  Mn2 nf  1 Aakash Educational Services Pvt. Ltd.

acidic

medium


Redox Reactions 185


Number of eq. of Mohr’s salt = Number of eq. of KMnO 4 400

x × 1 = (0.5 × 5) ×

1000

x=1 Number of Moles of Fe 2(SO4)3 = 3 mole mol % of Mohr’s salt 1

=

1 3

1

 100 =

4

 100 = 25%

22.. Th 22 The e vol volum ume e of of 3M 3M HNO HNO3 required to oxidised 8 g Fe 2+ is (HNO3 + Fe2+  Fe 3+ + NO + H2O) (1) 8 ml

(2)

16 ml

(3)

32 ml

(4)

64 ml

3

(4)

4

Sol. Answer (2) Fe

2

nf

5 H N O3



32

2

 Fe 3  NO  N 2O

52

nf

1

3 8

Number of moles of Fe +2 =



56

1 7

Number of eq. of Fe+2 = Number of eq. of HNO 3 8 56

 1 = (3 × 3) × V (litre) 8

V (litre) =

56  9

1000

V (ml) =

63



1 63

= 15.87  16 ml

23.. The stoich 23 stoichiome iometric tric coeff coefficie icient nt of S in the the following following reacti reaction on H2S + HNO3  NO + S + H 2O is balanced (in acidic medium): (1) 1

(2)

2

(3)

Sol. Answer (3) 2 H2 S nf

2



5 H N O3 nf

2

0

 NO  S  H2O

3

3H2S  2HNO3

3 S  H 2O  2NO  3S

coefficient of S = 3 [ Hint : n factor of oxidising agent will be coefficient of reducing agent and vice-versa] 24.. 12.5 24 12.53 3 ml ml of of 0.0 0.051 51 M SeO SeO2 reacted with 25.52 ml of 0.1 M CrSO 4 solution. In the reaction Cr 2+ was oxidised to Cr 3+. The oxidation state at which selenium was converted is (1) 3

(2)


+2

(3)

1

(4)

0


186 Redox Reactions


Sol. Answer (4) 4 S e O2

2 Cr SO 4



 4x

nf

nf

x

Se  Cr 3  Se

3–2 1

Number of meq. of SeO 2 = Number of meq. of CrSO 4 [0.051 × (4 – x)] × 12.53 = (0.1 × 1) × 25.52 4x

0 .1 .1 1 2 5. 5.5 2



0.051 0.051 12.53 12.53

4 – x = 3.99 x



0

25. A sam sampl ple e of of NaH NaHCO CO3 + Na2CO3 required 20 ml of HCl using phenolphthalein as indicator and 35 ml more required if methyl orange is used as indicator. Then molar ratio of NaHCO 3 to Na2CO3 is (1)) (1

2

1

(2)) (2

2

(3)) (3

3

3 4

1

(4)) (4

3

Sol. Answer (3) Let there are x1 meq. of NaHCO3 and x2 meq. of Na 2CO3. 1

HPh :

2

1

meq. of Na 2CO3 = meq. of HCl

x2 = 20 × N

2

x2 = 40 N

…(i)

MeOH : meq. of NaHCO 3 + meq. of Na2CO3 = meq. of HCl x1 + x2 = 55 N x1 = 15 N n1 ×1 = 15 N n 1 = 15 N x2 = 40 N n 2 × 2 = 40 N n 2 = 20 N n1: n 2 = 15 N : 20 N =3:4 26. A sam sampl ple e of of FeS FeSO O4 and FeC2O4 is dissolved in H 2SO4. The complete oxidation of sample required 8/3 eq. of KMnO 4 . After oxidation, the reaction mixture was reduced by Zn. On again oxidation by 5

KMnO4 required (1)) (1

3

eq. The mole ratio of FeSO4 and FeC2O4 is 7

3

(2)) (2

7

(3)) (3

3

5 7

7

(4)) (4

5

Sol. Answer (2) n1 mole

FeSO4 nf 1

n2 mo mole

 FeC 2O 4  KMnO 4  Fe

3

4  CO 2  Mn 2

nf  3



Redox Reactions 187


Number of eq. of FeSO 4 + Number of eq. of FeC 2O4 = number of eq. of KMnO 4 8

n1 ×1 + n 2 ×3 =

3

8

n1 + 3n2 = n1  n2



…(1)

3

5

…(2)

3

By solving (1) and (2), 7

n2



n1



6

and

7/6

n2



1/ 2

n2



1 2

7 3

27.. Th 27 The e equiv equivale alent nt mas mass s of Mn MnSO SO4 is half of its molecular mass when it is converted to (1) Mn2O3

(2)

MnO

(3)

MnO2

(4)

MnO4–

(4)

All of these

Sol. Answer (3) 2  MnO2 Mn 

n= 2  Eq. wt. =

M 2

28.. In the tit 28 titrat ration ion of NaH NaHCO CO3 with HCl, indicator cannot be used (1) Methyl orange

(2)

Methyl red

(3)

Phenolphthalein

Sol. Answer (3) Phenolphthalein cannot be used with weak base. 29.. Equiv 29 Equivalent alent weigh weightt of Mohr Mohr salt salt in the the titratio titration n with KMnO4 is (M – Molecular weight) M

(1)) (1

(2)) (2

1

M 4

M

(3)) (3

3

(4)) (4

M 2

Sol. Answer (1) Fe

2

 Fe 3

Equivalent weight 

M 1

30.. 100 ml of each HCl 30 HCl solution having having pH = 5 and NaOH NaOH having pH = 8 is mixture. mixture. How How much volume volume of N 100

NaOH is

(1) 0.009 ml

required to neutralise to 20 ml of this mixture? (2)

9 ml

(3)

20 ml

(4)

15 ml

Sol. Answer (1) Milliequivalent of HCl  10 5  100 = 10–3 Milliequivalent of NaOH = 10 –6 × 100 = 10 –4 Meq. H+ left unused = 9 × 10 –4 Aakash Educational Services Pvt. Ltd.


188 Redox Reactions





Nmix



9  10

9  10


4

 9  106

100 6

1

 10 10 

100

 VNaOH

 VNaOH = 0.009 ml

31.. The 31 The volu volume me of 0.1 0.1 M Ag AgNO NO3 which is required by 10 ml of 0.09 M K 2CrO4 to precipitate all the chromate as Ag2CrO4 is (1) 9 ml

(2)

18 ml

(3)

20 ml

(4)

36 ml

Sol. Answer (2) Mill equivalent Equation of AgNO3 = K 2CrO4 0.1 × V = 2 × 0.09 × 10 

V AgNO

 18 ml

3

32.. One litre 32 litre of a solut solution ion conta contains ins 15.12 15.12 g of HNO3 and one litre of another solution contains 3.2 g of NaOH. In what volume ratio must these solutions be mixed to obtain a neutral solution? 1

(1)) (1

2

(2)) (2

3

3

8

(3)) (3

3

3

(4)) (4

8

Sol. Answer (1) Eq. of HNO3 = Eq. of NaOH 1 5 .1 2

V 

63.5



V V

3 .2

 V

40

 0.333

33. Consider a titration of potassium dichromate solution solution with acidified Mohr’s Mohr’s salt solution solution using diphenylamine as [IIT-JEE 2007] indicator. The number of moles of Mohr’s salt required per mole of dichromate is (1) 3

(2)

4

(3)

5

(4)

6

Sol. Answer (4) 34.. The reaction 34 reaction of white phosphorus phosphorus with with aqueous aqueous NaOH gives phosphine phosphine along along with another another phosphorus phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product [IIT-JEE 2012] are respectively (1) Redox reaction; – 3 and – 5

(2)

Redox reaction; + 3 and + 5

(3) Disproportionation reaction; – 3 and + 5

(4)

Disproportionation reaction; – 3 and + 1

Sol. Answer (4) 35.. Hyd 35 Hydroge rogen n peroxi peroxide de in its reac reaction tion with KIO4 and NH2OH respectively, is acting as a [JEE(Advanced)-2014] (1) Reducing agent, oxidising agent

(2)

Reducing agent, reducing agent

(3) Oxidising agent, oxidising agent

(4)

Oxidising agent, reducing agent

Sol. Answer (1) H2O2 acting as a reducing agent with KIO 4 and H2O2 acting as an oxidising agent with NH 2OH. Aakash Educational Services Pvt. Ltd.


