Lecture No. 7. E_h-pH Diagrams

Constructing E_h-pH Diagrams

1. Identify the components of the system. (There can be hundreds.)
2. Get K_sp, acid and base dissociation constants, redox potentials, etc.
3. Decide on concentrations, temperature, pressure, etc.

Example of constructing a diagram: Aluminum in water

The aluminum species are Al (s), Al³⁺ (aq), Al(OH)₃ (s), Al(OH)₄^– (aq). We know that high and to the left (high E_h and low pH) there will be Al³⁺ in solution. Al (s) will be at the bottom (low E_h). Al(OH)₃ (s) will be high in E_h and toward the middle pH range. Al(OH)₄^– will be high and toward the right (high E_h, high pH).

Remember that pe = (5040.1/T) E_h, or E_h = (T/5040.1) pe. At 25¡ C (298.15 K), this works out to pe=16.9 E_h, or E_h = 0.05916 pe.

Now to plug in the governing equations:

Al (s) ó Al³⁺ (aq) + 3 e^-
E⁰ = –1.66 V

So this boundary is a horizontal line at E_h = –1.66 V.

Al(OH)₃ ó Al³⁺ + 3 OH^–
K_sp = {Al³⁺} {OH^–}³ = 5« 10^-33

If we set {Al³⁺} ó 1 M, then {OH^–} = 1.7 « 10^–11 M. Then we apply the water ionization equilibrium:

H₂O ó H⁺ + OH^–
K_w = {H⁺} {OH^–} = 1.0 « 10^–14

Therefore

{H⁺} = 1.0 « 10^–14/{OH^–} = 5.9 « 10^–4

pH = –log₁₀ {H⁺} = 3.2

So there is a vertical boundary between Al³⁺ and Al(OH)₃ at pH 3.2.

Now for the hard part:

Al(OH)₃ + 3 e^- ó Al (s) = 3 OH^–
E_h = E⁰ – (RT/nF) ln Q = E⁰ – (0.05916/n) log₁₀Q
And E⁰ = -2.3 V

Therefore

E_h = – 2.3 – (0.05916/3) log₁₀ [{Al (s)} {OH^–}³/Al(OH)₃}}

= – 2.3 – (0.05916/3) log₁₀[ (1) {OH^–}³/(1)]

= – 2.3 – (0.05916/3) « 3 log₁₀ {OH^–}

= – 2.3 – (0.05916/3) « 3 (-14+pH)

because the activity of a solid is taken as unity.

E_h = – 2.3 –0.05916 (-14+pH) = – 2.3 + 0.8302 – 0.05916 pH

E_h = –1.47 –0.05916 pH

We now have an equation for the boundary line between Al(OH)₃ and Al (s).

Note that this can also be written as

pe = – 24.8 – pH

And the slope of the line is – 1 pe/pH.

Next equation:

Al(OH)₃ + OH^– ó Al(OH)₄^–

K_eq = 40

{Al(OH)₄^–}/{OH^–} = 40

We have set the soluble aluminum concentration at 1 M; therefore,

1/{OH^–} = 40

Since {OH^–} = 1« 10^-14/{H⁺},

{H⁺}/1« 10^-14 = 40

{H⁺} = 4« 10^-13

pH = 12.4

The boundary between Al(OH)₃ and Al(OH)₄^– is therefore a vertical line at pH 12.4.

Finally, put in the water instability lines.

¸ O₂ + 2 H⁺ + 2 e^- ó H₂O E⁰ = + 1.229

E_h = 1.229 – (0.05916/2) log₁₀ [{H₂O}/{O₂}^¸ {H⁺}²]

= 1.229 – (0.05916/2) log₁₀ [1/{H⁺}²]

Since liquid water and atmospheric oxygen are also taken as being at unit concentration,

= 1.229 – (0.05916/2) « 2 log₁₀ [1/{H⁺}]

= 1.229 – (0.05916/2) « [-2 log₁₀ {H⁺}]

= 1.229 – (0.05916/2) « (2 pH)

E_h = 1.229 – 0.05916 pH

And

H⁺ + e^- ó ¸ H₂ E⁰ = 0.00 V

E_h = 0.00 – (0.05916/1) « log₁₀ [{H₂}^¸ /{H⁺}]

= 0.00 –0.05916 log₁₀ [–{H⁺}

E_h = 0.00 – 0.05916 pH

When we put these lines in, we find that Al (s) is not thermodynamically stable in the presence of water; or, to put it another way, water is not stable in the presence of Al (s). If water is allowed to come in contact with aluminum metal, they react:

Al + 3 H⁺ ¨ Al³⁺ + 3/2 H₂

or

Al + 3 H₂O ¨ Al(OH)₃ +3/2 H₂

(We knew that already from freshman chemistry, but it’s nice to see it confirmed.)

