Introduction to Inorganic Chemistry/Redox Stability and Redox Reactions

Chapter 4: Redox Stability and Redox Reactions
In redox reactions, one element or compound is reduced (gains electrons) and another is oxidized (loses electrons). In terms of everyday life, redox reactions occur all of the time around us. For example, the metabolism of sugars to CO2, which stores energy in the form of ATP, is a redox reaction. Another example of redox is fire or combustion, such as in a car engine. In a car engine, hydrocarbons in the fuel are oxidized to carbon dioxide and water, while oxygen is reduced to water. Corrosion (i.e. the formation of rust on iron) is a redox reaction involving oxidation of a metal.

Oxidation-reduction reactions are important to understanding inorganic chemistry for several reasons:


 * Transition metals can have multiple oxidation states
 * Main group elements (N, halogens, O, S...) also have multiple oxidation states and important redox chemistry
 * Many inorganic compounds catalyze redox reactions (which are especially useful in industrial and biological applications)
 * Energy conversion and storage technologies (solar water splitting, batteries, electrolyzers, fuel cells) rely on inorganic redox reactions and catalysis
 * Electrochemistry provides a way to measure equilibrium constants for dissolution/precipitation, complexation, and other reactions.


 * Reaction mechanisms in organometallic chemistry (oxidative addition, reductive elimination) involve changes in the oxidation states of metals.

Not all oxidizers and reducers are created equal. The electrochemical series ranks substances according to their oxidizing and reducing power, i.e., their standard electrode potential. Strong oxidizing agents are typically compounds with elements in high oxidation states or with high electronegativity, which gain electrons in the redox reaction. Examples of strong oxidizers include hydrogen peroxide, permanganate, and osmium tetroxide. Reducing agents are typically electropositive elements such as hydrogen, lithium, sodium, iron, and aluminum, which lose electrons in redox reactions. Hydrides (compounds that contain hydrogen in the formal -1 oxidation state), such as sodium hydride, sodium borohydride and lithium aluminum hydride, are often used as reducing agents in organic and organometallic reactions.

Learning goals for Chapter 4:


 * Balance complex oxidation-reduction reactions by the ion-electron method.
 * Understand periodic trends in the activity series and electrochemical series.
 * Use the Nernst equation to determine half-cell and cell potentials.
 * Derive the stability field of water and use this to rationalize aqueous redox chemistry.
 * Construct and be proficient with Latimer diagrams, using them to determine unknown reduction potential values and to quickly identify stable and unstable species.
 * Construct and be proficient with Frost diagrams, using them to identify stable and unstable species, as well as those that are strong oxidizers.
 * Construct and be proficient with Pourbaix diagrams, using them to identify redox and non-redox reactions, reactions that are and are not pH-dependent, and ultimately to predict and rationalize stability, reactivity, corrosion, and passivation.

4.1 Balancing redox reactions
In studying redox chemistry, it is important to begin by learning to balance electrochemical reactions. Simple redox reactions (for example, H2 + I2 → 2 HI) can be balanced by inspection, but for more complex reactions it is helpful to have a foolproof, systematic method. The ion-electron method allows one to balance redox reactions regardless of their complexity. We illustrate this method with two examples.

Example 1:

I&minus; is oxidized to IO3&minus; by MnO4&minus;, which is reduced to Mn2+.

How can this reaction be balanced? In the ion-electron method we follow a series of four steps:

 Step 1A : Write out the (unbalanced) reaction and identify the elements that are undergoing redox.


 * Mn O4&minus; + I&minus; → I O3&minus; + Mn2+ (The elements undergoing redox are Mn and I)

 Step 1B : Separate the reaction into two half reactions, balancing the element undergoing redox in each.
 * Mn O4&minus; → Mn2+
 * I&minus; → I O3&minus;

 Step 2A : Balance the oxygen atoms by adding water to one side of each half reaction.
 * Mn O4&minus; → Mn2+ + 4 H2 O
 * 3 H2 O + I&minus; → I O3&minus;

 Step 2B : Balance the hydrogen atoms by adding H+ ions.
 * 8 H+ + MnO4&minus;→ Mn2+ + 4 H2 O

The left side has a net charge of +7 and the right side has a net charge of +2
 * 3 H2 O + I&minus; → IO3&minus; + 6 H+

The left side has a net charge of &minus;1 and the right side has a net charge of +5

 Step 2C : Balance the overall charge by adding electrons
 * 8 H+ + 5 e&minus; + MnO4&minus;→ Mn2+ + 4 H2O

The left side has a charge of +2 while the right side has a charge of +2. They are balanced.
 * 3 H2O + I&minus; → IO3&minus; + 6 H+ + 6 e&minus;

The left side has a charge of &minus;1 while the right side has a charge of &minus;1. They are balanced.

