Redox Reaction of class 11 cbse isc.pptx

Redox Reaction of class 11 cbse

1. 11C08 Redox Reaction

2. 11C08.1 Introduction to Redox Reaction

3. Battery Combustion Rusting Refining of Impure metal Examples of Redox Reaction

4. Learning Objectives Classical and electronic concept of Oxidation and Reduction Oxidation number 11C08.1 Introduction to Redox Reaction

5. 11C08.1 CV 1 Classical and Electronic Concept of Oxidation and Reduction

6. Redox = Red uction + Ox idation T he reaction in which oxidation and reduction reactions occur simultaneously . Redox Reaction

7. Oxidation Addition of oxygen/electronegative element to a substance Removal of hydrogen/electropositive element from a substance             Classical Concept

8. Addition of hydrogen/electropositive element to a substance Removal of oxygen/electronegative element from a substance         Reduction Classical Concept

9.     Transfer of electron Electronic Concept

10.   Transfer of electron       Loss of electron   Electronic Concept

11. Oxidation Increase in positive charge Decrease in negative charge     Loss of electron Electronic Concept

12. Decrease in positive charge Increase in negative charge     Reduction Gain of electron Electronic Concept

13. Example-       Electronic Concept

14. Intensity of blue colour is reduced Initial stage Intermediate stage Final stage Cu +2 deposited as Cu Zn rod             Electron Transfer Reactions

15. Intensity of blue colour is reduced Initial stage Intermediate stage Final stage Cu +2 deposited as Cu   R elease of 2e - G ain of 2e - Zn rod       E lectron releasing tendency of the metals : Zn > Cu         Electron Transfer Reactions

16. Blue colour starts developing Initial stage Intermediate stage Final stage Ag deposited Cu rod             Intensity of blue colour increases Electron Transfer Reactions

17. Blue colour starts developing Initial stage Intermediate stage Final stage Ag deposited Cu rod       E lectron releasing tendency of the metals : Cu > Ag       R elease of 2e - G ain of 2e - Intensity of blue colour increases     Electron Transfer Reactions

18. Oxidation Reduction Addition of oxygen/electronegative element to a substance Removal of hydrogen/electropositive element from a substance Removal of oxygen/electronegative element from a substance Addition of hydrogen/electropositive element to a substance Loss of electron Gain of electron

19. 11C08.1 CV 2 Oxidation Number

20. R eal or imaginary charge on an atom in the compound     . .     . . High E.N. -1 + 1 Oxidation Number (O.N.)

21. In elements, in the free or the uncombined state , each atom bears an oxidation number of zero                                                       . . . . . . . . . . Rules for Calculating Oxidation Number . .

22. For ions composed of only one atom; O.N. = charge on the ion.   +2   +1   - 1 O.N.= +1 O.N.= +2 O.N.= -1 O.N. of Alkali metal in their compounds= +1 O.N. of Alkaline earth metal in their compounds = + 2 O.N. of Aluminum in its all compounds = +3 NOTE: Rules for Calculating Oxidation Number

23. 3. O.N. of oxygen : In peroxide   In super peroxide           O.N.= -2 O.N.= -1       +1 +1 +1 -1 -1 +1 - 1 - 1 . . . .       . . . . . . Mostly =   Rules for Calculating Oxidation Number

24. In   O.N.= +1 O.N.= +2 In                 -1 +1 -1 -1 +1 -1 +1 . . +1 . . . . . . . . Rules for Calculating Oxidation Number

25. O.N. of Hydrogen :       O.N.= +1 Exception- When H is bound to metal, O.N. = -1 e.g. - LiH , NaH , CaH 2 +1 -1 -1 + 1   O.N.= -1 . . . . Mostly +1 Rules for Calculating Oxidation Number

26. 5.           O.N. = -1, halide ion Cl, Br and I when combine with oxygen ( oxoacid and oxoanion ) P ositive oxidation number For ,   For   O.N=+7 O.N=+7 Rules for Calculating Oxidation Number