Redox Reactions 189

Solution of Assignment (Set-2) SECTION - B

Objective Type Questions (More than one options are correct) 1.

Rega Re gard rdin ing g th the e co comp mpou ound nd Cr CrO O5 which of the following statement is/are correct? (1)) Oxid (1 Oxidatio ation n number number of of Cr is (+6) (+6) (2)) 4-oxy (2 4-oxygen gen atoms atoms are are present present in form form of perox peroxide ide (3)) Oxid (3 Oxidatio ation n number number of of Cr is + 10 (4)) 2-oxy (4 2-oxygen gen atoms atoms are are present present in form form of perox peroxide ide

Sol. Answer (1, 2) Structure of CrO5 is

O 2-peroxide linkage

O

O Cr || O

O

 4-oxygen atoms are present in peroxide linkage and oxidation number is + 6.

2.

Which Whi ch of the the fol follow lowing ing can ac acts ts as as a red reduci ucing ng age agent? nt? (1) H2S

(2)

HNO3

(3)

FeSO4

(4)

SnCl2

(4)

HNO2

Sol. Answer (1, 3, 4)

3.

S2–  S + 2e–

(oxidation)

Fe2+  Fe3+ + e –

(oxidation)

Sn2+  Sn4+ + 2e –

(oxidation)

Which Whic h of the the follow following ing can can act act as a reducin reducing g as well as an an oxidis oxidising ing agent agent? ? (1) HNO3

(2)

H2SO4

(3)

H2O2

Sol. Answer (3, 4) Disproportionation Disproportion ation reaction H2O2  H 2O +

1 2

O2

In HNO2 also N is in +3 O.S. which can oxidize to +5 as well as can reduce to minimum –3 4.

Which Whi ch of the the follow following ing stat stateme ements nts are are correc correctt regard regarding ing this this equa equatio tion n Br 2  BrO 3– + Br – ? (1) Bromine is oxidised

(2)

Bromine is reduced

(3) It is an ex exam ampl ple e of of dis dispr prop opor orti tion onat atio ion n rea react ctio ion n

(4)

Brom Br omin ine e is is nei neith ther er ox oxid idis ised ed no norr red reduc uced ed

Sol. Answer (1, 2, 3) oxidation 0

Br2

 BrO3  Br  reduction

When the same substance undergoes oxidation and reduction simultaneously it is termed as disproportionate. Aakash Educational Services Pvt. Ltd.


190 Redox Reactions

5.


Which Whi ch of of the the follow following ing is/ is/are are dis disprop proporti ortiona onatio tion n react reactions ions? ? (1) 6N 6NaO aOH H + 3C 3Cll2  NaClO3 + 5NaCl + 3H2O (2) 2N 2NaO aOH H + Cl2  NaOCl + NaCl + H 2O (3) H2O2  H 2O +

1 2

O2

(4) 2KClO3  2KCl + 3O2 Sol. Answer (1, 2, 3) reduction

reduction

(1)) (1

Cl2

1

0

5



OH

(2)) (2

  ClO3  Cl

Cl2

 OCl  Cl oxidation

oxidation

oxidation 1

(3)) (3

H2O2

2

 H2 O 

1 2

0

O2

reduction

6.

Whic Wh ich h of th the e follo followi wing ng sta state teme ment nts s rega regardi rding ng H2SO5 is/are correct? (1)) The oxidati (1 oxidation on number number of sulphur sulphur is +6 (2)) Tw (2 Two o oxygen oxygen atoms atoms are present present in form of peroxide peroxide (3)) Three oxygen (3 oxygen atoms are present present in form of oxide oxide (4)) The oxi (4 oxidati dation on stat state e of sul sulphu phurr is + 8

Sol. Answer (1, 2, 3) Structure is

7.

Which Whi ch of the the fol follow lowing ing ca can n act act as an oxid oxidisi ising ng agen agent? t? (1) H2SO4

(2)

HNO3

(3)

KMnO4

(4)

K2Cr 2O7

(4)

CHCl3

Sol. Answer (1, 2, 3, 4) In all options oxidation number of central metal can decrease. 8.

The Th e oxi oxida dati tion on nu numb mber er of ca carb rbon on is zer zero o in (1) HCHO

(2)

CH2Cl2

(3)

C12H22O11

Sol. Answer (1, 2, 3) It is based on simple application of common oxidation states of H and oxygen.

HCHO 

net charge = 0

Similarly in CH2Cl2 and C12H22O11 (sugar) Aakash Educational Services Pvt. Ltd.


Redox Reactions 191


9.

150 ml

M 10

Ba(MnO4)2 in acidic medium can oxidise completely

(1) 15 150 0 ml ml 1M 1M F Fe e2+

(2)

50 ml 1M FeC2O4

(3)

100 ml 1M C2O42–

(4)

75 ml 1M K2Cr 2O7

(3)

1.79 M

(4)

3.035%

Sol. Answer (1, 2) Alternative solution Valency factors n for Ba(MnO4)2 = 10 Fe+2 = 1 FeC2O4 = 3 C2O4–2 = 2 K2Cr 2O7 = 6 Now on calculating Meq. Ba(MnO4)2 =

150 

1

 10  150

10

Meq of Fe+2 = 150 × 1 × 1 = 150 Meq of FeC 2O4 = 50 × 1 × 3 = 150 Meq. of C2O4–2 = 100 × 1 × 2 = 200 Meq of K2Cr 2O7 = 75 × 1 × 6 = 450 10.. The str 10 streng ength th of "20 vol volume ume"" H2O2 is equal to (1) 60 60.86 g/L

(2)

3.58 N

Sol. Answer (1, 2, 3) 20 volume H2O2 1NH2O2 = 5.6 volume 1

1 volume H2O2 =

5.6

NH2O2

20

20 volume H2O2 =

5.6

NH2O2 = 3.57 N

Strength in g/L = NE = 3.57 × 17 = 60.71 g/L N = M × n factor 3.57 = M × 2 M = 1.786  1.79 M 1, 2, 3 are correct options. volume strength = 5.6 × N 

N



20

 3 .5 7

5.6

Volume strength = 11.2 × M 

M



20

 1.78

11.2

S = N × E = 3.58 × 17 = 60.86 g/L Aakash Educational Services Pvt. Ltd.


192 Redox Reactions


11. Wh When en 1 mol mole e of of KMn KMnO O4 is reacted with FeC2O4 in acidic medium, then the reacted amount of FeC 2O4 is 5

(1)) (1

mo m oles

3

(2)

5 equivalents

(3)

3 5

mo moles

(4)

3 equivalents

Sol. Answer (1, 2) 7



K Mn O4 nf  7  2

FeC2O 4  Fe3  CO2  Mn Mn2 nf  (3  2) 2)  2  ( 4  3) 3)  1 2  3

Number of eq. of KMnO 4 = Number of eq. of FeC 2O4 nKMnO

 nf  nFeC

2 O4

4

1×5=

nFeC FeC2O4

nFeC FeC2O4





 nf

5 3

5 3

Number of eq. of of FeC2O4 = 1 × 5 = 5 equivalents 12.. For the followi 12 following ng balance balanced d redox reacti reaction, on, 2M 2MnO–4 + 4H+ + Br 2 weight of MnO–4 and Br 2 are x & y respectively then (1)) Equ (1 Equiva ivalen lentt weigh weightt of MnO4– is

x

2Mn 2M n2+ + 2BrO–3 + 2H2O. If the molecular y

(2)

Equi Eq uiva vale lent nt we weig igh ht of of Br Br 2 is

(4)

n-fa nfact ctor or ra rati tio o of Mn MnO O4– and Br 2 is 2 : 1

(1)) n(1 n-fa fact ctor or fo forr FeS FeS2 i is s 11

(2)

The ratio of moles of a : b is 4 : 11

(3) The ratio of moles of a : b is 11 : 4

(4)

The ratio of moles of c : d is 1 : 4

(3)) Equ (3 Equiva ivalent lent wei weight ght of Br 2 is

5

y 10

5

Sol. Answer (1, 3) 7 

 2MnO4  8H



  B r2   

nf = 1 × (7 – 2) =5 E



MnO 4

EBr  2

5

0

2Mn

2

 2BrO3  2H2O

nf = 2 × (5 – 0) = 10

x



5

y 10

n factor of

 MnO4

and Br 2

= 5 : 10 =1:2 Option 1, 3 are correct. 13.. Co 13 Cons nside iderr the the re reac actio tion, n, aFeS2 + bO 2  cFe2O3 + dSO2 Which is correct for the above reaction?



