3830 lecture notes part4_2008_redox

7/23/2019 3830 Lecture Notes Part4_2008_redox

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Chemistry 3830 Lecture Notes Dr. M. Gerken Page 76

5. Redox chemistry and multiple oxidation statesSo far we have only mentioned the fact that transition elements have multiple oxidation states, but we have ignored the

impact of this on chemical behavior. We will do so now, and then we will continue to use redox principles to treat thechemistry of the p-block elements. Redox chemistry is probably the most interesting (and complex) aspect of the chemica

behavior of the elements, and careful study of this material will greatly reward the effort expended. The majority of thereactions which you do in Part II of the lab are in some sense redox reactions.

As background to this section, you should review how to balance redox reactions and other basic electrochemical

principles from General Chemistry.

5.1 Review of basic princip les from Chemistry 2000

Oxidation-Reductionreactions, better known by the short-form redoxreactions, involve the transfer of electronsbetweenatoms and molecules during the reaction. They represent one of the major classes of chemical reactions. We can detect redoxreactions by monitoring the oxidation statesof the atoms in the reactants and products. Any change in these formal oxidationstates during the reactions means that a redox reaction has taken place. Oxidation states were introduced in Chem2000, andyou should go through as many exercises on assigning oxidation states as possible. You may wish to review the section in yourChem2000 textbook.

5.1.1. Definition of Oxidation States

For reaction involving the transformation of a neutral atom into an ion, e.g., the oxidation of Na to Na +, the identification

of the electron transfer (and the assignment of oxidation states) is obvious. The assignment of oxidation states in covalentcompounds is far more complex. In order to see which atom in a compound looses/gains electrons, we formally (on paper)decompose a covalent compound into monatomic ions. This is done by formal heterolytic cleavage of bonds according to theelectronegativity difference. For example: the H-Cl bond is formally broken by assigning the bonding electron pair to the moreelectronegative bonding partner (Cl). This gives the fragments: H+ and Cl-. The charges of these formal fragments are theoxidation states of the corresponding atoms in the covalent compound. In the case of element-element bonds, homonuclear

bonds, the bonds are cleaved homolytically (one electron of a single bond (two electrons of a double bond) goes to each of thebonding partners. Oxidation states are usually denoted by Roman numerals.Example: H2O21.step: Lewis structure 2.step: Formal bond cleavage

..O OH H

.... ..

......

O OH H..

( ))(

H HO....

: . O....

:.

3. step: Charges of fragments 4. step: Indicate oxidation states in Lewis structure

H HO....

: . O....

:.+ - +-

..O OH H

+I -I +I......

-I

5.1.2 Guidelines for Determining Oxidation Numbers

Usually, you do not have to go through the formal bond cleavage, as shown above. Simple rules has been developed tocalculate oxidation states for most of the known compounds whithout drawing Lewis structures. However, by using these rulesyou assume a certain Lewis structure.

1. Each atom in a pure element has an oxidation number of 0. The oxidation number of Cu in metallic copper is 0

and is zero for each atom in I2or S8. [always valid]2. For ions consisting of a single atom, the oxidation number is equal to the charge on the ion. Elements of periodicGroups 1, 2 and 13 form monatomic ions with a positive charge and oxidation number equal to the group numberAluminum therefore forms Al3+, and its oxidation number is +III. (See Section 3.3.) [always valid]

3. Fluorine is always I in compounds with other elements. [always valid]4. Cl, Br, and I are always I in compounds except when combined with more electronegative elements, such as O

or F. This means that Cl has an oxidation number of I in NaCl (in which Na is +I, as predicted by the fact that itis an element of Group 1). In the ion ClO the Cl atom has an oxidation number of +I (and O has an oxidationnumber of II; see Guideline 5).

5. The oxidation number of H is +I and of O is II in most compounds. Although this statement applies to manymany compounds, a few important exceptions occur.


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When H forms a binary compound with a metal, the metal forms a positive ion and H becomes a hydride ionH. Thus, in CaH2the oxidation number of Ca is +II (equal to the group number) and that of H is I. [Thedeciding factor remains the electronegativity of the bonding partner]

Oxygen can have an oxidation number of I in a class of compounds called peroxides, compounds based onthe O2

2ion. For example, in H2O2, hydrogen peroxide, H is assigned its usual oxidation number of +I, andso O is I. [If O is bonded to fluorine you have to use the definition given in 7.1.1]

6. The algebraic sum of the oxidation numbers in a neutral compound must be zero; in a polyatomic ion, the summust be equal to the ion charge. Examples of this rule are the previous compounds and others found in Example4.8.

5.1.3 Oxidation and Reduction - a working definition

The following basic definition is used to describe redox processes:If X loses one or more electrons, it is oxidized, and is the reducing agent.

X Xn+ + n eIf Y gains one or more electrons, it is reduced, and is the oxidizing agent.

Y + n e YnThese are definitions that must be learned cold. They are the oui and non of electrochemistry. All the other ideas are

based on them. The origin of the term oxidationcomes from the fact that combination with oxygen (e.g. in combustion) is onecommon form of an oxidation reaction. The element oxidized loses electrons to the oxygen atom. For example, whenmagnesium is burnt:

Mg Mg2+ + 2e

while oxygen gains electrons: O2 + 2e

O2

and the net reactionis:Mg + O2 MgO

Reductionoriginates in the concept of reducingan ore, usually a metal oxide, to the elemental form, usually the pure metal.For example when iron is made from iron ore, the iron reaction is:

Fe2+ + 2 Fe3+ + 8 e 3 Feand the carbon monoxide reaction is:

4 CO 4 CO2+ + 8 eHere the oxide ions are spectators, transferring from the iron to the oxidized carbon. The net reaction therefore becomes:

Fe3O4 + 4 CO 3 Fe + 4 CO2

5.1.4 Balancing redox equations in solution

These examples illustrate the common technique of separating redox reactions into two complementary redox halfreactions. In the above examples, the Mg/Mg2+ reaction is an oxidation half reaction. In such reactions, electrons arealways among the products. The CO/CO2+ reaction is also an oxidation half reaction. The other two are reduction halfreactions, for which electrons will always be reactants. Although simple reaction such as these are easy to balance at sightand we often dont stop to think that they are in fact redox reactions, more complicated redox reactions need a rigorousapproach for successful balancing. We will concentrate on reactions that take place in solution, and these will usually be eitherbasic or acidic solutions. Since electrochemistry is a branch of thermodynamics, we usually will be dealing with reactionunder conditions of standard state, which for solution chemistry means 1 Mconcentrations at 25C. Thus acidic solutionswill have [H3O

+] = 1.00 M, while basic solutions will have [OH] = 1.00 M. The usual Kwrelationship between hydroxide andhydronium ion will always hold.

