Reduction- Oxidation
Reactions
3rd lecture
Learning Objectives
What 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
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
2- Addition of 50 ml ceric,
E= 0.77 - 0.059/1 log 50/50 = 0.77v
50 ml Fe2+
50 ml Fe3+
50 ml Ce3+
1
90 ml Fe2+
10 ml Fe3+
10 ml Ce3+
3
3- Addition of 99.9 ml ceric,
E= 0.77 - 0.059/1 log 0.1/99.9 = 0.93v
4
4- Addition of 100 ml ceric
At the equivalence point,
0 ml Fe2+
100 ml Fe3+
100 ml Ce3+
Ee.p. = n1 E10 + n2 E20
n1 + n 2
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
0.1 ml Fe2+
99.9 ml Fe3+
99.9 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.
5
0.1 ml Ce4+
100 ml Fe3+
100 ml Ce3+
New
system
E
e.p.
ml of titrant
Detection of end point in redox titration
1- Potentiometric method: The potential of a cell
involving the solution titrated is followed.
The end point is known from the inflection point in 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
I 2.
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
Sample solu
color with ferricyanide.
 When all Fe2+ is oxidized by
K2Cr2O7, no more Fe2+ ions are Sample solu
available and the sample will not
give a blue color with ferricyanide.
Ferri
cyanide
Before e.p.
BB
Ferri
cyanide
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 -
0 . 059
log
n
[ In red ]
[ In oxid ]
 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 =
Eo
ind
-
0 . 059
n
log
1
10
Eind =
Eo
ind
+
0 . 059
n
 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 -
0 . 059
n
log
10
1
Eind = E0 ind –
0 . 059
n
 In other words a good redox indicator, must
change its color at a potential expressed by the
following formula:
Eind = E0 ind ±
0 . 059
n
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
colorless
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 E0 Fe3+/Fe2+ 0.77 and E0
Cr2O72-/Cr3+ 1.36.
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 (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