REDOX TITRATIONS - JSK NAGARAJAN

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REDOX TITRATIONS:

REDOX TITRATIONS Electron Transfer Titrations J.S.K.NAGARAJAN ASST. PROF OFF CAMPUS -JSS UNIVERSITY J.S.S. COLLEGE OF PHARMACY, OOTACAMUND – 643 001.

REDOX TITRATIONS:

Gravimetry REDOX TITRATIONS

PowerPoint Presentation:

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 TITRATIONS:

REDOX TITRATIONS Titrations Examples Acid-base Quantification of acetic acid in vinegar Complexometric Quantification of chloride ( Cl - ) in water Precipitation Water Hardness (Calcium & Magnesium) Redox Quantification of H 2 O 2

REDOX TITRATIONS:

REDOX TITRATIONS Titration example Analyte Titrant Indicator Acid-base Quantification of acetic acid in avinegar Acetic acid (CH 3 COOH) NaOH Phenolphthalein Complexometric Water Hardness Ca 2+ , Mg 2+ EDTA Eriochrome black T Murexide Precipitation Quantification of chloride ( Cl - ) in water Chlordie AgNO 3 Mohr, Volhard , Fajans Redox Quantification of hydrogen peroxide (H 2 O 2 ) Hydrogen peroxide (H 2 O 2 ) KMnO 4 No indicator

REDOX TITRATIONS:

REDOX TITRATIONS Titrations Example Type of Titrations Acid-base Quantification of acetic acid in vinegar ■ Direct □ Indirect □ Back Complexometric Water Hardness ( Ca 2+ & Mg2+) ■ Direct □ Indirect □ Back Precipitation Quantification of Cl in Water Mohr Method ■ Direct □ Indirect □ Back Fajans Method ■ Direct □ Indirect □ Back Volhard Method □ Direct □ Indirect ■ Back Redox Quantification of hydrogen peroxide (H 2 O 2 ) ■ Direct □ Indirect □ Back

REDOX TITRATIONS :

REDOX TITRATIONS Acid/Base reactions - involves donation /acceptance of protons Precipitation/ Solubility reactions : Involve donation / acceptance of negative charge REDOX TITRATION: what is being donated and accepted in a redox reaction? Electrons! Consider the reaction taking place in a disposable battery : 2Zn + 3MnO 2  Mn 3 O 4 + 2ZnO How can you tell that electrons are being donated and accepted? Which species is donating electron( s) and which is accepting electron (s)? Transfer leads to - D ecrease in ON of element = REDUCTION I ncrease in ON of element = OXIDATION SIMULTANEOUSLY

REDOX TITRATIONS:

LEO SAYS GER REDOX TITRATIONS Oxidation Loss of electrons Gain in oxygen Reduction Gain of electrons Loss of oxygen Sodium is oxidized Chlorine is reduced

REDOX TITRATIONS:

REDOX TITRATIONS O X I D A T I O N : Old Definition: Combination of substance with oxygen C (s) + O 2 (g) CO 2 (g) REDUCTION Current definition: L oss of E lectrons is O xidation ( LEO ) Na Na + + e - Positive charge represents electron deficiency Old definition: Removal of oxygen from a compound WO 3 (s) + 3H 2 (g) W(s ) + 3H 2 O(g) Current definition : Gain of E lectrons is R eduction ( GER ) Cl + e - Cl - Negative charge represents electron richness ONE NEGATIVE CHARGE MEANS RICH BY ONE ELECTRON Positive OS reflects the tendency atom to loose electrons N egative OS reflects the tendency atom to gain electrons

REDOX TITRATIONS:

Mg + S → Mg 2+ + S 2- (MgS) REDOX TITRATIONS The sulfur atom (which has no charge) is changed to a sulfide ion by gaining 2 electrons, and is reduced to S2- OBJECTIVES State the characteristics of a redox reaction and identify the oxidizing agent and reducing agent . Many of reactions may not even involve oxygen Redox- electrons are transferred between reactants OBJECTIVES Define oxidation & reduction in terms of loss or gain of O2/Electrons electrons Magnesium atom (has zero charge) changes to a Mg ion by losing 2 electrons, and is oxidized to Mg2+ Active metals : Lose electrons easily- easily oxidized- strong reducing agents Active nonmetals : Gain electrons easily-easily reduced-strong oxidizing agents

PowerPoint Presentation:

Mg (s) + S (s) → MgS (s) Mg is oxidized : loses e - , becomes a Mg 2+ ion S is reduced : gains e - = S 2- ion Mg is the reducing agent S is the oxidizing agent Losing electrons is oxidation & substance that loses the electrons is called the reducing agent REDOX TITRATIONS Gaining electrons is reduction, & substance that gains the electrons is called the oxidizing agent Sodium is oxidized – it is the reducing agent 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 + → Mg 2+ + S 2- REDOX TITRATIONS

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REDOX TITRATIONS

Not All Reactions are Redox Reactions:

Reactions - no change in ON are NOT redox reactions . Examples: Not All Reactions are Redox Reactions REDOX TITRATIONS 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

Corrosion:

Corrosion REDOX TITRATIONS Damage 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 materials make electrically conductive solutions that make electron transfer easy Luckily, not all metals corrode easily Gold & 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:

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 assigned to an atom to indicate its degree of oxidation or reduction .

PowerPoint Presentation:

Rules for Assigning Oxidation Numbers 1. ON of any uncombined element = 0 2. ON of a monatomic ion equals its charge . 3. ON of oxygen in compounds is -2, except in peroxides , such as H 2 O 2 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 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H 6. Sum of ON in formula of polyatomic ion is equal to its ionic charge . X + 3(-2) = -1 N O thus X = +5 X + 4(-2) = -2 S O thus X =+ 6

Rules for Oxidation States (OS):

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 = 0 2.OS for monatomic ions are the same as their charge. Eg : Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 3. Oxygen is assigned an OS of -2 in its covalent compounds except as a peroxide. ON of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its ON is –1. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. S um of ON of all atoms in a molecule /ion is equal to charge on molecule /ion.

Balancing Redox Equations:

Balancing Redox Equations OBJECTIVES Describe how ON are used to identify redox reactions. OBJECTIVES Balance a redox equation using the ON-change method . OBJECTIVES Balance a redox equation by breaking the equation into oxidation and reduction half-reactions, and then using the half-reaction method.

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In general, all chemical reactions can be assigned to one of two classes: oxidation-reduction, in which electrons are transferred: Single-replacement, combination, decomposition, & combustion 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 N 2(g) + O 2(g) → 2NO (g) 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. YES!

Balancing Redox Equations:

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: Use oxidation-number changes Use half-reactions

Using Oxidation-Number Changes:

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:

Using half-reactions

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d- block p- block s- block f- block

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Group 1A Group 2A Tend to loose 1e - OS = +1 Tend to loose 2e - OS = +2 Has 1e - in the outermost shell Has 2e - in the outermost shell Alkali metals Alkaline-earth metals

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p- block Electronegativity Increases Electro- negativity decreses Electronegativity increases as we more left to right along a period. Electronegativity decrease as move top to bottom down a group.

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Group 3A 3e - in outermost shell Tend to loose 3e - OS=+ 3 Group 4A 4e - in outermost shell loose 4e - /gain 4e - Oxidation state -4,-3,-2,-1,0,+1,+2,+3,+ 4 Group 5A 5e - in outermost shell loose 5e - /gain 3e - Oxidation state -3, +5

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Group 6A 6e - in outermost shell Tend to gain 2e - Oxidation state - 2 Group number – 8 chalcogens Group 7A 7e - in outermost shell Tend to gain 1e - Oxidation state - 1 Group number - 8 Halogens Group 8A 8e - in outermost shell Tend to gain/loose 0 e - Oxidation state -- 0 Inert elements/Noble gases