Notice also that, if we use pe rather than E_h, the slopes of the pe vs. pH lines are integers, which makes the plotting much easier. In this course we will normally use E_h because many of the published diagrams we have available only have E_h scales, and because E_h is what we read off an oxidation reduction potential (ORP) electrode. But you should get comfortable with both E_h and pe units.

The E_h-pH diagram we just drew shows the equilibrium lines between various chemical species. At a point to one side or the other of such a line, the chemical species shown predominates. That does not mean the other species are nonexistent, though. If we are considering two dissolved species, the minority one will be at a concentration smaller (generally by several powers of ten) than that of the majority species. If we are considering an equilibrium between a solid and a dissolved species, the dissolved species may still be in solution at a very low concentration in the region where the solid predominates.

Another point to note is that the conclusions drawn from the diagram are based on an assumption of equilibrium. At equilibrium, we know Al metal cannot coexist with water. Yet we know we can put water in an aluminum container and it will still be there a year from now (unless the water evaporates). That is because the world is in general not at equilibrium. (In this specific case, an impervious oxide coating forms on the metal surface, and then the water is no longer in contact with the metal.)

This brings us to our next topic:

Limitations of E_h-pH Diagrams

1. E_h is actually hard to measure. Typically a Pt electrode is used. But most redox couples don’t react on a platinum surface at near-neutral pH. (Many of them do at low pH, however.)
2. Natural systems are rarely at equilibrium. Equilibration time for many environmentally important reactions is on the order of years to centuries.
3. There are many individual redox equilibria that govern pairs of half-reactions. It is hard to relate all of them to a concept like "system E_h".

Nevertheless, these diagrams have their uses.

1. They can show us what is likely in a particular system, or where that system would like to be if it could reach equilibrium.
2. They can afford us a really good view of potentially stable minerals and/or aqueous solution species.

Now let’s step back a moment and ask the question, "What exactly does an electrochemical potential mean?"

Let’s take the example of two solutions. One contains 1 M dichromate at pH 0 (i.e., in 1M acid). The other solution contains 1 M Fe²⁺ with an inert anion (say chloride). If the two solutions were mixed, the dichromate would take electrons from the iron (II), because it is the stronger oxidizing agent.

If a salt bridge is placed between the solutions (a tube filled with a gel rich in a salt like KCl), so that electric charge can travel between the two solutions. Now immerse an inert electrode into each solution (say a platinum or carbon electrode) and connect the electrodes to a wire. The dichromate now has a way to take electrons away from iron, even though the two solutions are not in direct contact. A DC voltage will exist along the wire between the two solutions. That is because, in the normal course of things, dichromate would react with iron (II).

Cr₂O₇^2– + 14 H⁺ + 6 e^- ó 2 Cr³⁺ + 7 H₂O
E⁰ = 1.33 V

6 Fe²⁺ ó 6 Fe³⁺ + 6 e^-
E⁰ = -0.771 V

Adding these two equations, we get

Cr₂O₇^2– + 14 H⁺ + 6 Fe²⁺ ó 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O
E⁰ = 0.56 V

A Wheatstone bridge placed in contact with the two solutions would measure a potential of 0.56 V between the solutions. If instead of a Wheatstone bridge potentiometer we hooked the wire up to a light or a motor, the set of two solutions would function as a battery. Or say we mixed the two solutions and plunged a platinum ORP electrode into the mixed solution, it would momentarily read 560 mV in the split second before the dichromate and iron (II) started to react.

In summary, a redox potential measures the free energy of reaction between two redox-active species by measuring the voltage they would generate if they could use this free energy to drive electric current through a circuit.

