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Redox titrations   jsk nagarajan
 

Redox titrations jsk nagarajan

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  • Slide 2:Titrations are one of the two types of Classical Quantitative Analysis. What is the other classical quantitative analysis?Exactly, the other classical quantitative chemical analysis is gravimetry. You will see gravimetry in other parts of the course.Let’s continue by asking about titrations.What are the four types of titrations?Yes, remember the four types are:Firstly, acid-base titrations, secondly complexometric titrations, thirdly precipitation titrations and fourthly redox titrations.Remember that we started working with acid-base titrations, then we moved on to complexometric titrations and finally we saw precipitation titrations. Today we will be looking at iodometric and iodimetric titrations, which are examples of redox titrations.We have left redox titrations until now, because you needed to be familiar with the other three type of titrations.So, let’s look at redox titrations in more detail. Do you remember other redox titrations that we have done in the laboratory?Yes, we have done other redox titrations like the determination of the percent of hydrogen peroxide and other ones.Now, we are going to look at the redox titrations involving iodine. Notice here that there are two types of redox titrations involving iodine.The most important thing in this presentation is for you to understand the differences between iodometric and iodimetric titrations. Both involve iodine, but as you will see there are some differences.The analysis that we will perform in the laboratory is the iodometric titration of cooper, which is a classical quantitative chemical analysis, a redox titration involving iodine. It is used because it is necessary to quantify copper in water, alloys, minerals and so on
  • Slide 4:Before we continue, let’s review what we mean by the word titration.A definition of the word titration is:A titration is a procedure in which volume increments of the known reagent solution-which is called the titrant- are added to the analyte until the reaction is complete. Does anybody remember the four types of reactions that we call fundamental analytical reactions? Exactly, the four types are acid-base, complexometric, precipitation and redox. How do we classify titrations?Remember, we classify titrations according to the type of reaction between the titrant and the analyte.As you can see from the diagram, the titrant is usually delivered from a buret. Do you remember in which titrations the titrant is in the Erlenmeyer flask?That’s right, the titrant is in the Erlenmyer flask in standardization titrations.

Redox titrations   jsk nagarajan Redox titrations jsk nagarajan Presentation Transcript

  • REDOX TITRATIONSElectron Transfer Titrations J.S.K.NAGARAJAN ASST. PROF OFF CAMPUS -JSS UNIVERSITY J.S.S. COLLEGE OF PHARMACY, OOTACAMUND – 643 001.
  • REDOX TITRATIONSGravimetry
  • What is a titration? Act of adding standard solution in small quantities to the test solution till the reaction is complete is termed titration.What is a standard solution?A standard solution is one whose concentration is precisely known.What is a test solution?A test solution is one whose concentration is to be estimated
  • REDOX TITRATIONSTitrations ExamplesAcid-base Quantification of acetic acid in vinegarComplexometric Quantification of chloride (Cl-) in waterPrecipitation Water Hardness (Calcium & Magnesium)Redox Quantification of H2O2
  • REDOX TITRATIONS Titration Analyte Titrant Indicator exampleAcid-base Quantification Acetic acid NaOH Phenolphthale of acetic acid (CH3COOH) in in avinegarComplexometric Water Ca 2+ , Mg 2+ EDTA Eriochrome Hardness black T MurexidePrecipitation Quantification Chlordie AgNO3 Mohr, of chloride Volhard, (Cl-) in water FajansRedox Quantification Hydrogen peroxide KMnO4 No indicator of hydrogen (H2O2) peroxide (H2O2)
  • REDOX TITRATIONSTitrations Example Type of TitrationsAcid-base Quantification of acetic acid in vinegar ■ Direct □ Indirect □ BackComplexome Water Hardness (Ca 2+ & Mg2+)tric ■ Direct □ Indirect □ BackPrecipitation Quantification of Mohr Method Cl in Water ■ Direct □ Indirect □ Back Fajans Method ■ Direct □ Indirect □ Back Volhard Method □ Direct □ Indirect ■ BackRedox Quantification of hydrogen peroxide (H2O2) ■ Direct □ Indirect □ Back
  • REDOX TITRATIONSAcid/Base reactions - involves donation /acceptance of protons SPrecipitation/ Solubility reactions : Involve donation/ acceptance of negativecharge IREDOX TITRATION: what is being donated and accepted in a redox reaction? M U L E l e c t r o n s ! TConsider the reaction taking place in a disposable battery: A2Zn + 3MnO2 Mn3O4 + 2ZnO N EHow can you tell that electrons are being donated and accepted? Whichspecies is donating electron( s) and which is accepting electron (s)? O UTr a n s f e r l e a d s t o - SIncrease in ON of element = O X I D A T I O N LDecrease in ON of element = R E D U C T I O N Y