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Balancing simple redox reactions Cu (s) + Ag + ( aq ) Ag(s ) + Cu 2 + ( aq ) Step 1: Pick out similar species from the equation Cu(s) Cu 2 + ( aq ) Ag + ( aq ) Ag (S) Step 2: Balance the equations individually for charges and number of atoms Cu 0 (S) Cu 2 + ( aq ) + 2e - Ag + ( aq ) + e Ag (S)

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Balancing simple redox reactions Cu 0 (S) Cu 2 + ( aq ) + 2e - Ag + ( aq ) + e - Ag (S) Cu 0 (S) becomes Cu 2 + ( aq ) by loosing 2 electrons -Cu 0 (S ) oxidized to Cu 2+ ( aq ) - oxidizing half reaction. Ag + ( aq ) becomes Ag 0 (S) by gaining 1 electron -Ag + ( aq ) reduced to Ag(S ) is the reducing half reaction. Final Balancing act: Making the number of electrons equal in both half reactions [Cu 0 (S) Cu 2+ ( aq ) + 2e - ] × 1 [Ag + ( aq ) + e - Ag (S)] × 2 So we have, Cu 0 (S) Cu 2+ ( aq ) + 2e - 2Ag + ( aq ) + 2e - 2Ag (S) Cu 0 (S) Cu 2 + ( aq ) + 2e - 2Ag + ( aq ) + 2e - 2Ag (S) Cu 0 (S) + 2Ag + ( aq ) + 2e- Cu 2 + ( aq ) + 2Ag (S) + 2e - Cu 0 (S) + 2Ag + ( aq ) Cu 2+ ( aq ) + 2Ag (S ) Number of e - s involved in the overall reaction is 2

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Balancing complex redox reactions Fe +2 ( aq ) + MnO 4 - ( aq ) Mn +2 ( aq ) + Fe +3 ( aq ) Fe +2 ( aq ) Fe +3 ( aq ) + 1e - MnO 4 - ( aq ) Mn +2 ( aq ) Oxidizing half: Reducing half: Balancing atoms: Mn O 4 - ( aq )+ Mn +2 ( aq ) + 4H 2 O Balancing oxygens : Balancing hydrogens: MnO 4 - ( aq )+ 8H + Mn +2 ( aq ) + 4H 2 O Reaction happening in an acidic medium Oxidation numbers : Mn = +7 , O = -2 Mn = +2 Balancing electrons: The left side of the equation has 5 less electrons than the right side MnO 4 - ( aq )+ 8H + + 5e - Mn +2 ( aq ) + 4H 2 O

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Balancing complex redox reactions Final Balancing act: Making the number of electrons equal in both half reactions [Fe +2 ( aq ) Fe +3 ( aq ) + 1e - ] × 5 [MnO 4 - ( aq )+ 8H + + 5e - Mn +2 ( aq ) + 4H 2 O] ×1 5Fe +2 ( aq ) 5 Fe +3 ( aq ) + 5e - MnO 4 - ( aq )+ 8H + + 5e - Mn +2 ( aq ) + 4H 2 O 5Fe 2+ +MnO 4 - ( aq )+ 8H + + 5e - 5Fe 3+ + Mn +2 ( aq ) + 4H 2 O + 5e - 5Fe 2+ +MnO 4 - ( aq )+ 8H + 5Fe 3 + + Mn +2 ( aq ) + 4H 2 O 5 Fe 2+ ions oxidized by 1 MnO 4 - ion to 5 Fe 3+ ions . Conversely 1 MnO 4 - is reduced by 5 Fe 2+ ions to Mn 2+ .

