7. Redox chemistry and multiple oxidation states

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1 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page Redox chemistry and multiple oxidation states So 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 the chemistry of the p-block elements. Redox chemistry is probably the most interesting (and complex) aspect of the chemical behavior of the elements, and careful study of this material will greatly reward the effort expended. The majority of the reactions 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. A short recap of the some of this material is found in section Review of basic principles from Chemistry 2000 Oxidation-Reduction reactions, better known by the short-form redox reactions, involve the transfer of electrons between atoms and molecules during the reaction. They represent one of the major classes of chemical reactions introduced in the text in section We can detect redox reactions by monitoring the oxidation states of the atoms in the reactants and products. Any change in these formal oxidation states during the reactions means that a redox reaction has taken place. Oxidation states were introduced in section 4.10, and the problems in chapter 4 include many exercises on assigning oxidation states. You may wish to review this section. A summary of these rules is given here: Guidelines for Determining Oxidation Numbers 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 I 2 or S For ions consisting of a single atom, the oxidation number is equal to the charge on the ion. Elements of periodic Groups 1, 2 and 13 form monatomic ions with a positive charge and oxidation number equal to the group number. Aluminum therefore forms Al 3+, and its oxidation number is +3. (See Section 3.3.) 3. Fluorine is always 1 in compounds with other elements. 4. Cl, Br, and I are always 1 in compounds except when combined with O or F. This means that Cl has an oxidation number of 1 in NaCl (in which Na is +1, as predicted by the fact that it is an element of Group 1). In the ion ClO the Cl atom has an oxidation number of +1 (and O has an oxidation number of 2; see Guideline 5). 5. The oxidation number of H is +1 and of O is 2 in most compounds. Although this statement applies to many, many compounds, a few important exceptions occur. When H forms a binary compound with a metal, the metal forms a positive ion and H becomes a hydride ion, H. Thus, in CaH 2 the oxidation number of Ca is +2 (equal to the group number) and that of H is 1. Oxygen can have an oxidation number of 1 in a class of compounds called peroxides, compounds based on the O 2 2 ion. For example, in H 2 O 2, hydrogen peroxide, H is assigned its usual oxidation number of +1, and so O is The algebraic sum of the oxidation numbers in a neutral compound must be zero; in a polyatomic ion, the sum must be equal to the ion charge. Examples of this rule are the previous compounds and others found in Example 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 X n+ + n e If Y gains one or more electrons, it is reduced, and is the oxidizing agent. Y + n e Y n These 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 oxidation comes from the fact that combination with oxygen (e.g. in combustion) is one common form of an oxidation reaction. The element oxidized loses electrons to the oxygen atom. For example, when magnesium is burnt: while oxygen gains electrons: and the net reaction is: Mg Mg e ½ O e O 2 Mg + ½ O 2 MgO

2 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page 107 Reduction originates in the concept of reducing an 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: Fe Fe e 3 Fe and the carbon monoxide reaction is: 4 CO 4 CO e Here the oxide ions are spectators, transferring from the iron to the oxidized carbon. The net reaction therefore becomes: Fe 3 O CO 3 Fe + 4 CO 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/Mg 2+ reaction is an oxidation half reaction. In such reactions, electrons are always among the products. The CO/CO 2+ reaction is also an oxidation half reaction. The other two are reduction half-reactions, for which electrons will always be reactants. Although simple reaction such as these are easy to balance at sight, and we often don t stop to think that they are in fact redox reactions, more complicated redox reactions need a rigorous approach for successful balancing. We will concentrate on reactions that take place in solution, and these will usually be either basic or acidic solutions. Since electrochemistry is a branch of thermodynamics, we usually will be dealing with reactions under conditions of standard state, which for solution chemistry means 1 M concentrations at 25 C. Thus acidic solutions will have [H 3 O + ] = 1.00 M, while basic solutions will have [OH ] = 1.00 M. The usual K w relationship between hydroxide and hydronium ion will always hold. The rules for balancing redox reactions are as follows. These are better learned by doing many examples than by actual memorization of the rules! Step 1. Recognize the reaction as an oxidation-reduction. Step 2. Separate the overall process into half-reactions. Step 3. Balance each half-reaction for mass. In acid solution, add H 2 O to the side requiring O atoms. Then add H + to balance any remaining unbalanced H atoms. In basic solution, add 2 OH to the side requiring O atoms, and H 2 O to the other side. Then add H 2 O to the side requiring H atoms, and one OH to the other side. Step 4. Balance the half-reactions for charge. Step 5. Multiply the balanced half-reactions by appropriate factors. Step 6. Add the balanced half-reactions. Step 7. Eliminate common reactants and products. Step 8. Check the final result for mass and charge balance Some examples of balancing redox reaction equations The following reaction takes place in standard acid solution (1 M H + ) MnO 4 (aq) + HSO 3 (aq) Mn 2+ (aq) + SO 2 4 (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: HSO 3 2 SO 4 Mass balance HSO 3 + H 2 O 2 SO H + Charge balance HSO 3 + H 2 O 2 SO H e Multiply by 5 5 HSO H 2 O 2 5 SO H e Reduction: MnO 4 Mn 2+ Mass balance MnO H + Mn H 2 O Charge balance MnO H e Mn H 2 O Multiply by 2 2 MnO H e 2 Mn H 2 O Overall reaction: 2 MnO 4 (aq) + 5 HSO 3 (aq) + H + (aq) 2 Mn 2+ (aq) + 5 SO 2 4 (aq) + 3 H 2 O Finally, check that the atoms and charges balance at the right and at the left The following reaction takes place in standard acid solution (1 M H + ) Fe(OH) 2 (s) + CrO 2 4 (aq) Fe 2 O 3 (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 Fe 2 O 3 Mass balance 2 Fe(OH) OH Fe 2 O H 2 O Charge balance 2 Fe(OH) OH Fe 2 O H 2 O + 2 e