  • REDOX TITRATIONSOxidation Reduction Loss of electrons Gain of electrons Gain in oxygen Loss of oxygen 0 1 Cl e ClSodium is oxidized Chlorine is reduced LEO SAYS GER
  • REDOX TITRATIONS REDUCTIONOXIDATION: Old definition: Removal of oxygen from a compoundOld Definition:Combination of substance with WO3 (s) + 3H2(g) W(s) + 3H2O(g)oxygenC (s) + O2(g) CO2(g) Current definition: Gain of Electrons is Reduction (GER)Current definition: Cl + e- Cl -Loss of Electrons is Oxidation (LEO)Na Na + + e- Negative charge represents electron richness ONE NEGATIVE CHARGE MEANS RICHPositive charge represents electron BY ONE ELECTRONdeficiencyPositive OS reflects the tendency atom to loose electronsNegative OS reflects the tendency atom to gain electrons
  • REDOX TITRATIONS:•Direct Titrations•Indirect Titrations•Back Titrations
  • REDOX TITRATIONSOBJECTIVESDefine oxidation & reduction in terms of loss or gain of O2/Electrons electronsOBJECTIVESState the characteristics of a redox reaction and identify the oxidizing agentand reducing agent.Many of reactions may not even involve oxygenRedox- electrons are transferred between reactants Mg + S→ Mg2+ + S2- (MgS)Magnesium atom (has zero charge) changes to a Mg ion by losing 2 electrons,and is oxidized to Mg2+The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2electrons, and is reduced to S2-Active metals: Lose electrons easily- easily oxidized- strong reducing agentsActive nonmetals: Gain electrons easily-easily reduced-strong oxidizing agents
  • REDOX TITRATIONSLosing electrons is oxidation & substance thatloses the electrons is called the reducing agentGaining electrons is reduction, & substance thatgains the electrons is called the oxidizing agentMg is thereducing Mg(s) + S(s) → MgS(s) agent Mg is oxidized: loses e-, becomes a Mg2+ ion S is the oxidizing agent S is reduced: gains e- = S2- ion0 1Na Na e Sodium is oxidized – it is the reducing agent0 1Cl e Cl Chlorine is reduced – is the oxidizing agent
  • REDOX TITRATIONS The reaction of a metal and non-metal All the electrons must be accounted for! Mg + S 2+ + 2- → Mg S
  • REDOX TITRATIONS In water, oxygen is highly electronegative, so:It is easy to see theloss and gain of • Oxygen gains electrons (is reduced & is oxidizingelectrons in ionic agent)compounds, butwhat aboutcovalent • Hydrogen loses electrons (is oxidized & is reducingcompounds? agent)
  • REDOX TITRATIONSNot All Reactions are Redox ReactionsReactions - no change in ON are NOT redox reactions.Examples: 1 5 2 1 1 1 1 1 5 2Ag N O 3 ( aq ) Na Cl ( aq ) Ag Cl ( s ) Na N O 3 ( aq ) 1 2 1 1 6 2 1 6 2 1 22 Na O H ( aq ) H 2 S O 4 ( aq ) Na 2 S O 4 ( aq ) H 2 O (l ) Not all oxidation processes that use oxygen involve burning: Elemental iron slowly oxidizes to iron (III) oxide- “rust” Bleaching stains in fabrics H2O2 - releases oxygen when decomposes
  • REDOX TITRATIONSCorrosionDamage done to metal is costly to prevent and repair.Iron- Metal, corrodes by being oxidized to ions of iron by oxygen.Corrosion is faster in presence of salts & acids, because these materialsmake electrically conductive solutions that make electron transfer easyL u c kily, n o t a l l m e t a ls c o r rode e a s ilyGold & platinum are noble metals - are resistant to losing their electrons by corrosion.Other metals lose their electrons easily, but protected by coating on surface, such as Al.Iron has an oxide coating, but is not tightly packed, so water & air can penetrate easily
  • Oxidation Numbers OBJECTIVES Determine: Oxidation number of an atom of any element in a pure substance. Define: Oxidation & Reduction in terms of a change in ON, & identify atoms being oxidized or reduced in redox reactions.An “oxidation number” is positive /negative number assignedto an atom to indicate its degree of oxidation or reduction. Generally, a bonded atom’s oxidation number is the charge it would have if the electrons in the bond were assigned to the atom of the more electronegative element
  • Rules for Assigning Oxidation Numbers1. ON of any uncombined element = 0 2.ON of a monatomic ion equals its charge. 0 0 1 12 Na Cl 2 2 Na Cl3. ON of oxygen in compounds is -2, except in peroxides, such asH2O2 where it is -1.4. ON of hydrogen in compounds is +1, except in metal hydrides,like NaH, where it is -1.5. Sum of ON of atoms in compound must equal 0 1 2 2 2 1 H 2 O Ca (O H ) 2 2(+1) + (-2) = 0 (+2) + 2(-2) + 2(+1) = 0 H O Ca O H6. Sum of ON in formula of polyatomic ion is equal to its ionic charge. thus ? 2 X + 3(-2) = -1 thus ? 2 X + 4(-2) = -2 S O4 2 X =+6N O3 N O X = +5 S O
  • Rules for Oxidation States (OS)• The charge the atom would have in molecule (or ionic compound) if electrons were completely transferred.1 ON of elements in their standard states is zero. Eg: Na, Be, K, Pb, H2, O2, P4 = 02.OS for monatomic ions are the same as their charge. – Eg: Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. Oxygen is assigned an OS of -2 in its covalent compounds except as a peroxide.4. ON of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its ON is –1.5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.6. Sum of ON of all atoms in a molecule /ion is equal to charge on molecule /ion.