27. For a comound : Algebraic sum of O.N. of all the atoms = 0 For polyatomic ion : A lgebraic sum of O.N. of atoms = Charge on the ion .         Rules for Calculating Oxidation Number

28. Metallic element Positive O.N. Non Metallic element Negative O.N. Transition metal Several positive O.N. Group 1 2 13 14 15 16 17 Element Na Mg Al Si P S Cl Compound NaCl MgSO 4 AlF 3 SiCl 4 P 4 O 10 SF 6 HClO 4 Highest O.N. state of the group element +1 +2 +3 +4 +5 +6 +7 OXIDATION NUMBER   Rules for Calculating Oxidation Number

29. . . . . . . - 2 . . . . . . +2 . . . . - 2         +2 Individual Oxidation Number

30. . . . . . . - 2 . . . . . . +2 . . . . - 2         = +2+0 = +2 + 2 Individual Oxidation Number

31. . . . . . . - 2 . . . . . . +2 . . . . - 2         = 0+0 = 0 +2 Individual Oxidation Number

32. . . . . . . - 2 . . . . . . . . . . - 2         = 0+(+2) = +2 +2 + 2 Individual Oxidation Number

33.         . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - 2 + 2 - 2 + 2 - 2 + 2 + 2 + 2 - 2 - 2 - 2 - 2 - 2 + 2 + 2 + 2 Individual Oxidation Number

34.       . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - 2 - 2 - 2 + 2 + 2 - 2 - 2 - 2 - 2 - 2 + 2 + 2 + 2 = +2+2+2+0 = +6 + 2 + 2 + 2   Individual Oxidation Number

35.       . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - 2 + 2 - 2 + 2 - 2 + 2 - 2 - 2 - 2 - 2 - 2 + 2 + 2 + 2 = +2+2+0+0 = +4 + 2 + 2   Individual Oxidation Number

36.       . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - 2 + 2 - 2 + 2 - 2 + 2 + 2 + 2 - 2 - 2 - 2 - 2 - 2 = +2+2+2+0 = +6 + 2 + 2 + 2   Individual Oxidation Number

37.         . . . . . . - 1 + 1 + 2 - 2 - 2 - 2 - 2 . . . . - 1 + 1 + 2 + 2 + 2 . . . . . . . . . . . . . . . . Individual Oxidation Number

38.       . . . . . . - 1 - 2 - 2 . . . . - 1 + 1 . . . . . . . . . . . . . . . . + 2 + 2 + 1 = +2+2+1+0 = +5 - 2 - 2 + 2 + 2   Individual Oxidation Number

39.       . . . . . . - 1 + 1 + 2 - 2 - 2 . . . . - 1 + 1 + 2 . . . . . . . . . . . . . . . . = 0+0 = 0 - 2 - 2 + 2 + 2   Individual Oxidation Number

40.       . . . . . . - 1 + 1 + 2 - 2 - 2 . . . . - 1 + 1 + 2 . . . . . . . . . . . . . . . . = 0+0 = 0 - 2 - 2 + 2 + 2   Individual Oxidation Number

41.       . . . . . . - 1 + 1 + 2 - 2 - 2 - 2 . . . . - 1 + 2 . . . . . . . . . . . . . . . . = +2+2+1+0 = +5 + 2 + 1 - 2 - 2 + 2   Individual Oxidation Number

42. Alfred Stock Example :– Stannous chloride : Sn(II)Cl 2 S tannic chloride : Sn(IV)Cl 4 Stock Notation

43. Oxidation - O.N. of the element in the given substance   Reduction - O.N. of the element in the given substance   Reducing agent (R.A.)/Reductants : A reagent which can O.N. of the element in the substance Oxidizing agent (O.A.)/Oxidants : A reagent which can O.N. of the element in the substance Redox reaction - Reactions which involve change in O.N.

