Redox Titration Oxidation- Reduction Titration
REDOX TITRATION OXIDATION- REDUCTION TITRATION: B.KIRUTHIGA LECTURER DEPT OF PHARMACEUTICAL CHEMISTRY
What are we doing in this experiment? Determine the % of Iron, in a sample by performing a redox titration between a solution of the iron sample and potassium permanganate (KMnO4).
What is redox titration? A TITRATION WHICH DEALS WITH A REACTION INVOLVING OXIDATION AND REDUCTION OF CERTAIN CHEMICAL SPECIES. What is a titration? The 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
What is oxidation? Old definition: Combination of substance with oxygen C (s) O2(g) CO2(g) Current definition: Loss of Electrons is Oxidation (LEO) Na Na e- Positive charge represents electron deficiency ONE POSITIVE CHARGE MEANS DEFICIENT BY ONE ELECTRON
What is reduction? Old definition: Removal of oxygen from a compound WO3 (s) 3H2(g) W(s) 3H2O(g) Current definition: Gain of Electrons is Reduction (GER) Cl e- Cl - Negative charge represents electron richness ONE NEGATIVE CHARGE MEANS RICH BY ONE ELECTRON
OXIDATION-REDUCTION Oxidation and reduction go hand in hand. In a reaction, if there is an atom undergoing oxidation, there is probably another atom undergoing reduction. When there is an atom that donates electrons, there is always an atom that accepts electrons. Electron transfer happens from one atom to another.
How to keep track of electron transfer? Oxidation number or oxidation state (OS): Usually a positive, zero or a negative number (an integer) A positive OS reflects the tendency atom to loose electrons A negative OS reflects the tendency atom to gain electrons
Rules for assigning OS The sum of the oxidation numbers of all of the atoms in a molecule or ion must be equal in sign and value to the charge on the molecule or ion. Sulfate anion Potassium Permanganate KMnO4 SO42- OS of K OS of Mn OS of S 4(OS of O) -2 4(OS of O) 0 Ammonium cation NH4 OS of N 4(OS of H) 1
Also, in an element, such as S8 or O2 , this rule requires that all atoms must have an oxidation number of 0. In binary compounds (those consisting of only two different elements), the element with greater electronegativity is assigned a negative OS equal to its charge as a simple monatomic ion. NaCl MgS Na Cl- Mg2 S2-
When it is bonded directly to a non-metal atom, the hydrogen atom has an OS of 1. (When bonded to a metal atom, hydrogen has an OS of -1.) H2O 2H O2- NH4 N3- 4(H ) HCl H Cl- Except for substances termed peroxides or superoxides, the OS of oxygen in its compounds is 2. In peroxides, oxygen has an oxidation number of -1, and in superoxides, it has an oxidation number of -½ . Hydrogen peroxide: H O 2H 2O2 2 Potassium superoxide: KO2 K 2O -1/2
Please Remember !! In a periodic table, Vertical columns are called GROUPS Horizontal rows are called PERIODS Electronegativity increases as we more left to right along a period. Electronegativity decrease as we move top to bottom down a group.
s- block p- block d- block f- block
Group 1A Has 1e- in the outermost shell Tend to loose 1eOS 1 Alkali metals Group 2A Has 2e- in the outermost shell Tend to loose 2eOS 2 Alkaline-earth metals
s- block p- block d- block f- block
p - block Electronegativity Increases Electronegativity decreses
Group 3A Has 3e- in the outermost shell Tend to loose 3eOS 3
Group 4A Has 4e- in the outermost shell Can either loose 4eOr gain 4eExhibits variable Oxidation state -4,-3,-2,-1,0, 1, 2, 3, 4
Group 5A Has 5e- in the outermost shell Can either loose 5eOr gain 3eOxidation state -3, 5
Group 6A Has 6e- in the outermost shell Tend to gain 2e- Chalcogens Oxidation state -2 Group number - 8
Group 7A Has 7e- in the outermost shell Tend to gain 1e- Halogens Oxidation state -1 Group number - 8
Group 8A Has 8e- in the outermost shell Tend to gain/loose 0 e- Inert elements or Noble gases Oxidation state 0 Group number - 8