  • Balancing Redox EquationsOBJECTIVES Describe how ON are used to identify redox reactions.OBJECTIVES Balance a redox equation using the ON-change method.OBJECTIVESBalance a redox equation by breaking the equation into oxidationand reduction half-reactions, and then using the half-reaction method.
  • In general, all chemical reactions can be assigned to one of two classes: 1) oxidation-reduction, in which electrons are transferred: • Single-replacement, combination, decomposition, & combustion 2) Second class has no electron transfer, & includes all others: • Double-replacement and acid-base reactions In an electrical storm, N & O react to form NO N2(g) + O2(g) → 2NO(g) YES!•Is this a redox reaction? •If the ON of an element in a reacting species changes, then that element has undergone either oxidation or reduction; therefore, the reaction as a whole must be a redox.
  • Balancing Redox Equations• It is essential to write a correctly balanced equation that represents what happens in a chemical reaction – Fortunately, two systematic methods are available, and are based on the fact that the total electrons gained in reduction equals the total lost in oxidation. The two methods: 1. Use oxidation-number changes 2. Use half-reactions
  • Using Oxidation-Number Changes• Sort of like chemical bookkeeping, you compare the increases and decreases in oxidation numbers. – start with the skeleton equation – Step 1: assign oxidation numbers to all atoms; write above their symbols – Step 2: identify which are oxidized/reduced – Step 3: use bracket lines to connect them – Step 4: use coefficients to equalize – Step 5: make sure they are balanced for both atoms and charge.
  • Using half-reactions Half-reaction is an equation showing -oxidation or reduction that takes place they are then balanced separately, and finally combined• Step 1: write unbalanced equation in ionic form• Step 2: write separate half-reaction equations for oxidation & reduction• Step 3: balance the atoms in the half-reactions• Step 4: add enough electrons to one side of each half-reaction to balance the charges• Step 5: multiply each half-reaction by a number to make the electrons equal in both• Step 6: add the balanced half-reactions to show an overall equation• Step 7: add the spectator ions and balance the equation
  • p- blocks- block d- block f- block
  • Group 1A Group 2AHas 1e- in the Has 2e- in theoutermost shell outermost shellTend to loose Tend to loose1e- OS = +1 2e-OS = +2 Alkali metals Alkaline-earth metals
  • Electronegativity Increases p- blockElectro-negativitydecreses Electronegativity increases as we more left to right along a period. Electronegativity decrease as move top to bottom down a group.