The rules for balancing redox reactions are as follows. These are better learned by doing many examples than by actuamemorization of the rules!

Step 1. Assign oxidation states and recognize the reaction as an oxidation-reduction.Step 2. Separate the overall process into half-reactions.Step 3. Balance each half-reaction for material (first balance oxidized/reduced element, then balance O, then balance H).

In acid solution, add H2O to the side requiring O atoms.Thenadd H+to balance any remaining unbalanced H atoms.

In basic solution, add H2O to the side requiring O atoms.Thenadd H2O to the side requiring H atoms, andoneOH

to the other side. Add H2O/OH-for every

unbalanced H atom.Step 4. Balance the half-reactions for charge by inserting the correct number of electrons. Consider the number of

oxidized/reduced atoms.Step 5. Multiply the balanced half-reactions by appropriate factors to obtain the same number of electrons in both half-

reactions.Step 6. Add the balanced half-reactions.


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Step 7. Simplify the equation by eliminating common reactants and products.Step 8. Check the final result for material and charge balance.

5.1.5 Some examples of balancing redox reaction equations

The following reaction takes place in standard acid solution (1MH+)MnO4

(aq) + HSO3

(aq) Mn

2+(aq) + SO4

2(aq)

We recognize that Mn changes oxidation state from +7 to +2, while S changes from +4 to +6. The two half reactions are:Oxidation: HSO3

SO42

Material balance HSO3

+ H2O SO42

+ 3 H+

Electron balance HSO3

+ H2O SO42 + 3 H+ + 2 e

Multiply by 5 5 HSO3 + 5 H2O 5 SO4

2 + 15 H+ + 10 e

Reduction: MnO4 Mn2+

Material balance MnO4 + 8 H+ Mn2+ + 4 H2O

Electron balance MnO4 + 8 H+ + 5 e Mn2+ + 4 H2O

Multiply by 2 2 MnO4 + 16 H+ + 10 e 2 Mn2+ + 8 H2O

Overall reaction: 2 MnO4

(aq) + 5 HSO3

(aq)+ H

+(aq) 2 Mn

2+(aq) + 5 SO4

2(aq)+ 3 H2O

Finally, check that the atoms andcharges balance at the right and at the left

The following reaction takes place in standard basic solution (1MOH-)Fe(OH)2(s) + CrO4

2(aq) Fe2O3(s) + Cr(OH)4

(aq)

We recognize that Fe changes oxidation state from +2 to +3, while Cr changes from +6 to +3. The two half reactions are:Oxidation: Fe(OH)2 Fe2O3

Material balance 2 Fe(OH)2 + 2 OH Fe2O3 + 3 H2O

Electron balance 2 Fe(OH)2 + 2 OH Fe2O3 + 3 H2O + 2 e

Multiply by 3 6 Fe(OH)2 + 6 OH

3 Fe2O3 + 9 H2O + 6 e

Reduction: CrO42 Cr(OH)4

Material balance CrO4

2 + 4 H2O Cr(OH)4 + 4 OH

Electron balance CrO42 + 4 H2O + 3 e

Cr(OH)4 + 4 OH

Multiply by 2 2 CrO42 + 8 H2O + 6 e

2 Cr(OH)4 + 8 OH

Overall reaction: 6 Fe(OH)2(s) + 2 CrO4

2(aq) 3 Fe2O3(s) + 2 Cr(OH)4

(aq) + 2 OH

(aq) + H2O

Finally, check that the atoms andcharges balance at the right and at the left

5.1.6 Spontaneous reactions

Gibbs Free Energy of reaction tells us whether a certain reaction is product favoured or reactant-favoured in the forwarddirection. The electrochemical analogue to this concept is called the cell potentialgiven by the symbolE. Eis expressed inthe common electrical unit of volts, but is really just the Gibbs free energy in disguise! The exact relationship between cel

potential and Gibbs energy is given by the relationship:

G nFE

= in general, and for standard conditions;

G nFE

0 0=

;n = number of electrons transferred, F = Faradays constant = 96485 C mol-1

Because of the minus sign in the equation, the cell voltageEhas the opposite sign conventionto that of G:

Product-Favoured Reaction: ve G or +veEReactant-Favoured Reaction: +ve G or veE

Why do we have such a confusing convention? The origin of this discrepancy comes from the way that physicists definecurrent flow: although electrical current is entirelydue to the migrations of electrons in a circuit, a forward current flow isdefined in standard theories of electricity as the direction of positive charge flow. It is when harmonizing standard electricaconventions with chemical definitions that these apparently contradictory conventions are generated. In electrochemistry, areaction which is product favouredproducesa voltage, and is therefore called a Voltaic cell, in honor of Alessandro Volta. A


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reaction which is reactant favoured can be drivenforward by the application of a greater opposite voltage; such reactions arecalled electrolytic cells, and the process is named electrolysis.

We now want to remember another aspect of the meaningof G, which is that it tells for any given reaction the maximumamount of the total energy change that may be harnessed for useful work. The electrochemical equivalent of this idea is themaximum voltage of the electrochemical cell. This maximum potential is obtainable only from afully charged cell runningunder zero load. This means that we can calculatethe maximum cell voltage from standard free energies for the reaction.

Consider the reaction:Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

fG

(kJ mol

1

) 0 65.52 147.0 0Since rG = {fG(products)} {fG(reactants)}, rG= (-147.0 kJ/mol) (65.52 kJ/mol) = 212.5 kJ mol1, and we

would predict this to be a spontaneous reaction. It is spontaneous, but frankly quite useless unless you need some finelydivided metallic copper and have only these reactants at hand. Let us now see if it is indeed possible to harnessthe usefuwork from this reaction.

5.1.7 The electrochemical cell

To do this we must perform exactly the samechemical reaction but using an electrochemical cell.A typical cell-design is shown in the figure below. Itdivides the redox reaction into to compartments, eachof which contains one of the two redox half-reactions. What voltage does our device measure?

What voltage should it measure? Well, rG = 212.5 kJ mol1, and the value of n= 2, as can be seen

by breaking the reaction into its constituentelectrochemical half-reactions:

Zn(s) Zn2+(aq) + 2 e Oxidation

Cu2+(aq) + 2 e Cu(s) Reductionand so:

V10.1molkC485.962

molkJ5.212

nF

GE

1

100

+=

=

=

Any discrepancy between the actual measured

potential of a given cell and this value is usually dueto non-standard conditions, primarily that theconcentrations are not exactly one Molar.

