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Reduction- Oxidation Reactions . 3 rd lecture. Learning Objectives What are some of the key things we learned from this lecture ?. Calculating redox titration curves D etection of end point in redox titration Redox indicators. -Requirements of a redox indicator

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learning objectives what are some of the key things we learned from this lecture
Learning ObjectivesWhat are some of the key things we learned from this lecture?
  • Calculating redox titration curves
  • Detection of end point in redox titration
  • Redox indicators.
  • -Requirements of a redox indicator
  • -Transition range of redox indicator
  • - Examples of redox indicators
  • - Determination of equilibrium constant (Keq) for
  • redox reactions
redox titration curves
Redox titration curves

It is a plot of the number of mls of titrant aganist potential in volts.

These curves show the change in potential during the progress of a redox titration.

Suppose the titration of 100 ml of 0.1 N solu. of ferrous sulphate with 0.1 N ceric sulphate.

1- Upon addition of 10ml ceric sulphate the ratio

of Fe3+/ Fe2+ becomes 10/90 (10 ml ceric reacts

with 10 ml ferrous to give 10 ml ferric)

E= 0.77 - 0.059/1 log 90/10 = 0.69v

2- Addition of 50 ml ceric,

E= 0.77 - 0.059/1 log 50/50 = 0.77v


90 ml Fe2+

10 ml Fe3+

10 ml Ce3+


50 ml Fe2+

50 ml Fe3+

50 ml Ce3+



0.1 ml Fe2+

99.9 ml Fe3+

99.9 ml Ce3+


3- Addition of 99.9 ml ceric,

E= 0.77 - 0.059/1 log 0.1/99.9 = 0.93v

4- Addition of 100 ml ceric

At the equivalence point,

Ee.p. = n1 E10 + n2 E20

n1 + n2

E10 Standard oxidation potential of Fe3+/ Fe2+

E20 Standard oxidation potential of Ce4+/ Ce3+

n1 number of electrons lost or gained by Fe3+/ Fe2+

n2 number of electrons lost or gained by Ce4+/ Ce3+

Ee.p. = 0.77 X 1 +1.44 x 1 / 1+1 = 1.1v

0 ml Fe2+

100 ml Fe3+

100 ml Ce3+





0.1 ml Ce4+

100 ml Fe3+

100 ml Ce3+

5- After the equivalence point

Addition of 100.1 ml ceric

E= 1.44 - 0.059/1 log 100 /0.1 = 1.27v

It is clear from the curve that there is a

sudden change in potential at the

equivalence point.



ml of titrant

detection of end point in redox titration
Detection of end point in redox titration

1- Potentiometric method: The potential of a cell involving the solutiontitrated is followed.

The end point is known from the inflection pointin the titration curve.

2- Miscellaneous methods:

A- Specific indicators:

substances which react specifically with one of the reagents in a titration to produce a color.

e.g. starch gives blue color with iodine

SCN- ion gives red color with Fe3+


B- No (self) indicators:

If the color of the titrating agent undergoes a sharp enough change in color at the equivalence point. E.g. KMnO4, I2

1- When KMnO4, is used as atitrant for reducing agents, during titration the purple color of permenganate disappears due to formation of colorless Mn2+

When all the reducing agent has been oxidized a slight excess of KMnO4 colors the solution pink.

2- Similarly, titration with iodine, at the e.p. There should be the brown color of exx iodine.

But the color of I2 is unstable, so, we add starch which forms an intense blue color (adsorption compound) with exx free I2.


C- Irreversible indicators (acid-base indicators):

These are highly colored organic compounds which undergo irreversible oxidation or reduction with little excess of a titrant.

e.g. Methyl orange and methyl red in titrations with BrO3-

  • M.O. and M.R. are decolorised irreversibly (destroyed) by first exx of strong oxidants e.g. BrO3-.
  • Another example is naphthol blue black which changes from pink to colorless by a slight excess of BrO3-
  • The irrevesible indicator must be added near the end point.

D-External indicators:

E.g.1- Spot test method: for the determination of Fe2+ with K2Cr2O7

Near the equivalence point, drops of ferrous sample are removed and brought in contact with freshly prepared solu of potassium ferricyanide on a spot plate.

  • Normally Fe2+ forms a blue

color with ferricyanide.

  • When all Fe2+ is oxidized by

K2Cr2O7, no more Fe2+ ions are

available and the sample will not

give a blue color with ferricyanide.

