Constructing Eh-pH Diagrams

4.1The Water Eh-pH Diagram

Half Reactions and Data

In order to be able to make use of Eh-ph diagrams for aqueous solutions, we need to know what the stability of water is with potential and pH. Water can be reduced (to form H₂) and it can be oxidized (to form O₂). In terms of reductions alone, which is what we will be plotting, O₂ can be reduced to water. First we have to write suitable half reactions. Consider water reduction to H₂. Here we must simply follow the rules for balancing reactions:

[latex]\begin{align*} \ce{H2O}&=\ce{H2} \\ \ce{H2O}&=\ce{H2}+\ce{H2O} \\ \ce{H2O}&+\ce{2H+}=\ce{H2}+\ce{H2O} \\ \ce{\cancel{H2O}}&+\ce{2H+}+\ce{2e-}=\ce{H2}+\ce{\cancel{H2O}} \\ \ce{2H+_{aq}}&+\ce{2e-}=\ce{H2_g}\tag{35} \end{align*}[/latex]

This means that water reduction is equivalent to H+ reduction. The reduction of O2 to form water is,

\[\ce{O2_g + 4H^{+}_{aq} + 4e^- = 2H2O_l} \tag{36}\]

We also need ΔG°f data: ΔG°f(H2O) = -237.15 kJ/mol. For all the other species, ΔG°f = 0 by definition. Then we obtain ΔG° for each half reaction:

\[\text{For }\ce{2H+_{aq} + 2e^- = H2_g}\quad \Delta G^{\circ} = 0 – 0 = 0\ \text{kJ/mol} \tag{[34]}\]

\[\text{For }\ce{O2_g + 4H+_{aq} + 4e^- = 2H2O_l}\quad \Delta G^{\circ} = 2 \times (-237.15) = -474.30\ \text{kJ/mol} \tag{[35]}\]

Now the gas pressures must be specified. This depends entirely on what conditions one is interested in; for an autoclave we might be dealing with 100 atm, and for an open tank, perhaps 1 atm. First we will set gas pressures to be 1 atm. The only ion involved is H⁺, and its activity is incorporated into the pH term.

Calculating the Eh Equations

Starting with the H⁺/H₂ half reaction and using the Eh equation,

\[\;E_h = -\frac{\Delta G^{\circ}}{nF} – \frac{2.303RT}{nF}\log Q_H – \frac{2.303RT\,m}{nF}\,pH \tag{[36]}\]

m = 2 and n = 2; T = 25°C = 298.15 K:

\[\;E_h = 0 – \frac{2.303RT}{nF}\log(P_{H_2})
– \frac{2.303 \times 8.314 \times 298.15 \times 2}{2 \times 96485}\,pH
\tag{[37]}\]

\[\;E_h = -0.02958\log(P_{H_2}) – 0.05917\,pH = -0.05917\,pH \text{ when } P_{H_2}=1\ \text{atm} \tag{[38]}\]

Thus Eh = 0 at pH = 0, as expected, based on E°H+/H2 = 0. For the O2/H2O couple,

\[\;E_h
= -\frac{474{,}300\ \text{J/mol}}{2\ \text{mol e}^-/\text{mol} \times 96485\ \text{C/mol e}^-}
– \frac{2.303RT}{nF}\log\!\left(\frac{1}{P_{O_2}}\right)
– \frac{2.303 \times 8.314 \times 298.15 \times 4}{4 \times 96485}\,pH
\tag{[39]}\]

Note that DG° must be converted to J/mol from kJ/mol.

\[\;E_h = 1.229 – 0.01479\log\!\left(\frac{1}{P_{O_2}}\right) – 0.05917\,pH
= 1.229 – 0.05917\,pH \tag{[40]}\]

when P_O2 = 1 atm.

These two equations can be plotted on a graph, as in Figure 4. The resulting diagram is the Eh-pH diagram for water at 25°C and 1 atm pressure. Below the H⁺/H₂ line, water will be converted into H₂ (e.g. if a reducing agent with Eh < EhH⁺/H₂ is added it will yield electrons to the oxidant, H⁺, and form H₂).

And, above the O₂/H₂O line, water will be converted into O₂. It is important to bear in mind that the lines are plots of half reaction potentials, relative to standard H⁺/H₂ versus pH. It tells us, for instance, that if we add a strong oxidant, like KMnO₄ (E°_MnO4-/Mn+2 = 1.51 V in acid solution) to water at pH 0, it should oxidize it to form O₂ and Mn+2. This does occur, although very slowly. It tells us that water becomes increasingly prone to oxidation as pH rises. And, it tells us that water becomes progressively more stable toward reduction as pH rises. Thus, anywhere between the lines water is thermodynamically stable. The diagram shows us the Eh-pH domain over which water is stable, i.e. is the dominant species.

Understanding the Diagram

There are some important subtleties to take note of. For one, pure liquid water has an activity of 1 by definition (as do all pure solids and liquids). Between and on the lines water has unit activity. This means that above the O₂/H₂O line or below the H+/H₂ line water will not exist at all, at least thermodynamically speaking, since it can have activity of either 1 (pure water) or 0 (no water left). Practically, this would require that the oxidant, for instance, react rapidly with water. But, thermodynamically speaking, water is completely unstable outside its region of stability. These considerations apply to any pure solid or liquid.

Second, H₂ and O₂ can have varying pressures, so they can in principle exist in contact with water, i.e. within the stability region for water. For instance, if we maintain a potential of 0.2 V at pH 0 in water (for example by introducing a suitable redox couple, such as Ru⁺³/Ru⁺²; E° = 0.24 V and with a suitable reaction quotient a_Ru+2/a_Ru+3) we can calculate what the H₂ pressure can be. Now we must allow for a P_H2 different from 1 atm:

\[\;E_h
= -\frac{2.303 \times 8.314 \times 298.15}{2 \times 96485}\log P_{H_2}
– \frac{2.303 \times 8.314 \times 298.15 \times 2}{2 \times 96485}\,pH
\tag{[41]}\]

\[\;E_h = 0.20\ \text{V} = -0.02958\log P_{H_2} – 0.05917\,pH \tag{[42]}\]

Solving for P_H2 at pH 0 gives P_H2 = 4.17 x 10^-4 atm. This is considerably less than 1 atm, which is what the pressure was on the line (which we used to calculate the line). So species that can have variable activity (like gases and solutes) do exist within the region of dominance for another species, but at lower activities than the dominant species. Therefore, a region on an Eh-pH diagram shows the Eh-pH domain over which that one species is the predominant one.