44.   +1 - 2 +1 - 2 +4 - 2 Reduction Oxidation     Example-

45. Oxidation Reduction Addition of oxygen/electronegative element to a substance Removal of hydrogen/electropositive element from a substance Removal of oxygen/electronegative element from a substance Addition of hydrogen/electropositive element to a substance Loss of electron Gain of electron O.N . of the element in the given substance   O.N . of the element in the given substance  

46. 11C08.1 PSV 1

47. Q.       Sol. Oxidation +3 +7 +2 +4 Reduction In this reaction act as a reductant so ans is (A)  

48. ConcepTest Ready for a Challenge

49. Pause the video Time duration : 1 minute Q. What is the O.N. of Cl atom in and ?  

50.     O.N=+7 . . . . -2 + 2 . . . . + 2 -2 + 2 . . -2 . . +1 -1 . . O.N=+7 . . + 1 -1 . . . . -2 + 2 . . . . + 2 -2 . . . . + 2 -2 . . + 1 -1 Sol. Q. What is the O.N. of Cl atom in and ?  

51. Summary Redox Reaction - The reaction in which oxidation and reduction reactions occur simultaneously. Oxidation Number(O.N.)- It represents the real or imaginary charge on an atom or element in the compound Oxidation Reduction Addition of oxygen/electronegative element to a substance Removal of oxygen/electronegative element from a substance Removal of hydrogen/electropositive element from a substance Addition of hydrogen/electropositive element to a substance Loss of electron Gain of electron O.N. of the element in the given substance O.N. of the element in the given substance

52. Reference Questions NCERT Exercises: 1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 13, 14, 16, 17, 22 NCERT In text : 1, 2, 3, 4 Workbook Questions : 1, 2, 3, 4, 11, 12, 13, 16 11C08.1 Introduction to Redox Reaction

53. 11C08 .2 Types of Redox Reaction

54. Learning Objectives Combination and Decomposition Reaction Displacement Reaction Disproportionation Reaction 11C08 .2 Types of Redox Reactions

55. 11C08.2 CV 1 Combination and Decomposition Reaction

56. Combination reaction may be denoted as: Either one must be in elemental form             Combustion Reactions                         + Combination Reaction

57. Combination reaction may be denoted as: Either one must be in elemental form             Combustion Reactions Which make use of elemental dioxygen, are redox reaction                         + Combination Reaction

58. Combination reaction may be denoted as: Either one must be in elemental form                         Combustion Reactions Which make use of elemental dioxygen, are redox reaction Other Reactions                         + Combination Reaction

59. Opposite of combination reactions Breakdown of a compound into two or more components One of them must be in elemental state                                         Decomposition Reaction

60. All decomposition reactions are not redox reaction                   No change in O.N. Decomposition Reaction

61. All decomposition reactions are not redox reaction                   No change in O.N. Decomposition Reaction

62. 11C08.2 CV 2 Displacement Reaction

63. A n ion (or an atom) in a compound is replaced by an ion (or an atom) of another element Two types Metal displacement Non-metal displacement             + + Replaces Displacement Reactions

64. A metal in a compound can be displayed by another metal in the uncombined state Finds application in purification of metals                   Metal Displacement

65. A metal in a compound can be displayed by another metal in the uncombined state Finds application in purification of metals                   Reduced Metal Displacement

66. A metal in a compound can be displayed by another metal in the uncombined state Finds application in purification of metals                   Oxidised Metal Displacement

67. A metal in a compound can be displayed by another metal in the uncombined state Finds application in purification of metals                                   Reduced Metal Displacement

68. A metal in a compound can be displayed by another metal in the uncombined state Finds application in purification of metals                                   Oxidised Metal Displacement

69. Zn > Cu > Ag Metal activity series

70. It includes hydrogen displacement and a rarely occurring reaction involving oxygen displacement                 Water Sodium Fire Non-metal Displacement