Sample problem Find the OS of each Cr in K2Cr2O7 Let the OS of each Cr be x OS of K 1 (Remember K belongs to Gp. 1A) OS of O -2 (Remember O belongs to Gp. 6A) Net charge on the neutral K2Cr2O7 molecule 0 So we have, 2(OS of K) 2 ( OS of Cr) 7 (OS of O) 0 2( 1) 2 ( x) 7 (-2) 0 2 2 ( x) (-14) 0
2 2 ( x) (-14) 0 2 ( x) (-12) 0 2 ( x) (12) x 6 Find the OS of each C in C2O42Let the OS of each C be x OS of O -2 (Remember O belongs to Gp. 6A) So we have, 2(OS of C) 4 ( OS of O) -2 2(x) 4 ( -2) -2 2 ( x) (-8) -2
2 ( x) (-8) -2 2 ( x) 6 ( x) 3 Find the OS of N in NH4 Let the OS of each N be x OS of H 1 (Remember H belongs to Gp. 1A) So we have, (OS of N) 4 ( OS of H) 1 (x) 4 ( 1) 1 ( x) (4) 1 ( x) -3
Balancing simple redox reactions Cu (s) Ag (aq) Ag(s) Cu2 (aq) Step 1: Pick out similar species from the equation Cu(s) Ag (aq) Cu2 (aq) Ag (S) Step 2: Balance the equations individually for charges and number of atoms Cu0(S) Cu2 (aq) 2eAg (aq) eAg (S)
Balancing simple redox reactions Cu0(S) Cu2 (aq) 2e- Cu0(S) becomes Cu 2 (aq) by loosing 2 electrons. So Cu0(S) getting oxidized to Cu2 (aq) is the oxidizing half reaction. Ag (aq) eAg (S) Ag (aq) becomes Ag 0 (S) by gaining 1 electron. So Ag (aq) getting reduced to Ag (S) is the reducing half reaction. LEO-GER
Balancing simple redox reactions Final Balancing act: Making the number of electrons equal in both half reactions [Cu0(S) [Ag (aq) eSo we have, Cu0(S) 2Ag (aq) 2e- Cu2 (aq) 2e-] 1 Ag (S)] 2 Cu2 (aq) 2e2Ag (S)
Balancing simple redox reactions Cu0(S) 2Ag (aq) 2e- Cu2 (aq) 2e2Ag (S) Cu0(S) 2Ag (aq) 2eCu2 (aq) 2Ag (S) 2eCu0(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) Oxidizing half: Fe 2(aq) Reducing half: MnO4-(aq) Mn 2(aq) Fe 3(aq) Fe 3(aq) 1eMn 2(aq) Balancing atoms: Balancing -(aq) MnO 4 oxygens: Mn 2(aq) 4H2O
Balancing complex redox reactions Reaction happening in an acidic medium Balancing hydrogens: MnO4-(aq) 8H Mn 2(aq) 4H2O Oxidation numbers: Mn 7, O -2 Mn 2 Balancing electrons: The left side of the equation has 5 less electrons than the right side MnO4-(aq) 8H 5e- Mn 2(aq) 4H2O Reducing Half
Balancing complex redox reactions Final Balancing act: Making the number of electrons equal in both half reactions [Fe 2(aq) [MnO4-(aq) 8H 5e5Fe 2(aq) MnO4-(aq) 8H 5e- Fe 3(aq) 1e- ] 5 Mn 2(aq) 4H2O] 1 5Fe 3(aq) 5eMn 2(aq) 4H2O 5Fe2 MnO4-(aq) 8H 5e5Fe3 Mn 2(aq) 4H2O 5e-
Balancing complex redox reactions 5Fe2 MnO4-(aq) 8H 5Fe3 Mn 2(aq) 4H2O 5 Fe 2 ions are oxidized by 1 MnO4- ion to 5 Fe3 ions. Conversely 1 MnO4- is reduced by 5 Fe2 ions to Mn2 . If we talk in terms of moles: 5 moles of Fe 2 ions are oxidized by 1mole of MnO4- ions to 5 moles of Fe3 ions. Conversely 1 mole of MnO4- ions is reduced by 5 moles of Fe2 ions to 1 mole of Mn2 ions.
Conclusion from the balanced chemical equation For one mole of MnO4- to completely react With Fe2 , you will need 5 moles of Fe2 ions. So if the moles of MnO4- used up in the reaction is known, then the moles of Fe2 involved in the Mathematically written: reaction will be 5 times the moles of MnO4 Moles of Fe 5 [moles of MnO4 ] 2
How does this relationship concern our experiment? Titration of unknown sample of Iron Vs KMnO4: The unknown sample of iron contains, iron in Fe2 oxidation state. So we are basically doing a redox titration of Fe2 Vs KMnO4 5Fe2 MnO4-(aq) 8H 5Fe3 Mn 2(aq) 4H2O Moles of Fe 5 [moles of MnO4 ] 2
Vinitial Vfinal- Vinital Vused (in mL) Important requirement: The concentration of KMnO4 should be known precisely. KMnO4 Vfinal End point: Pale Permanent Pink color 250mL 250mL 250mL 1L Vused (in mL) VKMnO4 Used (in L) 1 1000mL
Moles of MnO4 Molarity of MnO4 VKMnO4 Used (in L) Moles of Fe 5 [moles of MnO4 ] 2 2 55.85 g of Fe 2 Grams of Fe moles of Fe 2 1 mole of Fe 2 2 grams of Fe % Fe in sample 100% mass of sample in grams
Problem with KMnO4 Unfortunately, the permanganate solution, once prepared, begins to decompose by the following reaction: 4 MnO4-(aq) 2 H2O(l) Æ 4 MnO2(s) 3 O2(g) 4 OH-(aq) So we need another solution whose concentration is precisely known to be able to find the precise concentration of KMnO4 solution.