The lines shift if we change the pressures of the gases. If we increase P_H2, for instance, the effect in equation [41] will be to decrease the Eh. Likewise, increasing PO2 in equation [39] will increase the Eh. The net effect is to make water more stable, increasing the size of the water region. However, the effect is small; at 10 atm the H⁺/H₂ line shifts down by 0.03 V and the O₂/H₂O line shifts up by 0.015 V.

Why Other H-O Compounds Were Not Included

Finally, there are O, H chemical species that we might have considered, such as O₃ (ozone gas) and H₂O₂ (hydrogen peroxide). Consider the sequence of half reactions,

[latex]\ce{O2_{g} + 2H+^{ }_{aq} + 2e- = H2O2_{aq}}[/latex]

[latex]\ce{H2O2_{aq} + 2H+^{ }_{aq} + 2e- = 2H2O_{l}}[/latex]

Thus O₂ could be reduced first to H₂O₂, then H₂O₂ to water. This suggests the possibility of two different lines (couples) on the Eh-pH diagram: O₂/H₂O₂ and H₂O₂/H₂O. For this to actually occur, Eh for the O₂/H₂O₂ couple would have to be greater than that for the H₂O₂/H₂O couple. We’ll see why. The ΔG°f value for H₂O₂ aq is -133.68 kJ/mol at 25°C. The Eh equations can be calculated for the couples using equation [36] and plotted on an Eh-pH diagram. This is shown in Figure 5. Note that thermodynamically speaking H₂O₂ at unit activity is able to oxidize water, so it should not survive in water at any pH, and we could thus leave it off the diagram. (However, H₂O₂ has substantial kinetic stability in water, i.e. it reacts with water slowly, so we can make H₂O₂ solutions.) But, note that if we reduce H₂O₂ using a suitable reductant we will form water, and at a potential well above Eh for the O2/H2O2 couple. This means that we can’t reduce O₂ to H₂O₂ first. Second, if we do try to reduce O₂ to H₂O₂, this will occur at a potential well below the H₂O₂/H₂O couple.

This means that the potential to form H₂O is higher and the result will be that we form water rather than H₂O₂. In short, the couples involving H₂O₂ are inconsistent with the thermodynamics, and we have no good thermodynamic basis for including them. Finally, note too that the region between the H₂O₂/H₂O and O₂/H₂O lines suggests that both O₂ and H₂O are dominant species here. Since this cannot be, it is another clue that the arrangement is untenable. (For a given pair of elements, only one species can be dominant in any given region).

Now that we have the water Eh-pH diagram we can start to build other diagrams for aqueous solutions. The water diagram tells us what is thermodynamically allowable (or not) in aqueous solution.

4.2 The Zinc-Water Eh-pH Diagram

Chemical Species and Data

When we refer to the Zn-H2O diagram, or system, we mean to consider all possible species that contain Zn, O and H. A list of the known species and their corresponding ΔG°f data is provided in Table 2. It should be apparent then that such diagrams are only as good as our chemical knowledge of any given system. For this diagram, to start with, we will use unit activities for solutes, 1 atm pressure for gasses and a temperature of 25°C.

Assigning Oxidation States

At this stage we need to determine the oxidation states of the species. The reason is that we would expect more highly oxidized species to appear at higher reduction potentials, and more reduced species to appear at lower potentials. This will give us a rough sense of what to expect. In addition some species may have the same oxidation state. This will mean that they are related to each other by non-electron transfer reactions (acid-base processes). Oxidation states can be assigned by the traditional rules. This was presented in the Chemistry Review Part I course notes. Another simple way to do it is to write a half reaction for the reduction of the compound/ion to the principal element (Zn here). To illustrate:

[latex]\ce{HZnO2- = Zn}[/latex]

[latex]\ce{HZnO2- = Zn + 2H2O}[/latex]

[latex]\ce{HZnO2- + 3H+ = Zn + 2H2O}[/latex]

[latex]\ce{HZnO2-_{aq} + 3H+_{aq} + 2e- = Zn_{s} + 2H2O_{l}}\quad(39)[/latex]

Next, divide the number of electrons (2 here) by the number of metal atoms involved in the reaction (1 Zn here). This gives a value of 2. The oxidation state for Zn in this ion then is +2.* The oxidation state for each of Zn⁺² (obviously +2), ZnO, HZnO₂^– and ZnO₂^2- is +2. The oxidation state for the element is always 0 by definition. This means that we would expect a diagram with Zn at the bottom and the other four Zn(II) species above. But, what will be the order for the four Zn(II) species? That comes next.

Hydrolysis Reactions

When metal ions in aqueous solution exhibit acid-base behaviour (and almost all do), we call it hydrolysis, meaning splitting by water. This was introduced in the Chemistry Review Part II course notes. Now we need a more thorough treatment. Consider the triprotic acid H3AsO4. It has three acid dissociable protons.

[latex]\ce{H3AsO4_{aq} = H2AsO4-_{aq} + H+_{aq}}\qquad \mathrm{p}K_{a1} = 2.24 \qquad (40)[/latex]

[latex]\ce{H2AsO4-_{aq} = HAsO4^{2-}_{aq} + H+_{aq}}\qquad \mathrm{p}K_{a2} = 6.96 \qquad (41)[/latex]

[latex]\ce{HAsO4^{2-}_{aq} = AsO4^{3-}_{aq} + H+_{aq}}\qquad \mathrm{p}K_{a3} = 11.50 \qquad (42)[/latex]

All four species have oxidation state +5. As we might expect, H₃AsO₄ is the strongest acid, followed by H₂AsO_4^–, followed by HAsO₄^2-. If we were to depict these three reactions on an Eh-pH diagram, there would be three vertical lines; one for each dissociation (no electrons are involved, so the pH at some fixed solute activity would be a constant independent of Eh for each reaction, as we noted previously as a special case in the development the of the Eh equations.)

