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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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
ml of titrant
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+
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.
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-
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.
color with ferricyanide.
K2Cr2O7, no more Fe2+ ions are
available and the sample will not
give a blue color with ferricyanide.
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.
Inox + n H+ + ne Inred
color A color B
Eind = E0ind -
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 –
Eind = E0 ind ±
Diphenylamine:1% solu in conc H2SO4 is used
(E0 = 0.76, n = 2)
0.73 - 0.79 v.
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.
because it is
insoluble in water
(diphenylamine p-sulphonic acid) has the same mechanism of action as diphenylamine.
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.
is converted by oxidation
into the pale blue
ferric complex (Ferrin).
(Ph)3 Fe3+ + e (Ph)3Fe2+ (E0 = +1.06 V)
pale blue Red
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.
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