Diagram of an electrochemical cell to harness the Zn/Cu reaction

5.1.8 Electrode or Half-cell potentials

Although the reaction only requires acopper(II) salt and metallic zinc, we have

built a cell in which both metallic zincand copper are present, and copper andzinc sulfate. The reason for this is tomake the reactions reversible. It turns

out that accurate potentials can only bemeasured for reversible electrochemicalcells. Electrochemists have developed ashort-hand notation to indicate thecomplete make-up of a givenelectrochemical cell. It is based on the

balanced redox reaction, but includesboth the reactantsand the products. Forour cell the correct notation would be:

Zn(s)| Zn2+(1 M)|| Cu2+(1 M)| Cu(s)

In this notation, the cathodereaction is alwaysput on the right hand side, and the anode reaction at the left. The cathodicprocessis alwaysreduction, while the anodic processis always oxidation. Thus the anode half-cell will always be written a


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the left hand side of the notation. A convenient mnemonic device to remember this convention is Right Red Cat, i.e. that thecathode process is a reduction process and is placed at the right of the cell notation. The vertical lines, |, in the notationindicate a phase boundary, such as between a solid and a liquid. Double lines, ||, indicate double boundaries, such ascommonly occur when a salt bridge is placed between the two half cells. If the components of a redox reaction co-exist insolution without a phase interface, they are listed together with a comma separating oxidized and reduced forms. This is oftenthe case when the current is introduced into a solution via an inert electrode (usually platinummetal); an example is theoxidation of iron(II) to iron (III) for which the notation could be: Pt(s)| Fe2+(1 M), Fe3+(1 M).

The operation of a voltaic cell is summarized in Figure 21.4. It represents a complete electrical circuit. In the wiresexternal to the cell, the current is carried exclusively by a moving stream of electrons. However, within the solution, thecurrent must be carried by migrating ionic species, such as in our example cell Zn2+and Cu2+ions. Within the salt bridge, ifone is used, inert ions such as K+and Clcarry the charges and provide for current flow.

Standard Half-Cell Reduction PotentialsFrom the construction of our electrochemical demonstration cell, and from the cell notation, it should now be obvious to

you that electrochemical half-reactions, which we arbitrarily introduced as a tool to balance redox reactions, have somedegree of physical reality they can after all be built into electrochemical half-cells. It should, for example, be possible touncouple the copper/zinc cell and construct other voltaic cells from them, for example a copper/lithium or a silver/zinc cell, etcThis is indeed possible. It would also be nice to be able to assign a voltage to each half-cell, but this turns out to beimpossible, since the voltage depends on a complete electrical circuit being established. Nonetheless, this idea of a half-cel

potential is so attractive, that chemists have figured out a way to do so. We do it by arbitrarily setting oneredox half-reactionto zero, and measuring all other cells with respect to this arbitrary zero reference point. The universal reference standard foelectrochemistry is the standard hydrogen electrode, or SHE. For example, if the zinc half-cell is combined with the SHE, asshown in the Figure below, a voltage is measured as +0.76 V. We assign this cell voltageentirely to the zinc half cell, sinceSHE is zero. Alternatively, we combine the SHE with the copper half cell, and get +0.34 V. Then we can establish the

potential of the copper zinc cell asE= + 0.76 + 0.34 V = +1.10 V.

Since for any given combination of two half-cells, we can never be sure which half will act as the anode and which as the

cathode, electrochemists have a standardized way of expressing half-cell potentials, and that is always to depict them asreductions. This leads to the compilation of the standard reduction potentials in aqueous solutionwhich are compiled inthe Table below.

We can use this table in numerous ways. First of all, we can now construct an electrochemical cell out of anycombination of half cell reactions. Thus we can go and obtain potentials for the examples cited above:


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5.1.9 Constructing cell potentials from standard half-cellpotentials

copper/lithium cellFrom the table we get the half-reactions:

Cu2+(aq) + 2 e Cu(s) E= +0.337 V

Li+

(aq) + e

Li(s) E= 3.045 VWe now combine them in such a way as to get a positive overall cell potential. This means we must reversethe lithiumequation, making it the anode (where the oxidation will take place.)

Li(s) Li+(aq) + e E= +3.045 Vnow we have the cell potential for the overall reaction

Cu2+(aq) + 2 Li(s) Cu(s) + 2 Li+(aq) Ecell= +3.382 V

Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for thisreaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in thevalue of n.

Table of Standard Reduction Potentials in Aqueous Solution at 25 C

Reduction Half-Reaction E(V)

F2(g)+ 2 e 2 F(aq) +2.87

H2O2(aq)+ 2 H3O+

(aq)+ 2 e 4 H2O(l) +1.77PbO2(s) + SO4

2(aq)+ 4 H3O+(aq)+ 2 e PbSO4(s) + 6 H2O(l) +1.685

MnO4(aq)+ 8 H3O

+(aq)+ 5 e Mn2+(aq)+ 12 H2O(l) +1.52

Au

3+

(aq)+ 3 e

Au(s) +1.50Cl2(g)+ 2 e 2 Cl(aq) +1.360

Cr2O72(aq)+ 14 H3O

+(aq)+ 6 e 2 Cr3+(aq)+ 21 H2O(l) +1.33O2(g)+ 4 H3O

+(aq)+ 4 e 6 H2O(l) +1.229Br2(l)+ 2 e

2 Br (aq) +1.08

NO3 (aq)+ 4 H3O+(aq)+ 3 e NO(g)+ 6 H2O(l) +0.96

OCl(aq)+ H2O(l)+ 2 e Cl(aq)+ 2 OH(aq) +0.89

Hg2+(aq)+ 2 e Hg(l) +0.855Ag+(aq)+ e Ag(s) +0.80Hg2

2+(aq)+ 2 e 2 Hg(l) +0.789Fe3+(aq)+ e Fe2+(aq) +0.771I2(s) + 2 e

2 I(aq) +0.535O2(g)+ 2 H2O(l)+ 4 e

4 OH(aq) +0.40

Cu2+(aq)+ 2 e Cu(s) +0.337Sn4+(aq)+ 2 e Sn2+(aq) +0.15

2 H3O + (aq)+ 2 e H2(g)+ 2 H2O(l) 0.00

Sn2+(aq)+ 2 e Sn(s) 0.14Ni2+(aq)+ 2 e Ni(s) 0.25V3+(aq)+ e V2+ (aq) 0.255PbSO4(s) + 2 e

Pb(s) + SO42(aq) 0.356

Cd2+(aq)+ 2 e Cd(s) 0.40Fe2+(aq)+ 2 e Fe(s) 0.440Zn2+(aq)+ 2 e Zn(s) 0.7632 H2O(l)+ 2 e

H2(g)+ 2 OH(aq) 0.8277

Al3+(aq)+ 3 e Al(s) 1.66

Mg2+

(aq)+ 2 e

Mg(s) 2.37Na+(aq)+ e Na(s) 2.714K+(aq)+ e K(s) 2.925Li+(aq)+ e Li(s) 3.045 In volts (V) versus the standard hydrogen electrode.