Now, if we start with AsO₄^3- at high pH and lower the pH (add H⁺) we would form HAsO₄^2-. This means that AsO₄^3- appears at the highest pH and the species next to it to the left (lower pH) is HAsO₄^2-. At pH 11.50, the buffer point, we would have equal activities (approximately equal concentrations) of both. Since in our Eh-pH diagrams we specify that the solutes have some fixed activity, the pH we would calculate by equation [32] would correspond to pK_a3. The same considerations can be applied to lowering the pH and forming H₂AsO₄^– from HAsO₄^2-. Likewise for H₂AsO₄^– and H₃AsO₄. This means that the strongest acid appears at the lowest pH, followed by the next strongest acid, etc. This is diagrammed in Figure 6.

Hydrolysis of metal ions involve reactions such as,

\[\ce{[M(H2O)6]^{2+}_{aq} = [M(H2O)5(OH)]+_{aq} + H+_{aq}} \tag{43}\]

\[\ce{[M(H2O)5(OH)]+_{aq} = M(OH)2_{s} + H+_{aq} + 4H2O_{l}} \tag{44}\]

\[\text{or,}\ \ce{[M(H2O)5(OH)]+_{aq} = MO_{s} + H+_{aq} + 5H2O_{l}} \tag{45}\]

etc. Numerous such dissociations are possible in principle. A more convenient shorthand for the same reactions is:

\[\ce{M^{2+}_{aq} + H2O_{l} = MOH+_{aq} + H+_{aq}} \tag{46}\]

\[\ce{MOH+_{aq} + H2O_{l} = M(OH)2_{s} + H+_{aq}} \tag{47}\]

\[\text{or,}\ \ce{MOH+_{aq} = MO_{s} + H+_{aq}} \tag{48}\]

etc. In almost all cases the hydrated metal ion, Mn+aq is the strongest acid and appears at the lowest pH.

The figure below is a summary of sequential hydrolysis reactions. It can be seen that with increasing extent of hydrolysis (number of acid dissociations) the number of possible metal oxide/hydroxide species increases as well. With each successive dissociation more O or less H resides with the metal. We can call these sequential compounds or ions hydrolysis states, analogous to oxidation states. For any given step only one of the possible species may appear on an Eh-pH diagram. The reason for this is that transformations between species with the same hydrolysis state involve only water, e.g.

\[\ce{M(OH)2_s = MO_s + H2O_l} \tag{49}\]

This was special case 4 in the development of the Eh equations.

Calculating Hydrolysis States

The simplest way to assess the order in which the acids go on an Eh-pH diagram, for a given oxidation state, is to calculate the hydrolysis state. (This works mainly for metal oxo- and hydroxy-ions.) This is show below:

\[\mathrm{HS} = \frac{2 \times \text{no. O atoms} – \text{no. H atoms}}{\text{no. metal atoms}} \tag{[43]}\]

Consider some examples:

[latex]\begin{align*} \mathrm{M^{n+}} \qquad &\mathrm{HS} = 0 \; (\text{unhydrolyzed; no acid dissociation has occurred}) \\[0.6em] \mathrm{M(OH)^+} \qquad &\mathrm{HS} = \frac{2 \times 1 - 1}{1} = 1 \\[0.6em] \mathrm{M(OH)_2} \qquad &\mathrm{HS} = \frac{2 \times 2 - 2}{1} = 2 \\[0.6em] \mathrm{MO} \qquad &\mathrm{HS} = \frac{1 \times 2 - 0}{1} = 2 \\[0.6em] \mathrm{[Mo_7O_{24}]^{6-}} \qquad &\mathrm{HS} = \frac{24 \times 2 - 0}{7} = \frac{48}{7} = 6.857 \end{align*}[/latex]

The larger the hydrolysis state, the more H+ that have been released per metal ion, and the weaker an acid the compound/ion is, and the further it appears to the right (higher pH). Thus hydrolyzed metal ions can be arranged from left to right in order of increasing HS. Finally, most metal ions have a small number of hydrolysis states. Some exceptions include oxo/hydroxo ions of V, Mo and W.*

Tentative Eh-pH Diagram for the Zn-H₂O System

The hydrolysis states for Zn(II) species are:

[latex]\begin{align*} \ce{Zn^{2+}_{aq}} \qquad &\mathrm{HS} = 0 \qquad\qquad \ce{HZnO2^-_{aq}} \qquad &\mathrm{HS} = 3 \\[0.6em] \ce{ZnO_s} \qquad &\mathrm{HS} = 2 \qquad\qquad \ce{ZnO2^{2-}_{aq}} \qquad &\mathrm{HS} = 4 \end{align*}[/latex]

(HZnO₂-is probably better presented as Zn(O)OH^–.) This then suggests a preliminary Eh-pH diagram as shown in Figure 8.

