Page 172 - Modern Analytical Chemistry

1400-CH06 9/9/99 7:40 AM Page 155

Chapter 6 Equilibrium Chemistry 155

or
2 +
1 1 [ Zn ]
pNH 3 = log b 3 + log
3 3 [ ( ) 2 + ] 2+
Zn NH 3 3 Zn
2+
2+
The concentrations of Zn and Zn(NH 3 ) 3 , therefore, are equal when
1 1 3 log b = 2.29
3
6
pNH 3 = log b 3 = log( 72 ´ 10 ) = 229
.
.
3 3
2+
A complete ladder diagram for the Zn –NH 3 system is shown in Figure 6.8. pNH 3 Zn(NH ) 2+
3 3
6F.3 Ladder Diagram for Oxidation–Reduction Equilibria log K = 2.03
4
Ladder diagrams can also be used to evaluate equilibrium reactions in redox sys-
tems. Figure 6.9 shows a typical ladder diagram for two half-reactions in which
the scale is the electrochemical potential, E. Areas of predominance are defined by Zn(NH ) 2+
3 4
3+
the Nernst equation. Using the Fe /Fe 2+ half-reaction as an example, we write
[Fe 2 + ] [Fe 2 + ]
°
E = E – . 0 05916 log 3 + = +. 0 771V – . 0 05916 log 3 +
[Fe ] [Fe ] Figure 6.8
2+
2+
Ladder diagram for Zn , Zn(NH 3 ) 3 , and
2+
For potentials more positive than the standard-state potential, the predominate Zn(NH 3 ) 4 , showing how cumulative
3+
species is Fe , whereas Fe 2+ predominates for potentials more negative than E°. formation constants are included.
4+
When coupled with the step for the Sn /Sn 2+ half-reaction, we see that Sn 2+ can be
3+
used to reduce Fe . If an excess of Sn 2+ is added, the potential of the resulting solu-
tion will be near +0.154 V.
Using standard-state potentials to construct a ladder diagram can
present problems if solutes are not at their standard-state concentra-
Fe 3+
tions. Because the concentrations of the reduced and oxidized species
are in a logarithmic term, deviations from standard-state concentra-
tions can usually be ignored if the steps being compared are separated E ° Fe /Fe 2+ = +0.771
3+
1b
by at least 0.3 V. A trickier problem occurs when a half-reaction’s po- E
tential is affected by the concentration of another species. For example,
Fe 2+
the potential for the following half-reaction
4+
2+
+
–
UO 2 (aq)+4H 3 O (aq)+2e t U (aq)+6H 2 O(l) 4+
Sn
depends on the pH of the solution. To define areas of predominance in
this case, we begin with the Nernst equation E ° Sn /Sn 2+ = +0.154
4+
.
0 05916 [U 4 + ]
.
E = 0 327 – log
2 [UO 2 + ][H 3 O +4 Sn 2+
]
2
+
and factor out the concentration of H 3 O .
.
.
0 05916 0 05916 [U 4 + ] Figure 6.9
.
E = 0 327 + log[HO + 4 log
] –
3
3+
4+
2 2 [UO 2 + ] Ladder diagram for the Fe /Fe 2+ and Sn /SN 2+ half-
2
reactions.
2+
From this equation we see that the areas of predominance for UO 2
4+
and U are defined by a step whose potential is
.
0 05916
E = 0 327 + log[HO + 4 =0 327 – 0 1183 pH
.
.
.
]
3
2
4+
2+
Figure 6.10 shows how a change in pH affects the step for the UO 2 /U half-reaction.