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silver/zinc cellFrom the table we get the half-reactions:

Zn2+(aq) + 2 e Zn(s) E= 0.763 VAg+(aq) + e Ag(s) E= +0.80 V

We now combine them in such a way as to get a positive overall cell potential . This means we must reversethe zincequation, making it the anode (where the oxidation will take place.)

Zn(s) Zn2+(aq) + 2 e E= +0.763 Vnow we have the cell potential for the overall reaction

2Ag+(aq) + Zn(s) 2Ag(s) + Zn2+

(aq) Ecell= +1.563 V

These examples should be sufficient to show how cell potentials are determined for standard reduction potentials. Remembethat voltaic cellsmust have a positive cell potential.

5.1.10 Other uses for the table of electrode potentials

In this course, we will not be concerned much with constructing electrochemical cells. But we do want to discuss thechemical behaviour of the elements. We can use the data in the Table directly in the description of chemical properties.

a) Spontaneity of reaction

Combine any two standard reduction reactions by reversing one of the couples and adding the resulting potentials togetherIf the E

cellis positive, the reaction as written is spontaneous. If negative, the reverse reaction is spontaneous.

b) Identifying reducing agents

Any species with a negative Eelectrode will tend to be a reducing agent, and elements with large negative electrodepotentials are strong reducing agents.

c) Identifying oxidizing agents

Any species with a positive Eelectrode will tend to be an oxidizing agent, and elements with large positive electrodepotentials are strong oxidizing agents.


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5.2 Latimer diagrams

The main disadvantage of the Table of Redox Potentials is that it is organized by the voltage of the process, rather than bythe chemical species involved. This is great if you want to build a battery, or to use an electrochemical process in an analyticatechnique, e.g. to monitor the concentration of a given species. But it is not great if you want to understand the complex redoxchemistry of an element such as iron, copper or manganese. To address this issue, Latimer devised a very helpful way todisplay redox potentials for the elements. These are given in an Appendix in the text by Shriver, Atkins and Langford.

We illustrate both the need for, and the method of construction, of Latimer diagrams with the following example: Wha

happens when Fe(s) is reacted with a strong, non-oxidizing, acid (for example, hydrochloric or perchloric acid)? Now thereare two common oxidation states of Fe, 2+ and 3+. We thus need to decide which of these two species is predicted bythermodynamics to form in acid solution?

Reminder: thermodynamic data will predict which reactions ought to occur, but cannot determine whether they

happen at an observable rate or not. Most of the redox reactions of inorganic compounds are rapid reactions, but there aremany times when thermodynamics predicts more than one possible product, and where the actual product is selected by therate of reaction. We will say a bit more about this later on. The example dealt with here can be decided unambiguously bythermodynamics.

From the table of reduction potentials, we pick out the data that is available for iron. We then remember that the free energiesof chemical reactions, i.e. G, obey algebra (often called Hess' Law of Heat Summation). This allows us to combine the twoequations to obtain the unknown redox potential relating iron in the 3+ state to the element. The calculations are summarized:

Reaction Potential G nFE =

Fe2++ 2e- Fe -0.440 V -2 F -0.440Fe3++ e- Fe2+ +0.771 V -1 F +0.771Fe3++ 3e- Fe NOT +0.331V!!!! = +0.109 F = -3 F -0.036

Thus we have obtained the desired potential, the so-called skip potential, for the direct conversion of elemental iron toiron(III):

Fe3++ 3e- Fe E = -0.036 Vand we can answer the original question. When Fe is dissolved in acid, we consider the reverse of reactions 1 and 3. We seethat the potential for the formation of Fe(II) is greater thanfor the formation of Fe(III). Thus, even though theformation of Fe3+is overall favored, the reaction has an evengreater tendency to stop at the Fe(II) stage. We can diagramthis effectively by plotting the free energy terms (asmultiples of the Faraday constant) as a function of oxidationstate. In effect, the reaction will stop in the "well" at Fe2+,

because it would cost energyto rise up from the Fe(II) stateup to the Fe(III) state.

Latimer diagrams are convenient for the display of all redox potentials relating to a given element, including the skippotentials. Thus the three redox potentials we have considered so far for iron are displayed as follows:

Fe3+

Fe2+

Fe+0.771

-0.036

-0.440

For clarity I have added arrows to the Latimer diagram. Thus the numbers as written are standard reduction potentials, i.ethe sign of the redox potential refers to the reaction as it proceeds from left to right. If you need to consider the reversereaction, as in our question above, the sign must be reversed. Latimer diagrams do not normally have such arrows, so that youneed to remember this convention!

5.2.1. Construction and use of the Latimer diagram for copper

Let's look at another example of such reasoning. Consider the two half-reactions:Reaction Potential G nFE =

Cu+ + e-Cu +0.520 V -1 F 0.520Cu2+ + e-Cu+ +0.159 V -1 F 0.159Cu2+ + 2e-Cu NOT 0.679! = -0.679F -2 F +0.340

Fe3+

Fe2+

G

Fe3+

Fe2+

Fe

oxidation state0 +I +II +III

Fe

+0.036 V

+0.440 V


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The free energy diagram is plotted at right, and the Latimer diagram becomes:

Cu3+

Cu2+

Cu+ Cu

+1.8

+0.340

+0.159 +0.520

a) Oxidation of elemental copperWhat happens when Cu(s) is placed in water or acid? The reverse of the reactions leading to Cu+ and Cu2+ are both

negative! Under standard conditions, copper will not be oxidized by acid! Remember that in your lab experiment on thereactions of metals, you had to use HNO3 to dissolve copper wire. In other words, we needed to add a reagent capable osupplying the energy required to overcome the barrier depicted in the above graph. This depends on the additional redox halfreaction:

2 NO3- + 4 H+ + 2 e- N2O4 + 2 H2O E = +0.803 V

The overall reaction is:Cu + 2 NO3

- + 4 H+ Cu2+ + N2O4 + 2 H2O E = +0.463You may also ask why the product of the nitric acid oxidation is Cu(II) rather than Cu(I). We need to work out the

potential for an overall reaction to form Cu(I):Cu + NO3

- + 4 H+ Cu+ + N2O4 + H2O E = +0.283However, the overall reaction to Cu+has E > 0; since this is a less positive EMF, the reaction proceeds directly to Cu2+and

does not halt at Cu+

. This is of course also clear from the free-energy diagram above.