Calculating the pH of the Hydrolysis Reactions

It is often best to start by calculating the hydrolysis pH lines. There is no Zn(OH)+ species listed, so we go directly from Zn+2 to ZnO. The reactions are:

Step 1

[latex]\begin{align*} &\ce{Zn^{+2} = ZnO} \\ &\ce{Zn^{+2} + H2O = ZnO} \\ &\ce{Zn^{+2}_{aq} + H2O_{l} = ZnO_{s} + 2H+_{aq}} \tag{50} \\ &\Delta G^{\circ} = \Delta G_f^{\circ}(\ce{ZnO}) - \Delta G_f^{\circ}(\ce{H2O}) - \Delta G_f^{\circ}(\ce{Zn^{+2}}) \tag{[44]} \\ &= -318.3 - (-237.15) - (-147.2) = 66.05\ \text{kJ/mol} \tag{[45]} \end{align*}[/latex]

Step 2

[latex]\begin{align*} &\ce{ZnO = HZnO2^-} \\ &\ce{ZnO + H2O = HZnO2^-} \\ &\ce{ZnO_{s} + H2O_{l} = HZnO2^-_{aq} + H+_{aq}} \tag{51} \\ &\Delta G^{\circ} = \Delta G_f^{\circ}(\ce{HZnO2^-}) - \Delta G_f^{\circ}(\ce{H2O}) - \Delta G_f^{\circ}(\ce{ZnO}) \tag{[46]} \\ &= -463.9 - (-237.15) - (-318.3) = 91.55\ \text{kJ/mol} \tag{[47]} \end{align*}[/latex]

Step 3

[latex]\begin{align*} &\ce{HZnO2^- = ZnO2^{2-}} \\ &\ce{HZnO2^-_{aq} = ZnO2^{2-}_{aq} + H+_{aq}} \tag{52} \\ &\Delta G^{\circ} = \Delta G_f^{\circ}(\ce{ZnO2^{2-}}) - \Delta G_f^{\circ}(\ce{HZnO2^-}) \tag{[48]} \\ &= -389.1 - (-463.9) = 74.8\ \text{kJ/mol} \tag{[49]} \end{align*}[/latex]

The activity of solutes = 1 (molal scale). From equation [33] for pH:

\[\;pH = -\frac{\Delta G^{\circ}}{2.303RT\,m} – \frac{1}{m}\log Q_H \tag{[50]}\]

Step 1

For Zn⁺²/ZnO with Q_-H = 1/a_Zn+2 and m = -2 (H⁺ is a product):

\[\;pH
= -\frac{66{,}050}{2.303 \times 8.314 \times 298.15 \times (-2)}
– \frac{1}{(-2)}\log\!\left(\frac{1}{a_{Zn^{2+}}}=1\right)
= 5.785
\tag{[51]}\]

Step 2

For ZnO/HZnO₂^– with Q_-H = a_HZnO2- (a_ZnO = 1 by definition) and m = -1:

\[ \mathrm{pH}= -\frac{91{,}550}{2.303 \times 8.314 \times 298.15 \times (-1)} – \frac{1} {(-1)} \log(a_{\ce{HZnO2^-}} = 1) = 16.037 \tag{[52]}\]

Step 3

For HZnO₂^–/ZnO₂^2- with Q_-H = a_ZnO22-/_aHZnO2- and m = -1

\[\mathrm{pH}= -\frac{74{,}800}{2.303 \times 8.314 \times 298.15 \times (-1)} – \frac{1}{(-1)} \log\!\left(\frac{a_{\ce{ZnO2^{2-}}}}{a_{\ce{HZnO2^-}}}\right) = 13.103 \tag{[53]}\]

Looking at the values we see that something seems amiss. The numbers should increase from one step to the next. Instead 5.785 < 16.037 > 13.103. This means that before ZnO converts to HZnO₂^– (by raising pH), HZnO₂^– has already converted to ZnO22-. This suggests that one of ZnO/HZnO₂^– or HZnO₂^–/ZnO₂^2- does not occur; that one of HZnO₂^– or ZnO₂^2- cannot have unit activity. (There does not appear to be any reason to reject the Zn⁺²/ZnO transformation.) The question is, which species will not appear on the diagram? The choices we are left with are HZnO₂^– and ZnO₂^2-. The only remaining possible reaction is for ZnO to go to ZnO₂^2-. (We have already determined the pH for ZnO/HZnO₂^– and HZnO₂^–/ZnO₂^2-.)

[latex]\begin{align*} &\ce{ZnO = ZnO2^{2-}}\\ &\ce{ZnO + H2O = ZnO2^{2-}}\\ &\ce{ZnO_{s} + H2O_{l} = ZnO2^{2-}_{aq} + 2H+_{aq}}\tag{53}\\ &\Delta G^{\circ} = \Delta G_f^{\circ}(\ce{ZnO2^{2-}}) - \Delta G_f^{\circ}(\ce{H2O}) - \Delta G_f^{\circ}(\ce{ZnO})\tag{[54]}\\ &= -389.1 - (-237.15) - (-318.3) = 166.35\ \text{kJ/mol}\tag{[55]} \end{align*}[/latex]

Step 4

For ZnO/ZnO₂^2- with Q-H = a_ZnO22- and m = -2:

\[\mathrm{pH}= -\frac{166{,}350}{2.303 \times 8.314 \times 298.15 \times (-2)} – \frac{1}{(-2)} \log(a_{\ce{ZnO2^{2-}}}=1) = 14.570 \tag{[56]}\]

Now we have to decide on the final arrangement. The options are shown below:

DIAGRAM

Now we can see that if we raise the pH we will convert ZnO into ZnO22- before we form HZnO_2-. Thus the arrangement on the diagram will be Zn⁺²/ZnO/ZnO₂^2-. *

Calculating the Eh Lines

Once the hydrolysis lines are known, start at one corner and work from there. The tentative diagram suggests we need a Zn⁺²/Zn line:

\begin{align*}
&\ce{Zn^{+2}_{aq} + 2e- = Zn_{s}} \tag{54}\\
&n = 2,\ m = 0\\
&\Delta G^{\circ} = \Delta G_f^{\circ}(\ce{Zn}) – \Delta G_f^{\circ}(\ce{Zn^{+2}}) = 0 – (-147.2) = +147.2\ \text{kJ/mol} \tag{[57]}
\end{align*}

from the Eh equation,

\[\mathrm{Eh} = -\frac{\Delta G^{\circ}}{nF} – \frac{2.303RT}{nF}\log Q_H – \frac{2.303RT\,m}{nF}\,\mathrm{pH} \tag{[58]}\]

