Redox-rnx class 11 science..notes fro pw

Hope it will help

1. FormulasheetofREDOXREACTIONS
BY
:SUHANSHUSIR
RedoxReactions
->Reactions
involving
OxidationandReduction
--
both
=
-
Reduction Oxidation
-
-
①additionof
H ①Removalof
H
3
old/ClassicalConcept
②
Removal
of
0 ②Addition
of
0
③gain
of
eQ ③
lossof
eQ
3
modern/electronic
concept
④
Decrease
inOxidation④
Increase
inOxidation
Number Number
⑤RemovalofEnelement
Addition
ofEnelement
old/Classical
Addition
ofEp
element
ORemovalof
Epelement3
concept

2. #Oxidation
Number/State(0
.
N/O
.
S)
->
imaginary/redchargedeveloped
overanatominits
combined
state.
-
Relative
charge
-
dependent
onbanded
atums/state
.
"¥
Rulesto
find
OxidationNumber
Rule
-I
->
OxidationMo
.
of
Alkalimetals=+1(L,
Na,,
Rb
,25)
OxidationNo.
of
alkalineearthmetals=+2
(Be,Mg,
C
,
Sr
,Ba)
Rule-
->
Hydrogen
metal
Hydrogen
+non-metal
"¥show
Rule
-
3
->
OxidationNo.
of
atominitsstandardstate/elemental
state
S
0=0
,=
,
,2
,Na
0.=0
is
zero
.

3. Rule-y=>OxidationNo
.
ofIons/chargedatom-charge
-
Rule-5=ForP-blockelements
,
showO
.N
from(9-18to
G-10)
where
-
(exceptinent
pair
effect,andPeriod-2
elements)
C=
Group
No
.
Group-13
=>O
.N
range
=-5to+3
-
Group-14
=0
.
Nrange
=
-4to+Y
-
Group-15=O.M
range
=
-3to+5--
&roup-16
=>0.N
.
range
=
-2to+6
Group-17=O
.N
range
=
-1to+7
--
&roup-18 =>0..
range
=
0to+8
O
Coxidei m)In
general,
O
.N
of
Oxygen
atom=-2
But
Oxygen
canshow
(
-1
,
-
2
,
0
,+,
+
2)
0"or02
=Peroxide
O2F2 OF2 0-"2
or
0
=Superoxide-
Is
+1"¥E

4. O
.N
.
of
Halogens(ingeneral)
=-It
But
C
,
Br
,
Icanshow+ 1
,
+3
,
5
,
7
Rule-gE
->Sumof0.
Notallatomsin
aneutralmolewle=0
Sum
of0
.
Mofallatomsin
a
chargedmolecule=charge
Rule-9[
-OxidationNo
.
of
somecompoundsC
,
P
,
S
,
Cretc)canbe
calculated
by
their
structures.
i+
(EN)
A
>B
/
A-
B
A=B
A=B
(
-
1](1)
(2)(
+
2)(3)(3)
Rule-10
->FractionalOxidationNumberrepresents
Oxidation
F average-
--
Number .
-
Rule-11
->E
For
Complex/CoordinationCompounds
,
sameabove
roles
-
applies
.
Rule-12-Incase
ofalloys/Amalgam,
O
.
ofmetal
=0

5. someimportantCompounds
for
OxidationNo
·Calculation
①MOU ⑬CrOJ ⑤[P
>
22(u)243]
-
-
② 2 MnOn
⑭H205 K2[2n(0)]
③2207
⑮30S
E3
④
2Cron
⑯
H2nOb ⑧H3
- ⑰
CH3COOn
--
⑤
20 y ⑱
M25208
③
#Hy
--
⑥282 ⑮
By010
⑳
Na-leg
⑦KO2
⑳
302
⑧⑳2
⑨HOC
④
BVOg
-
-
⑩
22
HC02 OF3Ou
-
⑪
HCO3 ⑬aH2
-
-
⑫CLO
Y
⑳ N92S203
-

6. #
Sck
Notations
=>Representationofelementswiththeir
OxidationNo
·inRomanletters
.
#ent/oxidant
=>Reactantwhichundergoes
reductionitself
=>
oxidising
"¥
Elementswith
high"¥elements
"¥
intheirhighest
oxidationstate
compoundswhichcaneasily
donate
oxygen
=>
ReducingAgent/Reductant
=Reactant
which
undergoes
Oxidationitself
.
-
"¥
elementswith
highelectropositive
nature
"¥
elementintheirlowest
OxidationstateLe
compoundswhichcaneasily
donateHydrogen
Highest
Oxidationstate=
Behaveas
oxidisingagent
LO
.
A)
E-&
inblwOxidationstate
=Both0
·A&R
.A
Elements
"¥
lowestOxidationstate
=Behaveas
Reducingagent
(RA)