  • Group 3A Group 4A Group 5A3e- in outermost shell 4e- in outermost shell 5e- in outermost shellTend to loose 3e- loose 4e-/gain 4e- loose 5e- /gain 3e-OS=+3 Oxidation state Oxidation state -3, +5 -4,-3,-2,-1,0,+1,+2,+3,+4
  • Group 6A Group 7A6e- in outermost shell 7e- in outermost shell Group 8ATend to gain 2e- Tend to gain 1e- 8e- in outermost shell Oxidation state -1 Tend to gain/loose 0 e- Oxidation state -2 Oxidation state -- 0 Group number - 8Group number – 8 Inert elements/Noblechalcogens Halogens gases
  • Balancing simple redox reactions Cu (s) + Ag +(aq) Ag(s) + Cu2+(aq)Step 1: Pick out similar species from the equation Cu(s) Cu2+(aq) Ag +(aq) Ag (S)Step 2: Balance the equations individually for charges and number of atoms Cu0(S) Cu2+(aq) + 2e- Ag +(aq) + e Ag (S)
  • Balancing simple redox reactions Cu0(S) Cu2+(aq) + 2e-Cu0(S) becomes Cu 2+(aq) by loosing 2 electrons -Cu0(S) oxidized to Cu2+(aq) -oxidizing half reaction. Ag +(aq) + e- Ag (S)Ag+(aq) becomes Ag 0 (S) by gaining 1 electron -Ag+(aq) reduced to Ag(S) is thereducing half reaction.Final Balancing act: Making the number of electrons equal in both half reactions [Cu0(S) Cu2+(aq) + 2e-] 1 [Ag +(aq) + e- Ag (S)] 2So we have, Cu0(S) Cu2+(aq) + 2e- 2Ag +(aq) + 2e- 2Ag (S) Cu0(S) Cu2+(aq) + 2e- 2Ag +(aq) + 2e- 2Ag (S)Cu0(S) + 2Ag +(aq) + 2e- Cu2+(aq) + 2Ag (S) + 2e- Cu0(S) + 2Ag +(aq) Cu2+(aq) + 2Ag (S) Number of e-s involved in the overall reaction is 2
  • Balancing complex redox reactions Fe+2(aq) + MnO4-(aq) Mn+2(aq) + Fe+3(aq) Oxidizing half: Fe+2(aq) Fe+3(aq) + 1e- Reducing half: MnO4-(aq) Mn+2(aq) BalancingBalancing atoms: MnO4-(aq)+ Mn+2(aq) + 4H2O oxygens: Balancing hydrogens: MnO4-(aq)+8H+ Mn+2(aq) + 4H2O Reaction happening in an acidic medium Oxidation numbers: Mn = +7, O = -2 Mn = +2Balancing electrons: The left side of the equation has 5 less electrons than the right side MnO4-(aq)+8H++ 5e- Mn+2(aq) + 4H2O
  • Balancing complex redox reactions Final Balancing act:Making the number of electrons equal in both half reaction [Fe+2(aq) Fe+3(aq) + 1e- ] 5[MnO4-(aq)+8H++ 5e- Mn+2(aq) + 4H2O] 1 5Fe+2(aq) 5Fe+3(aq) + 5e- MnO4-(aq)+8H++ 5e- Mn+2(aq) + 4H2O5Fe2++MnO4-(aq)+8H++ 5e- 5Fe3+ +Mn+2(aq) + 4H2O + 5e-5Fe2++MnO4-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2O 5 Fe 2+ ions oxidized by 1 MnO4- ion to 5 Fe3+ions. Conversely 1 MnO4- is reduced by 5 Fe2+ ions to Mn2+.
  • Choosing a Balancing Method 1.The oxidation number change 2.The half-reaction method works method works well if the oxidized best for reactions taking place in and reduced species appear only acidic or alkaline solution. once on each side of the equation, and there are no acids or bases.
  • Displacement Reaction a.k.a Single Hydrogen Displacement ReactionReplacementA + BC AC + BSr+2H2O Sr(OH)2 +H2Hydrogen Displacement TiCl4 + 2Mg Ti + 2MgCl2 Metal Displacement Cl2 + 2KBr 2KCl + Br2 The Activity Series fo Halogen Displacement M + BC AC + B M is metal, BC is acid or H2O B is H2 Ca + 2H2O Ca(OH)2 + H2
  • Types of Oxidation-Reduction Reactions 0 +1 +2 0+4 0 0 +20 -1 -1 0
  • C3H8O + CrO3 + H2SO4 Cr2(SO4)3 + C3H6O + H2OBalancing equations usingoxidation numbers
  • Review: balancing chemical equationsBalance the following chemical reaction: CuCl2 + Al Cu + AlCl3• Balanced equations by “inspection”.• Balancing equations-equal numbers of atoms on each side of the equation• Balance equations using oxidation #s.• Relies on the idea that -number of electrons lost by- element must be equal to number gained by different element.• Total gain in oxidation numbers -- equal to total lost.
  • Using Oxidation Numbers total +2 -2 0 0 +3 -3 oxidation # +2 -1+ CuCl Al 0 Cu + 0AlCl +3 -1 2 3• Notice: Cu has gained 2e– (oxidation # by 2)• Notice: Al has lost 3e– (oxidation # by 3)• But, number of e– gained must equal e– lost• Multiply Cu by 3, Al by 2: change is 6 for both -6 +6 change +6 0 0 +6 total +2 -2 0 0 +3 -3 oxidation # +2 -1 0 0 +3 -1 3CuCl2 + 2Al 3Cu + 2AlCl3
  • Steps to balancing equations Write the skeleton equation Assign oxidation numbers to all atoms Identify which atoms change oxidation numberMake the number of atoms that change oxidationnumber the same on both sides by insertingtemporary coefficients Compute the total change in oxidation number Make total increase in oxidation number equal the total decrease by multiplication using appropriate factors Balance the remainder by inspection. Do not change what has been balanced.Compounds with elements that have changed in onecase but not in another are considered twice.
