redox titration oxidation- reduction titration: b

REDOX TITRATION

OXIDATION- REDUCTION TITRATION:

B.KIRUTHIGALECTURER

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 deficiencyONE 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 richnessONE NEGATIVE CHARGE MEANS RICH BY ONE ELECTRON

OXIDATION-REDUCTION

Oxidation and reduction go hand in hand. In a reaction, if there is an atom undergoingoxidation, 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 toanother.

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 OSThe sum of the oxidation numbers of all of the atomsin a molecule or ion must be equal in sign and value to the charge on the molecule or ion.

KMnO4 SO42-

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

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 twodifferent elements), the element with greaterelectronegativity is assigned a negative OS equalto its charge as a simple monatomic ion.

NaClNa+ Cl-

MgSMg2+ 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.)

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 HCl2H+ O2-

NH4+

N3- 4(H+) H+ Cl-

Hydrogen peroxide: H2O2= 2H+ 2O-

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 rightalong a period.

Electronegativity decrease as we move top to bottomdown a group.

d- block

p- block

s-bl

ock

f- block

Group 1A Group 2A

Tend to loose 1e-

OS = +1Tend to loose 2e-

OS = +2

Has 1e- in the outermost shell Has 2e- in the

outermost shell

Alkali metals Alkaline-earthmetals

d- block

p- block

s- block

f- block

p - blockElectronegativity Increases

Electro-negativitydecreses

Group 3A

Tend to loose 3e-

OS = +3

Has 3e- in the outermost shell

Group 4A

Can either loose 4e-

Or gain 4e-

Has 4e- in the outermost shell

Exhibits variable Oxidation state

-4,-3,-2,-1,0,+1,+2,+3,+4

Group 5AHas 5e- in the

outermost shell

Can either loose 5e-

Or gain 3e-

Oxidation state

-3,+5

Group 6AHas 6e- in the

outermost shell

Tend to gain 2e-

Oxidation state-2

Group number - 8

Chalcogens

Group 7AHas 7e- in the

outermost shell

Tend to gain 1e-

Oxidation state-1

Group number - 8

Halogens

Group 8AHas 8e- in the

outermost shell

Tend to gain/loose 0 e-

Oxidation state0

Group number - 8

Inert elementsor

Noble gases

Sample problemFind 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 = 0So we have,2(OS of K) + 2 ( OS of Cr) + 7 (OS of O)= 0

2(+1) + 2 ( x) + 7 (-2)= 02+ 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 C2O42-

Let the OS of each C be = xOS of O = -2 (Remember O belongs to Gp. 6A)So we have,

2(OS of C) + 4 ( OS of O) = -22(x) + 4 ( -2) = -2

2 ( x) +(-8)= -2

2 ( x) +(-8)= -22 ( x) = +6

( x) = +3

Find the OS of N in NH4+

Let the OS of each N be = xOS 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 redoxreactionsCu (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 forcharges and number of atomsCu0(S) Cu2+(aq) + 2e-

Ag +(aq) + e- Ag (S)

Balancing simple redox reactionsCu0(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 theoxidizing half reaction.

Ag+(aq) becomes Ag 0 (S) by gaining 1 electron.So Ag+(aq) getting reduced to Ag (S) is thereducing half reaction.

LEO-GER

Balancing simple redoxreactionsFinal Balancing act:

[Cu0(S) Cu2+(aq) + 2e-] × 1

[Ag +(aq) + e- Ag (S)]× 2

Making the number of electrons equal in bothhalf reactions

So we have,Cu0(S) Cu2+(aq) + 2e-

2Ag +(aq) + 2e- 2Ag (S)

Cu0(S) Cu2+(aq) + 2e-

2Ag +(aq) + 2e- 2Ag (S)

Balancing simple redoxreactions

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 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:

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

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-

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.

Conclusion from the balanced

chemical equationFor 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 reactionis known, then the moles of Fe2+ involved in the reaction will be 5 times the moles of MnO4

-

Mathematically written:

][5 42 −+ ×= MnOofmolesFeofMoles

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 redoxtitration of Fe2+ Vs KMnO4

5Fe2++MnO4-(aq)+8H+ 5Fe3+ +Mn+2(aq) + 4H2O

][5 42 −+ ×= MnOofmolesFeofMoles

250mL 250mL 250mL

Vinitial

Vfinal

End point:Pale PermanentPink color

KMnO4

Vfinal- Vinital= Vused (in mL)

mLLmLinVLinV used

UsedKMnO 10001

1)()(

4×=

Important requirement:The concentration of KMnO4 should be known precisely.

)(444 LinVMnOofMolarityMnOofMoles UsedKMnO×= −−

][5 42 −+ ×= MnOofmolesFeofMoles

+

+

++ ×= 2

2

22

185.55 Feofmoles

FeofmoleFeofgFeofGrams

%100%2

×=+

gramsinsampleofmassFeofgramssampleinFe

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 concentrationis precisely known to be able to find the preciseconcentration of KMnO4 solution.

Titration of Oxalic acid 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 C2O42- ions are oxidized by 2 MnO4

- ions to 10 CO2molecules. Conversely 2 MnO4

- is reduced by 5 C2O42-

ions to 2Mn2+ ions.

Titration of Oxalic acid 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 C2O42- ions are oxidized by 2 moles MnO4

-

ions to 10 moles of CO2 molecules. Conversely 2 molesof MnO4

- is reduced by 5 moles of C2O42- ions to

2 moles of Mn2+ ions.

Conclusion from the balancedchemical 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 C2O4

2-, we will need 5 moles of C2O42- ions

−− ≡ 4

2

42 25 MnOofmolesOCofMoles−− ×≡ 4

2

42 521 MnOofmolesOCofMoles

250mL 250mL 250mL

Vinitial

Vfinal

End point:Pale PermanentPink color

KMnO4

Vfinal- Vinital= Vused (in mL)

mLLmLinVLinV used

UsedKMnO 10001

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

)(.

)(2

42

molgacidOxalicofWtMol

ginacidoxalicofWeightOCofMoles =−

422

2

422

42

44 1

152

OCNamolOCmolOxofMoles

OCmolMnOmolMnOmol

−

−

−− ××=

)(][

4

44 LinV

MnOmolMnOUsedKMnO

−− =

When preparing 0.9 M H2SO4

1.Wear SAFETY GOGGLES AND GLOVES2.Use graduated cylinder to dispense the acid

from the bottle3. Please have about 100 mL of water in

500 mL volumetric flask, before addingacid in to it.

4. Add acid to the flask slowly in small aliquots.

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