Titration of Oxalic acid Vs KMnO4 Primary standard Secondary standard 16 H (aq) 2 MnO4-(aq) 5 C2O4-2(aq) Æ 2 Mn 2(aq) 10 CO2(g) 8 H2O(l) 5 C2O42- ions are oxidized by 2 MnO4- ions to 10 CO2 molecules. Conversely 2 MnO4- is reduced by 5 C2O42ions to 2Mn2 ions.
Titration of Oxalic acid Vs KMnO4 16 H (aq) 2 MnO4-(aq) 5 C2O4-2(aq) Æ2 Mn 2(aq) 10CO2(g) 8 H2O(l) If we talk in terms of moles: 5 moles of C2O42- ions are oxidized by 2 moles MnO4ions to 10 moles of CO2 molecules. Conversely 2 moles of MnO4- is reduced by 5 moles of C2O42- ions to 2 moles of Mn2 ions.
Conclusion from the balanced chemical equation For 5 moles of C2O42- ions to be completely oxidized by MnO4- we will need 2 moles of MnO4- ions. Conversely for 2 moles of MnO4- to be completely reduced by C2O42-, we will need 5 moles of C2O42- ions 2 5 Moles of C2O4 2 moles of MnO4 2 2 1 Moles of C2O4 moles of MnO4 5
Vinitial Vfinal- Vinital Vused (in mL) Important requirement: The concentration of KMnO4 should be known precisely. KMnO4 Vfinal End point: Pale Permanent Pink color 250mL 0.15 g OXALIC ACID 100 mL of 0.9 M H2SO4.Heated to 80 C 250mL 250mL 1L Vused (in mL) VKMnO4 Used (in L) 1 1000mL
Moles of C2O4 2 Weight of oxalic acid (in g ) ) Mol.Wt of Oxalic acid ( g mol 2 2 mol MnO4 1mol C2O4 mol MnO4 2 Moles of Ox 5 mol C2O4 1mol Na2C2O4 mol MnO4 [ MnO4 ] VKMnO4Used (in L)
When preparing 0.9 M H2SO4 1.Wear SAFETY GOGGLES AND GLOVES 2.Use graduated cylinder to dispense the acid from the bottle 3. Please have about 100 mL of water in 500 mL volumetric flask, before adding acid in to it. 4. Add acid to the flask slowly in small aliquots.
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What is redox titration? A TITRATION WHICH DEALS WITH A . REACTION INVOLVING OXIDATION AND . REDUCTION OF CERTAIN CHEMICAL . SPECIES. What is a titration? The act of adding standard solution in small quantities to the test solution till the reaction is complete is termed titration.
ii. acid–base neutralization reactions iii. oxidation–reduction or redox reactions. Q.3. What are the important aspects of redox reactions? Ans: Almost every element participate in redox reactions. The important aspects of redox reactions are as follows: i. Large number of natural, biological and industrial processes involve redox reactions .
CHAPTER 12: Redox Reactions . Goals of Chapter: - Understand redox reactions in detail - Review oxidation numbers - Learn electrochemical techniques Application of Redox Chemistry - extracting metals from ores, e.g. Need to learn to balance tricky redox reactions Cu 2CO 3(OH) 2(s) C(s) Æ2Cu(s) 2CO 2(g) H 2O(g .
Scaffold : Redox Reactions Example Worksheet ATTENTION: This worksheet will be used to play the Redox relay and is modified into cards To determine if a reaction is a REDOX reaction: 1. Assign oxidation numbers to each element in the reaction. 2. If any of the oxidation numbers change in the reaction then the reaction is REDOX. Practice:
UNIT 6 – REDOX REACTIONS 6 The oxidation number of an atom is the charge that would exist on an individual atom if the bonding were completely ionic In simple ions, the oxidation number of the atom is the charge on the ion: - Na , K , H all have an oxidation number of 1 - Mg2 , Ca2 , Pb2 all have an oxidation number of 2 - Cl-, B
OCR Chemistry A H432 Redox and Electrode Potentials p. 3 Method 2: Constructing redox equations using oxidation numbers We can balance a redox equation using oxid
A) Oxidation-reduction reactions require oxygen as a reactant. B) Oxidation-reduction reactions are also called redox reactions. C) Reduction is the gain of electrons. D) Oxidation-reduction reactions involve the transfer of electrons from one substance to another. E) If one substance is oxidized, then another substance must be reduced. F .
Oxidation-Reduction Reactions 6.2 Oxidation Numbers. 6.3 Types of Chemical . Reactions. 6.4 Voltaic Cells. n many important chemical reactions, electrons are transferred from atom to atom. We are surrounded by these reactions, commonly called oxidation‑reduction (or . redox) reactions, inside and out.
Anatomy 2-5 Indications 5 Contra-indications 5 General preparation 6 Landmarks 6-7 Performing the block 7-8 Complications 8 Trouble shooting 9 Summary 9 References 10 Appendix 1 11. 6/10/2016 Fascia Iliaca Compartment Block: Landmark Approach 2 FASCIA ILIACA COMPARTMENT BLOCK: LANDMARK APPROACH INTRODUCTION Neck of femur fracture affect an estimated 65,000 patients per annum in England in .