Choosing a Balancing Method:

Choosing a Balancing Method

Displacement Reaction a.k.a Single Replacement A + BC AC + B Sr+2H2O Sr(OH)2 +H2 Hydrogen Displacement :

Displacement Reaction a.k.a Single Replacement A + BC AC + B Sr+2H 2 O Sr (OH) 2 +H 2 Hydrogen Displacement TiCl 4 + 2Mg Ti + 2MgCl 2 Metal Displacement Cl 2 + 2KBr 2KCl + Br 2 Halogen Displacement Hydrogen Displacement Reaction M + BC AC + B M is metal, BC is acid or H 2 O B is H 2 Ca + 2H 2 O Ca (OH) 2 + H 2 The Activity Series for Metals

Types of Oxidation-Reduction Reactions:

0 +1 +2 0 0 +4 0 +2 0 -1 -1 0 Types of Oxidation-Reduction Reactions

Balancing equations using oxidation numbers :

Balancing equations using oxidation numbers C 3 H 8 O + CrO 3 + H 2 SO 4  Cr 2 (SO 4 ) 3 + C 3 H 6 O + H 2 O

Review: balancing chemical equations:

Review: balancing chemical equations Balance the following chemical reaction: CuCl 2 + Al  Cu + AlCl 3 Balanced equations by “inspection”. Balancing equations-equal numbers of atoms on each side of the equation Balance equations using oxidation #s. R elies 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:

Using Oxidation Numbers total oxidation # CuCl 2 + Al  Cu + AlCl 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 change total oxidation # 3CuCl 2 + 2Al  3Cu + 2AlCl 3 -1 -2 +2 +2 0 0 0 0 +3 +3 -3 -1 -1 -2 +2 +2 0 0 0 0 +3 +3 -3 -1 +6 0 0 +6 +6 -6

Steps to balancing equations:

Steps to balancing equations Write the skeleton equation Assign oxidation numbers to all atoms Identify which atoms change oxidation number Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficients Compute the total change in oxidation number Make total increase in oxidation number equal the total decrease by multiplication using appropriate factors Compounds with elements that have changed in one case but not in another are considered twice. Balance the remainder by inspection. Do not change what has been balanced.

Example 1:

Step 2: Assign oxidation numbers Step 3: Identify which atoms change ox. # S (+6 to +4) and Al (0 to +3) Step 4: Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficients Step 5: Compute the total change in oxidation number Step 6: Make the total increase in oxidation number equal the total decrease by multiplication using appropriate factors Step 7: Balance the remainder by inspection. Note: only compounds that have not already been balanced need to be balanced here. Example 1 Balance the following equation : change total ox. # H 2 SO 4 + Al  Al 2 (SO 4 ) 3 + SO 2 + H 2 O +6 +6 +2 +1 0 0 +3 +6 +6 +18 -24 -2 Step 1: Write equation: already done for us -2 -8 +4 +4 -2 -4 +2 +1 -2 -2 2 +6 0 +6 +4 +6 -2 x 3 = -6 3 3 6 H 2 SO 4 + 3

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Step 4: Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficients Step 6: Make the total increase in oxidation number equal the total decrease by multiplication using appropriate factors Step 2: Assign oxidation numbers Step 7: Balance the remainder by inspection. Note: only compounds that have not already been balanced need to be balanced here. Step 3: Identify which atoms change ox. # Zn (0 to +2) and N (+5 to -3) Step 5: Compute total change in oxidation number Step 1: Write equation: already done for us Balance the following equation : Zn + HNO 3 + HNO 3 Zn(NO 3 ) 2 + NH 4 NO 3 + H 2 O +5 +5 +1 +1 0 0 +2 +2 +5 +10 -12 -2 -2 -6 -3 -3 +1 +4 +2 +1 -2 -2 +5 0 +2 -3 -8 +2 x 4 = +8 4 4 3 +5 +5 -2 -6 +5 +5 +1 +1 -2 -6 9