  • Example 1 Balance the following equation: -2 x 3 = -6 +6 change total +6 0 +6 +4 ox. # +2+6 -8 0 +6 +18-24 +4 -4 +2 -2 H2SO4 + Al +1+6 -2 0Al2(SO4)3 ++6 -2+ H2O -2 +3 SO2 +4 +1 -23H2SO4+3 2 3 6Step 6:7:Identify whichremainderin oxidation Step 4: Make the number of atoms that change Step3:Make the total increase done ox. usStep 1: Write equation: already by inspection. Balance the atoms change for #Step 5: Compute to +4) andthat have oxidation Step 2: Assigncompoundschange in+3) already oxidationS (+6the totalsame on tobynot Al (0 both sides by Note: equal oxidationdecrease number only the totalnumbers number the inserting temporary coefficients number balanced need to be balanced here. been multiplication using appropriate factors
  • Balance the following equation: +2 x 4 = +8 -8 0 +5 +2 -3 0 +1+5 -6 +1+5-6 +2+10-12 -3 +4+5 -6 +2 -2 0 +1+5-2 +1+5-2 +2 +5 -2 -3 +1+5 -2 +1 -24Zn+ HNO3+ 9 HNO3 4Zn(NO3)2 + NH4NO3+ H2O 3Step 7: Balance the remainder by inspection. Step 6: Make the total increase in oxidationStep 4: Make the number of atoms that change Note:Identify which oxidation numbers# us number equal totaltotal thatin oxidation number Step only compounds decreasedone Write equation: already not already the change have byStep 3: Step 2: Assign atoms change ox. for by 1: number the same on both sides oxidationStep 5: Compute need to be balanced here. been balanced +2) appropriate factors inserting Zn (0 to coefficients to -3) multiplication using and N (+5 temporary
  • Balance the following equation: -5 x 2 = -10 +2 x 5 = +10 +7 +4 +2 +6 1 7 -8 2 6 -8 2 6 -8 2 6 -8 2 6 -8 6 18 -24 2-2 1 7 -2 2 6 -2 1 6 -2 1 6 -2 2 6 -2 3 6 -2 1-22KMnO4+2FeSO4+ H2SO4 8 K2SO4+2MnSO4+ 5Fe2(SO4)3+8H2O 10Step 6: Make the total increase in oxidationStep 7: Balance the remainder by inspection.Step 4: Make the number of atoms that change number equal the total decrease byStep 3: Identify which atoms change ox. # Step 1: Write Assign oxidation numbers us Note:Step 2: equation: already done for oxidation number the same on both sides by only compounds that have not already Step 5: Compute total appropriate factorsnumber multiplication using change Feoxidation Mn (+7 to +2) and in (+2 to +3) inserting temporary coefficients been balanced need to be balanced here.
  • Balance the following equation: -2 x 3 = -6 +6 oxidation numbers Step 2: Assign +6 0 +6 +4 1 7 -8 2 6 -8 0 +6 +18 -24 +4 -4 +2 -2 1 7 -2 1 6 -2 0 +3 +6 -2 +4 -2 +2 -2 KMnO4+ H2C2O4+ H2SO4 2 CO2+ K2SO4+3 MnSO4+ H2O 3 Step 3: Identify which atoms change ox. #Step 4: Make (+6 the+4)increaseby oxidationStep 6: Make the to remainder(0 tothat change 7: Balance number of Al in inspection. S the total and atoms +3)Step 5: Compute thethe same inhaveby sides by oxidationequal total change on both already Note: only compoundsdecrease not number that oxidation number number total already done for us Step 1: Write equation: inserting temporary coefficients factors been balanced need to be balanced here. multiplication using appropriate
  • Oxidizing agents Reducing agentsExamples: Examples:permanganate (MnO4-), chromate Active metals(CrO42-), and dichromate (Cr2O72-)ions, sodium hypochlorite (bleach) sodium, magnesium,nitric acid (HNO3), aluminum & zinc,perchloric acid (HClO4), and NaH, CaH2 and LiAlH4.sulfuric acid (H2SO4) Oxidizing agents used Reducing agents used as as redox titrants redox titrants Sodium thiosulfateI3- (iodimetry) -most commonKMnO4, pot. permanganate -Stable in oxygen -reducing agents are oxidized byK2Cr2O7, potassium dichromate dissolved oxygen. -Stronger reducing agents areCerium(IV) solutions required, must work in oxygen-Titrations that create/consume I2 free environment
  • Redox Titrations- POTASSIUM PERMANGANATE • POWERFUL OXIDISING AGENT • 1ST Introduced by Margueritte for Iron • Acidic conditions – Strong Oxidising Agent. • Sulfuric Acid – No action on permanganate in dilute solution. • But with HCl – Chlorine liberates. • MnO4- + 8H++5e= Mn2++4H2O • With alkaline condition other reactions • Cant use as Primary standard • Standard solution and stability • Permanganate ion used often because - its own indicator. • MnO4- is purple, Mn+2 is colorless. When reaction solution remains clear, MnO4- is gone.