71. It includes hydrogen displacement and a rarely occurring reaction involving oxygen displacement                 Water Sodium Fire               Non-metal Displacement

72. It includes hydrogen displacement and a rarely occurring reaction involving oxygen displacement All alkali metals and some alkaline earth metals and will displace hydrogen from cold water                                   Water Sodium Fire               Non-metal Displacement

73. Less active metals such as and react with steam to produce gas                                     Production of dihydrogen gas Non-metal Displacement

74. Many metals are capable of displacing hydrogen from acids                         Used to prepare g as in the laboratory   Metals Acids Dihydrogen gas Salt + +     Non-metal Displacement

75. Note- and do not react with steam but are capable of displacing hydrogen from acids       Non-metal Displacement

76. 9 F 17 Cl 35 Br 53 I Halogens OXIDISING POWER         Displacement reactions of Cl, Br and I using fluorine are not carried out in aqueous solution     Halogen Activity Series

77.                             Chlorine can displace bromide and Iodide ions in an aqueous solution F or identifying and   Gives colour Layer Test Halogen Activity Series

78. Bromine can displace Iodide ion in solution   Halogen displacement reaction have industrial application Halides Halogen Oxidation Process Recovery   strongest oxidising agent oxidise electrolytically   Fluorine cannot recover from this method Halogen Activity Series

79. 11C08.2 CV 3 Disproportionation Reaction

80. A n element is simultaneously oxidised and reduced One of the reacting elements should exist in atleast three oxidation state             Disproportionation Reaction

81.                   P, S and Cl undergo disproportionation in alkaline medium :     Disproportionation Reaction

82.                   P, S and Cl undergo disproportionation in alkaline medium :                       Disproportionation Reaction

83.                   Oxidises colour bearing stains of the substance to colourless compounds P, S and Cl undergo disproportionation in alkaline medium :                                         Disproportionation Reaction

84. Fluorine reacts with alkali as: Being most E.N. element, it cannot exhibit any positive oxidation state F does not exhibit disproportionation tendency                   Exception of Disproportionation Reaction

85. ConcepTest Ready for a Challenge

86. Q . Pause the video Time duration: 2 minute Identify Disproportionation Reaction        

87. Sol.   - 4 +1 +4 -2 +1 -2 - 1   - 4 +1 +4 +1 - 1   -2 +1 - 1 +1 -2   +4 - 2 +1 -2 +3 -2 +5 -2 -2 +1 Ans -(D) Q . Identify Disproportionation Reaction        

88. Summary All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reaction Decomposition reaction leads to the breakdown of a compound into two or more components In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or atom) of another element

89. Reference Questions NCERT Exercises : 8.4, 8.15, 8.24 and 8.28 NCERT In text : 5, 6 and 7 Workbook Questions : 5 and 19 11C08 .2 Types of Redox Reactions

90. 11C08 .3 Balancing of Redox Reaction

91. Learning Objectives Oxidation number method Half reaction method 11C08 .3 Balancing of Redox Reaction

92. 11C08.3 CV 1 Oxidation number method

93. Lets take an example of unbalanced equation in acidic medium Step 1 : Assign the oxidation number to all elements in the reaction.   +6 - 2 +4 - 2 +3 +6 -2     Oxidation Number Method  

94. Lets take an example of unbalanced equation in acidic medium Step 1 : Assign the oxidation number to all elements in the reaction.   +6 - 2 +4 - 2 +3 +6 -2           Oxidation Number Method

95.   +6 - 2 +4 - 2 +3 +6 -2 Step 3: Find the total change in O.N. for entire molecule. For , -3x2 = -6       For , +2x1 = +2   Oxidation Number Method Step 2: Balance atoms undergoing change in O.N.  

