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![Balancing simple redox reactions
Final Balancing act:
[Cu0
(S) Cu2+
(aq) + 2e-
] × 1
[Ag +
(aq) + e-
Ag (S)]× 2
Making the number of electrons equal in both
half reactions
So we have,
Cu0
(S) Cu2+
(aq) + 2e-
2Ag +
(aq) + 2e-
2Ag (S)](https://image.slidesharecdn.com/redoxtitrationsoxidationstate-180123062739/75/Redox-titrations-amp-oxidation-state-28-2048.jpg)
![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
[MnO4
-
(aq)+8H+
+ 5e-
Mn+2
(aq) + 4H2O]×1
5Fe+2
(aq) 5Fe+3
(aq) + 5e-
MnO4
-
(aq)+8H+
+ 5e-
Mn+2
(aq) + 4H2O
5Fe2+
+MnO4
-
(aq)+8H+
+ 5e-
5Fe3+
+Mn+2
(aq) + 4H2O + 5e-](https://image.slidesharecdn.com/redoxtitrationsoxidationstate-180123062739/75/Redox-titrations-amp-oxidation-state-32-2048.jpg)
![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
reaction will be 5 times the moles of MnO4
-
Mathematically written:
][5 4
2 −+
×= MnOofmolesFeofMoles](https://image.slidesharecdn.com/redoxtitrationsoxidationstate-180123062739/75/Redox-titrations-amp-oxidation-state-34-2048.jpg)
![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
][5 4
2 −+
×= MnOofmolesFeofMoles](https://image.slidesharecdn.com/redoxtitrationsoxidationstate-180123062739/75/Redox-titrations-amp-oxidation-state-35-2048.jpg)
![)(444
LinVMnOofMolarityMnOofMoles UsedKMnO
×=
−−
][5 4
2 −+
×= MnOofmolesFeofMoles
+
+
+
+
×= 2
2
2
2
1
85.55
Feofmoles
Feofmole
Feofg
FeofGrams
%100%
2
×=
+
gramsinsampleofmass
Feofgrams
sampleinFe](https://image.slidesharecdn.com/redoxtitrationsoxidationstate-180123062739/75/Redox-titrations-amp-oxidation-state-37-2048.jpg)
![)(.
)(2
42
mol
gacidOxalicofWtMol
ginacidoxalicofWeight
OCofMoles =
−
422
2
42
2
42
4
4
1
1
5
2
OCNamol
OCmol
OxofMoles
OCmol
MnOmol
MnOmol
−
−
−
−
××=
)(
][
4
4
4
LinV
MnOmol
MnO
UsedKMnO
−
−
=](https://image.slidesharecdn.com/redoxtitrationsoxidationstate-180123062739/75/Redox-titrations-amp-oxidation-state-43-2048.jpg)
Uploaded byAnand Addagalla
Redox titrations & oxidation state
This document describes how to determine the percentage of iron in a sample through a redox titration with potassium permanganate (KMnO4). It provides background on redox titrations and oxidation-reduction reactions. KMnO4 is first standardized against oxalic acid, since its concentration decreases over time. This allows the moles of KMnO4 used to be determined. The reaction of KMnO4 with Fe2+ is then used to find the moles of Fe2+ in the sample. From this, the mass and percentage of iron can be calculated.
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Redox titrations & oxidation state
- 1. REDOX TITRATION OXIDATION- REDUCTIONTITRATION: IRON & PERMANGANATE
- 2. What are wedoing 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).
- 3. What is redoxtitration? 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.
- 4. What is astandard 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
- 5. What is oxidation? Olddefinition: 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
- 6. What is reduction? Olddefinition: 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
- 7. OXIDATION-REDUCTION Oxidation and reductiongo 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.
- 8. How to keeptrack 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
- 9. Rules for assigningOS 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. KMnO4 SO4 2- Potassium Permanganate Sulfate anion OS of K + OS of Mn + 4(OS of O) = 0 OS of S + 4(OS of O) = -2 NH4 + Ammonium cation OS of N + 4(OS of H) = +1
- 10. Also, in anelement, 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 Na+ Cl- MgS Mg2+ S2-
- 11. When it isbonded 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.) 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 . H2O HCl 2H+ O2- NH4 + N3- 4(H+ ) H+ Cl- Hydrogen peroxide: H2O2= 2H+ 2O- Potassium superoxide: KO2= K+ 2O -1/2
- 12. Please Remember !! Ina 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.
- 13.