  • Redox Titrations- POTASSIUM PERMANGANATE• Standard solution and stability• Permanganate ion used often because - its own indicator.• MnO4- is purple, Mn+2 is colorless.• When reaction solution remains clear, MnO4- is gone.• Aqueous solution of MnO4- are not entirely stable because the ions tend to oxidize water:• 4MnO4- + 2H2O 4MnO2(s)+3O2(g)+4OH-• Decomposition reaction-catalysed by light, heat,acids, MnO2• Stored in dark glass bottles & Kept away high temp.• Not a Primary Standard.• Standardised = Std. Oxalic acid• 2MnO4- + 5H2C2O4 +6H 2Mn2+ +10CO2+8H2O
  • Redox Titrations- POTASSIUM PERMANGANATEBACK TITRATION TECHNIQUE: Eg. Glycerol Determination• C3H8O3 + 14MnO4 - 20 OH- +3CO3• Ba2+ is added to ppt manganate, to mask its green color.• Acidify with sulfuric acid & back titrate xs St. MnO4- = std. Oxalic acid. 2MnO4- + 5H2C2O4 + 6H ------ 2Mn2+ +10CO2+8H2O Potassium mangante(VII) -in acid reduced to manganese(II) MnO4- + 8H+ + 5e- Mn2+ + 4H2O
  • Titration of unknown sample of Iron Vs KMnO4:The unknown sample of iron contains, iron in Fe2+ oxidation state. So we are basicallydoing a redox titration of Fe2+ Vs KMnO45Fe2++MnO4-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2OProblem with KMnO4 Unfortunately, the permanganate solution, once prepared, begins to decompose by the following reaction:4 MnO4-(aq) + 2 H2O(l)  4 MnO2(s) + 3 O2(g) + 4 OH-(aq)So we need another solution whose concentration is precisely known to be able to findthe precise concentration of KMnO4 solution.
  • Vinitial KMnO4 Titration of Oxalic acid Vs KMnO4 Vfinal Primary secondary standard standard End point: Pale Permanent Pink color 250mL 250mL 250mL16 H+(aq) + 2 MnO4-(aq) + 5 C2O4-2(aq) 2 Mn+2(aq) +10CO2(g) + 8 H2O(l)
  • Redox Titrations- POTASSIUM DICHROMATE Slightly Weaker Oxidising agents than MnO4 Its reactions are slow(Back titration) It does not oxidise Chloride Primary Std. Its solution need not be standardised (pure grade) Titrations are carried out in acidic condition In basic solution, Cr2O72- converted to yellow chromate ion CrO4 2- (oxidising power is nil) Potassium dichromate(VI) oxidising agent -reduced to chromium(III) Cr2O7- + 14H+ + 6e- 2Cr3+ + 7H2OPotassium dichromate acts as oxidizing agent in acidic medium only:The neutral aqueous solution of Potassium dichromate is 1:1 equilibrium mixture ofdichromate and chromate, a consequence of hydrolysis of dichromate ions.Cr2O7 2– + H2O = 2 CrO42– + 2H+Orange yellowChromate ions are weaker oxidizing agent than dichromate. Thus oxidizing strength ofdichromate is reduced in neutral solution. The above hydrolysis reaction however can bereversed by adding acid to the solution and this explains the necessity of acidic mediumfor the reaction.
