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Experiment #3. REDOX TITRATION. OXIDATION- REDUCTION TITRATION: IRON & PERMANGANATE. 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

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Presentation Transcript
slide1

Experiment #3

REDOX TITRATION

OXIDATION- REDUCTION TITRATION:

IRON & PERMANGANATE

slide2

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).

slide3

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.

slide4

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

slide5

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

slide6

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

slide7

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.

slide8

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

slide9

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 +

4(OS of O) = 0

OS of S + 4(OS of O) = -2

Ammonium cation

NH4+

OS of N + 4(OS of H) = +1

slide10

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-

slide11

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.)

NH4+

HCl

H2O

H+ Cl-

2H+ O2-

N3- 4(H+)

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: H2O2= 2H+ 2O-

Potassium superoxide: KO2= K+ 2O -1/2

slide12

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.

slide13

p- block

s- block

d- block

f- block

slide14

Group 1A

Group 2A

Has 1e- in the outermost shell

Has 2e- in the outermost shell

Tend to loose 1e-

Tend to loose 2e-

OS = +1

OS = +2

Alkali metals

Alkaline-earth

metals

slide15

s- block

p- block

d- block

f- block

slide16

p - block

Electronegativity Increases

Electro-

negativity

decreses

slide17

Group 3A

Has 3e- in the outermost shell

Tend to loose 3e-

OS = +3

slide18

Group 4A

Has 4e- in the outermost shell

Can either loose 4e-

Or gain 4e-

Exhibits variable

Oxidation state

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

slide19

Group 5A

Has 5e- in the outermost shell

Can either loose 5e-

Or gain 3e-

Oxidation state

-3,+5

slide20

Group 6A

Has 6e- in the outermost shell

Tend to gain 2e-

Oxidation state

Chalcogens

-2

Group number - 8

slide21

Group 7A

Has 7e- in the outermost shell

Tend to gain 1e-

Oxidation state

Halogens

-1

Group number - 8

slide22

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

slide23

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

slide24

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 = 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

slide25

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

slide26

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)

slide27

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) + e- Ag (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

slide28

Balancing simple redox reactions

Final Balancing act:

Making the number of electrons equal in both

half reactions

[Cu0(S) Cu2+(aq)+ 2e-] × 1

[Ag +(aq) + e- Ag (S)]× 2

So we have,

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

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

slide29

Balancing simple redox reactions

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

slide30

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)

Balancing atoms:

Balancing

oxygens:

MnO4-(aq)+Mn+2(aq) + 4H2O

slide31

Balancing complex redox reactions

Balancing hydrogens:

Reaction happening in an acidic medium

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

slide32

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-

slide33

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.

slide34

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:

slide35

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

slide36

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

slide38

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.

slide39

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 CO2

molecules. Conversely 2 MnO4- is reduced by 5 C2O42-

ions to 2Mn2+ ions.

slide40

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 moles

of MnO4- is reduced by 5 moles of C2O42- ions to

2 moles of Mn2+ ions.

slide41

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

slide42

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

0.15 g OXALIC ACID

+ 100 mL of 0.9 M

H2SO4.Heated to 80C

slide44

When preparing 0.9 M H2SO4

  • Wear SAFETY GOGGLES AND GLOVES
  • 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.