- 14. Group 1A Group2A 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
- 15. d- block p- block s-block f- block
- 16. p - block ElectronegativityIncreases Electro- negativity decreses
- 17. Group 3A Tend toloose 3e- OS = +3 Has 3e- in the outermost shell
- 18. Group 4A Can eitherloose 4e- Or gain 4e- Has 4e- in the outermost shell Exhibits variable Oxidation state -4,-3,-2,-1,0,+1,+2,+3,+4
- 19. Group 5A Has 5e- inthe outermost shell Can either loose 5e- Or gain 3e- Oxidation state -3,+5
- 20. Group 6A Has 6e- inthe outermost shell Tend to gain 2e- Oxidation state -2 Group number - 8 Chalcogens
- 21. Group 7A Has 7e- inthe outermost shell Tend to gain 1e- Oxidation state -1 Group number - 8 Halogens
- 22. Group 8A Has 8e- inthe outermost shell Tend to gain/loose 0 e- Oxidation state 0 Group number - 8 Inert elements or Noble gases
- 23. Sample problem Find theOS 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
- 24. 2+ 2 (x) +(-14)= 0 2 ( x) +(-12) = 0 2 ( x) = (12) x = 6 Find the OS of each C in C2O4 2- Let 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
- 25. 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
- 26. Balancing simple redoxreactions 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)
- 27. Balancing simple redoxreactions Cu0 (S) Cu2+ (aq) + 2e- Ag + (aq) + e- Ag (S) 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) becomes Ag 0 (S) by gaining 1 electron. So Ag+ (aq) getting reduced to Ag (S) is the reducing half reaction. LEO-GER
- 28. Balancing simple redoxreactions Final Balancing act: [Cu0 (S) Cu2+ (aq) + 2e- ] × 1 [Ag + (aq) + e- Ag (S)]× 2 Making the number of electrons equal in both half reactions So we have, Cu0 (S) Cu2+ (aq) + 2e- 2Ag + (aq) + 2e- 2Ag (S)
- 29. Cu0 (S) Cu2+ (aq) +2e- 2Ag + (aq) + 2e- 2Ag (S) Balancing simple redox reactions 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
- 30. Balancing complex redoxreactions Fe+2 (aq) + MnO4 - (aq) Mn+2 (aq) + Fe+3 (aq) Fe+2 (aq) Fe+3 (aq) + 1e- MnO4 - (aq) Mn+2 (aq) Oxidizing half: Reducing half: Balancing atoms: MnO4 - (aq)+ Mn+2 (aq) + 4H2O Balancing oxygens:
- 31. Balancing complex redoxreactions MnO4 - (aq)+8H+ Mn+2 (aq) + 4H2O Balancing hydrogens: 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 Reaction happening in an acidic medium
- 32. Balancing complex redoxreactions Final Balancing act: Making the number of electrons equal in both half reactions [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) + 4H2O 5Fe2+ +MnO4 - (aq)+8H+ + 5e- 5Fe3+ +Mn+2 (aq) + 4H2O + 5e-
- 33. Balancing complex redoxreactions 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.
- 34. Conclusion from thebalanced 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 reaction will be 5 times the moles of MnO4 - Mathematically written: ][5 4 2 −+ ×= MnOofmolesFeofMoles
- 35. How does thisrelationship 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 ][5 4 2 −+ ×= MnOofmolesFeofMoles
- 36. 250mL 250mL 250mL Vinitial Vfinal Endpoint: Pale Permanent Pink color KMnO4 Vfinal- Vinital= Vused (in mL) mL LmLinV LinV used UsedKMnO 1000 1 1 )( )(4 ×= Important requirement: The concentration of KMnO4 should be known precisely.
- 37. )(444 LinVMnOofMolarityMnOofMoles UsedKMnO ×= −− ][5 4 2−+ ×= MnOofmolesFeofMoles + + + + ×= 2 2 2 2 1 85.55 Feofmoles Feofmole Feofg FeofGrams %100% 2 ×= + gramsinsampleofmass Feofgrams sampleinFe
- 38. Problem 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 find the precise concentration of KMnO4 solution.
- 39. Titration of Oxalicacid Vs KMnO4 Primary standard Secondary standard 16 H+ (aq) + 2 MnO4 - (aq) + 5 C2O4 -2 (aq) 2 Mn+2 (aq) + 10 CO2(g) + 8 H2O(l) 5 C2O4 2- ions are oxidized by 2 MnO4 - ions to 10 CO2 molecules. Conversely 2 MnO4 - is reduced by 5 C2O4 2- ions to 2Mn2+ ions.
- 40. Titration of Oxalicacid Vs KMnO4 16 H+ (aq) + 2 MnO4 - (aq) + 5 C2O4 -2 (aq) 2 Mn+2 (aq) +10CO2(g) + 8 H2O(l) If we talk in terms of moles: 5 moles of C2O4 2- ions are oxidized by 2 moles MnO4 - ions to 10 moles of CO2 molecules. Conversely 2 moles of MnO4 - is reduced by 5 moles of C2O4 2- ions to 2 moles of Mn2+ ions.
- 41. Conclusion from thebalanced chemical equation For 5 moles of C2O4 2- 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 C2O4 2- , we will need 5 moles of C2O4 2- ions −− ≡ 4 2 42 25 MnOofmolesOCofMoles −− ×≡ 4 2 42 5 21 MnOofmolesOCofMoles
- 42. 250mL 250mL 250mL Vinitial Vfinal Endpoint: Pale Permanent Pink color KMnO4 Vfinal- Vinital= Vused (in mL) mL LmLinV LinV used UsedKMnO 1000 1 1 )( )(4 ×= Important requirement: The concentration of KMnO4 should be known precisely. 0.15 g OXALIC ACID + 100 mL of 0.9 M H2SO4.Heated to 80°C
- 43.
- 44. When preparing 0.9M 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.