REDOX REACTIONS

CHAPTER OUTLINE
Page
* Important Terms 131

Exerise
-.. 132
61 Introduction
62 Oidation number 133
63 Balancing of redox reactions 137
-- 41
64 Redox reaction and electrode potential
Maltiple Choice Questions 144

IMPORTANT TERMS

Loss of electron (M’M- - e )

Incease in the oxidation number of tthe element
L Oidation
Addition of electrongetive element
L Removal of electropositive element
Gain of electron (A +e’A)
Decrease in oxidation number of the element
2 Reduction
Addition of electropositive element
Removal of electronegative element
3. Oxidant:Oidising agent
4 Reductant : Reducing agernt
3. Redox reaction: A reaction in which simultaneous oxidation of one species and the reduction of the
other species takes place.
Combustion

Bleaching
Batteries
6. Applications of redox reaction
Metallurgy
Corrosion
Respiration

6. REDOX REACTIONS 131

 EXERCISE
the textbook
* Indicates question from
6.1 Introduction Q. 3. In the following reactions,
species undergoing oxidation and mention the
e Can vou tell?
(1) Why does cut apple
(Textbook page 81) (1) 2Fe,Osw +3Cs 4Fets +
3CO. redductio
turn brown when (2) 2H,Si + O:(g ’ 2S4 -2H.0 (2
exposed to air ? Ans.
marks
Ans. Catechol present in apple (1) In the given reaction, Fe undergoes
tion when exposed to air andundergoes oxida
imparts brOWn since oxygen is removed while C rundereductgn
colour. oxidation since it combines with oxvgen
(2) Why does old car
bumper changes (2) In the given reaction, S undergoes
colour?
since hydrogen is removed
while oxidator
Ans. Metallic bumper undergoes
air and changes colour.
oxidation by undergoes oxidation since it hydroge
combines wt
Oxygen.
(3) Why do new batteries
become useless after 0. 4. Write classical definition of oxidant/oxidisine
some days?
Ans. The batteries involve agent. Give suitable example. (2 marks)
redox reactions
during their use and consume the chemical Ans. Oxidant/oxidising agent : According
like H,SO.. Hence they become useless classical definition oxidant is a substanc:
after some days. which itself undergoes reduction and causE
Q. 1. What does word 'redox'
oxidation of another species.
represent? Give For example,
examples. (2 marks) Mgs + F2g ’MgF29)
Ans. Redox is an abbreviation used for the terms
In the reaction fluorine is oxidant whi
'oxidation and reduction'. For example, the
phenomenon such as respiration, combustion oxidises Mg.
of fuel, rusting, etc. involve redox reactions. Q. 5. Write classical definition of reductant
Q. 2. What are classical ideas of redox reactions? reducing agent? (1 mark)
(2 marks) Ans. Reductant/reducing agent : According t
Ans. Classically oxidation refers to combination of classical definition reductant is a substantt
an element or a substance with oxygen. which itself undergoes oxidation and
caue
the reduction of
For example, C + Ozg) ’ COzg another substance.
2Mgs + Ozig)’ 2MgOg For example,
In the reactions, Cy and Mgsy are oxidised to Mgs, + Ss)
their oxides, CO () and MgOs, respectively. In the reaction Mg is
Classically reduction refers to combination Q. 6.
reductant.
Explain oxidation and reduction with
of an element or a substance with hydrogen or help of
elements. electropositive
removal of oxygen from its compound.
and electronegative
(2 marks)
2H2 () t O2 (g) - 2H,0g Ans. Oxidation :Combination of a substance wu
In this reaction, hydrogen is oxidised and
Oxygen is reduced.
electronegative element or removal of e
tropositive element is called oxidation.
NAVNEET CHEMISTRY DIGEST
132 :STANDARD XI (PART 1)
 Reduction: Combination of a substance
electropositive element or removal of
(ii) Cu + 2Aga9) ’ Cug + 2Ags
with In the reaction Cu is a reducing agent and
electronegativeelement is called reduction.
Ag is an oxidising agent
Forexample,
(ii) 2Cop) +3 Ni +3 Nis
2Mg + 0, ’ 2MO
In the reaction Co is a reducing agent and
In this Mg undergoes oxidation while O
undergoes reduction. Ni is an oxidising agent.
Q7. Define oxidation and reduction considering
different aspects. (2 marks) 6.2 Oxidation number
defined as :
Ans, Oxidation is Q. 9. What is an oxidation number or
oxidation
(a) addition of oxygen. state? How is it useful to identify redox reac
(b) addition of electronegative element. tion,oxidantand reductant ? (2 marks)
(c) removal of hydrogen. Ans. Oxidation number or oxidation
state : The
(d) removed of electropositive element. Oxidation number (or oxidation state) of
(e) loss of electrons by any species. ion is defined as
an atom in a molecule or an
Reduction is defined as :
the number of charges it would carry if the
(a) removal of oxygen. electrons were completely transferred.
(b) removal of electronegative element.
(c) addition of hydrogen. Explanation :
state does
(1) The oxidation number or oxidation
(d) addition of electropositive element.
(e) gain of electrons by any species. not always imply ionic charges on the species.
mark) (2) The oxidation number of monoatomic ion is
0. 8. What is a displacement reaction ? (1
redox reac equal to charge of the ion. For example, Na has
Ans. Displacement reaction : Thisis a
in a +1charge while CI has -1charge, Ca? has
tion in which an ion (or an atom)
atom) of +2 charge and so on (But the oxidation states
Compound is replaced by an ion (or an
another element.
are 1 +,1-,2+ respectively).
(3) In case of a neutral molecule, the sum of the
For example, oxidation numbers of all the constituent atoms
X+ YZ XZ +Y
Fe + CuSO, ’ FeSO + Cu is always zero.
(4) The charge on a polyatomic ion is equal to the
" Try this (Textbook page 82) algebraic sum of the oxidation numbers of all
table of displace
Q. Complete the following oxidising and the constituent atoms of the ion.
ment reactions. Identify (5) From the change in the oxidation numbers in a
reducing agents involved.
redox reaction, an oxidant and a reductant can
Products
Reactants be identified.
(aq) + Cu) [Note : Oxidation number or oxidation state is represen
Znt (aq)
+......... ted as 1 +,2+, etc. while charge is represented as +1,
Cus + 2Ag (aq) +2, etc.]
Co?t,(aq) + Nis)
.... +
Q. 10. Explain the rules to assign Oxidation
Ans. number of an element? (3 marks)
Zn (ag) +Cus)
(i) Zns + Cu[ (aq) Ans. Following are the rules to assign the oxidation
agent and
In the reaction Zn 1s a reducing number (or state) to an atom in any molecule
agent.
Cu?t is an oxidising or an ion:

6. REDOX REACTIONS 133

 Rule 1: The oidation number of an atom in
(d) K,C,0, (e) H,S,0, () Cr,0
tre clemental state is alwavs ze0. Por (g) NaH,'O,. (1 mark each
example, cach atom in I,, O, Ca, P, Ihas Ans.
molecule
Ovidation unmber zero. (a) H,SO, is a neutral
Rule 2: The oxidation number may be zero, Sum of oxidation number of all atoms
Positive, negative, integral or fractional. H,SO, =0
Rule 3: The oidation number of an atomin .. 2 (oxidation number of H) +
monoatomic ion is equal to its charge.
For exanmple : Na', Ca'', cr,CI, S ,etc.
number of S + 4(oxidation number of
.'. 2(+ 1) +x + 4(-2) =0
oxidO)=0atio,
Rule 4 : The oxidation number of H atom Oxidation number of S = +6
(b) HNO, is a neutral molecule.
bonded to nonmetals is 1+,e.g. HCIWhile that
Sum of oxidation number of all
|+ HNO, =0 atoms of
of H atom bonded to electropositive elements Oxidation number of H + oxidation
or metals is 1-,e.g., NaH, CaH, of N +3 (oxidation number of O) =0 number
.. (+ 1) +x+3( 2) = 0
.. N= +5
Rule 5: The oxidation number of oxygen in
. ,Oxidation number of N= +5
Oxidesis 2- .e.g, MgO while in peroxide it is 1-, (c) H,PO, is a neutral molecule.
.. Sumn of oxidation number of all atoms ot

e-g. H,O,' Na,O,. (Only in OF,, the oxidation H,PO, =0

.. 3(Oxidation number of H) + oxidation
number of O is 2 +)
number of P +3 (oxidation number of O) =0
Rule 6 : The oxidation number of
() F is always 1-; ..3 x (+1) +x+3 x(- 2)
. . N= +3
(ii) alkali metals is always 1+,e.g. Na, K; Oxidation number of P= +3
(d) K,C,0, is a neutral molecule.
(Gii) alkaline earth metals is always 2 +, Sum of oxidation number of all atonms t
e.g. Mg, Ca K,C,0, =0
1 ‘. .. 2(oxidation number of K)+2(oxidati
2 + 2 +

number of C) + 4(oxidation number of O)=0

Rule 7: In a neutral molecule, the algebraic sum of .. 2(+ 1) + 2r + 4(- 2) =0
the oxidation numbers of all atoms of the molecule .. N +3
is zero.