Q-H = 1/a_Zn+2, solutes at unit activity

\[\mathrm{Eh} = -\frac{147{,}200}{2 \times 96485} – \frac{2.303 \times 8.314 \times 298.15}{2 \times 96485}\log Q_H – \frac{2.303RT \times 0}{nF}\,\mathrm{pH} \tag{[59]}\]

\[\mathrm{Eh} = -0.7628 – 0.02958 \log\!\left(\frac{1}{a_{\ce{Zn^{+2}}}}\right) = -0.7628\ \mathrm{V} \tag{[60]}\]

The reduction potential for Zn⁺²/Zn is a constant, independent of pH. It will be a horizontal line on the Eh-pH diagram. Assume a hypothetical redox couple that will supply electrons. As long as the reduction potential >-0.7628 V, Zn⁺² will persist. Once the reduction potential becomes ≤-0.7628 V, Zn⁺² is a strong enough oxidant to accept the electrons and be reduced. Thus as we lower the reduction potential we form Zn metal. Hence, as per our tentative diagram (Figure 8) the Zn(II) species lie above Zn metal. The line separates the Zn⁺² and Zn regions on the diagram. Above the line Zn⁺² is predominant; below it Zn is. The line will run from low pH to the pH where the next dominant Zn(II) species occurs, i.e. to pH 5.785. Above that pH ZnO is dominant and not Zn⁺². Hence the line stops at pH 5.785.

Next we need the ZnO/Zn couple line. Write the balanced half reaction:

[latex]\begin{align*} &\ce{ZnO = Zn} \\ &\ce{ZnO = Zn + H2O} \\ &\ce{ZnO + 2H+ = Zn + H2O} \\ &\ce{ZnO_{s} + 2H+_{aq} + 2e- = Zn_{s} + H2O_{l}} \tag{55} \\ &n = 2,\ m = 2,\ Q_H = 1\ (\text{there are no solute species; } 2.303RT \log Q_H/nF = 0) \\ &\Delta G^{\circ} = \Delta G_f^{\circ}(\ce{H2O}) - \Delta G_f^{\circ}(\ce{ZnO}) = -237.15 - (-318.3) = 81.15\ \text{kJ/mol} \tag{[61]} \\ &\mathrm{Eh} = -\frac{81{,}150}{2 \times 96485} - \frac{2.303RT}{nF}\log 1 - \frac{2.303 \times 8.314 \times 298.15 \times 2}{2 \times 96485}\,\mathrm{pH} \tag{[62]} \\ &= -0.4205 - 0.05917\,\mathrm{pH} \tag{[63]} \end{align*}[/latex]

Note: at pH 5.785 where the Zn⁺²/Zn line stops Eh = -0.7628 V, the same value as for the Zn⁺²/Zn couple. At pH 5.785 and Eh -0.7628 there are three species in equilibrium: Zn⁺², ZnO and Zn. The potentials then for the Zn⁺²/Zn and ZnO/Zn couples must be equal, otherwise there would be a driving force for a reaction (Zn⁺² or ZnO) « Zn. This is true in general. The point at which one line stops must be the point where the next starts.* The Zn⁺²/ZnO vertical line also stops at the same point; Zn(II) species cannot extend into the Zn metal region. The Zn⁺²/ZnO vertical line runs up from Eh = -0.7628 V to the top of the diagram. The diagram to this point is shown in Figure 9.

The ZnO/Zn line runs to pH 14.570, beyond which ZnO22- becomes the dominant Zn(II) species. Then we need a ZnO22-/Zn line:

\[\ce{ZnO2^{2-}_{aq} + 4H+_{aq} + 2e- = Zn_{s} + 2H2O_{l}} \tag{56}\]

\[\;n = 2,\ m = 4,\ Q_H = 1/a_{\ce{ZnO2^{2-}}}\]

\[\Delta G^{\circ} = 2\Delta G_f^{\circ}(\ce{H2O}) – \Delta G_f^{\circ}(\ce{ZnO2^{2-}}) = 2 \times (-237.15) – (-389.1) = -85.20\ \text{kJ/mol} \tag{[64]}\]

\[\mathrm{Eh} = -\frac{85{,}200}{2 \times 96485} – \frac{2.303 \times 8.314 \times 298.15}{2 \times 96485}\log Q_H – \frac{2.303 \times 8.314 \times 298.15 \times 4}{2 \times 96485}\,\mathrm{pH} \tag{[65]}\]

\[\mathrm{Eh} = 0.4415 – 0.02958 \log\!\left(\frac{1}{a_{\ce{ZnO2^{2-}}}}\right) – 0.1183\,\mathrm{pH} = 0.4415 – 0.1183\,\mathrm{pH} \tag{[66]}\]

The ZnO/ZnO₂^2- vertical line runs up from the Eh at pH 14.57. The Eh at the point of intersection is easily calculated from the ZnO/Zn or ZnO_22-/Zn Eh equations at pH 14.57. Plotting this yields the diagram in Figure 10. Note that the relative slopes of the lines are governed by the ratio m/n. The water Eh lines are included in such diagrams to remind us of the thermodynamic constraints of water stability.

Reading the Diagram

Now that we have the diagram we can start to use it. Zinc oxide is a solid and it is clearly quite highly soluble in dilute acid (pH 5.79 can sustain something on the order of 1 M (unit activity and thus somewhat less than 1 M) of a soluble zinc salt (e.g. ZnSO₄). However, by far the most common zinc mineral in nature is sphalerite, ZnS. One way to treat sphalerite is to roast it. The resulting product is ZnO (zinc calcine). This then can be leached in dilute acid. (Usually around pH 4 to improve the reaction rate.*) We can also see from the diagram that leaching ZnO in base is impractical. A pH of >14.6 would be required; probably on the order of 10 M NaOH. Since NaOH is a costly reagent, and there is no simple way to produce Zn metal from such a strongly basic solution, leaching in base is obviated.