A lot of metal corrosion happens by electrochemical mechanisms. That is how copper pipe commonly corrodes, even though copper is stable in the presence of water. A bacterial colony on one part of the pipe excludes oxygen from the surface. Oxygen is available in another spot on the surface. Normally oxygen would not attack metallic copper, because it has an inert coating of CuO. But the copper under the bacterial mat is oxygen-deprived. On the clean copper surface, we get the reaction

O₂ + 4 e^- + 2 H₂O ¨ 4 OH^–

Meanwhile, under the bacterial mat, we get

2 Cu ¨ 2 Cu²⁺ + 4 e^-

Since copper is a good electrical conductor, the electrons are transferred through the copper pipe itself. The net reaction is

2 Cu + O₂ + 2 H₂O ¨ 2 Cu²⁺ + 4 OH^–

Now let’s look at some E_h-pH diagrams that are of interest in environmental geochemistry.

In the Overheads

The two-page list of redox potentials I handed out previously lists most of the significant potentials we will be concerned with. The overheads are E_h-pH diagrams for various inorganic systems.

(Study these examples!)

The first diagram is for iron at 25¡ C and at atmospheric pressure. Total dissolved carbonate species = 10^–2 M. Total dissolved iron species = 10^–6 M. Let’s look at the various reactions shown in this diagram.

¸ O₂ + 2 H⁺ + 2 e^- ó H₂O

H⁺ + e^- ó ¸ H₂

Fe³⁺ + e^- ó Fe²⁺

FeOH²⁺ + H⁺ + e^- ó Fe²⁺ + H₂O

Fe(OH)₃ (s) + 3 H⁺ + e^- ó Fe²⁺ + 3 H₂O

FeCO₃ (s) ó Fe²⁺ + CO₃^2–

Fe(OH)₂ (s) + CO₃^2– ó FeCO₃ (s) + 2 OH^–

Fe(OH)₂ (s) + OH^– ó Fe(OH)₃^–

Fe(OH)₃ (s) + e^- ó Fe(OH)₃^–

Fe(OH)₃ (s) + e^- ó Fe(OH)₂ (s) + OH^–

Fe(OH)₃ (s) + CO₃^2– + e^- ó FeCO₃ + 3 OH^–

One important species and two important reactions have been left out of this diagram:

Fe(OH)₃(s) + OH^– ó Fe(OH)₄^–

Fe(OH)₄^– + e^- ó Fe(OH)₃^– + OH^–

The tetrahydroxyferrate (III) ion exists above pH 13.5.

As usual we have the water instability lines. Their slopes are –1 pe/pH, since there is one hydrogen ion per electron exchanged in these reactions. As expected, the lines between Fe³⁺ and FeOH²⁺, and between FeOH²⁺ and Fe(OH)₃ (s) are vertical, since these equilibria depend only on pH. The same is true for the pH-dependent boundaries between Fe²⁺ and FeCO₃, FeCO₃ and Fe(OH)₂, and between Fe(OH)₂ and HFeO₂^– (or Fe(OH)₃^–), which are all governed strictly by pH. The boundary between Fe²⁺ and Fe³⁺ is horizontal, since this equilibrium is purely a redox one. So is the boundary between Fe(OH)₃^– and Fe(OH)₃ (s). The boundary between Fe²⁺ and Fe(OH)₃ (s) has a slope of –3 pe/pH, since there are three hydrogen ions exchanged per electron. To be more precise,

where n=1. Since E_h = (2.303 RT/F) pe, we get

pe = pe⁰ – log₁₀ {Fe²⁺} +log₁₀ {Fe(OH)₃ (s)} + 3 log₁₀ {H⁺}

pe = pe⁰ – log₁₀ {Fe²⁺} + log₁₀ {Fe(OH)₃ (s)} – 3 pH

And the activity of Fe(OH)₃ (s) is unity, so its logarithm is zero.

pe = pe⁰ – log₁₀ {Fe²⁺} – 3 pH

So the slope of pe vs. pH is –3.

By similar reasoning, we see that the slope of the siderite-ferric hydroxide line is –2, and that of the ferrous hydroxide-ferric hydroxide line is –1.

Note one important consequence of the redox system shown in this diagram: iron (III) is more stable at higher pH. That is one of the main driving forces behind iron oxidation and reduction in nature.