Rule 8:The charge on a polyatomic ion is equal to .. Oxidation number of C= +3

the algebraic sum of the oxidation numbers of all (e) H,S,0, is aneutral
constituent atoms of the ion. molecule. atomsot
..Sum of oxidation of
number all
*O. 11, Calculate the oxidation number of under H,S,O, =0
lined atoms.
.. 2 (oxidation
(a) H,SO4 (b) HNO, (c) H,PO, (oxidation number of H) + 4
O) =0
number of S) + 6 (oxidation number of
134 NAVNEETCHEMISTRY DIGEST :STANDARD XL(PART 1.
 2(+1)+4(1) +6(--2) =0 .. X= +2
. : = t2.5 Oxidation numbers of S = +2.
Oidationnumber of S= +2.5 (c) HBO, is aneutral molecule.
Sum of oxidation number of all atoms of
a Cr.0: is an i0nic species carrying net
charge -2; HBO, = 0
+(oxidation
Cunm of oxidation number of all atoms of .. 3 (oxidation number of H)
0
Cr,0: = -2 number of B)+3 (oxidation number of O) =
..3(+1) +x+3(-2) =0
2 (oxidation number of Cr) +7 (oxidation
.. X= +3
number of O) = -2
:2r+7(-2) =-2 . .Oxidation number of B= +3.
the under
. =+6 Q. 13. Find the oxidation numbers to
Oxidation number of Cr = +6 lined species in the following compounds
or ions :
molecule.
(g) NaH,P0, is a neutral V,o (5) I;
Sum of oxidation number of all atoms of (1) PF; (2) CIF, (3) SbF, (4)
(7) CrOs. (1 mark each)
NaH,PO, =0 (6) K, [Fe(CN),]
Ans.
:.Oxidation number of Na + 2 (oxidation
(1) PF, :
number of H)+ oxidation number of P+4
Oxidation number of F =1
(oxidation number of O) = 0
Let the oxidation number of P=x
.. (+,1)+2(+ 1) +x+4(2) =0
Since the ion carries -1 charge,
. . x= +5
x(P) +6(1-)(F)= -1
..Oxidation number of P= +5
X-6=-1
*Q. 12. Assign oxidation number to
each atom in . . x=5
the following species. Hence the oxidation number of P =5+.
(a) Cr(OH); (b) Na,S,O, (c) H,BO,. (2) ClF, :
(1 mark each)
Oxidation number of F = 1
Ans.
Let the oxidation number of Cl=x
charge
la) Cr(OH), is an ionic species carrying net .. x (Cl)+3(1 -)(F) =0
-1. X-3=0 .. x=3
number of all atoms of
.. Sum of oxidation Hence the oxidation number of Cl =3t.
Cr(OH), = -1
O)+ 4 (oxidation (3) SbF :
.. x+4(oxidation number of Oxidation number of F=1
number of H) =-1
Let the oxidation number of Sb =x
.. x+4(-2) + 4(+1) = -1 Since the ion SbF carries -1 charge
. . X= +3
Cr= +3 x(Sb) +6(1-)(F) = -1
.. Oxidation number of
molecule. ..N-6= -1
(0) Na,S,0, is a neutral
number of all atoms of X=5
.. Sum of oxidation
Hence the oxidation number of Sb =5+.
Na,S,O,=0
Na) +2 (oxidation (4) V,0;:
.. 2(oxidation number of
of O) =0
number of S) +3 (oxidation number Oxidation number of O =2 -
=0
..2(+1) + 2(x) +3(2) Let the oxidation number of V =x
6. REDOX REACTIONS 135
 Since the ion V,0; ,carries 4 charge, 0. 15. Define (i) Oxidation and (ii) reduction in
2r(V) + 7(2 )(O) = 4 terms of oxidation number. (2 marks)
2r- 14 = 4 Ans.
.. X=5 (i) Oxidation : An increase in the oxidation
Hence the oxidation number of V=5+. number of an element in a given substance is
(5) I; : Tri-iodide ion Ii has two different oxida called oxidation.

tion numbers for I, namely zero and 1 (ii) Reduction : A decrease in the oxidation
number of an element in agiven substance is
(6) K, [Fe(CN)] : called reduction.
Oxidation number of K=1+
Oxidation number of C=2+ Q. 16. Define (i) oxidising agent and (ii) reduc
Oxidation number of N=3 ing agent in terms of oxidation number.
Let the oxidation number of Fe =x (2 marks)
K[Fe(CN),] has 3K, IFe, 6C and 6N atoms. Ans.
3(1 + )(K) + x(Fe) + 6(2 + )(C) + 6(3-)(N) =0 (i) Oxidising agent : A substance which
3+x+ 12 -18 =0 increases the oxidation number of an element
X=3 in a given substance and itself undergoes
Hence the oxidation number of Fe =3+. decrease in oxidation number of a constituent
(7) CrO,: In CrO,the oxidation states of Oare -1 element is called an oxidising agent.
and - 2. The oxidation state also depends upon (ii) Reducing agent : A substance which decreases
chemical bonding in the molecule, the oxidation number of an element in a given
0 2 substance and itself undergoes an increase in
the oxidation number of a constituent element
in it is called a reducing agent.
" Do you know? (Textbook page 85)
Remember (Textbook page 83) Some elements in a particular compound mav
Q. What are possible oxidation numnbers of an
possess fractional oxidation number.
element?
For example :C,O,, Br,Og, Na,S,O,, CH ett.
Ans. Oxidation number of an elenment can be
In these compounds oxidation nunmber of C,Br, S,
positive, negative, awhole number zero, or a Care 4/3, 16/3, 2.5, 9/4, respectivelv. These
fraction. Oxidation numbers are actually the average
Q. 14. Arrange the following in order of increas oxidation number of all the atoms of elements in
ing oxidation number ofC atom : that compound. Different atoms of the elenment in
such species exhibit different oxidation states.
CaCz, CO;, C,0, C,H,, CH,. (2 marks) For example : Tertrathionate ion has twoS atoms
Ans. The oxidation number of Catom in given
with oxidation number +5 and two with zer
compounds are as follows:
(0). Therefore, the average oxidation number ot 5
CaC, CO; C,o C,H, CH, in these species is 10/4 = 2.5
I.
|+5 0, 0

The increasing order of oxidation number of -O-S-S-s'-`-0 structure of S,O,

Catom is (tetrathionate ion)