NOTE:be sure that you can obtain such balanced redox reactions from the three redox half-reactions involved in this questionPast experience has shown that even the best students consistently fail this task!

b) Disproportionation of Copper (I)

Copper(I) in acid solution has an interesting property. Consider the possible reactions of Cu+: it can be reduced to Cu oroxidized to Cu2+. Let us consider both of these processes.

Reaction Potential G nFE =

Cu+ + e- Cu +0.520 V -1 F +0.520Cu+ Cu2+ + e- -0.159 V -1 F -0.1592Cu+ Cu + Cu2+ +0.3611 = - 0.361F -1 F +0.361

NOTE: The combination of these two redox half reactions describe a balanced redox reaction, i.e. effectively the final reactiondescribes an electrochemical cell. Whenever this is the case, the potentials obtained via the free energy calculation is the sameas just adding the redox potentials (with the correct sign!) together.

What we have demonstrated is that Cu+ is unstable towards disproportionation, the process in which a compoundundergoes an autoredoxreaction to produce forms of the element with higher and lower than the original oxidation state. Thereverse of this equation, the conversion of Cu and Cu2+to 2 Cu+is called comproportionation. Thermodynamics tells us thafor copper, the disproportionation reaction is product-favored, whereas comproportionation is not product-favored.

5.2.2 Generalized Latimer diagrams

Consider the Latimer diagram for iron in acid solution shown below in the form that it is found in SAL. It is extended by a

branch describing iron in the presence of both 1 M [CN-

] and 1 M acid. The potentials change because coordination with CNligands alters the free energy of both the Fe(II) and Fe(III) species.

Fe3+ Fe2+

[Fe(CN)6]3-

[Fe(CN)6]2-

Fe0.771 -0.44

-0.04

+III +II 0Acidic solution

0.361 -1.16

We remember from our discussion of acid/base chemistry that Fe3+in acid solution is a short-hand representation for the aquacomplex, [Fe(OH2)6]

3+, while Fe2+is really [Fe(OH2)6]2+. Thus the two branches of the diagram actually reflect identical redox

G

Cu2+

Cu

+

0 +I +II

Cu


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processes for iron in the presence of two ligand systems. We can now see at a glance that the hexacyano complex of Fe(III) ismore stable than the aqua ion with respect to reduction to Fe(II). Also, the oxidation of Fe(0) to Fe(II) is considerably morefavorable in the cyanide/acid mixture than in aqueous acid. A Latimer diagram can thus be extended to include any number orelated redox systems. We could just as well construct a branch in which iron was coordinated to ammonia or chloride ionsfor example.

In general, anything which alters the free energy of the system will change the redox potentials. The following factors alaffect the size of G : (1) Concentration

(2) Temperature(3) Other reagents which are not inert(4) pH (a special case of (3))

In practice, the most important of these two examples for aqueous element chemistry is pH changes. It has becomeconventional to construct Latimer diagrams for the two extremes of pH = 0 and pH = 14 (respectively 1 Macid and 1Mbase)This is shown for the element manganese below:

0.146 -0.234 -1.560.27 0.930.56

0.95 1.5MnO2 Mn

3+ Mn2+

MnO2 Mn2O3 Mn(OH)2

-1.18Mn

Mn

1.28 2.9HMnO4

- (H3MnO4)

MnO42- MnO4

3-

0.90MnO4

-

MnO4-

acidic solution

basic solution

+VII +VI +V +IV +III +II 0

Note that dramatic differences in redox potentials occur for these two sets of conditions. The reason for this is that largenumbers of H+and OH ions are usually involved in the redox half reactions. Anything that affects the concentration of theseions will therefore have a dramatic effect on the redox potentials. This is such an important consideration that it has led to thewide-scale use of a graphical presentation of the free energy changes that accompany redox reactions. We now consider suchFrost diagrams.

5.3 Frost diagrams

Frost diagrams are essentially the same as the graphs of free energy against oxidation state, where G is given in units of nF(i.e. in Volts). They are simply quantitative versions of the graphs we have already been considering. We demonstrate theconstruction of complex Frost diagrams for the element manganese, which has as complicated a redox chemistry as anyelement known.

5.3.1 Constructing the Frost diagram for Manganese

A Frost diagram relates the free energy of any given redox state to the energy of the elemental form. We calculate thevalue nE for the overall redox couple:

X(N)/X(0)and plot each couple against the oxidation state N. The oxidation state scale mustbe linear, even if the element does not exisin all possible oxidation states. We therefore need to calculate the value nE for each species in the Latimer diagram. This ismost economically done step-wise, as shown in the table below. Thus we start by calculating the two-electron process betweenMn2+and Mn, which works out to 2.36 V. For the next step, we recognize that Mn+3to Mn2+is an additionalone-electronstep, so that +1.5 V must be added to the previous step to get the value that relates the energy of Mn 3+to that of elementamanganese.

0.146 -0.234 -1.560.27 0.930.56

0.95 1.5MnO2 Mn

3+ Mn2+

MnO2 Mn2O3 Mn(OH)2

-1.18Mn

Mn

1.28 2.9HMnO4

- (H3MnO4)

MnO42- MnO4

3-

0.90MnO4

-

MnO4-

acidic solution

basic solution

+VII +VI +V +IV +III +II 0


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OxSt Species Calculation value of nE(V)

0 Mn 0 0.0

+II Mn2+ 2 1.18 2.36

+III Mn3+ 1 1.5 + 2.36 0.86

+IV MnO2 1 0.95 + 0.86 0.09

+V H3MnO4 1 2.9 + 0.09 2.99

+VI HMnO4 1 1.28 + 2.99 4.27

+VII MnO4 1 0.90 + 4.27 5.17

The results of our calculation are plotted on the graphshown at the right. In addition, the graph also includesthe results of a similar type of calculation using theLatimer diagram for basic solution above. This allowsus to directly compare the redox behavior of the elementin acid and basic solution.

