Page 351 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:13 PM Page 334

334 Modern Analytical Chemistry

E eq by combining the two Nernst equations. To do so we recognize that the poten-
tials for the two half-reactions are the same; thus,
2+
[ Fe ]
°
E eq = E Fe / Fe 2+ -0 05916. log
3
+
3+
[ Fe ]
3+
[ Ce ]
°
E eq = E Ce 4+ / Ce 3+ - . 0 05916 log 4+
[ Ce ]
Adding together these two Nernst equations leaves us with
3+
2+
[ Fe ][ Ce ]
°
°
.
2E eq = E Fe / Fe 2+ + E Ce 4+ / Ce 3+ - 0 05916 log 9.18
3+
4+
3+
[ Fe ][ Ce ]
At the equivalence point, the titration reaction’s stoichiometry requires that
2+
4+
[Fe ] = [Ce ]
3+
3+
[Fe ] = [Ce ]
The ratio in the log term of equation 9.18, therefore, equals one and the log term is
zero. Equation 9.18 simplifies to
E ° 3+ Fe 2+ + E Ce ° 4+ / Ce 3+ . 0 767 V + .70 V
1
Fe /
E eq = = = .123 V
2 2
3+
After the equivalence point, the concentrations of Ce and excess Ce 4+ are easy
to calculate. The potential, therefore, is best calculated using the Nernst equation
for the titrant’s half-reaction.
3+
[ Ce ]
E = E ° 4+ / Ce 3+ - . 0 05916 log 4+ 9.19
Ce
[ Ce ]
For example, after adding 60.0 mL of titrant, the concentrations of Ce 3+ and Ce 4+
are
intial moles Fe 2+ MV
Fe Fe
[Ce 3+ ] = =
total volume V Fe +V Ce
(. 0 100 M)(50.0 mL) -2
.
= =455 ´ 10 M
50.0 mL + 60.0 mL
moles excess Ce 4+ MV Fe Fe
Ce Ce -M V
[Ce 4+ ] = =
total volume V Fe +V Ce
(. 0 100 M)(60.0 mL) - (. 0 100 M)(50.0 mL) -3
= = . 909 ´ 10 M
50.0 mL + 60.0 mL
Substituting these concentrations into equation 9.19 gives the potential as
455 ´10 -2
.
.
.
E =+1 70 V -0 05916 log =166 V
.
.
909 ´10 -3
Additional results for this titration curve are shown in Table 9.17 and Figure 9.34.