Redox Reactions 193


Sol. Answer (1, 2, 4) 2 1 a Fe S 2  bO 2

3

4

  c Fe 2 O 3  d S O 2

n factor of FeS2 =

1 (3 ( 3  2)  2  [ 4  (  1) 1)]     

    

Fe

S

= 1 + 10 = 11 n factor of O 2 = 2 × 2 = 4 For balancing n factors are interchanged 4FeS2

a=4

 11O 2  2Fe 2O 3  8SO 2

b = 11

c= 2

d=8

a : b = 4 : 11 c:d=2:8=1:4 So option 1, 2, 4 are correct. 14. When Na2S2O3 is reacted with I 2 to form Na 2S4O6 and NaI then which statement is correct? (1)) n(1 n-fa fact ctor or fo forr Na Na2S2O3 is one (2)) n(2 n-fa fact ctor or fo forr I2 is two (3) 2 mol moles es of Na2S2O3 is reacted with one mole of I 2 (4)) n(4 n-fa fact ctor or fo forr Na Na2S4O6 is one Sol. Answer (1, 2, 3) 2 2 Na2S 2O 3

0



I2

Na 2

2.5 S4 O6

1

 2N aI

nf = 2 × (2.5 – 2) nf = 2 × 1 = 2 × 0.5

=2

=1 n factor of Na2S4O6 = 4(2.5 – 2) = 4 × 0.5 = 2 Hence 1, 2, 3 are correct. 15. Choo Choose se the the correct correct statem statement ent regardin regarding g following following react reaction ion HNO2  HNO3 + NO  (1)) It is an (1 an example example of dispro disproportion portionation ation reacti reaction on (2)) Equ (2 Equiva ivalen lentt weight weight of of HNO3 =

3M 2

(3)) Equ (3 Equiva ivalen lentt weight weight of of HNO3 = M (4)) It is an examp (4 example le of intramole intramolecular cular redox redox reaction reaction Sol. Answer (1, 2) i

HNO2

H NO 2   HN  HNO 3  N O

This is an example of disproportionation reaction. Aakash Educational Services Pvt. Ltd.


194 Redox Reactions


16.. 31.26 16 31.26 ml ml of a 0.165 0.165 M soluti solution on of of Ca(OH) Ca(OH)2 is required to just neutralise 25 ml of citric acid H 3C6H5O7. Then correct regarding this is/are (1) n-factor of citric acid is 3

(2)

Molarity of citric acid is 0.138 M

(3) Molarity of citric acid is 0.029 M

(4)

n-factor of citric acid is 2

Sol. Answer (1, 2) Equivalent of acid = equivalents of base 3 1 .2 6

25

 2  0 .1 6 5 

1 000

N ⇒ N 

3 1 .2 2 6

1000

 2  0 .1 6 5 

0 .4 1 2

25

17.. With 17 With 4 mole mole of KI KI one one mole mole of of Cl2 is treated to yield a solution, which is then treated with hypo solution, then correct regarding this is/are (1)) Equ (1 Equiva ivalen lentt weigh weightt of I2 is

M 1

(2)) To react completely (2 completely with with a solution, 1 mole mole of hypo solution solution is required required (3)) n-fa (3 n-facto ctorr of hypo hypo solu solution tion is 1 (4) nn-fa fact ctor or of of Cl2 is 2 Sol. Answer (3, 4) 2KI  Cl2 I2

 K I

 2KCl  I2

    

...(i)

KI3

...(ii)

2Na2S2O3 + I 2  Na2S4O6 + 2NaI 2NaI ... ...(iii (iii)) As per above reactions n-factor of Cl 2 (as per reaction (i), I 2 (as per reaction (iii) and Na 2S2O3 (as per reaction (iii)) is 2, 2 and 1 respectively. 18.. Cho 18 Choose ose the the corre correct ct regar regardin ding g indicat indicator or Titration

Indicator

(1)) Na (1 NaOH OH vs vs.. CH CH3COO OOH H

Phen Ph enol olph phth thal alei ein n

(2) KMnO4 vs. FeC2O4

KMnO4

(3) I2 vs. Na2S2O3

Starch

(4) K2Cr 2O7 vs. FeSO4

K3[Fe(CN)6] as external indicator

Sol. Answer (1, 2, 3, 4) KMnO4 acts as self indicator. 19. Reduct Reduction ion of the metal metal centre centre in aqueous permang permanganate anate ion involves involves (1) 3 electrons in neutral medium

(2)

5 electrons in neutral medium

(3) 3 electrons in alkaline medium

(4)

5 electrons in acidic medium

[IIT-JEE 2011]

Sol. Answer (1, 4) aq  MnO2 MnO  4 MnO

– 4

acidic

  Mn2

Therefore in aqueous and in acidic mediums 3 and 5 electrons will transfer respectively. Aakash Educational Services Pvt. Ltd.


Redox Reactions 195


20.. Fo 20 Forr th the e re reac acti tion on

[JEE(Advanced)-2014]

I– + ClO3– + H2SO4  Cl – + HSO4– + I2 The correct statements(s) in the balanced equation is/are (1)) St (1 Stoic oichiom hiometri etric c coeffici coefficient ent of HSO4– i is s6

(2)

Iodide is oxidized

(3) Su Sulphur is reduced

(4)

H2O is one of the products

Sol. Answer (1, 2, 4) Balanced chemical equation is 

6I

 ClO3  6H2SO 4  

Cl



 6HSO 4  3I 2  3H 2O

Here, H2O is produced and I – is oxidized. SECTION - C

Linked Comprehension Type Questions Comprehension-I Redox is a reaction in which both oxidation and reduction will take place simultaneously. It is obvious that if one substance gives electron there must be another substance to accept these electrons. In some reactions same substance is reduced as well as oxidised, these reactions are termed as disproportionation reactions. For calculating equivalent mass in redox reaction change in oxidation number is related to n-factor which is reciprocal of molar ratio. 1.

This Th is re reac acti tion on is an ex exam ampl ple e of Br 2 + OH–  BrO 3– + H 2O + Br – (1) Oxidation reaction only

(2)

Reduction reaction only

(3) Neutralization reaction

(4)

Disproportionation reaction

Sol. Answer (4) reduction

Br2



OH

  BrO3  H2O  Br  oxidation

∵

Br 2 is oxidized and reduced both.

 It is disproportionation reaction.

2.

When Wh en P reac reacts ts wit with h NaOH NaOH,, the the pr prod oduc ucts ts ar are e PH3 and NaH2PO2, which of the following statement is correct? (1) P is oxidised only

(2)

P is reduced only

(3) P is oxidised as well as reduced

(4)

P is neither reduced nor oxidised

Sol. Answer (3) oxidation



OH

3

 PH3 P 

 H2PO2

reduction



196 Redox Reactions

3.


How many mole les s of KM KMnO4 are reacted with one mole of ferrous oxalate oxalat e in acidic medium? 2

(1)) (1

1

(2)) (2

5

3

(3)) (3

5

5

(4)) (4

5

3

Sol. Answer (3) MnO4– + FeC2O4  Mn 2+ + Fe3+ + CO2 3 × MnO4– + 5e – + 8H+  4Mn 2+ + H2O 5 × FeC2O4  Fe 3+ + 2CO2 + 3e– ____ __ _____ _____ ____ ____ _____ _____ ____ ____ ____ _____ _____ ____ ____ _____ _____ ____ ____ _____ _____ ____ ____ _____ ____ _ 3MnO4– + 5FeC2O4 + 24H+  3Mn2+ + 5Fe3+ + 12H2O + 10CO2 ∵

5 moles of FeC2O4 reacts with 3 moles of KMnO 4 3

 1 mole of FeC2O4 reacts with

4.