NOTE: you should test your ability to construct Frostdiagrams by doing the base calculation and therebyconfirming the points on the graph for basic solution.

5.3.2 Reading the Frost diagram formanganese

Most reactions in aqueous solution occur between the extremes of 1 Macid and 1 M base. For neutral solution, a lineapproximately intermediate between the two lines plotted will be observed. It is rarely necessary to actually calculate theintermediate state in order to explain what happens in a neutral-solution reaction. There are several characteristic features o

Frost diagrams that you should be aware of in order to be able to interpret chemical behavior quickly from the graphs. (NOTEthere is unfortunately no universal agreement as to whether the oxidation state in such diagrams are plotted as increasingfromleft to right, or vice versa. The two text books used for Chemistry 2810, Shriver-Atkins-Langford and Rayner-Canham, diffein this regard. In these notes I will consistently use the most-positive-at-right convention (that used in Shriver-Atkins-Langford), and I will provide alternatives to all the reversed diagrams presented in Rayner-Canham's book.

a) Identifying oxidizing and reducing agents

Species lying high on the diagram are oxidizing agents towards species on their left, while species high on the diagram acas reducing agents to other oxidizing agents on their right. Another way to visualize this is by considering the lines connectingthe higher and lower lying species. If the line has a postitive slope, the higher-lying species is an oxidizing agent. If the linehas a negative slope, the higher-lying species is a reducing agent.

Thus for Mn in both acid and base solution, MnO4 is an oxidizing agent (the line has a positive slope), being reduced to

several possible manganese species of lower oxidation state. In acid solution, the remaining forms of manganese down to Mn

3+

are all potential oxidizing agents, while in base the lowest oxidizing agent is MnO2. Elemental manganese, (i.e. manganesemetal) is a reducing agent (the line has a negative slope), being itself oxidized most readily to Mn2+in acid solution, and Mn2Oin base. In base only, Mn(II) can also act as a reducing agent.

b) Identifying strong and weak agents

Once we have identified potential oxidizing and reducing agents from the slopes of the lines, their relative strength can bedetermined by the steepness of the slope of the lines. The steeper the slope, the stronger the agent. Thus MnO4

is a muchmore powerful oxidizing agent in acidic solution than in basic solution. On the other hand, metallic manganese is a slightlystronger reducing agent in basic solution.

Frost Diagram for Manganese

Mn2+

Mn3+

MnO2

H3MnO4

(HMnO4)-

(MnO4)-

(MnO4)-(MnO4)2-

(MnO4)3-

MnO2

Mn2O3Mn(OH)2

Mn

-4

-3

-2

-1

0

1

2

3

4

5

6

O I II III IV V VI VII

Oxidation State

nE(

V) Aci

Bas


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c) Identifying redox products (unreactive redox states)

Species at the bottom of the graph have low free-energy, thus little tendency to react. The lowest species on the graph arethe thermodynamic final product(s) of the redox reactions involving that element. Note that many Frost diagrams display athermodynamic well. This is the case for manganese, for which the well is Mn2+in acid and Mn2O3(i.e. Mn(III)) in base.

d) Identifying species likely to undergo disproportionation

If a species lies abovethe line connecting its neighbors, it is thermodynamically unstable towards disproportionation. Thishas been described as a point lying along a concave line. For example, in basic solution MnO4

3lies on a point which is abovethe line connecting MnO42and MnO2. This means that the reaction:

2 MnO43 + 2 H2O MnO4

2 + MnO2 + 4 OH

is predicted to be product-favored.

e) Identifying species likely to undergo comproportionation

Species likely to undergo comproportionation to a third species are located to left and right of a point which lies belowtheline connecting the two species. Thus in acid solution, a mixture of MnO2and Mn are expected to react together to form Mn

2+

The rate of this reaction may be hindered by the insolubility of both species, but thermodynamically it is favored. Similarly, inbase, a mixture of MnO2 and Mn(OH)2 should comproportionate to Mn2O3. Note an important difference betweendisproportionation and comproportionation. The former process renders a single species unstable, in that an autoredox reactioncan occur at any time. Comproportionation, on the other hand, requires both reactants to be present at the same time, and ione is missing the other species remains stable.

5.3.3 An overview of transition metal redox behavior in the first period

One of the great advantages of the Frost concept is the ability to compare the relative behavior of different species. We arenow finally ready to take a more detailed look at the redox behavior of the transition element, which was one of the goals weset for ourselves when we introduced redox tools. Consider the following Frost diagram for all the first-row transitionelements in acid solution. Despite the large number of compounds, the voltages of the redox processes are quite different, sothat the points do not on the whole obscure each other.


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We see that allthe metals are potent reducing agents, with the exceptionof copper, for which the oxidized forms have ahigher free energy than the element. The reducing strength of the metals goes down smoothly from calcium to nickel,across the period, with nickel being only a mild reducing agent. The jump to copper is fairly large, but its behavior isconsistent with the trend towards weaker reducing power - copper simply has none.

The earlier transition elements favor the +3 oxidation state as the most stable form (bottom of the diagram), while for thelatter elements +2 is more stable, sometimes (as for cobalt) considerably more stable.

The elements in the middle of the series - Mn and Fe - have the largest range of accessible oxidation states. But the freeenergies of the highest oxidation states of these elements are extremelyhigh, and they are all potent oxidizing agents. Infact, only Ti in its highest state (+4) has virtually no oxidizing power (if we ignore the calcium and scandium ions, whichimmediately form stable noble-gas configurations forms and are extremely stable and totally non-oxidizing.)

5.3.4 Trends in the redox behavior down Group 6

The Frost diagram at the right picks on one of the elements in the abovediagram, still in acid solution, and compares its behavior to the remainder of thetransition elements in this group. This may be taken as fairly representative of theredox trends down any d-block group. Note that this diagram has not includedthe chemical form of the species involved. It isThe highest oxidation state for Cr, +6, is strongly oxidizing. For Mo, +6 is mildlyoxidizing, but for W it is completely non-oxidizing.The most stable oxidation states are +3 for Cr, +4 for Mo, and +6 for W. Higheroxidation states become more favorable in the 2ndand especially the 3rdtransition

series.In the +5 state, the energy of Mo and W species are the same, which leads to verysimilar behavior for the two elements in this oxidation state.


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5.4 Kinetics of redox processes

A detailed consideration of the kinetics of redox processes is beyond the scope of this course. However, in order to makesense of redox reactions, particularly those of the non-metals, we will require to take account of a few kinetic principles.