Note: We did not need to explicitly determine the oxidation states of Mn or I to arrive at the correct number of electrons in each half reaction.

 Step 3 : Combine the half reactions so that there are equal numbers of electrons on the left and right sides
 *  6  (8 H+ + 5 e&minus; + MnO4&minus; → Mn2+ + 4 H2O)
 *  5  (3 H2O + I&minus; → IO3&minus; + 6 H+ + 6 e&minus;)

48 H+ + 30 e&minus; + 15 H2O + 6 MnO4&minus; + 5 I&minus; → 5 IO3&minus; + 6 Mn2+ + 24 H2O + 30 e&minus; + 30 H+

Cancel the H+, electrons , and water :

48 H+ + 30 e&minus; + 15 H2 O + 6 MnO4&minus; + 5 I&minus; → 5 IO3&minus; + 6 Mn2+ + 24 H2 O + 30 e&minus; + 30 H+

The overall balanced reaction is therefore:

18 H+ + 6 MnO4&minus; + 5 I&minus; → 5 IO3&minus; + 6 Mn2+ + 9 H2O

Check your work by making sure that all elements and charges are balanced.

 Step 4 : If the reaction occurs under basic conditions, we add OH&minus; to each side to cancel H+

18 H+ + 18 OH&minus; + 6 MnO4&minus; + 5 I&minus; → 5 IO3&minus; + 6 Mn2+ + 9 H2O + 18 OH&minus;

The 18 H+ + 18 OH&minus; will become 18 H2O so the overall balanced reaction is:

9 H2O + 6 MnO4&minus; + 5 I&minus; → 5 IO3&minus; + 6 Mn2+ + 18 OH&minus;

Again, it is a good idea to check and make sure that all of the elements are balanced, and that the charge is the same on both sides. If this is not the case, you need to find the error in one of the earlier steps.

Example 2:

Redox reaction of S2O32&minus; and H2O2

S2O32&minus; + H2O2 →  S4O62&minus; + H2O

Which elements are undergoing redox? S and O

 Step 1 : Write out half reactions, balancing the element undergoing redox
 * 2 S2O32&minus; → S4O62&minus;
 * H2O2 → 2 H2O

 Step 2A : Balance oxygen (already balanced)

 Step 2B : Balance hydrogen:
 * 2 S2O32&minus; → S4O62&minus;
 * H2O2 + 2 H+ → 2 H2O

 Step 2C : Balance charge by adding electrons:
 * 2 S2O32&minus; → S4O62&minus; + 2 e&minus;
 * H2O2 + 2 H+ + 2 e&minus; → 2 H2O

 Step 3 : Combine the half reactions so that there are equal numbers of electrons on the left and right sides (already equal)

Overall balanced reaction:

2 S2O32&minus; + H2O2 + 2 H+ → S4O62&minus; + 2 H2O

Note that again, we did not need to know the formal oxidation states of S or O in the reactants and products in order to balance the reaction. In this case, assigning the oxidation states would be rather complex, because S2O32&minus; and S4O62&minus; both contain sulfur in more than one oxidation state.

4.2 Electrochemical potentials
In electrochemical cells, or in redox reactions that happen in solution, the thermodynamic driving force can be measured as the cell potential. Chemical reactions are spontaneous in the direction of -ΔG, which is also the direction in which the cell potential (defined as Ecathode – Eanode) is positive. A cell operating in the spontaneous direction (for example, a battery that is discharging) is called a galvanic cell. A cell that is being driven in the non-spontaneous direction is called an electrolytic cell. For example, let us consider the reaction of hydrogen and oxygen to make water:


 * 2 H2(g) + O2(g) → 2 H2O(&#8467;)

Thermodynamically, this reaction is spontaneous in the direction shown and has an overall standard free energy change (ΔG°) of &minus;237 kJ per mole of water produced.

When this reaction occurs electrochemically in the spontaneous direction (e.g., in a hydrogen-air fuel cell), the two half cell reactions that occur are:


 * Anode:    H2(g) → 2 H+(aq) + 2 e&minus;
 * Cathode: O2(g) + 4 H+(aq) + 4 e&minus; → 2 H2O(&#8467;)

Here the anode is the negative electrode and the cathode is the positive electrode; under conditions of very low current density (where there are minimal resistive losses and kinetic overpotentials), the potential difference we would measure between the two electrodes would be 1.229 V.