5

mole KMnO4

The Th e eq equi uiva vale lent nt we weig ight ht of Cu2S in the following reaction is Cu2S + O2  Cu +2 + SO3 (1)) (1

M. wt

M. wt

(2)) (2

1

M. wt

(3)) (3

10

8

(4)) (4

M. wt 11

Sol. Answer (2) 1 2 Cu2 S  O 2

 2Cu

2

6  1 0e  SO 3  10

 n-factor = 10  Equivalent mass =

5.

M 10

Which Whi ch of the the foll followi owing ng is is an an examp example le of of redox redox rea reacti ction? on? (1) 2NO2  N2O4

(2)

NH4OH  NH4+ + OH–

(3) 2NO2 + H2O  HNO3 + HNO2

(4)

N2O5 + H2O  2HNO3

Sol. Answer (3) NO2 + H2O  HNO3 + HNO2 This is an example of disproportionation reaction.  It is a redox reaction.

6.

NO2

NH2 –

+ ne (i)

(ii)

For converting one mole of nitrobenzene to aniline, how many moles of electrons are transferred? (1) 2

(2)

3

(3)

6

(4)

8

Sol. Answer (3) O

O N (+3) + 6e


H

H N (–3)

–


Redox Reactions 197


Comprehension-II The strength of H 2O2 is expressed in many ways like molarity, normality, % strength and volume strengths. But out of all these form of strengths, volume strength has great significance for chemical reactions. The decomposition of H 2O2 is shown as under : H2O2(l )  H2O(l ) +

1 2

O2(g).

'x' volume strength of H 2O2 means one volume (litre or ml) of H 2O2 releases x volume (litre or ml) of O 2 at NTP. 1 litre H2O2 release x litre of O 2 at NTP =

x 22.4

moles of O 2

From the equation, 1 mole of O2 produces from 2 moles of H 2O2. x 22.4

=

moles of O2 produces from x 11.2

2

x 22.4

moles of H 2O2

moles of H 2O2 x

So, molarity of H 2O2 =

11.2

x



1

11.2

M

Normal Nor mality ity = n-facto n-factorr × molarity molarity =

1.

2

x



1 1 .2

x

5 .6

N

What Wh at is is the the perc percen enta tage ge stre streng ngth th of of “15 “15 volu volume me”” H 2O2? (1) 6.086%

(2)

4.55%

(3)

3.03%

(4)

1.5%

Sol. Answer (2) 1 volume = 0.303% 15 volume = 15× 0.303 = 4.55% 2.

30 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 ml of “28 volume” strength of H2O2 in acidic medium then what will be the percentage purity of Ba(MnO 4)2 in the sample? (Ba = 137, Mn = 55, O = 16) (1) 10%

(2)

40%

(3)

62.5%

(4)

80%

Sol. Answer (3) NH O  2

2

Volu Volume me stre streng ngth th 5.6



28 5.6

5



198 Redox Reactions


Eq. of H2O2 = Eq. of Ba(MnO 4)2 100

5

W



100 0

 10

375

 WBa (MnO4)2 = 18.75 g  % purity =

3.

18.7 18.75 5  100 100

 62.5

30

What volume of H2O2 solution of “11.2 volume” strength is required to liberate 2240 ml of O 2 at NTP? (1) 300 ml

(2)

500 ml

(3)

100 ml

(4)

200 ml

Sol. Answer (4) 11.2 volume H 2O2 means 1 vol H2O2 gives 11.2 vol O2  11.2 ml O2 obtained from 1 ml H 2O2  2240 ml O2 obtained from =

2240  1 11.2

= 200 ml

Comprehension-III 2 g of brass containing Cu and Zn only reacts with 3 M HNO 3 solution. Following are the reactions taking place Cu(s) + HNO3 (aq)  Cu2+ (aq) + NO2(g) + H2O(l) Zn(s) + H+(aq) + NO3– (aq)  NH4+ (aq) + Zn2+ (aq) + H2O(l) The liberated NO2(g) was found to be 1.04 L at 25°C and 1 atm [Cu = 63.5, Zn = 65.4] 1.

The Th e per perce cent ntag age e by by mas mass s of of Cu Cu in in bras brass s was was (1) 67%

(2)

70%

(3)

80%

(4)

90%

(4)

10.5 ml

(4)

0.358 g

Sol. Answer (1) Cu  4H 4HNO3  Cu(NO3 )2  2NO2  2H2O nNO



2

PV



RT

1  1 .0 4 0.0821 2 98 98

 0 .0 4 2

 nCu = 0.021  WCu = 1.335 g 

2.

% Cu



1.33 1.335 5  100 100

 66.67 %

2

The volume of HNO3 consumed during the reaction with brass is (1) 9. 9.52 ml

(2)

14.5 ml

(3)

16.25 ml

Sol. Answer (2) 3.

How many gra ram ms of NH 4NO3 will be obtained in the above reaction? (1) 0. 0 .405 g

(2)

0.0428 g

(3)

0.2018 g

Sol. Answer (3) [H]

  NH3 NO3  NH3

 H  NH4



Redox Reactions 199

Solution of Assignment (Set-2) SECTION - D

Assertion-Reason Type Questions 1.

STATEMENT-1 :

acidic medium

      Mn KMnO 4 

2

, n factor of KMnO 4 is 5.

and STATEMENT-2 : Equivalent mass of KMnO4 in acidic medium is

M 5

(M = molecular mass of KMnO 4).

Sol. Answer (1) Equivalent Equivalent mass 

Molar mass n  fact factor or

MnO4– + 5e– + 8H+  Mn 2+ + 4H2O  n-factor = 5 M

and equivalent mass =

2.

5

STATEMENT-1 : H2O2  H2O +

1 2

O2. This is an example of disproportionation reaction.

and STATEMENT-2 : H2O2 can act as a oxidising as well as reducing agent. Sol. Answer (1) oxidation –1

H — O— O— H

–2

0

O

H

H

+ O2

reduction ∵

H2O2 is oxidized as wall as reduced.

 Its an example of disproportionation reaction.

3.

STA ST ATE TEME MENTNT-1 1 : Oxida Oxidatio tion n numb number er of of oxyg oxygen en in OF OF2 compound is +2. and STATEMENT-2 : An element has a fixed oxidation state.

Sol. Answer (3) In OF2, x – 2 = 0 x = +2 and an element can have variable oxidation state. 4.

STA ST ATEME TEMENT-1 NT-1 : Oxid Oxidatio ation n state state of of carbon carbon in in its com compoun pound d is alway always s +4. and STATEMENT-2 : An element can show variable oxidation numbers.

Sol. Answer (4) Carbon can show –4 and many more oxidation states also e.g ., ., CH4, CO, CH2Cl2 etc. Aakash Educational Services Pvt. Ltd.


200 Redox Reactions

5.


STA ST ATEM TEMEN ENT-1 T-1 : Equ Equiva ivale lent nt ma mass ss of KM KMnO nO4 in different mediums are different. and STATEMENT-2 : KMnO4 can act as a oxidising agent.

Sol. Answer (2) In acidic medium KMnO 4 is reduced to Mn2+.  It is an oxidising agent.

6.

STA ST ATE TEME MENT NT-1 -1 : In In the the re reac acti tion on : Cl Cl2 + OH–  Cl – + ClO4– chlorine is oxidised only. only. and STATEMENT-2 : Oxidation and reduction cannot take place alone.

Sol. Answer (4) 0



Cl2  OH OH

1

7

  Cl Cl O 4

In above reaction, Cl 2 is oxidized as well as reduced both because for redox reaction both oxidation and reduction reactions are required. M

7.

STA ST ATEM TEMENT ENT-1 -1 : The eq equiv uivale alent nt mas mass s of KMn KMnO O4 in acidic medium is

5

where M = Molecular mass of KMnO 4.

and STATEMENT-2 : Equivalent mass is equal to product of molecular mass and change in oxidation number. Sol. Answer (3) 7 acidic K Mn O 4   Mn2 medium

nf = 7 – 2 = 5 E



M



nf

M 5

Equivalent Equivalent mass 

8.