5.4.1 Overpotentials for gas formation

Reactions which produce gases, especially H2, O2and N2 often have an overpotentialassociated with them, and tend tobe kinetically slow. Overpotentials are simply the extra energy required to overcome activation energies for reactions to occuat appreciable rates. Consider the reaction to initiate the reduction of H3O+to H2:

H3O+ + e- 4H2 + H2O

Thus the overpotential is simply the voltage needed to sustain a particular rate of electron transfer.

Overpotential (V) for gas evolution at 25 C10 A/m2 100 A/m2 1000 A/m2 10000 A/m2

Electrode H2 O2 H2 O2 H2 O2 H2 O2

PlatinizedPt 0.015 0.398 0.030 0.521 0.040 0.638 0.048 0.766Smooth Pt 0.024 0.721 0.068 0.85 0.288 1.28 0.676 1.49Cu 0.479 0.422 0.584 0.580 0.801 0.660 1.254 0.793Ag 0.475 0.588 0.762 0.729 0.875 0.984 1.089 1.131

Au 0.241 0.673 0.390 0.963 0.588 1.244 0.798 1.63Graphite 0.599 0.778 0.977 1.220Sn 0.856 1.076 1.223 1.230Pb 0.52 1.090 1.179 1.262Zn 0.716 0.746 1.064 1.229Cd 0.981 1.134 1.216 1.254Hg 0.9 1.0 1.1 1.1Fe 0.403 0.577 0.818 1.292

Ni 0.563 0.353 0.747 0.519 1.048 0.726 1.241 0.853

Source:International Critical Tables, 6, 339 (1929).

For the simple reaction of hydrogen evolution by a metal dissolving in water or acid, the overpotential is ~0.6 V. Thisexplains why Zn, Fe, Ni and Pb do not evolve hydrogen gas when placed in water. Although the Efor Zn is 0.76 V in 1 M

acid, this is reduced considerably when the [H3O+

] is 1

10-7

in neutral solution. [HINT: use the Nernst equation!] Thus wefound in the lab that Zn did not react with boiling water, but produced H2rapidly from 6 M H3O+.

ASIDE: The presence of overpotentials is often exploited in the design of electrochemical experiments and industrial cellsThus redox reactions can be studied at voltages where H2or O2evolution from water should be occurring, but does not due tothe overpotential. This is where the table from Harris becomes useful in finding an electrode material with in this case thehighest possible overpotentials. On the other hand, there is a great deal of interest in photoelectrochemistry, in whichsunlight is harnessed directly in a suitable electrochemical cell to split water into H2and O2. Such cells require electrodes withthe lowest overpotential, and thus they often use platinized Pt (Pt covered with a colloidal deposit of fresh Pt on the surface) asthe electrode material.

5.4.2 Outer-sphere electron transfer

If a redox process can occur with no change in theatomic composition of the reacting species, electron transfer is oftenvery fast, and the rates of reaction closely follow the thermodynamic predictions of G. For example, consider the reaction:

[Fe(phen)3]3+ + [Fe(CN)6]

4- [Fe(phen)3]2+ + [Fe(CN)6]

3-This is an example of a reaction which proceeds by an outer-sphere electron transfer, in which the electron is transferred

between the two species much as a baton is between two participants in a relay race.

5.4.3 Inner-sphere electron transfer

If a redox reaction requires the transfer of atoms from one reactant to the other, reaction tends to be much slower dueto significant activation energies associated with the atom transfer process. The mechanism of such reactions are inner-sphereelectron transfers, in which the composition of the coordination sphere of a complex changes during the reaction:

[CoCl(NH3)5]2+ + [Cr(OH2)6]

2+ + 5 H2O + 5 H+ [Co(OH2)6]

2+ + [CrCl(OH2)5]2+ + 5NH4

+


15/18


The inner-sphere mechanism is common for redox reactions involving oxoanions. For example, the reduction ofoxoanionsby NO2

-occurs by attack of N on the oxygen atom of the oxoanion:NO2

- + OCl- NO3 + Cl-

The rate of reduction are found to be:ClO4

-< ClO3-< ClO2

-< ClO-and

ClO4-< SO4

2-< HPO42-

Thus the lowerthe oxidation state of the central atom, the faster the reaction is found to be. Why? Because the O-E bond istrongestfor the highest oxidation state, and this bond must be broken for the reaction to proceed. Further evidence in favoof this hypothesis is the effect of size:

ClO4-< BrO4

-< IO4-

The strength of the bond is reduced as the central atom size is increased.

5.4.4 Noncomplementary redox reactions

Very slow kinetics are observed when the change in oxidation states in the oxidizing and reducing agent are not the same,e.g.:

2 Fe3+ + Tl+ 2 Fe2+ + Tl3+If this reaction is to proceed stepwise, then one Fe3+reacts with one Tl+to give a very unfavorable Tl2+ion. Alternatively thereaction must be fully termolecular, requiring an activated complex containing two Fe3+ ions and a Tl+ ion. This is also veryunlikely, so the reaction tends to be slow.

5.5 Pourbaix diagrams

We now wish to discuss the fate of the metallic elements in aqueous solution in somewhat more detail. This ties in with a loof the work in the lab so far, and has important implications particularly for environmental chemistry.

5.5.1 Oxidation of the elements by water

The basic reaction we will consider is:M + H2O M

+ + H2 + OH-

This is the reaction involved in oxidation of metals by water, i.e. the process commonly known as rusting. The tables ostandard reduction potentials tells us for which metals this reaction goes in 1 M H 3O

+. We must allow for the overpotential~0.6 V. We must also correct for pH if we want to discuss neutral solutions!

Consider Mg: E = -2.37 V is the standard reduction potential in 1Macid, i.e.

Mg + 2H3O+

Mg2+

+ 2H2O + H2 E = +2.37Now apply the Nernst equation:

2

220

]H[

]Mg][H[log

2

0591.0EE

+

+

=

For pH 7 this becomes: 956.1)10(

1log

2

0591.037.2E

27 +==

So here is a reaction that has enough voltage to overcome the overpotential, so long as a fresh surface is kept exposed tothe water such that the oxidation can proceed. This latter condition is what protects Al: a tough coating of Al2O3protects

bulk aluminum from air and water oxidation. This is also why the Mg had to be put in boiling water in the lab for it to react aappreciable rates. A good rule of thumb then is a metal with a reduction potential of ~ {0.4 + 0.6} = 1V or greater wiloxidize in water at appreciable rates in the absence of air. NB: dissolved O2will change this picture, of course! We all knowthat iron rusts in aerated water.)