In an electrolytic cell, this reaction is run in reverse. That is, we put in electrical energy to split water into hydrogen and oxygen molecules. In this case, the half reactions (and their standard potentials) reverse. O2(g) bubbles form at the anode and H2(g) is formed at the cathode. Now the anode is the positive electrode and the cathode is negative. Electrons are extracted from the substance at the anode (water) and pumped into the solution at the cathode to make hydrogen. An animation of the cathode half reaction is shown at the right.

In both galvanic and electrolytic cells, oxidation occurs at the anode and reduction occurs at the cathode.

Half-cell potentials. As noted above, the equilibrium voltage of an electrochemical cell is proportional to the free energy change of the reaction. Because electrochemical reactions can be broken up into two half-reactions, it follows that the potentials of half reactions (like free energies) can be added and subtracted to give an overall value for the reaction. If we take the standard hydrogen electrode as our reference, i.e., if we assign it a value of zero volts, we can measure all the other half cells against it and thus obtain the voltage of each one. This allows us to rank redox couples according to their standard reduction potentials (or more simply their standard potentials), as shown in the table on the right.

Note that when we construct an electrochemical cell and calculate the voltage, we simply take the difference between the half cell potentials and do not worry about the number of electrons in the reaction. For example, for the displacement reaction in which silver ions are reduced by copper metal, the reaction is:


 * 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)

The two half-cell reactions are:
 * Ag+(aq) + e&minus; → Ag(s)      +0.80V
 * Cu2+(aq) + 2 e&minus; → Cu(s)   +0.34V

and the standard potential Eo = +0.80 – 0.34 V = +0.46V

The reason we don't need to multiply the Ag potential by 2 is that E° is a measure of the free energy change per electron. Dividing the free energy change by the number of electrons (see below) makes E° an intensive property (like pressure, temperature, etc.).

Relationship between E and ΔG. For systems that are in equilibrium,  ΔG° = &minus;nFE°cell, where n is number of moles of electrons per mole of products and F is the Faraday constant, ~96485 C/mol. Here the o symbol indicates that the substances involved in the reaction are in their standard states. For example, for the water electrolysis reaction, the standard states would be pure liquid water, H+ at 1 M concentration (or more precisely, at unit activity), and O2 and H2(g) at 1 atmosphere pressure.

More generally (at any concentration or pressure), ΔG = &minus;nFE, where


 * $$ E = E^o-\frac{RT}{nF} * ln Q$$,

or at 298 K
 * $$ E = E^o-\frac{0.0592}{n} * log Q$$

where Q is the reaction quotient, calculated as the concentration ratio of products over reactants, raised to the powers of their coefficients in the reaction. This equation (in either form) is called the Nernst equation. The second term in the equation, when multiplied by &minus;nF, is RT*lnQ. This is the free energy difference between ΔG and ΔG°. We can think of this as an entropic term that takes into account the positive entropy change of dilution, or the negative entropy change of concentrating a reactant or product, relative to its standard state.

Using the Nernst equation

Example 1:

For the half reaction 2 H+ + 2 e&minus; → H2, E°1/2 = 0.000 V (by definition)

What is E1/2 at pH 5 and PH 2 = 1 atm?

pH = &minus;log [H+] = 5, so [H+] = 10&minus;5 M


 * $$ E = E^o -\frac{0.0592}{2} * log$$ $$ \frac{P_{H2}}{(H^+)^2} = E^o -\frac{0.0592}{2} * log (10^5)^2 = E^o -\frac{0.0592}{2} (10)$$


 * $$ = 0.000 - 0.296 = -0.296V$$

Example 2:

What is the potential of a fuel cell (a galvanic H2/O2 cell) operating at pH 5?

Overall reaction: 2 H2(g) + O2(g) → 2 H2O(&#8467;)

In this reaction, H2 is oxidized to H+ and O2 is reduced to H2O. According to our convention, we write out and balance both half-cell reactions as reductions. For convenience, we do this in acid. (It is left as an exercise to the interested reader to try to work the problem in base)

Half cell reactions:

2 H2(g) → 4 H+(aq) + 4 e&minus;

O2(g) + 4 H+(aq) + 4 e&minus; → 2 H2O(&#8467;)

To solve this problem we need to find the difference between the H2/ H+ and O2/H2O half cell potentials at pH 5.