One of the most common ways of recovering pure metal from aqueous solutions is electrolysis (electrowinning). This forces an otherwise thermo-dynamically unfavourable reduction. However, it is evident from the diagram that at all pH where Zn⁺² is dominant, the Zn⁺²/Zn couple lies far below the H⁺/H₂ line, i.e attempting to reduce Zn⁺² in aqueous solution should produce hydrogen gas first; H+ is a much stronger oxidant than Zn⁺² at all relevant pH. Nevertheless, Zn⁺² electrowinning (EW) is practiced all over the world. This seems paradoxical. It is another case of where kinetics apparently trumps thermodynamics. It turns out that on the surface of very pure zinc metal, reduction of H⁺ to H2 is very slow, while the rate at which Zn⁺² can be reduced to Zn is very fast. Thus in Zn EW most of the electricity goes to plate Zn metal, while a small fraction goes to form H₂. The driving force for hydrogen evolution increases as pH decreases (the Eh gap between the H⁺/H₂ and Zn⁺²/Zn couples increases, meaning that H⁺ becomes an increasingly stronger oxidant relative to Zn⁺².)

The Effects of Varying Ionic Solute Activities

Changing the solute activities will shift the lines where Eh is a function of solute activities. The effects are shown in the Eh-pH diagram in Figure 11. At 10-4 m solute activities a region for HZnO₂^– is present. The Zn⁺² region has expanded to lower Eh and higher pH. The lower Eh for Zn⁺² reduction occurs due to the –log(1/a_Zn+2) term. The shift in pH can be rationalized on the basis of the reaction,

\[\ce{ZnO_s + 2H^{+}_{aq} = Zn^{2+}_{aq} + H2O_l} \tag{57}\]

\[\;K = \frac{a_{Zn^{2+}}}{(a_{H^+})^2} \tag{[67]}\]

If a_Zn+2 drops, so must aH+, i.e. pH increases. Other changes can be similarly understood. Finally, the ZnO region shrinks with decreasing solute activities, while the ZnO₂^2- region expands.

4.3 The Sulfur-Water Diagram

Sulfur plays a large role in hydrometallurgy (and pyrometallurgy). Many metals of commercial interest occur in sulfide minerals. Hence it is important that we be able to develop and understand the sulfur-water Eh-pH diagram. Once that is in place we can combine it with metal-water diagrams to get information about aqueous chemistry associate with metal sulfides.

Chemical Species and Data

Again, we need species with the elements of sulfur and water (S, H, and O). Sulfur forms a wide range of oxoanions and solid elemental sulfur has three common forms that occur in hydrometallurgy. These are shown in Table 3.

The oxidation states can be easily determined by the method outlined earlier in these notes. There are a lot of species to consider. It turns out, however, that only the S(-2), elemental sulfur and S(+6), i.e. sulfate species are thermodynamically stable. Many of the others have substantial kinetic stability (decompose slowly) under certain conditions, but in time they will revert to one of the three stable compounds. What this usually means is that only these three will appear on an Eh-pH diagram that considers the most thermodynamically stable species. The reasons for this will be demonstrated later. (However, because many other sulfur species have kinetic stability, and for other reasons, Eh-pH diagrams can be drawn to include the relevant ones. This requires some additional considerations.) Thermodynamic data for the three sulfur species are given in the table below.

Tentative Diagram

Again, we might expect the most oxidized species to appear at the highest Eh (have the highest reduction potentials). Of the S(+6) species, H₂SO₄ is a strong acid and fully dissociates into HSO₄^– and H⁺. HSO₄^– is a weak acid. We should expect HSO₄^– to be dominant at lower pH. Likewise the order for the S(-2) species should be, from low to high pH: H₂S, HS^–, S^2-. Then our diagram should look somewhat like that in Figure 12. Based on our experience with the Zn–H₂O diagram we might expect negatively Eh sloping lines.

Hydrolysis (Acid Dissociation) Lines

The pH values for the vertical lines can be calculated as before:

Step 1

[latex]\begin{align*} \ce{H2S_g} &= \ce{HS^-_{aq}} + \ce{H+_{aq}} \tag{58} \\ \Delta G^{\circ} &= 11.44 - (-33.56) = 45.00\ \text{kJ/mol} \tag{[68]} \end{align*}[/latex]

Step 2

[latex]\begin{align*} \ce{HS^-_{aq}} &= \ce{S^{2-}_{aq}} + \ce{H+_{aq}} \tag{59} \\ \Delta G^{\circ} &= 117 - 11.44 = 105.56\ \text{kJ/mol} \tag{[69]} \end{align*}[/latex]

Step 3

[latex]\begin{align*} \ce{HSO4^-_{aq}} &= \ce{SO4^{2-}_{aq}} + \ce{H+_{aq}} \tag{60} \\ \Delta G^{\circ} &= -744.55 - (-755.91) = 11.36\ \text{kJ/mol} \tag{[70]} \end{align*}[/latex]

The pH values for the acid dissociation reactions can be calculated next. We will use unit activities for solutes and 1 atm pressure for gases.

Step 1

For H₂S/HS- with Q_-H = a_HS-/P_H2S and m = -1 (H+ is a product):

\[\;pH = -\frac{45{,}000}{2.303 \times 8.314 \times 298.15 \times (-1)} – \frac{1}{(-1)}\log\!\left(\frac{a_{HS^-}}{P_{H_2}}\right) \tag{[71]}\]

\[\;pH = 7.883 + \log\!\left(\frac{a_{HS^-}=1}{P_{H_2}=1}\right) = 7.883 \tag{[72]}\]

Step 2

For HS^–/S^2- with Q_-H = a_S2-/a_HS– = 1 (at all activities) and m = -1:

\[\;pH = -\frac{105{,}560}{2.303 \times 8.314 \times 298.15 \times (-1)} = 18.5 \tag{[73]}\]

Note that because of the form of Q_-H, the log term falls away. If we use ΔG°f(S2-) = 92.2 kJ/mol, the pH becomes 14.15. This is quite commonly seen in hydrometallurgy literature, though probably incorrect.