7. #
ApplicationsofOxidationNumber
①Strengthof
Aid&OxidationNumber
Strength
ofBase
I
--
OxidationNumber
⑳EquivalentMars
(E)
Equivalent
mansof
X
=
Marman
of
X
n-factor
For
oxidis,
agent
=>
n-factor
=no
.ofegaineding
Lea
vivalentman
of
0.A=
Normans of
o
e
ForReducingagent
=>
n-factor=noofeflost
"¥
quivalent
manofRA=
Marmanofoa

8. #esofRedoxReadine
ion- atleasttwodifferent
species
①Redox
combine
to
form
athirdcompound
A+B->C with
change
inOxidation
No.
->eitherAorBorBothA&Bmustbeinelemental
form
.
②
Redox
DecompositionReactionAcompound
breakstoform
morethan
oneproduct
with
change
inO
.N
-
A-
B+C
->eitherBorCmustbeinelementalform
③Redoxdisplacement
reaction
-
change
inOxidationNo.
-
d
"¥
metal
displacement
x
=>
Ahigh
activemetal
non-metaldisplacement
-
+
displaces
anothermetal
Hydrogendisplacement
+
2n+
Sou->
Ins0yflu
Halogendisplacement
.

9. ④isproportionationreactionsspecialtypeof
RedoxRXR.
->sameelement
undergoes
oxidationaswellasReductionin
the2x4
.
-Reverse
of
disproportionation
=
Comproportionation
.
-Toshow
disproportionation
,any
elementmusthaveatleastthree
--
Some
importantdisproportionation
"¥
Oxidation
Number
XMS .
02
-40+400 Highest
&lowestOxidation
Pu
->
PH3
+H2PO
Statecannotshow
disproportional
H202
->120
+
02
-
-
A
*
-O
->S-
2
+5203
*
↓I
Showdispropontimate
Mn0,
2
->Mo
+
MnO2"¥
+
33
Sg C
"¥
HMOz-HMOz
+NO
OX

10. Redox
Rx
canbe
of
two
types
->Intermolecular
Redox
-
to
WhenOxidation&Reductiontakes
L
IntramolecularRedox
place
separately
in
different
-
↓ compounds
.
WhenOxidationandreductiontake
place
ina
singlecompound
. Snd2
+
Fed-Sndu
+Fede
KC0z
-KU+
02

11. #
Balancing
ofRedoxReactions
-Half-rx"method/im-electron
"¥
acidic/neutralmedium
-
OxidationMo
.method
Lo "¥
asicmedium.
atomsa
chargesareconserved
=>xidationNo.method(usedinexams)
Step-1-
writeIonic
equation
andbalanceatomsexcept⑪ inQ
O-S
⑧in
Step
-It
CalculateO.N
.
of
eachelementand
also
calculateincicance
-
anddecreaseino
.Mand
makethem
equalby
multiplying
a
suitableNo.
#
-Balancecharge
by
adding
no
inacidic
medium/neutral
Ontin
Basicmedium
Step-I-Balance0byadding
suitableno
of H
20
-

12. OfApplications
RedoxRXin
Chemistry
On-factor
acidic2
-
->
in E=
medium
5
&
Mnot weakalkaline
+Y
Y-
Mr02E
=M
-
or
<
neutral 3
O
.N=⑰
"¥
Basic
MnO?
E
=
+medium
-2
acidic
C207-Crt3n-factor
=5
medium
during
xidation
-
>n-factor
=no
.of
20lost/increase
inOxidation
No-
Reduction
=>
n-factor
=no
·
ofegained/decrease
inOxidation
during-
No.
Redox
Rx
=n-factor
=(increas=decrease)
inOxidationNo
.

13. ②no
.
ofequivalents
=noof
molsxn-factor
=
Molarity
xVolume(sxn-factor
=
Normality
xVolume(1)
Lawofequivalence
-
During
achemicalwxn
,
no
of
movesof
reactants
or
productsmay
or
may
notbe
equal
.Buttheirnoof
equivalents
are
alwaysequal
A+bB
-C+dD
no
of
mols
of
AB+C+D
But
(e)
A
=B=c
=D
③
rain-selfindicator
iodine
gives
BluecolourwithStarch
.