2 H+ + 2 e&minus; → H2                 E°1/2 =   0.000 V

Like all standard potentials, this is written as a reduction. We need to reverse it and change the sign of E° since H2 is being oxidized:

H2 → 2 H+ + 2 e&minus; E°1/2 = &minus;0.000 V

and add the standard potential of the substance being reduced at the cathode:

O2 + 4 H+ + 4 e&minus; → 2 H2O         E°1/2 = +1.229 V

The difference between the two standard half cell potentials is +1.229 – 0.000 = +1.229 V

E°cell = +1.229 Volts

We now use the Nernst equation to account for the fact that H+ is not in its standard state:


 * $$ E_{cell} = E^o -\frac{0.0592}{4} * log$$ $$ \frac{[H^{+}]^4}{P_{O2}P_{H2}^2[H^+]^4} = E^o -\frac{0.0592}{4} * log(1)  = E^o = +1.229 V$$

Note that the value of Ecell does not change with pH since both couples shift -59.2 mV/pH according to the Nernst equation. This is the consequence of the fact that the number of electrons equals the number of protons in each of the half cell reactions. Another way to rationalize this result is to remember that the overall reaction (2 H2 + O2 → 2 H2O) does not involve H+ as a reactant or product, so ΔG and E should be independent of pH.

We can plot the shift in the H2/H+ and O2/H2O half-cell potentials with pH on a potential-pH diagram (also called a Pourbaix diagram) as shown at the left. The pH-dependent potentials of the H2 and O2 couples are shown as dotted lines. Notice that the potential difference between them is always 1.23 V. The dark circles represent the standard potentials.

Pourbaix diagrams are essentially electrochemical phase diagrams, which plot regions of thermodynamic stability for redox-active substances. As in other kinds of phase diagrams, the lines represent conditions under which two phases coexist in equilibrium. The shaded area in the water Pourbaix diagram represents the conditions of potential and pH where liquid water is stable relative to hydrogen or oxygen. Outside the shaded region, water is thermodynamically unstable and is reduced to H2(g) or oxidized to O2(g). Although these processes are spontaneous in the thermodynamic sense (for example, water is unstable in the presence of Pb4+, Cl2, Fe, Zn, or Al), they are kinetically slow and require catalysis to proceed.

4.3 Latimer and Frost diagrams
In addition to Pourbaix diagrams, there are two other kinds of redox stability diagrams known as Latimer and Frost diagrams. Each of these diagrams contains similar information, but one representation may be more useful in a given situation than the others.

Latimer and Frost diagrams help predict stability relative to higher and lower oxidation states, usually at one fixed pH. Pourbaix diagrams help understand pH-dependent equilibria, which are often coupled to solubility equilibria and corrosion (which will be talked about more later).

Latimer diagrams:

Latimer diagrams are the oldest and most compact way to represent electrochemical equilibria for substances that have multiple oxidation states. Electrochemical potential values are written for successive redox reactions (from highest to lowest oxidation state), typically under standard conditions in either strong acid ([H+] = 1 M, pH 0) or strong base ([OH&minus;] = 1 M, pH 14). The oxidation states of successive substances in a Latimer diagram can differ by one or more electrons. Oxidation states for the element undergoing redox are typically determined by difference; we assign the oxygen atoms an oxidation state of -2 and the hydrogen atoms an oxidation state of +1.

Example:
 * Mn in Acid



The Latimer diagram compresses into shorthand notation all the standard potentials for redox reactions of the element Mn. For example, the entry that connects Mn2+ and Mn gives the potential for the half-cell reaction:

E1/2° = &minus;1.18V
 * Mn2+(aq) + 2 e&minus; → Mn(s)

and the entry connecting Mn4+ and Mn3+ represents the reaction:

E1/2° = +0.95V
 * MnO2(s) + 4 H+(aq) + e&minus; → Mn3+(aq) + 2 H2O(&#8467;)

We can also calculate values for multi-electron reactions by first adding ΔG°(=&minus;nFE°) values and then dividing by the total number of electrons

For example, for the 5-electron reduction of MnO4&minus; to Mn2+, we write


 * $$E^{o} = \frac{1(0.564)+1(0.274)+1(4.27)+1(0.95)+1(1.51)}{5} = +1.51 V$$

and for the three-electron reduction of MnO4&minus;(aq) to MnO2(s),
 * $$E^{o} = \frac{1(0.564)+1(0.274)+1(4.27)}{3} = +1.70 V$$

Remember to divide by the number of electrons involved in the oxidation number change (5 and 3 for the above equations).