5.5.2 Reduction of elements by waterThe redox reaction involved in acid solution is:

O2 + 4 H+ + 4 e- 2 H2O E = +1.23 V

In basic solution it becomes:O2 + 2 H2O + 4 e

- 4 OH- E= +0.40 VSome strong oxidizing agents, e.g. Co3+, are reducedby water. The overall reaction in acid is:

4 Co3+ + 2 H2O 4 Co2+ + O2 + 4 H

+ E = 0.59 VThis reaction is thus at the overpotential boundary. It becomes fully favored in basic solution, with E = 1.42 V


16/18


5.5.3 Stability field of water

We can combine the reduction and oxidation of water with the pHdependence and construct a diagram which represents the stability fieldfor water. This is shown in the diagram at right.

The region boxed in the middle is the normal range found fornatural waters, in which water is not oxidized or reduced, and those arethe pH ranges found in the various types of natural waters. SpecificE/pH zones for different kinds of environments are indicated by the

circles drawn into the stability field diagram.Well aerated natural waters near the surface contain enough

dissolved oxygen to get close to the oxidation of water to O2.Eutrophic lake water contains sufficient dissolved organic matter toapproach closely the H+ reduction line. Ocean waters are relatively

basic, and may be oxidizing if saturated in dioxygen, or reducing ifsaturated in organic matter (i.e. in stagnant lagoons, etc.)

Fresh waters are considerably more acidic (because of dissolvedcarbon dioxide), but again they can be oxidizing if saturated in oxygen,or reducing if too much organic matter is consuming all the oxygen.This acidity is greatly enhanced in bogs and organic-laden soils, due tohigh humic acid content. (Humic acids are complex organic acidsoccurring in the soil and in bituminous substances formed by the

decomposition of dead vegetable matter.)These are strongly reducing conditions, and explain the formationof CH4in marshes. Methane was first discovered from this source; Alessandro Volta was one of the first to identify this gasAmmonia and hydrogen sulfide as well as the inflammable phosphine, PH3, can also emanate from swamps. (There are manyrumors of eerie glows emanating in misty swamps; such tales are likely rooted in phosphine being released and burning abovethe surface of the waters.) If you remember that bogs and marshes release such compounds, in other words very reducedchemical compounds, it is should help you to remember the acidic, reducing character of bog water.

5.5.4 Pourbaix diagrams

The final type of diagram we want to consider for redox chemistry is a type of combination redox/pH predominancediagram developed by the French electrochemistry Pourbaix. The diagrams are usually named after him. These diagrams areclosely related to the stability field for water just discussed. Indeed, most Pourbaix diagrams include either the main E/pHlines from the water diagram, or both the main and overpotential lines. What these diagrams add are the relationships between

the redox activity and the Brnsted acidity of the elements. They are thus related both to Latimer diagrams, and topredominance diagramsfor acid/base reactions.

Consider as an example the Pourbaix diagram foriron. Note that some of the lines from the stability fieldof water are drawn into this diagram, since this diagramapplies to aqueous solutions of iron compounds.

The vertical axis plots the standard reductionpotential, and the horizontal the pH. Remember whata predominance diagram was: simple vertical

boundaries where the most abundant speciesaltered.

The bottom of the diagram refers to reduced species,

i.e. Fe(s), or to the type of conditions that lead toreduction.

The top of the diagram to oxidized species and/oroxidizing conditions.

Vertical lines indicate changes in acid base chemistryindependent of E, e.g. Fe3+/Fe(OH)3.

Horizontal lines indicate redox changes unaffectedby pH, e.g. Fe/Fe2+below pH 6.

In the more general case, the lines slope, since both Eand [H+] or [OH-] affect the redox process.

+1.6

+1.2

+0.8

+0.4

-0.4

-0.8

0

2 4 6 8 10

pH

pH = 9

pH = 4

O2/H2O

H2O/H2

With overpotential

With

overpotential

The stability field for water

Fresh surface water

Ocean water

Organic-richlake water

Organic-richwaterloggedsoils

Organic-richsaline water

Bogwater

E(V)


17/18


To test your ability to read Pourbaix diagrams, see if you can find answers to the following:(a) The form of iron which is the strongest oxidizing agent: FeO4

2-at [H+] 1 M(b) The form of iron which is the strongest reducing agent: Fe(0), elemental iron(c) The predominant form at pH = 7 and E = 0.0 V: Fe(OH)3predominates, but close to Fe

2+/Fe(OH)2(d) Efor (acid) reduction of FeO4

2-to Fe3+: 1 M, 0 pH = 0 so E = 2.2 V(e) Efor reduction of Fe

2+to Fe(s): E = 0.5 V (This MUST be an acid process. Why?)

On the diagram, dashed lines dand erepresent respectively the normal and overpotentialfor oxygen evolution according to:

2 H2O

4 H

+

(aq) + O2 + 4 e

-

E

= +1.229 V.The actual E for hydrogen evolution is given byf, while the overpotential is given by line g:2H++ 2e- H2 E = 0.00 V

5.5.6 How elements behave in natural waters

We can now take any of the other Pourbaix diagrams, and compare them to the natural water limits, and predict whatforms may exist in variousenvironments. Severallanthanide and actinideelement Pourbaix diagramsare shown in the figure atright. What form will Yb take in

natural lake water?Answer: Yb(OH)3

Can uranium besolubilized in sea water?Answer: Yes as UO2

2+. Would you expect to find

cerium metal free innature?Answer:

Plutonium is highly toxic,as well as being stronglyradioactive. What wouldhappen if plutonium was

released into a lake or astream?Answer:

Imagine that plutoniumoxides from a nuclearweapons processing centrehad been dumped into asmall, well aerated lakewhere the pH = 6-8 and E= 0.0-0.5 V. Over thecourse of 20 years, thelake converted to a bog,where the pH = 4 and E =

0.1 V. Discuss theenvironmental concerns inthe initial stages and thefinal stages of the lake"storage", rememberingthat plutonium would be aserious toxic hazard if itentered the food stream.Answer:


18/18


Note that there are some limitations to this approach. First of all, the concentration of the elements in natural waters isoften much lower than standard conditions. Furthermore, these data are only valid for pure water plus the element inquestion. For example, on this basis we would say gold cannot exist in sea water. However, it does, as a chloro complexwhich is soluble. This is due to a complexation equilibrium. Nevertheless the combined E/pH diagrams provide a verycomprehensive insight into the behavior of the elements in aqueous solution, and this is clearly the most important set ofconditions relevant to the terrestrial environment.

3830 lecture notes part4_2008_redox

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