CH, < C,H, <CaC, < C,O; < co

136 NAVNEET CHEMISTRY DIGEST : STANDARD XI (PART 1)
 Balancing of redox reactions (2) Half reaction methods :
6.3
atoms
redox
Step 1: Assign oxidation numbers to all
Q. 17. What is a reaction ? (2 marks)
in the given reaction.
Ans, half
Step 2: Divide the equation into two
Redox reaction : A reaction in which oxidation the other for
andreduction reactions occursimultaneously is equations, one for oxidation and
reduction.
calledredox reaction. For example, gain of
Step 3: Find the loss of electrons and
Zn,, + Cu(aq) Znt Cu
electrons by atoms in the reaction.
(2) In this reaction, one species (Zn) loses electrons electrons.
and undergoes oxidation. Step 4: Balance the loss and gain of
For example, Zn, Zn;(aq) + 2e (oxidation) Step 5: Add both the reactions.
added
Step 6: For acidic medium Ht ions are
(3) The other species (Cu) accepts the donated
electrons and undergoes reduction. while for basic medium OH are added.
For example, Cu + 2e ’ Cu, (reduction) *Q. 19. Balance the reactions/equations by
oxidation number method :
(4) Hence the overall redox reaction involves simul
taneous reactions of oxidation and reduction. (a) Cr,0; bo) +SOlug) Cra + SO;4(aq) (acidic).
(1 mark)
In the above reaction, Zn gets oxidised and
Cu get reduced. |Note :Cr,0, given in the textbook is changed to Cr,0I
Q. 18. Explain balancing of redox reactions.
Ans.
(2 marks) Step 1: Assign oxidation number to each atom in
chemical
Ans. Two methods are used to balance the given reaction.
equation for redox processes : Oxidation + SO (aq) (aq) + SO (ag)

number method and Half reaction method or

+3 +6
lon electron method. +6 +4

(1) Oxidation number reihod : Step 2: Identify the atoms that have changed
Step 1: Balance the given equation for all atoms oxidation numbers.
except H and O. Identify the atoms undergoing Cr,0; n) + sO Cr3+(a) + SO? (aq)
(aq)
change in oxidation number.
oxidation number +6 +3
Step 2: Show an increase in
+ 6
Reduced
in oxidation
for oxidised species and a decrease Oxidised

number for reduced species. Balance total Step 3:Find the toal decrease in oxidation number
numbers.
increase and decrease in oxidation
to the of the reduced atom Cr and total increase in oxida
Step 3: Balance O atoms by adding H,O tion number of the oxidised atom S.
balance H atoms
side deficient ofO atoms and Total decrease in oxidation number
side deficient of
by adding H ions to the
by form =(+6)-(+3) =3
Hatoms and they are finally removed
Total increase in oxidation number
ing H,0.
ions are added =(+6)(+4) =2
Step 4 : In basic medium OH-
which are removed as H,0. Step 4: To balance the net change in oxidation
Step 5: Check the equation with respect to
numbers, multiply Cr,0 by l and Cr* by 2and
number of atoms of each element on
both the multiply S0; and SO} by 3.
sides. Cr,0} (a9) + 3S0: (aq) ’ 2Cr + 3S0 (aq
6. REDOX REACTIONS 137
Step :Fnd net chargs on lett hai ani ight (c) H.S0.-C -CO; -SO: - H,0
han side ot the quatOn. (acidic). (1 mark)
Totalcharge on let han siie Ans.
Step 1: Assign Oxidation number to each
= - ) - 3(-2= -8 atom
Total change on right hand side the given equation.
HSO. tC, CO + SO;e +H0.
=-3)-3(-2) =0
Sine there is net -S change on let hand side. add
SH to it since the etion is in acidic medium.
Hene balaned equation is Step 2: Identifv the atoms that have changed
oidation numbers.
Cr.0:
Cra -380;7 + 4H.0. HSO. +Co COsg t SO:g) + H,0.
(b) MnOBr - MnO,s- BrO,a (basic).
[Note : IKO schanged to BrO! (1 mark) Redue

Oidised
Ans.
Step 1 :Assign oidation number to each atom in Step 3 : Find the total decrease in oxidation number
the equation.
of reduced atom S and total increases in oxidation
number of the oxidised atom C.
MnO. + Br MnO - BrOs Iotal decrease in oxidation number
-

=(+6) -(+4)=2
Total increase in oxidation number
Step 2 : ldentify the atoms that have changed =(+4) -(0) =4
oxidation numbers.
Step 4 : To balance the net charge in oxidation
MnO numbers, multiply H,SO, and SO, by 2.
Reduce
2H,SO: g) +Cs COg + 2SO2g t H,O,
Since there are 4H on left side and 2H on right side,
Oricised
multiply H,0 by 2.
Step 3: Find the total decrease in oxidation number .. 2H,SO. ag) + Ca COzg t 2S021g +2H,O%
of reduced atom Mn and total increase in oxidation
Step 5: Since there is no net charge on left and right
number of oxidised atom Br. hand sides, the above equation is balanced.
Total decrease in oxidation number (d) Bi(OH);(9 + Sn(OH)3(ag) Bi + Sn(OH)
=(-7) -(+4)=3 (basic). (1mark)
Total increase in oxidation number Ans.
=(+5)-(-1) =6 Step 1:Assign oxidation number to each atom n
Step 4 : To balance the net change in oxidation the equation.
numbers, multiply MnO, and MnO, by 2. Bi(OH),s t Sn (OH)a0) Bi + Sn + (OH);4
’ 2MnO,(, + BrO, faqy
2MnO.4) + Br
Step 5: Since there are 3 negative charges on left
hand side and 1negative charge on right hand side Step 2 : ldentify the atoms that have change
of the equation add lH,0 on leftand 20H on right oxidation number.
hand side. Bi(OH)>) +Sn(OH), Bi + Sn(OH)eg
.2MnOot Br (2q) + H,O, +3 +2
1
+4