Thermodynamically stable and unstable oxidation states

An unstable species on a Latimer diagram will have a lower standard potential to the left than to the right.

Example:
 * 2 MnO43&minus; → MnO2 + MnO42-    (the MnO43&minus; species is unstable to disproportionation)
 * E° = +4.27 &minus; 0.274 = +3.997 V   ( spontaneous disproportionation )

Which Mn species are unstable with respect to disproportionation?
 * MnO43&minus;    5+ → 6+ and 4+
 * Mn3+       3+ → 4+ and 2+

So stable species are: MnO4&minus;, MnO42&minus;, MnO2, Mn2+, and Mn0.

But MnO42&minus; is also unstable. Why? This reaction can happen: MnO42- → MnO2. It undergoes a multi-electron transfer!

So overall, 2 MnO42&minus; → MnO4&minus; + MnO2     E° = 2.272 – 0.564 = +1.708 V    (spontaneous disproportionation)

Take Home Message: All possible disproportionation reactions (even those with more than one electron transferred) must be considered in order to determine stability (this is often more convenient with a Frost diagram).

Note: Thermodynamically unstable ions can be quite stable kinetically. For example, most N-containing molecules (NO2, NO, N2H4) are unstable relative to the elements (O2, N2, H2), but they are still quite stable kinetically.

Frost diagrams: In a Frost diagram, we plot ΔG°&frasl;F (= nE°) vs. oxidation number. The zero oxidation state of the element is assigned a nE° value of zero.

Stable and unstable oxidation states can be easily identified in the plot. Unstable compounds are higher on the plot than the line connecting their neighbors. Note that this is simply a graphical representation of what we did with the Latimer diagram to determine which oxidation states were stable and unstable.

The standard potential for any electrochemical reaction is given by the slope of the line connecting the two species on a Frost diagram. For example, the line connecting Mn3+ and MnO2 on the Frost diagram has a slope of +0.95, the standard potential of MnO2 reduction to Mn3+. This is the number that is written above the arrow in the Latimer diagram for Mn. Multielectron potentials can be calculated easily by connecting the dots in a Frost diagram.

A Frost diagram:


 * Contains the same information as in a Latimer diagram, but graphically shows stability and oxidizing power.
 * The lowest species on the diagram are the most stable (Mn2+, MnO2). The highest species on diagram are the strongest oxidizers (MnO4&minus;).

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&minus; ions:


 * Fe2+(aq) + 6 CN&minus;(aq) &#8652; [Fe(CN)6]4&minus;         (1)
 * Fe3+(aq) + 6 CN&minus;(aq) &#8652; [Fe(CN)6]3&minus;         (2)

Which oxidation state of Fe is more strongly complexed by CN&minus;? We can answer this question by measuring the standard half-cell potential of the [Fe(CN)6]3&minus;/4&minus; couple and comparing it to that of the Fe3+/2+couple:


 * Fe3+(aq) + e&minus; &#8652; Fe2+(aq)               E° = +0.77 V          (3)
 * [Fe(CN)6]3&minus; + e&minus; &#8652; [Fe(CN)6]4&minus;       E° = +0.42 V          (4)

Iron(III) is harder to reduce (i.e., E° is less positive) when it is complexed to CN&minus;

This implies that the equilibrium constant for complexation reaction (1) should be smaller than that for reaction (2). How much smaller?

We can calculate the ratio of equilibrium constants by adding and subtracting reactions:


 * Fe3+ + 6 CN&minus; &#8652; [Fe(CN)6]3&minus;         K = K2
 * [Fe(CN)6]4&minus; &#8652;  Fe2+ + 6 CN&minus;          K = 1/K1
 * ____________________________________
 * Fe3+ + Fe(CN)64&minus; &#8652; Fe2+ + Fe(CN)63&minus;

The equilibrium constant for this reaction is the product of the two reactions we added, i.e., K = K2/K1.

But we can make the same overall reaction by combining reactions (3) and (4):
 * Fe3+(aq) + e&minus; &#8652; Fe2+(aq)               E° = +0.77 V
 * [Fe(CN)6]4&minus; &#8652; [Fe(CN)6]3&minus; + e&minus;       E° = -0.42 V
 * ____________________________________
 * Fe3+ + Fe(CN)64&minus; &#8652; Fe2+ + Fe(CN)63&minus;

In this case, we can calculate E° = 0.77 – 0.42 = +0.35 V

It follows from nFE° = -ΔG° = RTlnK that


 * E° = RT/nF ln(K2/K1)


 * K2/K1 = exp(nFE°/RT) = exp[(1 equiv/mol)(96,500 C/equiv)(0.35 J/C)/(8.314 J/mol K)(298 K)] = exp(13.63) = 8 x 105

Thus we find that Fe(CN)63&minus; is about a million times more stable as a complex than Fe(CN)64&minus;.