Step 3

For HSO₄^–/SO₄^2- with Q-H = a_SO42-/a_HSO4- = 1 at all activities and m = -1:

\[\;pH = -\frac{11{,}360}{2.303 \times 8.314 \times 298.15 \times (-1)} = 1.990 \tag{[74]}\]

Calculating the Eh Lines

We will start at the lower left and work out from there. First then we need a S-H2S reduction line:

\[\ce{S_s + 2H+_{aq} + 2e^- = H2S_g} \tag{61}\]

\[n = 2,\; m = 2,\; Q_H = P_{H_2S} = 1\]

\[\Delta G^{\circ} = -33.56\ \text{kJ/mol} \tag{[75]}\]

\[\;E_h = -\frac{-33{,}560}{2 \times 96485} – \frac{2.303 \times 8.314 \times 298.15}{2 \times 96485}\log Q_H – \frac{2.303 \times 8.314 \times 298.15 \times 2}{2 \times 96485}\,pH \tag{[76]}\]

\[\;E_h = 0.1739 – 0.02958\log P_{H_2S} – 0.05917\,pH \tag{[77]}\]

\[\;E_h = 0.1739 – 0.05917\,pH \tag{[78]}\]

This line will run from low pH up to the H2S/HS- vertical line at pH 7.883. Beyond that point H2S is no longer dominant; HS- is. Hence we need a S/HS- line:

[latex]\begin{align*} \ce{S_s} &+ \ce{H+_{aq}} + \ce{2e^-} = \ce{HS^-_{aq}} \tag{62} \\[0.4em] n &= 2,\; m = 1,\; Q_H = a_{HS^-} = 1 \\[0.4em] \Delta G^{\circ} &= 11.44\ \text{kJ/mol} \tag{[79]} \\[0.4em] E_h &= -\frac{11{,}440}{2 \times 96485} - \frac{2.303 \times 8.314 \times 298.15}{2 \times 96485}\log Q_H - \frac{2.303 \times 8.314 \times 298.15 \times 1}{2 \times 96485}\,pH \tag{[80]} \\[0.6em] E_h &= -0.05928 - 0.02958\log(a_{HS^-}) - 0.02958\,pH \tag{[81]} \\[0.4em] E_h &= -0.05928 - 0.02958\,pH \tag{[82]} \end{align*}[/latex]
This line starts at pH 7.883. The diagram to this point is shown in Figure 13. At this point it is a good idea to begin to work up as well. (Looking at the diagram so far we might expect that an HSO₄^–/S line could intersect the S/HS^– line.) Thus we need an HSO₄^–/S line next.

\[\ce{HSO4^-_{aq} + 7H+ + 6e^- = S_s + 4H2O_l} \tag{63}\]

\[n = 6,\ m = 7,\ Q_H = 1/a_{HSO4^-} = 1\]

\[\Delta G^{\circ} = 4 \times (-237.15) – (-755.91) = -192.69\ \text{kJ/mol} \tag{[83]}\]

\[\;E_h = -\frac{-192{,}690}{6 \times 96485} – \frac{2.303×8.314×298.15}{6 \times 96485}\log Q_H – \frac{2.303×8.314×298.15 \times 7}{6 \times 96485}\,pH \tag{[84]}\]

\[\;E_h = 0.3329 – 0.009861\log\!\left(\frac{1}{a_{HSO4^-}}\right) – 0.06903\,pH \tag{[85]}\]

\[\;E_h = 0.3329 – 0.06903\,pH \tag{[86]}\]

This line runs from low pH to pH 1.990, after which SO₄^2- becomes dominant. Next we need the SO₄^2-/S line.

\[\ce{SO4^{2-}_{aq} + 8H+ + 6e^- = S_s + 4H2O_l} \tag{64}\]

\[n = 6,\ m = 8,\ Q_H = 1/a_{SO4^{2-}} = 1\]

\[\Delta G^{\circ} = 4 \times (-237.15) – (-744.55) = -204.05\ \text{kJ/mol} \tag{[87]}\]

\[\;E_h = -\frac{-204{,}050}{6 \times 96485} – \frac{2.303×8.314×298.15}{6 \times 96485}\log Q_H – \frac{2.303×8.314×298.15 \times 8}{6 \times 96485}\,pH \tag{[88]}\]

\[\;E_h = 0.3525 – 0.009861\log\!\left(\frac{1}{a_{SO4^{2-}}}\right) – 0.07889\,pH \tag{[89]}\]

\[\;E_h = 0.3525 – 0.07889\,pH \tag{[90]}\]

This line begins at pH 1.990. Plotting the two new lines yields the diagram below. It is apparent that the SO₄^2-/S and S/HS- lines intersect. Where the lines meet is where they stop. Continuing the lines past their point of intersection makes no sense as shown in Figure 15. This would create apparent regions with more than one S species as dominant. The pH and Eh of intersection need to be calculated using the Eh equations so that the point of intersection can be plotted.

Figure 15. S-H₂O Eh-pH diagram, detail around the intersection of the S/HS^– and SO₄^2-/S lines, 25°C, unit activities for solutes, 1 atm pressure.

The relevant equations are:

\[\text{S/HS:}\quad E_h = -0.05928 – 0.02958\,pH \tag{[91]}\]

\[\text{SO}_4^{2-}\text{/S:}\quad E_h = 0.3525 – 0.07889\,pH \tag{[92]}\]

Solving these two equations gives the pH, Eh point = 8.351, -0.3063 V.