2MnO, + BrO,n, t 20H(aq) Reduced

Oxidised
This is a balanced equation.
138 NAVNEET CHEMISTRY DIGEST : STANDARD XI (PART 1)
 total dewreasein ovidation number Mn change from -7to -1
3 :Findthe
Step
atomBi and total inerease in oxidation Hence erease in oiiztior umier s

number ot
Ovitised atom $n.
in oxidation number S changes from -4 to -6
lotalderres nuTCEr ot > s
=3 Hence inrease in ovdation
(+3) -(0)
oNidation number (+6)-(-4) = -2
rotal increase in deTEzSE nd Toese r
(+)-(+2) =2 Step 4: Io balance the net
\ o ri
Shen 4: To balance the net charge
in oxidation oudation numbers egual. muitpir
b Hene
numbers, multiply Bi(OH), and Bi
by 2 while Mn bv 2and SO: and SO:
by 3.
Sn(OH), and Sn(OH); hari nir t
’2Bi, +3Sn(OH) on lert
2Bi(OH);t 3Sn(OH)) Step 3 : Find the net charges
charges on left hand sides.
Step 5: Since there are 3 negative side:
ight hand side Total charge on left hand
hand side and 6 negative charges on +5x(-2S0)= -1
basic medium on left 2x(-1) (Mn0;)
ofthe equation,add 3OH tor side :
hand side. Total charge on right hand
2x(-1)(Mn²)+5 x(-2)SO)= -
2Bi(OH) t 3Sn(OH)t3OH
charge on et
2Bi + 3Sn(OH) Hence there is net -6extra
in adiic meiun
This is a balanced equation. side. Since the reaction is
handsiie balne
Q. 20. Balance the following
equation for the 6H (6positive harges) on lett
medium by oxidation 6 extra negative charges.
redox reaction in acidic
number method: +3S0; a-oti a
’ Mn+S0;. (1mark)
MnO 4aq)
+ SO;3 (aq)
SS n t
Ans. Since there are 6H nd 30 atoms
Assign oxidation number to each atomn in hand side, write 3H.0 on ighthand siie
balae
Step 1:
the given reaction. H and Oatoms.
Mn² + SO: Hence the inal equation is
MnO (aq) + SO; (aq)
1 +3S0;g -6H,
2MnO
+2
2Mn -80-H,0
atoms that have changed Q. 21. Balance the following redor equztions by
Step 2 : Identify the
Oxidation numbers. oidation number method. The reactions ow
has been
number of Mn in MnO7 in acidic medium.
The oxidation in S0; has
from 7+ to 2 + and that of S HO;a + Cr,0
Teduced
4+ to6+.
Deen increased from Ans.
’ Mn + SO:
MnO (aq)
+ SO; (aq Step 1: H,0,,
+7

Reduced
Oxidiset

number Step 2 :0 in H,0, is oidised nom 1 - to zer n

decrease in Oxidation
Step 3 : Find the total ncrease in oxidation O,, while Cr in Cr.0: is redue i r o - t
reduced atom Mn and total
of in Cr.
atom S.
number of oxidised
6. REDOX REACTIONS
Step 3 : lncrease in Oxidation number of Step 3 :
atom =0-(1-)=1. Loss of electrons by Catom =(+4) (+3) =1
Sine there are 20 atoms in H.O,, total increase in Gain of electrons by Mn atom =(+7)-(+2) =5
Hence we can write,
oidation number =2 xl=2.
Derease in oxidation number of Cr atom H,C,0, (a4)+ 5e 2C0, + 2e
’ Mn²
(oxidation)
= (6-)- (3+)=3+ MnO, (aq) (reduction)
Step 4 :To balance loss and gain of electrone
Total decrease in oxidation number for 2Cr
atoms =2 x (3 +)=6+.
multiply oxidation equation by 5 and reduction
equation by 2.
Step 4:To balance the increase and decrease ’ 10COze + 10e (oxidation)
5H,C,O(aq)
in oxidation numbers equal, multiply H,0, and O, + 10, + 2Mn (reduction)
br 3.
2MnO4(aq) aq

Step 5: By adding both the equations,

3H0:2q),+Cr,0 30e +2Cr SH,C,Oitag) t 2Mn, (aq) 2Mn + 10CO2g)
Step 5: Since the reaction is in acidic medium, to Step 6 : To balance the charges and H atoms add
balance charges, add SH on left hand side. 6H on left hand side of equation.
(Both the sides, have net charge + 6). 5H,C,0,(9) + 2MnO0) + 6H+
3029+ 2Cr3+ 2Mn t 8H,O, + 10 CO,
3H;0, +CrO-a0 + 8H ) The above equation is a balanced equation.
Step 6 : Add 7H,0 on right hand side to balance H
(b) Bi(OH)se + SnO SnOs + Bis, (basic)
and O atoms on both the sides. (1 mark)
SH,02 e9 +Cr,Oa +8H (aq) Ans.
302g) t 2Cragt 7H,O Step 1 : Assign oxidation numbers to all atoms.
The equation is balanced with respect to all atoms Bi(OH):(9) + SnOzg) ’ SnOs) + Bi
and charges. +3-2 +1 +2 -2 +4-2