  • Redox titrations -Iodine• I2, or I3- --Good oxidizing agent - used as a titrant in iodimetry – I2 not very soluble in water. Dissolved in aqs. KI solution. – In excess I-, forms I3-, very soluble, dark red• I2 in Equations should replaced by I3-, with addition I- on other side• Stability: Iodine solutions lack stability Volatility of I, Looses of Iodine from an open vessel occur in short time. Air oxidation of Iodide in M of Iodine solution. 4I- + O2(g) + 4H+ 2I2 + 2H2O This reaction is promoted by Acids, Heat, Light and Nitrogen oxides. Stored in closed dark glass stoppered bottles and from elevated temp.• I3- usually standardized against As2O3• Iodine is volatile, solution used immediately after standardization• Starch indicator detects appearance of excess I3-
  • Titration involing with Iodine IODIMETRIC IODOMETRIC DIRECT METHOD INDIRECT METHOD Analysing strong reducing species Analysing strong oxidising species Titration of I2 produced by analyte againstTitration of analyte with std. I2 soln. (I3-) sod. thiosulfate Add Xs of I- to soln of analyte. I2 isTitration in acidic/weakly alkaline solution produced in amount equivalent to oxi. agentRed. Power of red. Agents is increased in Lib I2 is titrated against Sod. Thio sulfate neutral solution pH is maintained neutral by adding Cr2O7 2- + 6I-(Xs)---14H + 2Cr3+ +3I2 + 7H2O NaHCO3 I2+2S2O3 2- --- 2I- + S4O6 2- Sodium thio sulfate is universal titrant for Iodine AsO3 3- + I2 + H2O ---- AsO4 3- +2I- + 2H+ in neutral/acidic soltuion. In neutral solution, at low acid: equilibirium is shited to right Titrations occur in acidic solutions Acidity promotes oxidising agent- iodide Potential of As(V)/As(III) couple is reaction decreased sufficiently that As(III) willreduce I2 and increase reducing power of Lib I2 is titrated against Sod. Thio sulfate AsO3 3-AsO4 3- + 2H+ +2e-------AsO33- + H2O How? Equiliibrium is shitted to right. MnO4 - + 8H+ + 5e ---- Mn2+ +4H2O H3AsO4 + 2H+ + 2e ---- H3AsO3 +4H2O
  • Titration involing with Iodine IODIMETRIC IODOMETRIC pH Effect: pH Effect: Very Little influence on electrode potential of Increase of acidity promotes oxidising I2/I couple because H+ does not participate in agent – iodide reaction the half reaction. Keep the solution neutral. Complexing agent: Fe3+/Fe2+ system Precipitating agent: Fe 3+ +e --- Fe 2+ E0=0.77V Determination of Cu 2+ I3-+2e -----3I- E0=0.536V Cu 2+ + e -------Cu+ E0 Fe 3+ /Fe+ is hr than E0 I /I- system 2 To force the reaction ----oxidation of I-by Under normal condition I 2 could not oxidise Cu2+ ---- you have to increase E of Fe 2+ into Fe 3+ Cu2+/Cu+ system to be >0.53V Addition of complexing agenta as F- or EDTA By adding SCN/I- reagent to ppt that form stable complex with Fe3+ shit to Cu+ ----- [Cu+] --- E of Cu2+/Cu+lowerr E (0.53v) that allows oxidation of Fe2+ with I2 Indicator: Starch Indicator: Starch Sources of error in Iodometry: Sources of error in Iodimetry: Decomposition of thiosulfate solutionDue to I2: lack of stability (vol, air oxi of Iodide) Premature addition ofstarch & its decomposition Due to Starch: Due to I2: lack of stability (vol, air oxi of Iodide)Aq. Starch decomposes –few days- bacterial action. Determination of oxidising agents, Decomposition products: Glucose, 2MnO4, Cr2O7, BrO3, IO3. Boric acid/formamide-as preservative 2Fe3+ , H3AsO4 Direct titrations Titrate with liberated I2 against sod. Thio sulfate Back titrations techniques.