Looking at the diagram to this point, it appears that we have a SO₄^2- region and a HS- region yet that need a boundary. This calls for a SO₄^2-/HS^– line.

\[\ce{SO4^{2-}_{aq} + 9H+ + 8e^- = HS^-_{aq} + 4H2O_l} \tag{65}\]

\[n = 8,\ m = 9,\ Q_H = a_{HS^-}/a_{SO4^{2-}} = 1\ (\text{regardless of solute activities})\]

\[\Delta G^{\circ} = 4 x (-237.15) + 11.44 – (-744.55) = -192.61\ \text{kJ/mol} \tag{[93]}\]

\[\;E_h = -\frac{192610}{8 x 96485} – \frac{2.303×8.314×298.15}{8 x 96485}\log Q_H – \frac{2.303×8.314×298.15 x 9}{8 x 96485}\,pH \tag{[94]}\]

\[\;E_h = 0.2495 – 0.06656\,pH \tag{[95]}\]

This line starts at pH 8.351 and would run to the HS^–/S^2- line at pH 18.5. However, such extremely high pH (on the order of 30,000 M OH^–) is not attainable. (A pH of 14 corresponds to ~ 1 M OH^–). Next we could calculate a SO₄^2-/S^2- line, though it is of little significance since the pH involved is so unrealistically high. However, if one takes ΔG^°_f(S^2-) to be 92.2 kJ/mol (as is common in the hydrometallurgical literature) then the HS^–/S^2- pH would be 14.14 and a SO₄^2-/S^2- line would be reasonable. The half reaction is:

\[\ce{SO4^{2-}_{aq} + 8H+_{aq} + 8e^- = S^{2-}_{aq} + 4H2O_l} \tag{66}\]

\[n = 8,\ m = 8,\ Q_H = a_{S^{2-}}/a_{SO4^{2-}} = 1\ (\text{regardless of solute activities})\]

\[\Delta G^{\circ} = 4 x (-237.15) + 92.2 – (-744.55) = -111.85\ \text{kJ/mol} \tag{[96]}\]

(IF we take ΔG°f(S^2-) to be to the conventional value of 92.2 kJ/mol, rather than the more likely 117 kJ/mol.)

\[\;E_h = -\frac{111{,}850}{8 x 96485} – \frac{2.303×8.314×298.15}{8 x 96485}\log Q_H – \frac{2.303×8.314×298.15 x 8}{8 x 96485}\,pH \tag{[97]}\]

\[\;E_h = 0.1449 – 0.05917\,pH \tag{[98]}\]

The final diagrams are shown in Figure 16 and Figure 17. The sulfur region is a downward sloping wedge that is thermodynamically stable to about pH 8.4. This influences the shape of many metal sulfide regions in corresponding Eh-pH diagrams (in many cases the metal sulfide region parallels the sulfur region). Changing the activities of the solutes or the gas pressure will change the diagram.

Less Stable S-H2O Species

We stated earlier that only S(-2), S and S(+6) species would appear on the diagram. Here we will see why that is so. Consider thiosulfate, S₂O₃^2-, ΔG^°_f = -532.21 kJ/mol. The S oxidation state is +2. We might expect that a region for S₂O₃^2- would lie above the S region and below the SO₄^2- region. Plausible half reactions and Eh equations (solute activities = 1) are shown below.

Step 1

\[\mathrm{SO_4^{2-}\ to\ S_2O_3^{2-}:}\]

\[\ce{2SO4^{2-}_{aq} + 10H+_{aq} + 8e^- = S2O3^{2-}_{aq} + 5H2O_l} \tag{67}\]

\[n = 8,\ m = 10,\ Q_H = a_{S2O3^{2-}}/(a_{SO4^{2-}})^2 = 1\]

\[\Delta G^{\circ} = 5 x (-237.15) + (-532.21) – 2 x (-744.55) = -228.86\ \text{kJ/mol} \tag{[99]}\]

\[\;E_h = -\frac{228{,}860}{8 x 96485} – \frac{2.303×8.314×298.15}{8 x 96485}\log Q_H – \frac{2.303×8.314×298.15 x 10}{8 x 96485}\,pH \tag{[100]}\]

\[\;E_h = 0.2965 – 0.007396\log\!\left(\frac{a_{S2O3^{2-}}}{(a_{SO4^{2-}})^2}\right) – 0.07396\,pH \tag{[101]}\]

\[\;E_h = 0.2965 – 0.07396\,pH \tag{[102]}\]

Step 2

S₂O₃^2- to S:

\[\ce{S2O3^{2-}_{aq} + 6H+_{aq} + 4e^- = 2S_s + 3H2O_l} \tag{68}\]

\[n = 4,\ m = 6,\ Q_H = 1/a_{S2O3^{2-}} = 1\]

\[\Delta G^{\circ} = 3 x (-237.15) – (-532.21) = -179.24\ \text{kJ/mol} \tag{[103]}\]

\[\;E_h = -\frac{179{,}240}{4 x 96485} – \frac{2.303×8.314×298.15}{4 x 96485}\log Q_H – \frac{2.303×8.314×298.15 x 6}{4 x 96485}\,pH \tag{[104]}\]

\[\;E_h = 0.4644 – 0.01479\log\!\left(\frac{1}{a_{S2O3^{2-}}}\right) – 0.08875\,pH \tag{[105]}\]

\[\;E_h = 0.4644 – 0.08875\,pH \tag{[106]}\]

These equations can be plotted on the Eh–pH diagram as shown in Figure 18. The diagram shows that the SO₄^2-/S₂O₃^2- line occurs within the S region, or within the HS^– region. So, if we had SO₄^2- and lowered the reduction potential, we would form the more reduced elemental S before we would form S₂O₃^2-. Or, S forms more easily (at higher potential) than does S₂O₃^2-. This shows that there cannot be a SO₄^2-/S₂O₃^2- line. Similarly, at higher pH, we would form HS^– by reduction of SO₄^2- before we form S₂O₃^2-. The S₂O₃^2-/S line lies above the S region and in the SO₄^2- region. Since S is not stable at this Eh (SO₄^2- is), this line is not possible either. These kinds of considerations would pertain to all the other sulfur species under these conditions, leaving S(-2), S and S(+6) species as the only possibilities to include on the diagram.

Kinetic Effects

The fact that other sulfur species don’t show up on the diagram does not mean they are unimportant. Indeed, thiosulfate is being studied intensively as a complexing agent for gold leaching, and SO_{2 aq}/HSO₃^–/SO₃^2- are important in some aspects of hydrometallurgy, etc. These species are not stable relative to S(-2), S and S(+6). This means that they will eventually decompose into more stable species by reactions with acid, base or oxygen, or they will disproportionate, e.g.