Gain of 3 e
*Q. 22. Balance the following redox equations by Loss of 2e
half reaction method :
Step 2 : Divide the equation into two half equations
(a) H.C,04 24) + MnOs(ag) ’ CO,g + Mn one for oxidation and the other for reduction.
(acidic). (1 mark)
SnO; (aq) SnO (Oxidation half equation)
Ans. Bi(OH),(s) + Bis (reduction half equation)
Step 1 : Assign oxidation numbers to all atoms. Step 3: Loss of electrons by Sn =(+4) -(+2) =
H,C,0a2q) t MnO,(aq) COg t Mnag) Gain of electrons by Bi =(+3) (0) =3
-1 -3-2 +7 -2 +4-2 Hence we can write,
Loss of le
Gain of 5 e
SnO;2 (aq) ’ SnOt2e (Oxidation)
Bi(OH),(s) t 3e Bi (reduction)
Step 2 : Divide the given equation into two half Step 4 : To balance loss and gain of electrons mult
equations, one for oxidation and the other for
ply oxidation equation by 3 and reduction equato
reduction.
by 2.
C+ CO;* (oxidation half equation) ’ 3SnO+6e
7* MnO, ’ Mn² (reduction half equation) 3SnO:aq)
2Bi(OH)s6) + 6e
3(a)
’ 2Bi

140 NAVNEET CHEMISTRY DIGEST: STANDARD XI (PART 1)

Step5: By adding both the vquatiom,
2Bi(O) 3SiO 215i Viett

Step6:To balance Hnd 0 atoms add 3|,0 on Salbdge

righthandside. elestte
Zine
clectrode tcathue)
3SnOa Bio, t 311,0, (mode)

this(Textbook Page 88) holuton

o Try
Classify the following unbalanced half letofe fextions

equations as oxidation and reduction. .Z1''+ 2

Peductim (yin of ele trts)
oxidation (loss of cetrons)
Exampleiei Type
Fig. 6.1 : Daniel Cell
(a) ClCl2) oxidation
Construction :
One half
|(b) OCI CI () Daniell cell consist of two half cells.
(c) Fe(OH), ’ Fe(OH), cell consists of a beaker containing Zn0, (1 M)
y! solution in which a polished strip of netallic
(d) Vo"
(aq)
zinc is immersed while the second half cell
consists of a beaker containing CuSO, (1 M)
Ans.
solution in which a polished strip of metallic
Example Type copper is immersed.
(a) Cl Cig Oxidation (ii) Two half cells are electrically connected by a
U-shaped salt bride containing a gel of KCl or
(b) OCI, ’CI
NH,NO, in agar-agar.
reduction
(ii) Working of Daniell cell: When the circuit is
2 3+

|(c) Fe(OH)9) ’ Fe(OH)4) Oxidation complete, following reactions take place :

At zinc electrode : Zinc atoms from zinc plate
(d) VO 0q) -v (aq) reduction lose electrons spontaneously which flow in the
external circuit from zinc plate tocopper plate
through wire.
6.4 Redox reaction and electrode potential Zn Zn + 2e (oxidation)
Q 23. Explain a redox reaction with the help of At copper electrode : Cu' from solution receive
Daniel cell. (3 marks) these electrons through copper plate and are
Ans. Consider the following displacement reaction, reduced to copper atoms which are deposited
Loss of 2e on the copper plate.
Zns) +Cu? Zn
Cu + 2e ’ Cu,, (reduction)
(aq) +Cus) The net reaction in the cell is an electro
Gain of 2e
chemical redox reaction represented as,
The electron transfer from Zn atom to Cu' ions Zno, +CuZn(aq) +Cus
Can be
demonstrated with the help of Daniel cell In the cell zinc electrode is an anode while
reactions. copper electrode is a cathode. In the external
circuit electrons flow from anode (Zn) to
cathode (Cu). An electrical potential called
6. REDOX REACTIONS
141
 clectrode potential is ostablished at vo cee (o) APwiivevalueottadardI
trodes of the cleetrohenical cwl. I" inieatenthat the todoN couple eductienl
Q. 24. What is an electrode potential ? mark) nidisiny ayent than ||/0, coupe
Ans. Anclectrical potentialestablished at the clee ()large ngative value of F" inicatey
trode ot an clectroehemical cell is called an Nronger
toducing power o the redos COuple while lary,
electrode potential. The magnitude ot the cev positive value ot "indicates sltongoroviiine
trode potential depends upon the nature of pOwer ot the redox couple,
metal and ions, oncentration of ions and
temperature. Q. 29, On the basis of standard electrode poten.
Q. 25. What is eleetrode reaction ? (I mark) tial explain : (a) the reducing power of alkali
Ans. The reaction associated with an clectrode metals and (b) oxidising power of luorine.
involving metal and ions is called electrode (2 marks)
reaction. Ans.