  • Titration involing with Iodine IODIMETRIC IODOMETRICDirect titration with only 1 reaction: Not direct titration because 2 reactions:analyte +titrant (I2)→product (iodide I-) analyte + I- →I2unknown known I2 + std. thiosulfate → producta) A reducing analyte b) One reaction a) An oxidizing analyte b) Two reactionsc) Standard solution: Iodine (I2) c) Standard solution: Sodium thisoufateAnalytical applications:Species analyzed : (reducing analytes), Species analyzed : (oxidizing analytes) HOCl,SO2, H2S, Zn2+, Cd2+, Hg2+,Pb2+ Cysteine, Br2, IO3- , IO4- , O2, H2O2, O3,glutathione, mercaptoethanol, Glucose NO2- , Cu 2+, MnO4-, MnO2(and other reducing sugars) Species Reaction Species Oxidation reaction SO2 SO2 + H2O < == > H2SO3 HOCl HOCl+H++3I- <=>Cl-+I3-+ H2O H2SO3 + H2O< == > SO24- + 4H+ + 2e- Br2 Br2 + 3 I- < == > 2 Br - + I3- IO3- 2 IO3-+16 I-+12 H+<=>6 I3- +6 H2O H2S H2S < == > S(s) + 2H+ + 2e- IO4- 2 IO4-+22 I-+16 H+<=>8 I3-+ 8 H2O Zn2+, Cd2+-- M2+ + H2S MS(s) + 2H+ O2 O2+4 Mn(OH)2+2H2O<=>4 Mn(OH)3 Hg2+, Pb2+ -- MS(s) < == > M2+ + S + 2e- 2Mn(OH)3+6H++6I-<=>2Mn2++2I3-+6H2O H2O2 H2O2+3I-+2H+<=>I3-+2H2O Cysteine, glutathione, 2RSH<=>RSSR+2H++2e- O O3+3I-+2H+<=>O2+ I3-+H2O mercaptoethanol NO2 - 2HNO2+2H++3I-<=>2NO+I3-+2H2O Aldehydes H2CO+3OH-<=>HCO2-+2H2O+2e- S2O 8 2- S O 2-+3I-<=>2SO 2- +I - 2 8 4 3 Glucose (and other reducing sugar) O Cu2+ 2 Cu2+ + 5 I - < == > 2 CuI(s) + I3- MnO4- 2MnO4-+16H++15 I-<=>2Mn2++5I3-+8 RCH + 3OH- < == > HCO2-+2H2O + 2e- H 2O Ascorbic acid MnO2 MnO2(s)+4H+ +3I-<=>Mn2++I3-+2H2O
  • Example: Quantification of Ascorbic Acid (Vitamin C)C6H8O6 + I2 → CçH6O6 + 2I- + 2H+Iodine rapidly oxidizes ascorbic acid, C6H8O6 , to producedehydroascorbic acid, C6H6O6 .Ascorbic acid Dehydroascorbic acid
  • Redox Titrations- Cerimetry Strong Oxidizing agent Eg: Determination of Hydrogen peroxide (direct), Glycerol (back titration)Cerric solution in H2SO4 does not oxidize chloride and can be used to titrate HCl solutions of analytes Its solultion need not be standardisedStandard solution and stability:Salt is dissolved in H2SO4 – to prevent pptn of basic salts.Solution in sulfuric acid is indefinitely stable.Solution in HNO3 undergoes photo-chemical decomposition but slowly.Ceric salts hydrolyses to ceric hydroxide if not in acid.
  • Introduction• Sodium Thiosulfate (Na2S2O3) is used in Iodometry to quantify the amount of iodine in solution, in the form of Triiodide (I3-)• Since Sodium Thiosulfate is rarely/never available as a primary standard, Na2S2O3 solutions must be standardized between preparation and use• To standardize an Na2S2O3 solution, KIO3 is usually employed as the primary standard• Two chemical equilibrium equations are involved in the process:(1) IO3- + 8I- + 6H+  3I3- + 3H2O(2) I3- + 2S2O3 2-  3I- + S4O6 2-
  • Preparation of Primary Standard The primary standard (KIO3) is accurately weighed, dissolved in dH2O, and chemically treated to obtain I3- in solution, according to Eq. (1): (1) IO3- + 8I- + 6H+  3I3- + 3H2OA slight excess of I- is supplied The H+ can be supplied by addingby dissolving solid KI… a non-reactive acid, such as H2SO4… In this manner, 3 moles of I3- can be obtained in the solution for every mole of KIO3 dissolved…
  • Standardization of Sodium Thiosulfate Solution The standardization of Na2S2O3 occurs according to Eq. (2): (2) I3- + 2S2O3 2-  3I- + S4O6 2- Starch indicator is used to accentuateAn exact amount of the treated the endpoint…primary standard soln. is placedin an Erlenmeyer flask… The unstandardized thiosulfate soln. is placed in the buret… Thus, 2 moles of S2O3 2- are required to neutralize each mole of I3-…
  • CERIMETRY VITAMIN c PERMANGANOMETRY FERROUS SULFATE FERROUS FUMARATE AMMONIUM CHLORIDE FERROUS GLUCONATE FAS NIFIDIPINE HYDROGEN PEROXIDE ACETAMENOPHEN OXALIC ACID VITAMIN CIODOMETRY REDOX DIIODOHYDROXY TITRATIONS QUINOLINEBENZALKONIUM Cl DIMERCAPROLCAPTOPRIL GLYCERYLMONOSTERATECEPHLORIDINE GUAPHENESINCETRIMIDE MANNITOLPHENINDIONE POVIDONE IODINESODIUM METABISULFITE SODIUM THIOSULFATE BROMOMETRYSODIUM DIATRIAZOATE IODINE IRON CHLOROCRESOL CHLOROXYLENOL PHENOL