Molecular Molecular mass Total change change in oxidation oxidation number

STA ST ATE TEME MENT NT-1 -1 : For For the the rea react ctio ion n NaOH NaOH + H2CO3  NaHCO3 + H 2O equivalent weight of H2CO3 is 62. and STATEMENT-2 : n factor of H2CO3 is 1 (in above reaction) and equivalent mass =

Molecular mass n factor

.

Sol. Answer (1) NaOH +H2CO3  NaHCO3 + H 2O Since only one H atom is being replaced from H 2CO3 in above reaction hence n factor of H2CO3 = 1 E



M 1



2  1  12 12  3  16 16 1

=

62 1

= 1 equivalent mass 


Molecular mass n factor


Redox Reactions 201

Solution of Assignment (Set-2) SECTION - E

Matrix-Match Type Questions 1.

Matc Ma tch h Colu Colum mnn-II with with Col Colum umnn-II II Column-I

Column-II

(A) CaOCl2

(p) + 6, + 6

(Oxidation state of Cl) (B) S2O32–

(q) + 1, –1

(Oxidation state of S) (C) NH4NO3

(r)

Pero Pe roxy xy li link nkag age e is pr pres esen entt

(Oxidation state of N) (D) H2SO5 and H2S2O8

(s) –3, + 5

(Oxidation state of S) (t)

–2, + 6

Sol. Answer A(q), B(t), C(s), D(p, r) (A)) Str (A Struct ucture ure of CaOC CaOCll2 is

(B) S2O32– , 2x – 6 = –2 x = +2 (C) NH4NO3 may written as NH4+, 

oxidation no. of N = – 3 and in NO 3–, oxidation no. of N = +5

(D)) H SO (D 2

5

⇒

O || HO — S — O — O — H || O peroxide linkage

and oxidation no. = +6 For H2S2O8, structure is

and oxidation no. = +6 Aakash Educational Services Pvt. Ltd.


202 Redox Reactions

2.


Match the foll llo owin ing g Column-I (A)

XI2

Column-II

 YNO3   IO3  

(B)

X ClO



(p) X > Y

NO

2 Acidic med

 YCr O2  Cl 





CrO 4

(q) X < Y

 alkaline 

(C)) (C

 YBrO3  Br  

X N2O4



(r)) (r

NO3

n-fact n-f actor or (oxid (oxidant ant)) > n-fac n-factor tor (Red (Reduct uctant ant))

 Acidic 

(D)) (D

3

XAsO3

 YMnO4 

3

AsO4

 MnO2

(s)) n-f (s n-fact actor or (Reduc (Reductan tant) t) > n-facto n-factorr (Oxidan (Oxidant) t)

 Acidic 

(t)) (t

Oxidant Oxida nt has hig highes hestt O.N. O.N. am amon ong g the the oxidants involved in the reactions

Sol. Answer A(q, s), B(p, s), C(p, r), D(p, r, t)

3.

 10NO3  8H   2IO3  10NO2  4H2O

(A)

I2

(B)

3ClO

(C)) (C

3N2O4

(D)) (D

3 AsO3



 2CrO2  2OH   3Cl   2CrO4   H2O  BrO3  3H2O   6NO3  Br   6H

3

 2MnO4   3 AsO43  2MnO2  2H2 O

Match the foll llo owin ing g Column-I

Column-II (n-factor)

(A) KMnO4 i in n acidic medium

(p) 10

(B) Ba Ba(M (MnO nO4)2 i in n acidic medium

(q) 6

(C) S2O32– i in n alkaline medium

(r)

(D) K2Cr 2O7 i in n acidic medium

(s) 8

5

Sol. Answer A(r); B(p); C(s); D(q)

4.

(A) KMnO4  Mn 2+

nf = 5

(B) Ba Ba(M (Mn nO4)2  Mn2+

nf = 10

2

(C)) (C

S 2O 5

(D)) (D

Cr 2 O 7



2



2

SO 4

3

Cr

nf = 8 nf = 6

Match the foll llo owin ing g Column-I

Column-II (n-factor of reactants)

(A) S2O32– i in n different medium

(p) 8

(B) KMnO4 i in n different medium

(q) 5

(C) FeC2O4 is converted into Fe3+ and CO2

(r)

(D) Cu2S is converted into Cu 2+ and SO2

(s) 1


3


Redox Reactions 203


Sol. Answer A (p, s); B (q, r, s); C (r); D (p) Acidic medium (nf = 1)

(A)

S2O3

2–

Basic medium (nf = 8) Acidic medium (nf = 5)

(B)

2–

S4O6

KMnO4

SO4

2–

2+

Mn

+4

Neutral medium (nf = 3)

MnO2

Basic medium (nf = 1)

MnO4

+6 2–

(C)) (C

So, total 3 e– are loosed by one molecule of FeC 2O4. Hence, n-factor for FeC 2O4 is 3.

(D)) (D

So, total 8 e – are loosed by one molecule of Cu 2S. Hence, n-factor for Cu 2S is 8. 5.

3.48 g of MnO2 is added to 500 ml of 0.1 M oxalic acid solution. The resulting solution is then titrated against either 0.02 M KMnO4 or 0.02 M K2Cr 2O7 solution. Column I

Column II

(A) KMnO4 s so olution

(p) N = 0.1

(B) MnO2

(q) nfactor = 2

(C) Oxalic acid

(r)

(D) K2Cr 2O7 s so olution

(s) Vo Vol = 200 ml (t)

Vol = 167 ml nfactor = 6

Sol. Answer A(p, s), B(q), C(q), D(r, t)  Mn

MnO2 2e



2

 2e



 H2C 2O 4   CO 2  H 2O

 nf (MnO2 and oxalic acid) = 2

Equivalent of

MnO2



3.48

 2  0.08

87

Equivalent of oxalic acid = 0.1 Equivalents of oxalic acid left = 0.02 Equivalents of oxalic acid = Equivalents of KMnO 4 0.02 = 0.02 × 5 × V V KMnO

4



1 5

litre = 200 ml.

Volume of K2Cr 2O7 required = 1/6 L = 167 ml Aakash Educational Services Pvt. Ltd.


204 Redox Reactions

6.


I with nature of the reaction/type of the products in Column II. II . Match th the re reactio ion ns in in Column I with [IIT-JEE 2007] Column I

Column II

 O 2  O 22

(A)

O2

(B)

CrO4

(C)) (C

MnO4

(D)) (D

NO3

2





(p) Re Redo dox x re reac acti tion on

 H 

(q)) One of the (q the produc products ts has trigo trigonal nal plana planarr structur structure e

 NO2  H 

(r)

 H2SO4  Fe 2 

Dimeri Dim eric c bridg bridged ed tetra tetrahe hedra drall metal metal ion ion

(s)) Di (s Disp spro ropo port rtio iona nati tion on

Sol. Answer A(p, s), B(r), C(p, q), D(p) SECTION - F

Integer Answer Type Questions 1.

0.144 g of pure FeC2O4 was dissolved in dilute H 2SO4 and the solution was diluted to 100 ml. What volume in ml of 0.1 M KMnO 4 will be needed to oxidise FeC 2O4 solution?

Sol. Answer (6) meq meq. of FeC FeC 2O 4  meq meq of KMnO KMnO 4 W

 1000 1000  N.V. .V.

E 0.144

Or,

 1000  0.1 5  V

144 144 / 3

Or V = 6 ml 2.

2.48 g of Na2S2O3. xH2O is dissolved per litre solution. 20 ml of this solution required 10 ml 0.01 M iodine solution. What is value of x?

Sol. Answer (5) Meq. of Na2S2O3 × H2O = meq of l 2 2.48

 20  10  0.01 2

M

 M = 248  248 = 158 + 18x  x=5

3.

1.245 g of CuSO4.xH2O was dissolved in water and H 2S was passed into it till CuS was completely precipitated. The H2SO4 produced in the filtrate required 10 ml of N-NaOH solution for complete neutralisation. What is value of x?

Sol. Answer (5) H S

2 CuSO4 .xH2O   CuS

 H2SO 4

H2SO4  NaOH Aakash Educational Services Pvt. Ltd.


Redox Reactions 205


Meq. of CuSO4 xH2O = meq. of H 2SO4 = meq. of NaOH W

 1000 1000  N.V. .V.

E 1 .2 .2 45 45  2

Or

4.

