4.4: Redox Reactions with Coupled Equilibria

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 Ksp, Ka, and Kd values.

As an example, we consider the complexation of Fe2+ and Fe3+ 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 Fe3+/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 = K1/K2.

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)63- is about a million times more stable as a complex than Fe(CN)64-.

Solubility Equlibria

We can use a similar procedure to measure Ksp values electrochemically.
For example, the silver halides (AgCl, AgBr, AgI) are sparingly soluble. We can calculate the Ksp 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 = Ksp = 9.7 x 10-11 M2. 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 Ksp 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 Châtelier's principle, it follows that removing protons (increasing the pH) should stabilize the reactant, MnO42-. 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 (FeS2) are exposed to the air and oxidized. The reaction produces aqueous Fe3+ and H2SO4. 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.