5.5: Redox Reactions with Coupled Equilibria
-
- Last updated
- Save as PDF
Redox Reactions with Coupled Equilibria
Coupled equilibria (solubility, complexation, acid-base, and other reactions) change the value of E°, effectively by changing the concentrations of free metal ions. We can use the Nernst equation to calculate the value of E° from the equilibrium constant for the coupled reaction. Alternatively, we can measure the half-cell potential with and without the coupled reaction to get the value of the equilibrium constant. This is one of the best ways to measure K sp , K a , and K d values.
As an example, we consider the complexation of Fe 2 + and Fe 3 + by CN - ions:
\[\ce{Fe^{2+}_{(aq)} + 6CN^{-}_{(aq)} -> [Fe(CN)6]^{4-}} \label{1}\]
\[\ce{Fe^{3+}_{(aq)} + 6CN^{-}_{(aq)} -> [Fe(CN)6]^{3-}} \label{2}\]
Which oxidation state of Fe is more strongly complexed by CN - ? We can answer this question by measuring the standard half-cell potential of the [Fe(CN) 6 ] 3-/4- couple and comparing it to that of the Fe 3 +/2+ couple:
\(\ce{Fe^{3+}_{(aq)} + e^{-}= Fe^{2+}_{(aq)}}\) E° = +0.77 V (3)
\(\ce{[Fe(CN)6]^{3-} + 3^{-} = [Fe(CN)6]^{4-}}\) E° = +0.42 V (4)
Iron(III) is harder to reduce (i.e., E° is less positive ) when it is complexed to \(\ce{CN^{-}}\)
This implies that the equilibrium constant for complexation reaction (Equation \ref{1}) should be
smaller
than that for reaction (Equation \ref{2}). How much smaller?
We can calculate the ratio of equilibrium constants by adding and subtracting reactions:
\(\ce{Fe^{3+} + 6CN^{-} -> [Fe(CN)6]^{3-}}\: \: K=K_{1}\)
\(\ce{[Fe(CN)6]^{4-} -> Fe^{2+} + 6CN^{-}} \: \: K=\frac{1}{K_{2}}\)
____________________________________
\(\ce{Fe^{3+} + Fe(CN)6^{4-} <-> Fe^{2+} + Fe(CN)6^{3-}}\)
The equilibrium constant for this reaction is the product of the two reactions we added, i.e., K = K 1 /K 2 .
But we can make the same overall reaction by combining reactions (3) and (4)
\(\ce{Fe^{3+}_{(aq)} + e^{-} = Fe^{2+}_{(aq)}} \: \: E^{o}= + 0.77V\)
\(\ce{[Fe(CN)6]^{4-} = [Fe(CN)6]^{3-} + e^{-}} \: \: E^{o} = -0.42V\)
____________________________________
\(\ce{Fe^{3+} + Fe(CN)6^{4-} <-> Fe^{2+} + Fe(CN)6^{3-}}\)
In this case, we can calculate E° = 0.77 - 0.42 = +0.35 V
It follows from nFE° = -ΔG° = RTlnK that
\(E^{o} = \frac{RT}{nF} ln \frac{K_{1}}{K_{2}}\)
\(\frac{K_{1}}{K_{2}} = exp\frac{(nFE^{o}}{RT} = exp[(1 \: equiv/mol)(96,500 C/equiv)(0.35J/C)(8.314 J/molK)(298)] = exp(13.63) = 8 \times 10^{5}\)
Thus we find that Fe(CN) 6 3- is about a million times more stable as a complex than Fe(CN) 6 4- .
Solubility Equlibria
We can use a similar procedure to measure K
sp
values electrochemically.
For example, the silver halides (AgCl, AgBr, AgI) are sparingly soluble. We can calculate the K
sp
of AgCl by measuring the standard potential of the AgCl/Ag couple. This can be done very simply by measuring the potential of a silver wire, which is in contact with solid AgCl and 1 M Cl
-
(aq), against a hydrogen reference electrode. That value is then compared to the standard potential of the Ag
+
/Ag couple:
\(\ce{AgCl_{(s)} + e^{-} -> Ag_{(s)} + Cl^{-}_{(aq)}} \: \: E^{o} +.207V\)
\(\ce{Ag^{+}_{(aq)} + e^{-} -> Ag_{(s)}} \: \: E^{o}= + 0.799V\)
Subtracting the second reaction from the first one we obtain:
\(\ce{AgCl_{(s)} = Ag^{+}_{(aq)} + Cl^{-}_{(aq)}} \: \: E^{o} = + 0.207 - 0.799 = -0.592V\)
and again using nFE° = RTlnK, we obtain K = K sp = 9.7 x 10 -11 M 2 .
Because the solubility of the silver halides is so low, this would be a very difficult number to measure by other methods, e.g., by measuring the concentration of Ag+ spectroscopically, or by gravimetry. In practice almost all K sp values involving electroactive substances are measured potentiometrically.
Acid-Base Equilibria
Many electrochemical reactions involve H + or OH - . For these reactions, the half-cell potentials are pH-dependent .
Example: Recall that the disproportionation reaction \(\ce{3MnO4^{2-}_{(a)} -> 2MnO4^{-}_{(aq)} + MnO2_{(s)}}\) is spontaneous at pH=0 ([H + ] = 1M), from the Latimer diagram or Frost plot.
However, when we properly balance this half reaction we see that it involves protons as a reactant :
\[\ce{3MnO4^{2-}(aq) + 4H^{+}(aq) <=> 2MnO4^{-}(aq) + MnO2(s) + 2H2O(l)}\]
By Le Chatelier's principle, it follows that removing protons (increasing the pH) should stabilize the reactant, MnO 4 2 - . Thus we would expect the +6 oxidation state of Mn, which is unstable in acid, to be stabilized in basic media. We will examine these proton-coupled redox equilibria more thoroughly in the context of Pourbaix diagrams below.
|
Acid mine drainage occurs when sulfide rocks such as pyrite (FeS 2 ) are exposed to the air and oxidized. The reaction produces aqueous Fe 3+ and H 2 SO 4 . When the acidic effluent of the mine meets a stream or lake at higher pH, solid Fe(OH) 3 precipitates, resulting in the characteristic orange muddy color downstream of the mine. The acidic effluent is toxic to plants and animals. |