159. 159.5 5  18x

 10  1

or x



5

In th the e re rea acti tio on VO + Fe2O3  FeO + V2O5, what is the n-factor for V 2O5 ?

Sol. Answer (6) 2

VO



Fe 2 O 3

  FeO 

5

V2O 5

    

6e  involve involved d per mole mole

5.

, nF (V2O5) = 6

 x   AO3  Mn    I n th th e re re ac ti on A  MnO4 

A

x

1 2

 O2 , if one mole of MnO oxidises 1.67 moles of 4

  to AO3 , then what will be the value of x?

Sol. Answer (2) 

MnO4

 5e   Mn 

These 5 moles of e – are lost by A+x ion & gained by

Θ

MnO4

.

 1 mole A+x will lose e– = 5/1.67 = 3 moles (approx.)

Now

A

x

 AO3

 Oxidation state of A,

x = +5 – 3 = +2 6.

A 0.1 mol mol of a metal metal is burnt burnt in air air to form form oxide. oxide. The The same same oxide oxide is then then reduced reduced by by 0.05 0.05 M, 4 litre litre

2

S 2 O3

(acidic medium) to +3 oxidation state of metal. What is the oxidation state of metal in oxide ? Sol. Answer (5) M1

 O2  

x

M

0.1mol

0.1mol

0.1 × n = 0.05 × 1 × 4 n= 2  x=+5

7.

100 ml of

N a2 S2O 3 solution

is divided into two equal parts A and B. A part requires 12.5 ml of 0.2 M I 2 solution

(acidic medium) and part B is diluted x times and 50 ml of diluted solution requires 5 ml of 0.8 M I 2 solution in basic medium. What is value of x? Sol. Answer (5) Part A 1 × M × 50 = 2 × 12.5 × 0.2 Aakash Educational Services Pvt. Ltd.

…(1) Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

206 Redox Reactions


Part B 8 × M 1 × 50 = 2 × 5 × 0.8

…(2)

Dividing (1) by (2) we get M



8 M1

12.5  2

M  5M1

58

 M is 5 times diluted.

8.

Forr a give Fo given n redu reduct ctan ant, t, rat ratio io of of volu volume mes s of 0. 0.2 2 M KMnO KMnO 4 and 1 M K2Cr 2O7 in acidic medium will be

Sol. Answer (6) V1M1n1(KMnO4) = V2M2n2 (K2Cr 2O7) V1 V2

9.



M2n 2



M1n1

1 6 0.2  5

6

The diffe differenc rence e in the the oxidat oxidation ion numb numbers ers of of the two two types types of of sulphu sulphurr atoms atoms in Na 2S4O6 is

[IIT-JEE 2011]

Sol. Answer (5) Structure of S 4O62– is

 Difference in oxidation state is +5 – 0 = +5.

10.. Reac 10 Reacti tion on of Br Br 2 with Na2CO3 in aqueous solution gives sodium bromide and sodium bromate with evolution of CO 2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is [IIT-JEE 2011] Sol. Answer (5)

Br2  2Br –  5 –    5 Br2  2Br  10 e  2 e  Br 2 + 5Br 2  10NaBr + 2NaBrO3 + 6CO2 + 6Na2CO3 3Br 2 + 3Na2CO3  5NaBr + NaBrO3 + 3CO2 11.

Conside Con siderr the foll followin owing g list list of reag reagent ents s:

[JEE(Advanced)-2014]

Acidified K2Cr 2O7, alkaline KMnO4, CuSO4, H 2O2, Cl2, O3, FeCl3, HNO3 and Na2S2O3. The total number of reagents that can oxidise aqueous iodide to iodine is Sol. Answer (7) K2Cr 2O7 + I– + H+  I 2 CuSO4 + I–  I 2 H2O2 + I–  I 2 Cl2 + I–  I 2 O3 + I–  I 2 FeCl3 + I–  I 2 HNO3 + I–  I 2 Aakash Educational Services Pvt. Ltd.


Redox Reactions 207

Solution of Assignment (Set-2) SECTION - G

Multiple True-False Type Questions 1.

STA ST ATEM EME ENT-1 : KM KMnO4 acts as a powerful oxidising agent in acidic, alkaline and neutral medium. STATEMENT-2 : Equivalent weight of KMnO 4 in acidic medium is M/5 and in strongly alkaline medium is M/3. STATEMENT-3 : KMnO4 solution is used as a primary standard and acts as self indicator during its titration with Mohr salt. (1) T F T

(2)

F T T

(3)

F F T

(4)

F F F

Sol. Answer (1) 

 Mn MnO4  or,



MnO4



(Acidic)

 8H  5e   Mn   4H2O



2

 MnO4 MnO4  2.

(Alkaline)

STA ST ATEMENT TEMENT-1 -1 : In dispropor disproportionat tionation ion reaction, reaction, 50% 50% of the the substance substance is is oxidised oxidised and remainin remaining g 50% is reduce reduced. d. STATEMENT-2 : Decomposition of H2O2 is not a disproportionation reaction. STATEMENT-3 : Both HNO2 & H3PO3 can undergo disproportionation reaction. (1) T T T

(2)

F F T

(3)

F F F

(4)

T F T

Sol. Answer (2) 2H2O2  H2O + O 2  1

 1

H2 O2  H 2 O2

or,

3.

|

|

2 

  H2 O 



0 

O2



STATE STA TEME MENT NT-1 -1 : In the the re reac actio tion n Zn(s Zn(s)) + C Cu u2+(aq)  Zn2+(aq) + Cu(s). Cu 2+ ions act as oxidising agent and Zn atoms act as a reducing agent. STATEMENT-2 : Every redox reaction cannot be splitted into two reactions one being oxidation and the other being reduction. STATEMENT-3 : The oxidation numbers are artificial and are useful as a book keeping device of electrons in reactions. (1) T T T

(2)

F F T

(3)

F F F

(4)

T F T

Sol. Answer (4) 4.

STATE STA TEME MENT NT-1 -1 : N/1 N/10, 0, 10 100 0 ml ml KMnO KMnO4 solution is sufficient to oxidise M/10, 50 ml FeC 2O4 solution in acidic medium. STATEMENT-2 : The left solution of statement-1 is sufficient to react with 8.33 ml of M/10 K 2Cr 2O7 solution in acidic medium. STATEMENT-3 : 1.06 g Na2CO3 will require 100 ml 0.1 M HCl solution with phenolphthalein. (1) FTT

(2)


FTF

(3)

TFT

(4)

TTF


208 Redox Reactions


Sol. Answer (1) 1 10

1

 100  10 milli equivalent of KMnO 4,

10

 50  3  15 milli equivalent of FeC2O4

The left over FeC 2O4 = 5 milli equivalent  K2Cr 2O7 required = 8.33 ml

M 10

 nfactor  6 

With phenolphthalein nfactor for Na2CO3 = 1 5.

To ox oxidis ise e M/10 Fe FeSO4 (100 ml solution) STATEMENT-1 : Volume required for 0.1 M solution is in order

VKMnO

4

STATEMENT-2 : The number of equivalents required will be in order STATEMENT-3 : The nfactor is in order (1) T T F

(2)

nH

2

O2



VK

H2O2

2

Cr2O7



VH

2

O2

.

 KMnO4  K 2 Cr2 O7 .

 nKMnO  nK Cr O . 4

2

TFT

2

7

(3)

FFT

(4)

FTF

Sol. Answer (3) Volume required  1/n Number of equivalents are same for all. Number of equivalents of all reactants used must be equal. SECTION - H

Aakash Challengers Questions 1.

During Durin g the the ox oxida idatio tion n of of arse arsenit nite e ion, ion, As AsO O 3–3 to arsenate AsO43– in alkaline medium, the numbers of moles of hydroxide ions consumed per mole of arsenite ion are (1) 2

(2)

3

(3)

2/3

(4)

3/2

Sol. Answer (1) 2.

When coppe copperr is treated treated with a certain certain conce concentrat ntration ion of nitric nitric acid, acid, nitric nitric oxide oxide and nitrogen nitrogen dioxid dioxide e are liberate liberated d in equal volumes according to the equation xCu + yHNO 3  Cu(NO3)2 + NO + NO2 + H 2O The coefficients x and y are (1) 2 and 6

(2)

4 and 12

(3)

1 and 3

(4)

3 and 8

Sol. Answer (2) 4Cu  12 12HNO3  4Cu(NO3 )2  2NO 2  2NO  6H2O 3.