96. Step 5: Charge balance using H + (acids) , OH - (basic)   Step 4: Balance the or O.N. by multiplying some suitable number   Oxidation Number Method   +6 - 2 +4 - 2 +3 +6 -2            

97. Step 6: Add H 2 O to balance H & O atom   Step7: Verification:   Oxidation Number Method Atoms of each element Total Charge  

98. Q. Balance this equation by oxidation method in basic medium Sol.     +7 -1 +4 +5 Step 1 : Assign the oxidation number to all elements in the reaction.    

99. Q. Balance this equation by oxidation method in basic medium Sol.     +7 -1 +4 +5 Step 1 : Assign the oxidation number to all elements in the reaction.        

100.   +7 -1 +4 +5 Step 2: Balance atoms undergoing change in O.N.     Step 3: Find the total change in O.N. for entire molecule. For , -3x1 = -3   For , +6x1 = +6  

101. Step 4: Balance the or O.N. by multiplying some suitable number     +7 -1 +4 +5      

102. Step 5: Charge balance using OH - (basic)   Step 4: Balance the or O.N. by multiplying some suitable number     +7 -1 +4 +5          

103. Step 6: Add H 2 O to balance H & O atom   Atoms of each element Total Charge   Step7: Verification:

104. 11C08 .3 PSV 1

105. Q. Balance the following equation in basic medium by oxidation number method. Sol .     - 1 +5 -2 +2 Step 1 : Assign the oxidation number to all elements in the reaction.    

106. Q. Balance the following equation in basic medium by oxidation number method. Sol .     - 1 +5 -2 +2 Step 1 : Assign the oxidation number to all elements in the reaction.        

107. Step 2: Balance atoms undergoing change in O.N.     - 1 +5 -2 +2

108. Step 2: Balance atoms undergoing change in O.N.     - 1 +5 -2 +2    

109. Step 2: Balance atoms undergoing change in O.N.   -2 Step 3: Find the total change in O.N. for entire molecule. For , +4x2 = +8   For , -6x1 = -6       - 1   +5 +2

110. Step 4: Balance the or O.N. by multiplying some suitable number     - 1 +5 -2 +2  

111. Step 4: Balance the or O.N. by multiplying some suitable number         +5 -2   - 1 +2

112. Step 4: Balance the or O.N. by multiplying some suitable number           +5 -2   - 1 +2

113. Step 4: Balance the or O.N. by multiplying some suitable number             - 1 +5 -2 +2    

114. Step 4: Balance the or O.N. by multiplying some suitable number   Step 5: Charge balance using OH - (basic) Charge is already balanced           - 1 +5 -2 +2    

115. Step 6: Add H 2 O to balance H & O atom   Atoms of each element Total Charge     Step7: Verification:

116. 11C08.3 CV 2 Half reaction Method

117. Lets take an example   Step-1 Separate the equation into half – reactions Oxidation half reaction:   +3   + 2 + 6 +3 I on-electron method Half Reaction Method

118. Lets take an example   Step-1 Separate the equation into half – reactions Oxidation half reaction:   Reduction half reaction:   I on-electron method Half Reaction Method +3   + 2 + 6 +3

119. Half Reaction Method +3   + 2 + 6 +3 Oxidation half reaction Reduction half reaction Step-2 Balance the atom other than O and H in each half reaction individually.   (Fe is already balanced)    

120. Half Reaction Method +3   + 2 + 6 +3 Oxidation half reaction Reduction half reaction       Step-3 For acidic medium, add H 2 O to balance O atoms and H + to balance H atoms  

121.   Half Reaction Method +3   + 2 + 6 +3 Oxidation half reaction Reduction half reaction   Step-4 Add e - to one side of the half reaction to balance the charge    

122.     Half Reaction Method +3   + 2 + 6 +3 Oxidation half reaction Reduction half reaction   Step-5 Multiply with suitable number to equalize e - in both half reaction      

123. Half Reaction Method +3   + 2 + 6 +3 Oxidation half reaction Reduction half reaction Step-6 : Now add the two half reactions and cancel the e - on each side.          