The next diagram shows the aqueous iron system without carbonate and with total iron at 10^–5 M. Note that the diagram extends outside the water stability field, so that metallic iron also is in the picture. Both of the ferrate ions — trihydroxyferrate (II) and tetrahydroxyferrate (III) are shown. There is no siderite field because there is no carbonate.

The next overhead shows an E_h-pH diagram for carbon. In this diagram the 8-electron reduction of carbon dioxide to methane is the main redox reaction. Depending on the pH, the reaction is one of the following three:

H₂CO₃ + 8 H⁺ + 8 e^- ó CH₄ + 3 H₂O

HCO₃^– + 6 H₂O + 8 e^- ó CH₄ + 9 OH^–

CO₃^2– + 7 H₂O + 8 e^- ó CH₄ + 10 OH^–

In the first case, there are 8 electrons per 8 hydrogen ions, so the pe/pH slope of the line between the carbonic acid field and the methane field is – 1. In the second case there are 9 hydroxides gained, and therefore 9 hydrogen ions lost, per 8 electrons, so the slope of the line between bicarbonate and methane fields is –9/8. In the third case, the slope of the line between the carbonate and methane fields is –10/8, since 10 hydrogen ions are involved per 8 electrons.

There seems to be no stability field for graphite. It forms under pressure and temperature conditions very different from those in this diagram. Always keep in mind that a given E_h-pH diagram is good for just one temperature, pressure, and set of concentrations.

Note that methane is stable at just slightly above the lower water stability line for water. This is consistent with the observation that some methane-generating bacteria also reduce water to liberate hydrogen gas, and hydrogen is very important to the metabolism of methanogens.

The diagram for nitrogen is a little more complex, because nitrogen exists in several stable forms, depending on pe and pH. Note that nitrite does not have a stability field, even though we know that there are organisms in soil and water that liberate nitrite. What is going on?

The answer is that nitrite is metastable in water; it is unstable with respect to disproportionation, i.e., self-oxidation and reduction, in water, according to the reaction

5 NO₂^– + 2 H⁺ ¨ 3 NO₃^– + N₂ + H₂O

The fact that nitrite solutions exist serves to remind us that nature is frequently not at equilibrium. E_h-pH diagrams always show equilibrium conditions. This means we can actually miss some important phenomena by relying on them too much.

In terms of thermodynamically stable species, nitrogen can exist as nitrate, zero-valent nitrogen (dissolved N₂ gas), or ammonia or ammonium ion, depending on pH. Nitrate is stable just under the upper stability line for water, which makes nitrate look like a pretty good electron acceptor (oxidizing agent) if there is no free oxygen. This is in fact the case. There are quite a few facultative anaerobes, bacteria that can make use of nitrate for their respiration if there is no available oxygen.

Moving to another overhead, we see that, if N₂ is not redox-active, there is a finite stability field for nitrite. And in fact nitrogen is usually not redox-active. The kinetics of most geochemical systems are such that ammonia and nitrite do not convert easily to nitrogen gas.

Now let’s take a look at sulfur. We see that there is a very wide stability field for sulfate. Sulfide is stable below the sulfate-sulfide line, which is at lower E_h than the nitrogen-ammonia line at acidic pH, but slightly higher than that line at high pH. It is also higher than the carbonic acid-methane line. Also note that there is a small stability field for zero-valent sulfur at low pH. This is where much of our mineral sulfur comes from. Finally, note that there is no stability field for sulfite or sulfurous acid; like nitrite, sulfite is unstable with respect to disproportionation.

What these three diagrams tell us is that sulfur reduction will happen at a higher E_h than methane production. It also tells us that N₂ reduction will set in at about the same E_h as sulfate reduction give or take a little. That would be true if denitrifiers and nitrogen fixers were always present to reduce nitrate to nitrogen and to reduce nitrogen to ammonia. But N₂ is almost never redox-active! More commonly, we deal with dissimilatory nitrate reducers, which take nitrogen all the way from nitrate to ammonia by way of nitrite. In that case (which is the usual one in the real world) nitrite reduction takes place at a much higher E_h than sulfate reduction.

