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Chapter 9 Titrimetric Methods of Analysis 337

9D.2 Selecting and Evaluating the End Point
The equivalence point of a redox titration occurs when stoichiometrically equiva-
lent amounts of analyte and titrant react. As with other titrations, any difference be-
tween the equivalence point and the end point is a determinate source of error.

Where Is the Equivalence Point? In discussing acid–base titrations and com-
plexometric titrations, we noted that the equivalence point is almost identical
with the inflection point located in the sharply rising part of the titration curve.
If you look back at Figures 9.8 and 9.28, you will see that for acid–base and com-
plexometric titrations the inflection point is also in the middle of the titration
curve’s sharp rise (we call this a symmetrical equivalence point). This makes it
relatively easy to find the equivalence point when you sketch these titration
curves. When the stoichiometry of a redox titration is symmetrical (one mole
analyte per mole of titrant), then the equivalence point also is symmetrical. If the
stoichiometry is not symmetrical, then the equivalence point will lie closer to the
top or bottom of the titration curve’s sharp rise. In this case the equivalence
point is said to be asymmetrical. Example 9.12 shows how to calculate the equiv-
alence point potential in this situation.

9
EXAMPLE .12
Derive a general equation for the electrochemical potential at the equivalence
2+
–
point for the titration of Fe with MnO 4 ; the reaction is
2+
3+
+
2+
–
5Fe (aq) + MnO 4 (aq)+8H 3 O (aq) t 5Fe (aq)+Mn (aq) + 12H 2 O(l)
SOLUTION
The redox half-reactions for the analyte and the titrant are
2+
3+
Fe (aq) t Fe (aq)+e –
2+
–
+
–
MnO 4 (aq)+8H 3 O (aq)+5e t Mn (aq) + 12H 2 O(l)
for which the Nernst equations are
2+
[ Fe ]
E eq = E ° 2+ - . 0 05916 log
3+
Fe / Fe 3+
[ Fe ]
+
2
. 0 05916 [ Mn ]
E eq = E ° 2+ - log
-
MnO 4 / Mn 5 - + 8
[MnO 4 ][ H 3 O ]
Before adding together these two equations, the second equation must be
multiplied by 5 so that the log terms can be combined; thus
2+
2+
[ Fe ][ Mn ]
°
°
.
6E eq = E Fe / Fe 2+ + 5E MnO 4 / Mn 2+ - 0 05916 log 3+ + 8
3+
-
-
[ Fe ][ MnO 4 ][ H 3 O ]
At the equivalence point, we know that
2+
–
[Fe ]=5 ´[MnO 4 ]
3+
2+
[Fe ]=5 ´[Mn ]