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&minus;(aq), against a hydrogen reference electrode. That value is then compared to the standard potential of the Ag+/Ag couple:


 * AgCl(s) + e&minus; &#8652; Ag(s) + Cl&minus;(aq)       E° = +.207V
 * Ag+(aq) + e&minus; &#8652; Ag(s)                     E° = +0.799 V

Subtracting the second reaction from the first one we obtain:


 * AgCl(s) &#8652; Ag+(aq) + Cl&minus;(aq)             E° = +0.207 – 0.799 = -0.592 V

and again using nFE° = RTlnK, we obtain K = Ksp = 9.7 x 10&minus;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&minus;. For these reactions, the half-cell potentials are pH-dependent.

Example: Recall that the disproportionation reaction 3 MnO42&minus;(a) → 2 MnO4&minus;(aq) + MnO2(s) is spontaneous at pH = 0 ([H+] = 1 M under standard conditions), from the Latimer diagram or Frost plot.

However, when we properly balance this half reaction we see that it involves protons as a reactant:


 * 3 MnO42&minus;(aq) + 4 H+(aq) &#8652; 2 MnO4&minus;(aq) + MnO2(s) + 2 H2O(&#8467;)

By Le Châtelier's principle, it follows that removing protons (increasing the pH) should stabilize the reactant, MnO42&minus;. 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.

4.5 Pourbaix diagrams
Pourbaix Diagrams plot electrochemical stability for different redox states of an element as a function of pH. As noted above, these diagrams are essentially phase diagrams that map the conditions of potential and pH (most typically in aqueous solutions) where different redox species are stable. We saw a simple example of such a diagram in section 4.2 for H2O. Typically, the water redox reactions are plotted as dotted lines on these more complicated diagrams for other elements.

The lines in Pourbaix diagrams represent redox and acid-base reactions, and are the parts of the diagram where two species can exist in equilibrium. For example, in the Pourbaix diagram for Fe below, the horizontal line between the Fe3+ and Fe2+ regions represents the reaction Fe3+(aq) + e&minus; → Fe2+(aq), which has a standard potential of +0.77 V. While we could use standard potentials for all these lines, in practice Pourbaix diagrams are usually plotted for lower ion concentrations (often 1 mM) that are more relevant to corrosion and electrochemical experiments. Example: Iron Pourbaix diagram
 * Areas in the Pourbaix diagram mark regions where a single species (such as Fe2+(aq) or Fe3O4(s)) is stable. More stable species tend to occupy larger areas.

Lines mark places where two species exist in equilibrium.


 * Pure redox reactions are horizontal lines – these reactions are not pH-dependent
 * Pure acid-base reactions are vertical lines – these do not depend on potential
 * Dividing lines for reactions that are both acid-base and redox have a slope of &minus;0.0592 V/pH × (# H+&frasl;# e&minus;). In an oxidation, the overall slope of the dividing line is negative, and the number of electrons transferred (the denominator) is positive. In a reduction, the slope of the dividing line is positive, and the number of electrons transferred (the denominator) is negative.


 * Examples of equilibria in the iron Pourbaix diagram (numbered on the plot):

1. Fe2+ + 2 e&minus; → Fe(s)  (pure redox reaction – no pH dependence)

2. Fe3+ + e&minus; → Fe2+  (pure redox reaction – no pH dependence)

3. 2 Fe3+ + 3 H2O → Fe2O3(s)+ 6 H+  (pure acid-base, no redox)

4. 2 Fe2+ + 3 H2O → Fe2O3(s)+ 6 H+ + 2 e&minus;  (slope = &minus;59.2 x 6/2 = -178 mV/pH; oxidation, so the number of electrons transferred is positive)

5. 2 Fe3O4(s) + H2O → 3 Fe2O3(s) + 2H+ + 2 e&minus;  (slope = &minus;59.2 x 2/2 = &minus;59.2 mV/pH; oxidation, so the number of electrons transferred is positive)

The water redox lines have special significance on a Pourbaix diagram for an element such as iron. Recall that liquid water is stable only in the region between the dotted lines. Below the H2 line, water is unstable relative to hydrogen gas, and above the O2 line, water is unstable with respect to oxygen. For active metals such as Fe, the region where the pure element is stable is typically below the H2 line. The areas where the two substances are stable do not overlap. This means that iron metal is unstable in contact with water, undergoing reactions:


 * Fe(s) + 2 H+ → Fe2+(aq) + H2  (in acid)
 * Fe(s) + 2 H2O → Fe(OH)2(s) + H2 (in base)

Iron (and most other metals) are also thermodynamically unstable in air-saturated water, where the potential of the solution is close to the O2 line in the Pourbaix diagram. Here the spontaneous reactions are:
 * 4 Fe(s) + 3 O2 + 12 H+ → 4 Fe3+ + 6 H2O (in acid)
 * 4 Fe(s) + 3 O2 → 2 Fe2O3(s) (in base)

Corrosion and passivation. It certainly sounds bad for our friend Fe: unstable in water, no matter what the pH or potential. Given enough time, it will all turn into rust. But iron (and other active metals) can corrode, or can be stabilized against corrosion, depending on the conditions. Because our civilization is dependent on the use of active metals such as Fe, Al, Zn, Ti, Cr... for practically everything, it is important to understand this, and we can do so by referring to the Pourbaix diagram.

The corrosion of iron (and other active metals such as Al) is indeed rapid in parts of the Pourbaix diagram where the element is oxidized to a soluble, ionic product such as Fe3+(aq) or Al3+(aq). However, solids such as Fe2O3, and especially Al2O3, form a protective coating on the metal that greatly impedes the corrosion reaction. This phenomenon is called passivation.

Draw a vertical line through the iron Pourbaix diagram at the pH of tap water (about 6) and you will discover something interesting: at slightly acidic pH, iron is quite unstable with respect to corrosion by the reaction:


 * Fe(s) + 2 H+ → Fe2+(aq) + H2

but only in water that contains relatively little oxygen, i.e., in solutions where the potential is near the H2 line. Saturating the water with air or oxygen moves the system closer to the O2 line, where the most stable species is Fe2O3 and the corrosion reaction is:


 * 4 Fe(s) + 3 O2 → 2 Fe2O3(s)

This oxidation reaction is orders of magnitude slower because the oxide that is formed passivates the surface. Therefore iron corrodes much more slowly in oxygenated solutions.

More generally, iron (and other active metals) are passivated whenever they oxidize to produce a solid product, and corrode whenever the product is ionic and soluble. This behavior can be summed up on the color-coded Pourbaix diagram below. The red and green regions represent conditions under which oxidation of iron produces soluble and insoluble products, respectively.

In the yellow part of the diagram, an active metal such as iron can be protected by a second mechanism, which is to bias it so that its potential is below the oxidation potential of the metal. This cathodic protection strategy is most frequently carried out by connecting a more active metal such as Mg or Zn to the iron or steel object (e.g., the hull of a ship, or an underground gas pipeline) that is being protected. The active metal (which must be higher than Fe in the activity series) is also in contact with the solution and slowly corrodes, so it must eventually be replaced. In some cases a battery or DC power supply – the anode of which oxidizes water to oxygen in the solution – is used instead to apply a negative bias.

Another common mode of corrosion of iron and carbon steel is differential aeration. In this case, part of the iron object – e.g., the base of a bridge, or the drill in an oil rig – is under water or in an anoxic environment such as mud or soil. The potential of the solution is close to the H2 line in the Pourbaix diagram, where Fe can corrode to Fe2+ (aq). Another part of the iron object is in the air, or near the surface where water is well oxygenated. At that surface oxygen can be reduced to water, O2 + 4 H+ + 4 e&minus; → 2 H2O. The conductive iron object completes the circuit, carrying electrons from the anode (where Fe is oxidized) to the cathode (where O2 is reduced). Corrosion by differential aeration can be rapid because soluble ions are produced, and the reaction has a driving force of over 1 V. Iron or carbon steel that is subjected to frequent weathering, such as the cast iron bridge and lamppost shown at the right, is corroded on the surface by differential aeration.

Differential aeration is involved in the formation of a rust ring around wet areas of cast iron, e.g., an iron frying pan left partially submerged in water for a day or more. (You may have seen this mechanism of corrosion in action when you did not get to the dirty dishes right away). Under the water, Fe is oxidized to soluble Fe2+, and at the water line O2 is reduced to H2O. As Fe2+ ions diffuse towards the water surface, they encounter oxygen molecules and are oxidized to Fe3+. However Fe3+ is insoluble at neutral pH and deposits as rust, typically just below the water line, forming the rust ring.