Next, consider the diagram for manganese. In the water stability field, we see that under acidic conditions soluble Mn²⁺ will predominate. At more alkaline pH there are a number of oxyhydroxides and oxides in which manganese has an oxidation state between +2 and +4. (Note the typo on pyrochroite, which should be Mn(OH)₂. Moving up in E_h, we encounter hausmannite, Mn₃O₄, which has an apparent oxidation number of +8/3. This makes more sense if we consider it a manganous manganate, MnOáMn₂O₃, in which 1/3 of the manganese atoms are in the +2 state, and 2/3 of them are in the +3 state.

Moving farther up we see bixbyite, and manganite, which are Mn (III) compounds, and pyrolusite and nsutite, which are Mn (IV) minerals. Not all of the minerals are strictly stable in water; note the dashed lines between bixbyite and manganite, and between pyrolusite and nsutite. Also note that this diagram is for total dissolved manganese activity of 1 micromolar. At higher Mn activities or at different temperatures and pressures, the situation might be different.

The next overhead shows diagrams for copper + sulfur and for copper + sulfur + iron. The dissolved sulfur species are not shown. Notice the large field for native copper that extends from moderately acidic to very basic pH. Copper exists in three oxidation states: +2, +1, and 0. Interestingly, the Cu (II) sulfide mineral, covellite, is sandwiched between two fields where Cu (I) sulfides are stable. Also note, by comparison with the sulfur diagram we saw, that sulfur in the –2 state is much more stable in the presence of copper than in its absence; this is due to the large negative free energy (i.e., large stability) of the Cu₂S crystal lattice. Another thing to note: there is no stable field for dissolved Cu (I). Cuprous ion disproportionates very quickly to Cu ⁰ and Cu²⁺:

2 Cu⁺ ¨ Cu²⁺ + Cu

This is true at low copper concentrations and in the absence of complexing agents such as chloride. However, in the presence of the right anions or chelating agents, the reaction is reversible.

In the copper-iron-sulfur diagram there is another curious thing: None of the iron monosulfides are stable, but only pyrite, an iron disulfide. All of the common iron monosulfides, such as troilite, pyrrhotite, and mackinawite, (approximate formulas FeS) are thermodynamically unstable vs. oxidation to pyrite. Also note the reaction path shown for copper, moving from CuO in the presence of iron (II) to Cu₂S in the presence of pyrite.

The next overhead shows arsenic’s stability fields in the presence of sulfur. The main oxidation states are +5 and +3. There is also a stability field for realgar, which is written as AsS, but which is really a molecular compound As₄S₄, in which As is in the +2 oxidation state.

We know there are negative-valent arsenide minerals, such as skutterudite, (Co,Ni)As₃. The diagram tells us that they are not stable at 25¡ C and 1 atmosphere. Stability fields are different at higher temperatures and pressures, which is presumably where such minerals form.

If there is sufficient sulfur present, note that As (III) oxide is not very stable except under acidic or neutral conditions. However, if there is not much sulfur, arsenite salts can form.

In the As + S diagram, the only As (III) oxide species shown is As₂O₃, arsenolite. In solution, arsenous acid (sometimes called arsenious acid), H₃AsO₃, is stable as a very weak acid. Under basic conditions it can form arsenite salts, such as sodium arsenite, Na AsO₂. Note that the acid loses one molecule of water in forming the salt.

H₂AsO₃^– + Na⁺ ó NaAsO₂ + H₂O

There probably is also a monohydrogen arsenite ion in very basic solutions, HAsO₃^2–, but there is no tribasic arsenite ion, AsO₃^3–. (Note that the arsenic oxyanions are shown in the As-no S diagram.) Arsenic acid, on the other hand, has 3 separate ionizations and forms salts analogous to phosphates, e.g., NaH₂AsO₄, Na₂HasO₄, and Na₃AsO₄.

A set of compounds not shown in either diagram is the thioarsenites. Under high sulfide concentrations (> about 10 ppm), As₂S₃ redissolves, forming a set of anions such as H₂AsS₃^–, H₃As₂S₅^–, and others. These do not show on the As+S diagram, presumably because the sulfide concentration is too low.

Next: More E_h-pH relationships

To previous lecture: Lecture No 6. Grain Size Controls

To next lecture: Lecture No 8. Eh-pH Diagrams II

Back to list of lecture notes