The Th e coeff ffic icie ient nts s of I , IO IO3 and H in the following reaction,

I

 IO I2  H2O in a balanced state IO3  H 

would be respectively (1) 5,1,6

(2)

1,5,6

(3)

6,1,5

(4)

5,6,1

Sol. Answer (1) 6H+ + 5I– + IO3–  3I 2 + 3H2 Aakash Educational Services Pvt. Ltd.


Redox Reactions 209


4.

Which Whi ch of the the follow following ing comp compoun ounds ds does does not not decolo decolouris urise e acidifi acidified ed KMnO KMnO4 solution? (1) FeCl3

(2)

FeSO4

(3)

SO2

(4)

H2O2

Sol. Answer (1) Fe+3 cannot be oxidized by KMnO 4. 5.

Number of of mo mole les s of of KM KMnO 4 that will be needed to completely react with 2 moles of ferrous oxalate in acidic solution. (1) 6/5

(2)

3/5 3/

(3)

2/5 2/

(4)

1/3 1/

(4)

–2, +1 and –2

(4)

Cr2 O3

Sol. Answer (1) n factor of FeC2O4 is 3. 6.

The Th e oxi oxida dati tion on st stat ate e of of sul sulph phur ur in S 8, S 2F2 and H2S respectively are (1) 0, +1 and –2

(2)

0, +2 and –2

(3)

+2, +1 and –1

Sol. Answer (1) S8 (Zero) S2F2(+1) H2S(–2) 7.

Cr(OH)3  ClO  OH  ......  Cl  H2 O . The missing ion is 2 Cr2 O7

(1)) (1

(2)

Cr3+

(3)) (3

CrO 24 

Sol. Answer (3) Cr +3 is oxidized to Cr +6 and in alkaline medium Cr +6 exists as CrO 4–2. 8.

The chrom chromate ate ion ion present present in water water samp sample le is reduce reduced d to insolub insoluble le chromiu chromium m hydroxi hydroxide, de, Cr(OH) Cr(OH)3 by dithionation, in basic solution. S2O24  CrO24  2H2O    2SO32  Cr(OH)3  OH 100 L of water requires 387 g of Na 2S2O4. The molarity of CrO 42– in waste water is (1) 0.0448

(2)

4.448

(3)

0.0148

(4)

0.0224

Sol. Answer (3) Balanced redox reaction is 2

2 H2O  3S 2O 4 MNa

2

Now 

 2CrO 42   6 SO 32  2 Cr 3  4 OH 

S2O4  387

2n

2

S2 O 4

n



CrO CrO 42 (in (in 100 100  )

3 n

2

CrO4



2  1 74 38 7



1

 1 .4 8

3

 nCO4–2 in 1 L H2O = 0.0148

9.

A sampl sample e which which con contai tains ns exac exactly tly 0.5 g of ura uranium nium in the the form form of U 4+. The total uranium is allowed to oxidized by 50 ml of KMnO 4. The reaction taking place is U4 

 KMnO 4  H2O  UO 22  Mn 2  H3O 

Find the concentration of KMnO 4 required for the above purpose [U = 238] (1) 0.0336 M

(2)


0.0084 M

(3)

0.0168 M

(4)

0.0672 M


210 Redox Reactions


Sol. Answer (3) On balancing redox reaction 5U+4 + 2MnO4–  5UO2+2 + 2Mn+2 n=2

n=5

Equivalents of KMnO4 = Equivalents of U +4 50

N



1000

0.5

 238   2   

 N = 0.084

Hence

MKMnO

 4

N



0 .0 8 4

5

 0.0168 M

5

10. A 50 ml of a 20% 20% (w/w) solution solution of density density 1.2 1.2 g/ml is diluted diluted until until its strength strength becomes becomes 6% (w/w). (w/w). Determine Determine the mass of water added (1) 88 g

(2)

120 g

(3)

140 g

(4)

180 g

Sol. Answer (3) %

w w

11.

W t. of o f so solute



 100

W t. t. of solution

1.245 1.24 5 g of Cu CuSO SO4.xH2O was dissolved in water and H 2S gas was passed through it till CuS was completely precipitated. The H 2SO4 produced in the filtrate required 100 ml of 0.1 M NaOH solution. Calculate x (approximately) (1) 5

(2)

6

(3)

7

(4)

8

I in KI3

Sol. Answer (3) Eq. of H2SO4  Eq. of NaOH Eq. of H2SO4  Eq. of CuSO4 � xH x H2 O

12. Determ Determine ine the oxidati oxidation on numbers numbers of the the following following elements elements as indicated. indicated. (i)

(ii)

Mn in K2MnO4

(iii)

I in KIO3

(iv)

(v) Fe in in Fe3O4

(vi)

C in CH3OH

(vii)

S in Na2S4O6

(viii) Cr in in CrO CrO2Cl2

(ix)) Cr in (ix in [Cr( [Cr(NH NH3)6] Cl3

(x)

Fe in K4[Fe(CN)6]

(ii)

+ 6

(iii)

+5

(iv)

(vi)

–2

(vii)

+ 2.5

(viii) + 6

(x)

+2

Sol. (i) (v)

P in NaH2PO4

+5 



1 3

8 3

(ix) + 3

13. Balanc Balance e the following following chemic chemical al equations equations by the oxidati oxidation on number number method. method. (ii)

P + HNO3  HPO 3 + NO + H2O

(iiii) FeS2 + O2  Fe 2O3 + SO2 (i

(iv)

Zn + HNO3  Zn (NO3)2 + N2O + H 2O

(v) SnO2 + C  S Sn n + CO

(vi)

FeCl3 + H2S  FeCl2 + S + HCl

(vii) H2O2 + PbS  PbSO4 + H 2O

(viii) MnO2 + HCl  MnCl2 + Cl2 + H2O

(ix) I2 + HNO3  HIO3 + NO2 + H2O

(x)

(i)

CuO + NH3  Cu + N2 + H2O


K2Cr 2O7 + HCl  KCl + CrCl3 + Cl2


Redox Reactions 211


(ii)

3P + 5HNO3  3HPO3 + 5NO + H2O

(iii (i ii)) 4F 4Fe eS2 + 11O2  2Fe2O3 + 8SO2

(iv)

4Zn + 10HNO3  4Zn(NO3)2 + N 2O + 5H2O

(v) SnO2 + 2C  S Sn n + 2CO

(vi)

2FeCl3 + H2S  2FeCl2 + S + 2HCl

(viii) 4H2O2 + PbS  PbSO4 + 4H2O (v

(viii) MnO2 + 4HCl  MnCl2 + 2H2O + Cl2

Sol. (i)

3CuO + 2NH3  3Cu + N2 + 3H2O

(ix) l2 + 10HNO3  2HIO3 + 10NO2 + 4H 2O (x) K2Cr 2O7 + 14HCl  2KCl + 2CrCl3 + 3Cl2 + 7H2O 14. Balan Balance ce the the following following equat equations ions by Ion-ele Ion-electron ctron meth method od MnO4– + Fe2+  Mn 2+ + Fe3+

(In acidic medium)

(ii) MnO4– + H2C2O4  Mn2+ + CO2

(In acidic medium)

(iii) Cu2+ + SO2  SO 42– + Cu+

(In acidic medium)

(iv) N2O4 + BrO3–  NO 3– + Br –

(In acidic medium)

(v) Br– + BrO3–  Br 2

(In acidic medium)

(vi) Cl2 + OH–  Cl – + ClO–

(In basic medium)

(viii) S + OH–  S 2– + S 2O32– (v

(In basic medium)

(viii (v iii)) Al + NO3–  Al(OH)4– + NH3

(In basic medium)

(ix) Cr 2O72– + Fe2+  Fe3+ + Cr 3+

(In acidic medium)

(x) Fe3O4 + MnO4–  Fe2O3 + MnO2

(In basic medium)

(i)

Sol. (i)

MnO4– + 5Fe2+ + 8H +  Mn 2+ + 5Fe3+ + 4H2O

(ii) 2MnO4– + 5H2C2O4 + 6H+  2Mn2+ + 10CO2 + 8H2O (iii) 2Cu2+ + SO2 + 2H2O  2Cu + + 4H+ + SO42– (iv) 3N2O4 + BrO3– + 3H2O  6NO3– + Br – + 6H+ (v) 5Br– + BrO3– + 6H+  3Br 2 + 3H2O (vi) Cl2 + 2OH–  Cl – + ClO– + H2O (viiii)) 4S + 6OH–  2S2– + S2O32– + 3H2O (v (viii) (vi ii) 8Al + 3NO3– + 18H2O + 5OH–  8Al(OH)4– + 3NH3 (ix) 6Fe2+ + Cr 2 O 72– + 14H+  6Fe 3+ + 2Cr 3+ + 7H2O (x) 6Fe3O4 + 2MnO4– + H2O  9Fe2O3 + 2MnO2 + 2OH– 15.. A solu 15 solutio tion n cont contain aining ing 2.68 2.68 × 10 10 –3 mol of An+ ions require 1.61 × 10 –3 mol of MnO4– for complete oxidation of An+ to AO3– in acidic medium. What is the value of n? Sol.