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Step 7: Balance the remainder by inspection. Note: only compounds that have not already been balanced need to be balanced here. Step 4: Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficients Step 5: Compute total change in oxidation number Step 1: Write equation: already done for us Step 6: Make the total increase in oxidation number equal the total decrease by multiplication using appropriate factors KMnO 4 + FeSO 4 + H 2 SO 4  K 2 SO 4 + MnSO 4 + Fe 2 (SO 4 ) 3 + H 2 O Step 2: Assign oxidation numbers Step 3: Identify which atoms change ox. # Mn (+7 to +2) and Fe (+2 to +3) Balance the following equation : 7 7 1 1 -2 -8 2 +7 +2 +6 +2 -5 x 2 = -10 2 8 2 6 6 2 2 -2 -8 6 6 2 1 -2 -8 6 6 2 1 -2 -8 6 6 2 2 -2 -8 6 18 6 3 -2 -24 +4 x 5 = +10 2 1 -2 -2 10 5 8

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Step 2: Assign oxidation numbers Step 3: Identify which atoms change ox. # S (+6 to +4) and Al (0 to +3) Step 4: Make the number of atoms that change oxidation number the same on both sides by inserting temporary coefficients Step 5: Compute total change in oxidation number Step 6: Make the total increase in oxidation number equal the total decrease by multiplication using appropriate factors Step 7: Balance the remainder by inspection. Note: only compounds that have not already been balanced need to be balanced here. Balance the following equation : 7 7 1 1 0 0 +3 +6 +6 +18 -24 -2 Step 1: Write equation: already done for us -2 -8 +4 +4 -2 -4 +2 +2 -2 -2 2 +6 0 +6 +4 +6 -2 x 3 = -6 3 3 3 KMnO 4 + H 2 C 2 O 4 + H 2 SO 4  CO 2 + K 2 SO 4 + MnSO 4 + H 2 O 6 6 2 1 -2 -8

Oxidizing agents Examples: permanganate (MnO4-), chromate (CrO42-), and dichromate (Cr2O72-) ions, sodium hypochlorite (bleach) nitric acid (HNO3), perchloric acid (HClO4), and sulfuric acid (H2SO4):

Reducing agents Examples: Active metals sodium, magnesium, aluminum & zinc, NaH , CaH 2 and LiAlH 4 . Oxidizing agents Examples : permanganate (MnO 4 - ), chromate (CrO 4 2- ), and dichromate (Cr 2 O 7 2- ) ions, sodium hypochlorite (bleach) nitric acid (HNO 3 ), perchloric acid (HClO 4 ), and sulfuric acid (H 2 SO 4 ) Oxidizing agents used as redox titrants I 3 - ( iodimetry ) KMnO 4 , pot . permanganate K 2 Cr 2 O 7 , potassium dichromate Cerium(IV) solutions Titrations that create/consume I 2 Reducing agents used as redox titrants Sodium thiosulfate -most common -Stable in oxygen -reducing agents are oxidized by dissolved oxygen. - Stronger reducing agents are required, must work in oxygen-free environment

Redox Titrations- POTASSIUM PERMANGANATE:

Redox Titrations- POTASSIUM PERMANGANATE POWERFUL OXIDISING AGENT 1 ST 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. MnO 4 - is purple, Mn +2 is colorless. When reaction solution remains clear, MnO 4 - is gone.

Redox Titrations- POTASSIUM PERMANGANATE:

Redox Titrations- POTASSIUM PERMANGANATE Standard solution and stability Permanganate ion used often because - its own indicator. MnO 4 - is purple, Mn +2 is colorless. When reaction solution remains clear, MnO 4 - is gone. Aqueous solution of MnO 4 - 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 PERMANGANATE:

Redox Titrations- POTASSIUM PERMANGANATE BACK 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) MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O

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Titration of unknown sample of Iron Vs KMnO 4 : The unknown sample of iron contains, iron in Fe 2 + oxidation state. So we are basically doing a redox titration of Fe 2+ Vs KMnO 4 5Fe 2+ +MnO 4 - ( aq )+ 8H + 5Fe 3 + + Mn +2 ( aq ) + 4H 2 O Problem with KMnO 4 Unfortunately, the permanganate solution, once prepared, begins to decompose by the following reaction: 4 MnO 4 - ( aq ) + 2 H 2 O (l)  4 MnO 2 (s) + 3 O 2 (g) + 4 OH- ( aq ) So we need another solution whose concentration is precisely known to be able to find the precise concentration of KMnO 4 solution.