Q. 26. What is a redox couple ? (1 murk) (a) Since redox couple ot alkali metals has larger
Ans. The two chemical species which are linked by negative value tor standard reduction potential,
transfer of electronsrepresents aredox couple. they have a great tendency t0 give away cler:
For example, ZnlZn trons and forn1 cations,
Q. 27. What is a standard electrode potential ?
Nag Na' te 2.7 V
(1 mark)
Therefore
ENNa
alkali metals are strong reducing
Ans. Standard electrode potential :The potential
of astandard electrode in which the concentra agents.

tion of each species taking part in the electrode (b) Since redox ouple of tluorine has verv large
reaction is unity at temperature of 298 K is
positive value tor standard potential, it has a
called standard electrode potential and
denoted by E'.
great tendency to gain electron and form anion.
Fg + 2e 2F E +2.87 V
Q. 28. What is the significance of standard elec
trode potential, E°? (3 marks)
Therefore fluorine is a strong oxidising agent
Ans. *Q. 30. Which of the following redox couple is
(1) Generally electrode potential (E°) for reduction stronger oxidising agent?
reaction of the electrode is considered. (a) Cl,(E"=1.36 V) and Br, (E"=1.09 V)
(2) It may have positive or negative value. Stand (b) Mn0, (E =1.51 V) andCr.0² (E=1.33 \)
hydrogen electrode potential has zero values. (2 marks)
(3) From the value of standard
reduction potentials, Ans.
oxidising and reducing properties can be
explained. (a) Since the standard reduction potential, t
(4) The value and sign of the standard Cl, (+ 1.36 V) is higher than E of Br, (1.09\
electrode
potential is a measure of the tendency of the the stronger oxidising agent in the redoxcouple
constituent species to remain in the oxidised/ is Cl,.
reduced state. (b) Sinee standard reduction potential
(5) A negative value of E" (or E) indicates that
MnO, (I.51 V) is higher than E° tor Cr0:
the redox couple is a stronger
reducing agent (1.33 V), MnO, will be a stronger Oxidisig
than H/H,couple.
agent than Cr,0: in the redox couple.
142 NAVNEET CHEMISTRY DIGEST: STANDARD XI('ART )
 which acts as a uctant

(' marks cach)

Ans,

(i) Sis nitisnt tnm

() CuOis an vidant
() us is a mutant
*Q Wie the formula hor the ollowing

(A) Menuvl)chloride Sinr then is o change in the onidation

() Thalliumlsuhhate numrot anv lement it is not a rorNCtion.

() mium(llWmide

Compound Formula

(i) The Mtin is a NoN ction.

( Thalha() shat () is lued from en to-1.
( )S is oxdi fon +2 to + 25
(iv) l, is an oxidant.
() $O is a rnductant.
* i ln which chemial tion does carbon
exàibit variation ot onidation state trom *Q. 36 What is oridation ? Which one of the
land chemical rtion. tollowing pairs ot species is in its oxidized
to -Write state ?

(a) Mg/Mg (b) Cu/Cu'

C
Ans H () 0,/0 (3 arks)
Ans Fr deênitin ot nidatin eer o 7 .
The s i s in the nii stat:
*QH ln which ution does nitrogen exhitbit
variation of onidation state trom (a) M
( ar}
Ans NH th Q. 37. Justity the tollowing reartion as rrior
rution.

