When the activities (a_i) can be considered as equal to the
molar, or the
molal, concentrations (C_i) at sufficiently diluted concentrations when the
activity coefficients (\gamma_i) tend to one, the term regrouping all the activity coefficients is equal to one, and the
Nernst equation can be written simply with the concentrations (C_i) denoted here with square braces [ ]: : E_h = E_\text{red} = E^{\ominus}_\text{red} - \frac{0.05916}{z} \log\left(\frac{\left[C\right]^c\left[D\right]^d}{\left[A\right]^a\left[B\right]^b}\right) - \frac{0.05916\,h}{z} \text{pH} There are three types of line boundaries in a Pourbaix diagram: Vertical, horizontal, and sloped.
Vertical boundary line When no electrons are exchanged (
z = 0), the equilibrium between , , , and only depends on and is not affected by the
electrode potential. In this case, the reaction is a classical
acid-base reaction involving only
protonation/deprotonation of dissolved species. The boundary line will be a vertical line at a particular value of pH. The reaction equation may be written: : a \, A + b \, B + h \, \ce{H+} \quad \ce{} \quad c \, C + d \, D and the energy balance is written as \Delta G^\circ = -RT \ln K, where is the
equilibrium constant: : K = \frac{\left[C\right]^c \left[D\right]^d}{\left[A\right]^a \left[B\right]^b \left[\ce{H+}\right]^h} Thus: :\Delta G^\circ = -RT \ln\left( \frac{\left[C\right]^c \left[D\right]^d}{\left[A\right]^a \left[B\right]^b \left[\ce{H+}\right]^h} \right) or, in base-10 logarithms, :\Delta G^\circ = -RT \lambda \, \left( \log \left( \frac{\left[C\right]^c \left[D\right]^d}{\left[A\right]^a \left[B\right]^b} \right) + h\,\ce{pH} \right) which may be solved for the particular value of pH. For example The potential Eh is a function of temperature via the
thermal voltage V_T and directly depends on the ratio of the concentrations of the and ions: :E_h = {E^\circ} - V_T\lambda \log\left(\frac{\ce{[Fe^{2+}]}}{\ce{[Fe^{3+}]}}\right) For both ionic species at the same concentration (e.g., 10^{-6} \mathrm M) at STP, log 1 = 0, so, E_h = E^\circ=0.771\, \mathrm V, and the boundary will be a horizontal line at
Eh = 0.771 volts. The potential will vary with temperature.
Sloped boundary line In this case, both electrons and H+ ions are involved and the electrode potential is a function of pH. The reaction equation may be written: : a \, A + b \, B + h \, \ce{H+} + z \, e^{-} \quad \ce{} \quad c \, C + d \, D Using the expressions for the free energy in terms of potentials, the energy balance is given by a Nernst equation: : E_h = {E^\circ} - \frac{V_T\lambda}{z} \left( \log \left( \frac{\left[C\right]^c \left[D\right]^d}{\left[A\right]^a \left[B\right]^b} \right) + h\, \ce{pH} \right) For the iron and water example, considering the boundary line between the
ferrous ion Fe2+ and
hematite Fe2O3, the reaction equation is: : Fe2O3(s) + 6 H+(aq) + 2 e^- 2 Fe^{2+}(aq) + 3 H2O(l) : with E^\circ = 0.728 \mathrm V. The equation of the boundary line, expressed in base-10 logarithms is: : E_h = {E^\circ} - \frac{V_T \lambda}{2} \left ( \log\left ( \frac\ce{[Fe^{2+}]^2[H2O]^3}\ce{[Fe2O3]} \right ) + 6 \ \ce{pH}\right) As, the activities, or the concentrations, of the solid phases and water are always taken equal to unity by convention in the definition of the
equilibrium constant : [Fe2O3] = [H2O] = 1. The Nernst equation thus limited to the dissolved species and is written as: : E_h = {E^\circ} - \frac{V_T \lambda}{2} \left ( \log\ \ce{[Fe^{2+}]^2} + 6 \ \ce{pH}\right) For, [Fe2+] = 10−6 M, this yields: : E_h = {1.0826} - {0.1775} \ {pH} \quad (\text{in volts}) Note the negative slope (−0.1775) of this line in a Eh–pH diagram. == The stability region of water ==