7  Mn O 4

nf = 5–n

nf = 5

A

n

5

 A O3  Mn2

Number of eq. of A +n = Number of eq. of

 MnO 4

2.68 × 10–3 × (5 – n) = 1.61 × 10 –3 × 5 5–n=3 n=5–3=2 Aakash Educational Services Pvt. Ltd.


212 Redox Reactions


16.. Calcu 16 Calculat late e the the weigh weightt of Mn MnO O 2 and the volume of HCl of specific gravity 1.2 g/ml (and 4% by weight) needed to produce 1.78 litre of Cl2 at STP by the reaction: MnO 2 + 4HCl  MnCl2 + 2H2O + Cl2.  MnCl2 + Cl2 + 2H2O

Sol. MnO2 + 4HCl

Moles of MnO2 = moles of Cl2 W 55  32 W 87



volum e



m ol olar vo lu lum e

1.78 2 2 .4

W = 6.913 g 17. What is the the mass of of sodium bromate and and molarity molarity of the solution solution necessar necessary y to prepare prepare 85.5 ml ml of 0.672 N – + – solution when the half cell reaction is 2BrO 3 + 12H + 10e  Br 2 + 6H2O ? wt. wt. of NaB NaBrO rO3 (w) Sol. N =

Eq. Eq. wt. of NaBrO3 Volume Volume of solutio solution n ( ) w  5  1000

0.672 = 

w



151 151 85.5 85.5 0.67 0.672 2  151 151 85. 85.5 5000

 w = 1.735 g

18. 5.7 g of bleaching bleaching powder powder was suspended suspended in 500 500 ml of water. water. 25 ml of this solution solution on treatment treatment with with KI in the presence of HCl liberated iodine which reacted with 24.35 ml of

N 10

Na2S2O3. Calculate the % of available

chlorine in the bleaching powder. Sol.

∵

Weight of bleaching powder present in 500 ml = 5.7 g

 weight of bleaching powder present in 25 ml = 1

Normality of Na2S2O3 =

2 5  5 .7 500

= 0.285 g

 0.1

10

1 ml N Na2S2O3  0.0355 g Cl 0.1 N, 24.35 ml Na 2S2O3  0.1 × 24.35 × 0.0355 g Cl % of available chlorine in bleaching powder 100

= 0.1 × 24.35 × 0.0355 ×

0.285

= 30.33%

19.. Bo 19 Bora rax x in wat water er giv gives es : B4O72– + 7H2O  4H 3BO3 + 2OH– How many grams of borax (Na 2B4O7.10H2O) are required to (a)) Prep (a Prepare are 50 50 ml of 0.2 0.2 M soluti solution? on? (b)) Neutra (b Neutralise lise 25 ml of 0.1934 0.1934 M of HCl HCl and H2SO4 separately? Aakash Educational Services Pvt. Ltd.


Redox Reactions 213


Sol. B4O72– + 7H 2O  4H3BO3 + 2OH–

Average oxidation number of B in

Borax



4334

 3.5

4

H3BO3

(n factor = 3.5 × 4 – 12 = 2)

(a)) Let weight (a weight of of Borax Borax require required d is W gram. gram. 50

0.2 × 2 ×

100

W

=

382 2

MNa B O .10H2O 2 4 7



38 38.2

W = 3.82 g (b)) Number of equivalent (b equivalent of Borax = Number of equivalent equivalent of HCl W 382

= 0.1934 × 1 × 25

2

W = 0.9235 gm Number of Borax = Number of eq of H 2SO4 W 382

= 0.1934 × 2 × 25

2

W = 1.847 g 20.. On 20 One e gram gram of of comm commerc ercia iall AgNO AgNO3 is dissolved in 50 ml of H 2O. It was treated with 50 ml of KI solution. The M

silver iodide thus precipitated is filtered off. The excess KI in the filtrate is titrated with presence of 6N HCl till all the I – is converted to ICl. It requires 50 ml of

10

KIO3 solution in

M 10

KIO3. Under similar conditions,

M

20 ml of the same stock solution of KI required 30 ml of

10

KIO 3. Calculate the percentage of silver nitrate

in the sample. KIO 3 Aakash Educational Services Pvt. Ltd.

 2KI  6HCl    3ICl  3KCl  3H2O


214 Redox Reactions


Sol. KIO3 + 2KI + 6HCl  3ICl + 3KCl + 3H 2O 1 mole KIO3  2 moles KI Number of millimoles of KI in 20 ml of stock solution is 2 × 3 = 6 Number of millimoles of KI in 50 ml of the same solution = 1

Number of millimoles of KIO 3 is 50 ml of solution =

6

50

 15

20

 50  5

10

Number of millimoles of KI used with 50 ml of KIO 3 solution = 2 × 5 = 10 Number of millimoles of KI used with AgNO 3 = 15 – 10 = 5 AgNO3 + KI  AgI + KNO3 1 mole AgNO3 reacts with 1 mole KI. Number of millimoles of AgNO 3 = 5 Weight of AgNO3 = 5 × 10 –3 × 170 0.85 0.85  100 100

% AgNO3 =

1 0

 85 %

21.. A solu 21 soluti tion on con conta tain ins s Na2CO3 and NaHCO3. 20 cm3 of this solution required 5.0 cm 3 of 0.1 M H2SO4 solution for neutralisation using phenolphthalein as an indicator. Methyl orange is then added when a further 5.0 cm 3 of 0.2 M H 2SO4 was required. Calculate the masses of Na 2CO3 and NaHCO3 in 1 litre of this solution. Sol. Let mass of Na 2CO3 = a g Mass of NaHCO 3 = b g HPh = HPh = 1 2



1 2

a 106

number of Na2CO3 = number of H 2SO4

 0.1 2 

5.0 1000

 a = 0.106 g

2 ∵

20 cm3 contains 0.106 g

 1000 cm3 (1 L) contains =

0.10 0.106 6  1000 1000 20

 5.3g

Methylorange 1 2

× number of eq. of Na 2CO3 + 1 × Number of eq. of NaHCO 3

= number of eq. of H 2SO4 0.1 2 

5.0 1000

 1

b 84

=

5

0 .2  2 

1000

b

0.001 +

84

= 0.002

b = 0.084 g ∵

20 ml contains 0.084 g.

 1000 ml (1 L) contains = 4.2 g Aakash Educational Services Pvt. Ltd.


Redox Reactions 215


22.. A 1.0 22 1.0 g sam sampl ple e of of Fe Fe2O3 solid of 55.2 percent purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made upto 100 ml. An aliquot of 25 ml of this solution requires 17.0 ml of 0.0167 M solution of an oxidant for titration. Calculate the no. of electrons taken up by oxidant in the reaction of above titration. Sol. Weight of Fe2O3 in 25 ml solution =

1 55.2  25 100 100  100 100

 0.138g

Milliequivalents of Fe2O3 = Milliequivalents of Fe2+ = Milliequivalent of oxidant 0 .1 .1 38 38  2 160

 1000 = 17 × 0.0167 × n

n= 6 So, 6 electrons are taken up by the oxidant in the titration.









Redox Reactions

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