Sample solu




Before e.p.

Sample solu



After e.p.


E- Internal redox indicators:

Redox indicator changes its color when the oxidation potential of the titrated solution reaches a definite value.

Requirement of a redox indicator:

1- The color change should be intense so that small amount of titrant gives the color change.

2- The indicator action should be reversible so that back titration can be performed.

3- The 2 colors of the indicator (reduced form, and oxidised form) should be sufficiently different so that the color change would be sharp.

4- The transition potential of the indicator should be in between the Eo values of the 2 systems used in the titration.

i.e. Lower than the titrant but higher than the sample WHY?

5- The transition potential of the indicator should not be affected by changes in pH, other wise the pH must be controlled.


Transition range of redox indicator

  • Redox indicators are typical redox systems.
  • They can be represented by the general half reaction as follows:

Inox + n H+ + ne Inred

color A color B

  • At a potential E, the ratio of the 2 forms of the indicator are determined by the Nernst equation as follows:

Eind = E0ind -

  • A good redox indicator must show the colour of its oxidized form when the ratio of [Inred] / [Inoxd] is not more than 1/10. At this limit the potential becomes:

Eind = Eoind - Eind = Eoind +


And shows the colour of its reduced form when the ratio of [Inred] / [Inoxd] is not more than 10/1. At this limit the potential becomes:

Eind = E0 ind - Eind = E0 ind –

  • In other words a good redox indicator, must change its color at a potential expressed by the following formula:

Eind = E0 ind ±


Examples or redox indicators:

Diphenylamine:1% solu in conc H2SO4 is used

(E0 = 0.76, n = 2)

  • The range of diphenylamine Eind = 0.76 ± 0.059/2 =

0.73 - 0.79 v.

  • At potential below 0.73 v the color of the rduced form predominates (colorless).
  • At potential above 0.79v the color of the oxidized form predomintes (blue-violet).
  • Between 0.73-0.79v the color of the solution changed gradually from colorless to blue-violet.

The first reaction involving the formation of diphenylbenzidine is non-reversible, the second gives violet product which can be reversed and constitutes the actual indicator reaction.

  • It is prepared in

sulphuric acid

because it is

insoluble in water

  • sulphonic acid derivative of diphenylamine

(diphenylamine p-sulphonic acid) has the same mechanism of action as diphenylamine.

  • It is water soluble with sharp color change during oxidation (Transition potential 0.8V), and is independent on pH.



Diphenylamine is unsuitable indicator for the determination of ferrous with dichromate WHY?

Because E0 of Fe3/Fe2+ 0.77 and E0 of diphenylamine 0.76 which are very close.

So we have to lower the oxidation potential of Fe3+/Fe2+ by adding PO43 .

Phenylanthranilic acid is suitable indicator for the determination of ferrous with dichromate WHY?

Because E0 of phenylanthranilic acid 1.08 which is intermediate between E0Fe3+/Fe2+ 0.77 and E0 Cr2O72-/Cr3+ 1.36.

chelate of ferrous with 1 10 ortho phenanthroline ferroin
Chelate of ferrous with 1,10 ortho phenanthroline (Ferroin)
  • It is intensely red and

is converted by oxidation

into the pale blue

ferric complex (Ferrin).

(Ph)3 Fe3+ + e  (Ph)3Fe2+ (E0 = +1.06 V)

pale blue Red

(Ferrin) (Ferroin)

  • It is an excellent indicator for Ce4+.
determination of equilibrium constant k eq for redox reactions
Determination of equilibrium constant (Keq) for redox reactions
  • a Aox + n e a Ared E0 A cathode
  • B red b Box + n e E0 A anode
  • By combining the two equations:

a Aox+ b Bred a Ared + b Box

Keq = [ Ared ]a[ Box]b / [Aox]a[ Bred ]b

E cell = E cathode - E anode

Equilibrium constant , indicates reaction completeness.

High value of Keq indicate complete reaction while low value indicate incomplete reaction.


At equilibrium the current stops

  • Ecathode= E anode

EA = E B s

E0A - 0.o59/ n log [Ared]a / [Aoxd] a= E0B - 0.o59/ n log [Bred]b / [Boxd]b

E0A - E0B = 0.o59/ n log Keq

Log Keq = n (E0A - E0B ) / 0.059