*Q31 lustity that the following ractions ar Na(s) +S(s) Na,S

rior rution: identity the spies oi Find out the oxidizing and rducing gents
ìdtui which acts as an idant and
 Ans. 2Na Na,Ss "Q. 39. Complete the following table :
0
Assign oxidation number to the
()) The reaction is a redox reaction.
(ii) Ss is an oxidising agent.
species and write Stock notation of
pound
underlined
com-
(ii) Nas is areducing agent.
(iv) Nas is oxidising and Ss is reduced. Compound Oxidation Stock
number
notation
*Q. 38. Provide the stock notation for the (1) AuCI,
following compounds :
(2) SnCl,
(a) HAuCI, (b) TI,0, (c) FeO, (d) Fe,O,
(e) MnO and () CuO (1 mark each) (3) V,o
Ans.
(4) PrCI:
Compound Stock notation (5) H,AsO,
(a) HAuCl, HAu(II) CI,
-1+3-4 Ans. (1 mark each)
(b) Tl,0 TI,() o Oxidation
+1 -2 Compound number
Stock
notation
(c) FeO Fe(II) O (1) AuCl, +3
+2 -2
Au(lllCi,
(2) SnCl, +2
(d) Fe,O, Fe,(I) O, Sn(ll) C1,
-3 -2
(3) V,o +5
V,(VO
(e) MnO
-2 -2
Mn(I) O (4) PCI +4
Pt(|V)C1;
() CuO (5) H,AsO, +3
+2 -2
Cu(ll) O H,As(lllO,
" Activity
Performn redox reaction experiment with the
help of Daniel cell under teacher
guidance.
MULTIPLE CHOICE QUESTIONS
Q. 40. Select and write the
most appropriate *2. Which of the
answer from the given alternatives for each following is not an example t
redox reaction?
subquestion. (1 mark each) (a) CuO+H, Cu +HO
*1. Oxidation numbers of Cl atoms marked as (b) Fe,O, + 3C0,
CI ’2Fe +3C0,
and Cl in CaOCI, (bleaching (c) 2K + F, 2KF
powder) are
C (d) BaCl, + H,SO, ’ BaSO, + 2HCI
Ca O-Cb
*3. A compoundcontains atoms ofthre elemet
(a) zero in each A, B and C. lfthe oxidation state ot A is t "
(b) -1 in CP and + 1in C| is + 5 and that of Cis -2, the compouNt
(c) +1 in C and -1 in CIb possibly represented by
(d) 1in each (a) A,(BC,): (b) A, (BC);
(c) A, (B,C), (d) ABC:
144 NAVNEET CHEMISTRY DIGEST:STANDARD XI (PART 1)
 *4. The coefficients p, q, r, s in the reaction 14. Oxidation number of oxygen in O; ion is
pCr,0,- +qFe Cr +S Fe' + H,0 (a) -2 (b) +1
respectively are : (c) -1 (d) +2
(a) 1, 2, 6, 6 (b) 6, 1, 2, 4 15. The sum of oxidation states of all atomns in
(c) 1, 6, 2, 6 (d) 1, 2, 4, 6 CIO, ion is
*5. For the following redox reactions, find the (a) zero (b) 4 (c) -1 (d) +2
Correct statement.
16. The oxidation number of arsenic in sodium
Sn² +2Fe ’Snt +2Fe2+
arsenate is
(a) Sn' is undergoing oxidation (a) 5 (b) 3+ (c) 4+ (d) 5 +
(b) Fe is undergoing oxidation
17. C,H,OH on oxidation forms CH,COOH.
(c) It is not a redox reaction Hence the oxidation number of carbon
(d) Both Sn and Fe are oxidised
changes from
*6. Oxidation number of carbon in H,CO, is (b) 2- to 0
(a) 0to 2+
(a) +1 (b) +2
(c) 2- to 3 + (d) 1+ to 3 +
(c) +3 (d) +4
18. The oxidation number of P in MgNH,PO, is
*7. Which is the correct stock notation for
magenese dioxide ?
(a) 3+ (b) 5+ (c) 2+ (d) 7+
19. The oxidation number of S in S.O is
(a) Mn(l)O, (b) Mn(I)O,
(c) Mn(II)O, (d) Mn(1V)O, (a) 2.5 + (b) 3.5 + (c) 2+ (d) 4+
*8. Oxidation number of oxygen in superoxide is 20. The ratio of number of moles of KMnO, and
(a) -2 (b) -1 FeSO, in acidic redox reaction is
2 1 3 5
(c) -; (d) 0 (a) 5
(b) (c); (d) ;
*9. Which of the following halogens does always 21. In the reaction,
show oxidation state-1? 3H,A,0ag) + BrOsag) ’ Br +3H,A,O4
(a) F (b) C1 The elements undergoing reduction and oxi
(c) Br (d) I dation respectively are
*10. The process, SO, ’ S,C1, is (a) As and Br (b) Br and O
(a) Reduction (c) Br and As (d) As and O
(b) Oxidation 22. When ferrous chloride is reacted with potass
(c) Neither oxidation nor reduction ium dichromate, the products formed are
(d) Oxidation and reduction
(a) Fe?*, CI, (b) Fe?*, CIO,
11. Oxidation number of P is +3 in the
(c) Fet, Ci (d) Fe, CI,
compound 23. In the reaction,
(a) H,PO0, (b) H,PO,
(c) H,P04 (d) H,P,0, Ale + BiONO,3(s) Bi + NH, + Al(OH),
which atom/atoms
12. Oxidation number of Clin KCIO, is undergoes /undergo
(b) -5 reduction?
(a) -1
(c) +1 (d) +5 (a) N (b) Bi, N (c) Bi, O (d) Al, Bi
13. Oxidation number of H in AIH, is 24. In bleaching powder (CaOCl,) the oxidation
(a) +1 (b) -1 state / states of Cl is/ are
(c) +3 (d) zero (a) 1 (b) 0 (c) 1-,1l+ (d) 1-,0
6. REDOX REACTIONS 145
 25. In the reaction, 10. (a) Reduction
aMnO, +bC,0, +H' cMn' +dCO, 11. (b) H,PO,
(a) a =2 b=6 (b) b=5 c=8 12. (d) +5
(c) b=5 e=2 (d) a =5 b=10 13. (b) -1
26. In the reaction, xP,H, ’yPH, + zP,H, 14. (c) -1.
(a) x= 6, y =4 (b) y =8, z =2
15. (c) -1
(c) x= 5, z =2 (d) y =6, z =1
16. (d) 5+
27. In NHNO,, the oxidation number of N
17. (b) 2- to 0
is/ are
18. (b) 5+
(a) 3- (b) 5+ (c) 3-, 5+ (d) 3-,5
Answers 19. (a) 2.5 +
1. (b) -1 in C and + 1 in Clb
2. (d) BaCl, + H,SO4 ’ BaSO, + 2HCI 20. (b)
3. (b) A, (BC4)2
21. (c) Br and As
4. (c) 1, 6, 2, 6
22. (d) Fe3+,,Cl,
5. (a) Sn² is undergoing oxidation
23. (b) Bi, N
6. (d) +4
24. (c) 1-,1 +
7. (d) Mn(IV)O,
25. (c) b=5, c=2
8. (c)
26. (d) y=6, z =1
9. (a) F 27. (c) 3-,5+.

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NAVNEET CHEMISTRY DIGEST: XI