124. Half Reaction Method +3   + 2 + 6 +3 Step-7 Verification: Atoms of each element Total Charge   Note- For reaction in basic medium. Firstly, follow all the steps of acidic medium . For each H + ion , add equal no. of OH - ion to both sides of the equation .

125. Q. Balance the following redox reaction in basic medium by ion-electron method. Sol .   +7   +4 -1 Oxidation half reaction:   Step-1 Separate the equation into half – reactions

126. +7   +4 -1 Q. Balance the following redox reaction in basic medium by ion-electron method. Sol .   Oxidation half reaction:   Reduction half reaction:   Step-1 Separate the equation into half – reactions

127.   Step-2 Balance the atom other than O and H in each half reaction individually. Oxidation half reaction Reduction half reaction      

128.     Step-2 Balance the atom other than O and H in each half reaction individually. Oxidation half reaction Reduction half reaction     ( Mn is already balanced)

129.   Oxidation half reaction Reduction half reaction   Step-3 A dd H 2 O to balance O atoms and H + to balance H atoms  

130.   Oxidation half reaction Reduction half reaction     Step-3 A dd H 2 O to balance O atoms and H + to balance H atoms  

131.   Oxidation half reaction Reduction half reaction     Step-3 A dd H 2 O to balance O atoms and H + to balance H atoms    

132.     Oxidation half reaction Reduction half reaction   Step-4 Add e - to one side of the half reaction to balance the charge  

133.   Oxidation half reaction Reduction half reaction   Step-4 Add e - to one side of the half reaction to balance the charge      

134.     Oxidation half reaction Reduction half reaction Step-5 Multiply with suitable number to equalize e - in both half reaction              

135.   Oxidation half reaction Reduction half reaction     Step-6 : Now add the two half reactions and cancel the e - on each side.    

136.   Step-7 In basic medium, For each H + ion , add equal no. of OH - ion to both sides of the equation            

137.   Step-8 Verification: Atoms of each element Total Charge  

138. ConcepTest Ready for a Challenge

139. Q. Balance the following redox reaction in acidic medium by ion-electron method.   Pause the video Time duration: 2 minute

140. Q. Balance the following redox reaction in acidic medium by ion-electron method.   Sol . Step-1 Separate the equation into half – reactions Oxidation half reaction:     +7 +4 +2 +6

141. Q. Balance the following redox reaction in acidic medium by ion-electron method.   Sol . Step-1 Separate the equation into half – reactions Oxidation half reaction:   Reduction half reaction:     +7 +4 +2 +6

142.   Step-2 Balance the atom other than O and H in each half reaction individually. Oxidation half reaction Reduction half reaction     ( Mn is already balanced) ( S is already balanced)

143.   Oxidation half reaction Reduction half reaction Step-3 A dd H 2 O to balance O atoms and H + to balance H atoms        

144.   Oxidation half reaction Reduction half reaction Step-3 A dd H 2 O to balance O atoms and H + to balance H atoms            

145.       Oxidation half reaction Reduction half reaction Step-4 Add e - to one side of the half reaction to balance the charge  

146.       Oxidation half reaction Reduction half reaction Step-4 Add e - to one side of the half reaction to balance the charge    

147.   Oxidation half reaction Reduction half reaction Step-5 Multiply with suitable number to equalize e - in both half reaction          

148.       Oxidation half reaction Reduction half reaction Step-5 Multiply with suitable number to equalize e - in both half reaction            

149.   Oxidation half reaction Reduction half reaction     Step-6 : Now add the two half reactions and cancel the e - on each side.    