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250mL 250mL 250mL V initial V final End point: Pale Permanent Pink color KMnO 4 Titration of Oxalic acid Vs KMnO 4 Primary standard secondary standard 16 H + ( aq ) + 2 MnO 4 - ( aq ) + 5 C 2 O 4 -2 ( aq ) 2 Mn +2 ( aq ) + 10CO 2 (g) + 8 H 2 O (l)

Redox Titrations- POTASSIUM DICHROMATE:

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, Cr2O 7 2- converted to yellow chromate ion CrO4 2- ( oxidising power is nil) Potassium dichromate(VI) oxidising agent -reduced to chromium(III ) Cr 2 O 7 - + 14H + + 6 e -  2Cr 3+ + 7H 2 O Potassium dichromate acts as oxidizing agent in acidic medium only: The neutral aqueous solution of Potassium dichromate is 1:1 equilibrium mixture of dichromate and chromate, a consequence of hydrolysis of dichromate ions. Cr2O7 2 – + H2O = 2 CrO42 – + 2H+ Orange yellow Chromate ions are weaker oxidizing agent than dichromate. Thus oxidizing strength of dichromate is reduced in neutral solution. The above hydrolysis reaction however can be reversed by adding acid to the solution and this explains the necessity of acidic medium for the reaction.

Redox titrations -Iodine:

Redox titrations -Iodine I 2 , or I 3 - --Good oxidizing agent - used as a titrant in iodimetry I 2 not very soluble in water. Dissolved in aqs . KI solution. In excess I - , forms I 3 - , very soluble, dark red I 2 in Equations should replaced by I 3 - , 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. I 3 - usually standardized against As 2 O 3 Iodine is volatile, solution used immediately after standardization Starch indicator detects appearance of excess I 3 -

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Titration involing with Iodine IODIMETRIC IODOMETRIC DIRECT METHOD INDIRECT METHOD Analysing strong reducing species Analysing strong oxidising species Titration of analyte with std. I2 soln. (I3-) Titration of I2 produced by analyte against sod. thiosulfate Titration in acidic/weakly alkaline solution Add X s of I- to soln of analyte. I2 is produced in amount equivalent to oxi . agent Red. Power of red. Agents is increased in neutral solution Lib I2 is titrated against Sod. Thio sulfate pH is maintained neutral by adding NaHCO3 AsO 3 3- + I 2 + H 2 O ---- AsO 4 3- +2I - + 2H + In neutral solution, at low acid: equilibirium is shited to right Potential of As(V)/As(III) couple is decreased sufficiently that As(III) will reduce I2 and increase reducing power of AsO3 3- AsO4 3- + 2H+ +2e-------AsO33- + H2O Cr 2 O 7 2- + 6I - ( Xs )---14H + 2Cr 3+ +3I 2 + 7H 2 O I 2 +2S 2 O 3 2- --- 2I - + S 4 O 6 2- Sodium thio sulfate is universal titrant for Iodine in neutral/acidic soltuion . Titrations occur in acidic solutions Acidity promotes oxidising agent- iodide reaction Lib I2 is titrated against Sod. Thio sulfate How? Equiliibrium is shitted to right. MnO 4 - + 8H + + 5e ---- Mn 2+ +4H 2 O H 3 AsO 4 + 2H + + 2e ---- H 3 AsO 3 + 4H 2 O