4.6 Discussion questions

 * Explain the simplified Pourbaix diagram for nitrogen, shown below. Discuss the reactions that are implied by the lines, and explain why they have the slopes they do.


 * Use the information in the Pourbaix diagram to construct a Frost diagram for nitrogen at pH 7.

4.7 Problems
1.	Balance the following redox reactions, adding H2O and H+ (or OH&minus;) as needed. Predict for each one whether the reaction would become more or less spontaneous at higher pH.


 * (a) MnO4-(aq) + N2O(g) → Mn2+(aq) + NO3&minus;(aq)   (in acid)


 * (b) Cr2O72&minus;(aq) + S2O32&minus;(aq) → Cr2O3(s) + SO42&minus;(aq)  (in base)


 * (c) H2O2(aq) + HI(aq) → I3&minus;(aq)  (in acid)


 * (d) HOBr(aq) → HBr(aq) + HBrO2(aq) (in acid)


 * (e) C12H22O11 (sucrose, aq) + ClO3&minus;(aq) → HCO3&minus;(aq) + Cl&minus;(aq) (in base)

2.	Silver metal is not easily oxidized, and does not react with oxygen-saturated water. However, when excess NaCN is added to a suspension of silver particles, some silver dissolves. If oxygen is removed (e.g., by bubbling nitrogen through the solution), the dissolution reaction stops. Given the fact that Ag(I) tends to form linear two-coordinate complexes, write a balanced equation for the dissolution reaction (hint: it is a redox reaction).

3. The standard potentials for the Fe3+/Fe2+ and Cl&minus;/Cl2 couples are +0.77 and +1.36 V.  Calculate the cell potential of a redox flow battery that has Fe3+/Fe2+ and Cl&minus;/Cl2 solutions on the two sides (both containing 1.0 M HCl as the electrolyte). The pressure of Cl2 gas on the chlorine side is 0.2 atm, and the concentrations of Fe2+ and Fe3+ on the iron side are both 0.10 M.

4.	The Latimer potential diagram for iodine (in acidic solutions) is given below.

+1.60     +1.13     +1.44   +.535  H5IO6  →  IO3-    →    IO-  →  I2   →   I-

(a) What is the standard reduction potential for the IO3&minus;/I&minus; redox couple?

(b) Construct a Frost diagram and identify any species that are unstable with respect to disproportionation.

5.  The Latimer diagram for nitrogen in acidic solutions is shown below:

+5        +4       +3       +2       +1      0       -1        -2      -3      0.80      1.07      1.00     1.59    1.77   -1.87      1.41     1.28  NO3-   →   N2O4  →  HNO2  →   NO   →   N2O  →  N2  →  NH3OH+  →  N2H5+ →  NH4+


 * (a) Write a balanced half reaction for reduction of NO to NH4+ in acid.


 * (b) What is the value of Eo for the half reaction in part (a)?


 * (c) Is this half reaction more thermodynamically favorable in acid or in base? Explain.


 * (d) All the nitrogen-containing molecules and ions listed above are kinetically stable, but only three are thermodynamically stable with respect to disproportionation in acid. Which ones are they?

6.  Referring to the Pourbaix diagram for Mn below:


 * (a)	Write out the balanced reduction half reaction corresponding to the line separating Mn2+ and Mn2O3. Use your answer to calculate the slope of the line (give units).


 * (b)	Is Mn metal stable in water at any pH? If so, in what range of pH?


 * (c)	What spontaneous redox reaction would you expect for an aqueous solution of MnO4&minus; at pH 6?


 * (d)	Describe an electrochemical procedure (specifying pH and potential) for making Mn3O4(s) from aqueous Mn2+.


 * (e)	Label the regions of the diagram that correspond to corrosion and passivation of Mn metal.



7. The Pourbaix diagram for copper is shown below.


 * (a) Write a balanced half-reaction that corresponds to the boundary between the Cu2O(s) and Cu(OH)2(s) regions of the diagram.


 * (b) What is the slope of the line (do not try to measure it from the graph!) that connects the Cu2+(aq) and Cu2O(s) regions? Explain your reasoning.


 * (c) Over what pH range (if any) is copper metal stable in contact with pure water?


 * (d) Thermodynamically, the reaction between copper and oxygen is spontaneous at pH 7. Would you expect copper to corrode or to be passivated against corrosion in aerated water at neutral pH? Explain your reasoning.


 * (e) Some recent studies have suggested the use of Cu2O as a water splitting photocatalyst. Would you expect Cu2O to be stable in the presence of the oxygen formed in the reaction? Explain your reasoning.