150.   Step-7 Verification: Atoms of each element Total Charge  

151. Reference Questions NCERT Exercises : 8.18, 8.19, 8.23, and 8.25 NCERT In text : 8, 9 and 10 Workbook Questions : 17 and 18 11C08 .3 Balancing of Redox Reaction

152. 11C08.4 Applications of Redox Reactions

153. 11C08.4 Applications of Redox Reactions Learning Objectives Redox Reactions as the basis for Titrations Electrochemical cell Electrode Potential

154. 11C08.4 CV 1 Redox Reaction as the basis for Titrations

155. Redox Reaction as the basis of Titration Redox Titrations Oxidant Reductant Indicator Method to determine the strength of a reductant/oxidant using a redox sensitive indicator Substance that changes color in response to a chemical change

156. Oxidizing Agents used in Redox Titration Potassium Permanganate   Potassium Dichromate   Iodine  

157. Potassium Permanganate (oxidant) Self indicator   Reductant   End Point   Pink colour

158. Potassium Dichromate (oxidant) Not a self indicator   Reductant   Diphenylamine / Diphenylbenzidine Intense blue colour End Point  

159. Iodine     KI solution Oxidising Agent /   Iodine gives an intense blue colour with starch Starch

160. Iodine   solution   Sodium T hiosulphate Starch   End Point

161. 11C08.4 CV 2 Electrochemical Cell

162. Electrochemical Cell Devices which are used to convert electrical energy into chemical energy and vice versa Electrical energy is used to carry out a non-spontaneous redox reaction. Chemical energy of a spontaneous redox reaction is converted into electrical energy. Battery Anode Cathode Salt Solution Cathode Anode Salt Bridge

163. Galvanic Cell LOAN Left Oxidation Anode Negative     Zn Anode Cu Cathode Overall cell reaction   Solution   Solution   Voltmeter Anode Cathode Salt Bridge NaCl ( aq )

164. Salt Bridge U-shaped tube containing a chemically inert salt like Mobility of cation = Mobility of anion Completes the electrical circuit. Maintains the electrical neutrality of both electrodes. Maintains the spontaneity of cell reaction.   Salt Bridge

165. IUPAC representation of Galvanic Cell At anode i.e. oxidation half: IUPAC notation: At cathode i.e. reduction half: IUPAC notation: Overall cell reaction: IUPAC notation:   Zn Anode Cu Cathode Voltmeter Salt Bridge Galvanic Cell

166. 11C08.4 CV 3 Electrode Potential

167. Electrode Potential Potential difference developed between electrode and electrolyte. Anode: Solution potential < Metal potential   Cathode: Solution potential > Metal potential   According to IUPAC convention, standard reduction potentials are now called standard electrode potentials . Electrode Potential Electromotive Force      

168. Standard Hydrogen Electrode         solution   rod   surface   STP: and      

169. Determination of standard electrode potential of metals     solution   rod       Standard metal electrode Metal rod

170. Electrochemical Series Elements Electrode Reaction Elements Electrode Reaction (SHE) Oxidising Nature Reducing Nature

171. Selection of oxidising and reducing agents Higher the Reduction Potential higher will be the tendency to get reduced and better oxidising agent, So silver will undergo reduction. Lower the Reduction Potential higher will be the tendency to get oxidised and better reducing agent, So chromium will undergo oxidation.   (SHE)

172. Reactivity of Metal Lower the Reduction Potential (more negative) higher will be the reactivity of metal , So chromium more reactive metal.   (SHE)

173. 11C08.4 PSV 1

174. Q. Given the standard electrode potentials, Arrange these metals in their increasing order of oxidising power. Sol. We know that, Higher the electrode potential, higher will the oxidising power of metal. So the order will be,  

175. Summary Redox titrations Oxidizing agents: , ,   Electrochemical Cell (Galvanic cell) Salt bridge Electrode Potential- Potential difference developed between electrode and electrolyte Higher the Reduction Potential higher will be the tendency to get reduced and better oxidising agent

176. 11C08.4 Applications of Redox Reactions Reference Questions NCERT Exercises: 26 to 30 Workbook Questions : 6 to 10, 14, 15, 20