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Titration involing with Iodine IODIMETRIC IODOMETRIC pH Effect: Very Little influence on electrode potential of I 2 /I couple because H + does not participate in the half reaction. Keep the solution neutral. pH Effect: Increase of acidity promotes oxidising agent – iodide reaction Complexing agent: Fe3+/Fe2+ system Fe 3+ +e --- Fe 2+ E0=0.77V I3-+2e -----3I - E0=0.536V E0 Fe 3+ /Fe + is hr than E0 I 2 /I - system Under normal condition I 2 could not oxidise Fe 2+ into Fe 3+ Precipitating agent: Determination of Cu 2+ Cu 2+ + e -------Cu + To force the reaction ----oxidation of I-by Cu2+ ---- you have to increase E of Cu 2+ /Cu + system to be >0.53V By adding SCN/I- reagent to ppt Cu+ ----- [Cu+] --- E of Cu2+/Cu+ Addition of complexing agenta as F- or EDTA that form stable complex with Fe3+ shit to lowerr E (0.53v) that allows oxidation of Fe2+ with I2 Sources of error in Iodometry : Decomposition of thiosulfate solution Premature addition ofstarch & its decomposition Indicator: Starch Sources of error in Iodimetry : Due to I2: lack of stability ( vol , air oxi of Iodide) Indicator: Starch Due to I2: lack of stability ( vol , air oxi of Iodide) Determination of oxidising agents, 2MnO4, Cr2O7, BrO3, IO3. 2Fe3+ , H3AsO4 Titrate with liberated I2 against sod. Thio sulfate Due to Starch: Aq. Starch decomposes –few days- bacterial action. Decomposition products: Glucose, Boric acid/ formamide -as preservative Direct titrations Back titrations techniques .

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Titration involing with Iodine IODIMETRIC IODOMETRIC a) A reducing analyte b) One reaction c) Standard solution: Iodine (I 2 ) a) An oxidizing analyte b ) Two reactions c) Standard solution: Sodium thisoufate Analytical applications: Species analyzed : ( reducing analytes ), SO 2, H 2 S, Zn 2 + , Cd 2+ , Hg 2 + ,Pb 2+ Cysteine , glutathione, mercaptoethanol, Glucose (and other reducing sugars) Species analyzed : ( oxidizing analytes ) HOCl, Br2, IO3- , IO4- , O2 , H2O2, O3, NO2- , Cu 2 +, MnO4- , MnO2 Species Oxidation reaction SO 2 SO 2 + H 2 O < == > H 2 SO 3 H 2 SO 3 + H 2 O< == > SO 2 4- + 4H + + 2e - H 2 S H 2 S < == > S( s ) + 2H + + 2e - Zn 2+ , Cd 2 + -- M 2 + + H 2 S  MS( s ) + 2H + Hg 2+ , Pb 2+ -- MS( s ) < == > M 2+ + S + 2e - Cysteine , glutathione, 2RSH<=>RSSR +2H + +2e - mercaptoethanol Aldehydes H 2 CO+3OH - <=>HCO 2 - +2H 2 O+2e - Glucose (and other reducing sugar ) O RCH + 3OH - < == > HCO 2 - +2H 2 O + 2e - Ascorbic acid Species Reaction HOCl HOCl+H + +3I - <=> Cl - +I 3 - + H 2 O Br 2 Br 2 + 3 I - < == > 2 Br - + I 3 - IO 3 - 2 IO 3 - +16 I - +12 H + <=>6 I 3 - +6 H 2 O IO 4 - 2 IO 4 - +22 I - +16 H + <=>8 I 3 - + 8 H 2 O O 2 O 2 +4 Mn (OH) 2 +2H 2 O<=>4 Mn (OH) 3 2Mn(OH) 3 +6H + +6I - <=>2Mn 2+ +2I 3 - +6H 2 O H 2 O 2 H 2 O 2 +3I - +2H + <=>I 3 - +2H 2 O O O 3 +3I - +2H + <=>O 2 + I 3 - + H 2 O NO 2 - 2HNO 2 +2H + +3I - <=>2NO+I 3 - +2H 2 O S 2 O 8 2- S 2 O 8 2- +3I - <=>2SO 4 2- +I 3 - Cu 2+ 2 Cu 2+ + 5 I - < == > 2 CuI ( s ) + I 3 - MnO 4 - 2MnO 4 - +16H + + 15 I - <=> 2Mn 2 + + 5I 3 - +8 H 2 O MnO 2 MnO 2 ( s )+4H + +3I - <=>Mn 2+ +I 3 - +2H 2 O Direct titration with only 1 reaction: analyte +titrant (I 2 )→product (iodide I - ) unknown known Not direct titration because 2 reactions: analyte + I - →I 2 I 2 + std. thiosulfate → product

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Example: Quantification of Ascorbic Acid (Vitamin C) C 6 H 8 O 6 + I 2 → C ç H 6 O 6 + 2I - + 2H + Iodine rapidly oxidizes ascorbic acid, C 6 H 8 O 6 , to produce dehydroascorbic acid, C 6 H 6 O 6 . Ascorbic acid Dehydroascorbic acid

Redox Titrations- Cerimetry:

Redox Titrations- Cerimetry Strong Oxidizing agent Cerric solution in H2SO4 does not oxidize chloride and can be used to titrate HCl solutions of analytes Its solultion need not be standardised Standard 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. Eg : Determination of Hydrogen peroxide (direct), Glycerol (back titration)

Introduction:

Introduction Sodium Thiosulfate (Na 2 S 2 O 3 ) is used in Iodometry to quantify the amount of iodine in solution, in the form of Triiodide (I 3 -) Since Sodium Thiosulfate is rarely/never available as a primary standard, Na 2 S 2 O 3 solutions must be standardized between preparation and use To standardize an Na 2 S 2 O 3 solution, KIO 3 is usually employed as the primary standard Two chemical equilibrium equations are involved in the process: IO 3 - + 8I- + 6H+  3I 3 - + 3H 2 O I 3 - + 2S 2 O 3 2-  3I- + S 4 O 6 2-

Preparation of Primary Standard:

Preparation of Primary Standard IO 3 - + 8I- + 6H+  3I 3 - + 3H 2 O The primary standard (KIO 3 ) is accurately weighed, dissolved in dH2O, and chemically treated to obtain I 3 - in solution, according to Eq. (1): A slight excess of I- is supplied by dissolving solid KI… The H+ can be supplied by adding a non-reactive acid, such as H 2 SO 4 … In this manner, 3 moles of I 3 - can be obtained in the solution for every mole of KIO 3 dissolved…

Standardization of Sodium Thiosulfate Solution:

Standardization of Sodium Thiosulfate Solution The standardization of Na 2 S 2 O 3 occurs according to Eq. (2): (2) I 3 - + 2S 2 O 3 2 -  3I- + S 4 O 6 2 - An exact amount of the treated primary standard soln. is placed in an Erlenmeyer flask… The unstandardized thiosulfate soln. is placed in the buret… Starch indicator is used to accentuate the endpoint… Thus, 2 moles of S 2 O 3 2 - are required to neutralize each mole of I 3 -…

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IODOMETRY BENZALKONIUM Cl CAPTOPRIL CEPHLORIDINE CETRIMIDE PHENINDIONE SODIUM METABISULFITE SODIUM DIATRIAZOATE VITAMIN C DIIODOHYDROXY QUINOLINE DIMERCAPROL GLYCERYLMONOSTERATE GUAPHENESIN MANNITOL POVIDONE IODINE SODIUM THIOSULFATE IODINE IRON BROMOMETRY CHLOROCRESOL CHLOROXYLENOL PHENOL REDOX TITRATIONS CERIMETRY VITAMIN c FERROUS SULFATE FERROUS FUMARATE FERROUS GLUCONATE NIFIDIPINE ACETAMENOPHEN PERMANGANOMETRY AMMONIUM CHLORIDE FAS HYDROGEN PEROXIDE OXALIC ACID

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