3 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page 108 Multiply by 3 6 Fe(OH) OH 3 Fe 2 O H 2 O + 6 e Reduction: 2 CrO 4 Cr(OH) 4 Mass balance 2 CrO H 2 O Cr(OH) OH Charge balance 2 CrO H 2 O + 3 e Cr(OH) OH Multiply by CrO H 2 O + 6 e 2 Cr(OH) OH Overall reaction: 6 Fe(OH) 2 (s) + 2 CrO 2 4 (aq) + 2 H 2 O 3 Fe 2 O 3 (s) + 2 Cr(OH) 4 (aq) + 2 OH Finally, check that the atoms and charges balance at the right and at the left Spontaneous reactions In the last section, we learned to use the Gibbs Free Energy of reaction to tell us whether a certain reaction is product favored or reactant-favored in the forward direction. The electrochemical analogue to this concept is called the cell potential given by the symbol E. E is expressed in the common electrical unit of volts, but is really just the Gibbs free energy in disguise! The exact relationship between cell potential and Gibbs energy is given by the relationship: G = nfe in general, and for standard conditions G = nfe 0 0 Because of the minus sign in the equation, the cell voltage E has the opposite sign convention to that of G: Product Favored Reaction: ve G or +ve E Reactant Favored Reaction: +ve G or ve E Why do we have such a confusing convention? The origin of this discrepancy comes from the way that physicists define current flow: although electrical current is entirely due to the migrations of electrons in a circuit, a forward current flow is defined in standard theories of electricity as the direction of positive charge flow. It is when harmonizing standard electrical conventions with chemical definitions that these apparently contradictory conventions are generated. In electrochemistry, a reaction which is product favored produces a voltage, and is therefore called a Voltaic cell, in honor of Alessandro Volte. A reaction which is reactant favored can be driven forward by the application of a greater opposite voltage; such reactions are called electrolytic cells, and the process is named electrolysis. We now want to remember another aspect of the meaning of G, which is that it tells for any given reaction the maximum amount of the total energy change that may be harnessed for useful work. The electrochemical equivalent of this idea is the maximum voltage of the electrochemical cell. This maximum potential is obtainable only from a fully charged cell running under zero load. This means that we can calculate the maximum cell voltage from standard free energies for the reaction. Consider the reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) G f (kj mol 1 ) Since G rxn = Σ{ G f (products)} Σ{ G f (reactants)}, G rxn = (-147.0) (65.52) = kj mol 1, and we would predict this to be a spontaneous reaction. It is spontaneous, but frankly quite useless unless you need some finely divided metallic copper and have only these reactants at hand. Let us now see if it is indeed possible to harness the useful work from this reaction The electrochemical cell To do this we must perform exactly the same chemical reaction but using an electrochemical cell. A typical cell-design is shown in the figure below. It divides the redox reaction into to compartments, each of which contains one of the two redox halfreactions. What voltage does our device measure: What voltage should it measure? Well, G rxn = kj mol 1, and the value of n = 2, as can be seen by breaking the reaction into its constituent electrochemical half-reactions: Zn(s) Zn 2+ (aq) + 2 e Oxidation Cu 2+ (aq) + 2 e Cu(s) Reduction

4 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page 109 and so: E 0 0 G 2125kJ = =. = V nf kC Any discrepancy between the actual measured potential of a given cell and this value is usually due to non-standard conditions, primarily that the concentrations are not exactly one Molar Electrode or Half-cell potentials Diagram of an electrochemical cell to harness the Zn/Cu reaction Although the reaction only requires a copper(ii) salt and metallic zinc, we have built a cell in which both metallic zinc and copper are present, and copper and zinc sulfate. The reason for this is to make the reactions reversible. It turns out that accurate potentials can only be measured for reversible electrochemical cells. Electrochemists have developed a short-hand notation to indicate the complete make-up of a given electrochemical cell. It is based on the balanced redox reaction, but includes both the reactants and the products. For our cell the correct notation would be: Zn(s) Zn 2+ (1 M) Cu 2+ (1 M) Cu(s) In this notation, the cathode reaction is always put on the right hand side, and the anode reaction at the left. The cathodic process is always reduction, while the anodic process is always oxidation. Thus the anode half-cell will always be written at the left hand side of the notation. A convenient mnemonic device to remember this convention is Right Red Cat, i.e. that the cathode process is a reduction process and is placed at the left of the cell notation. The vertical lines,, in the notation indicate a phase boundary, such as between a solid and a liquid. Double lines,, indicate double boundaries, such as commonly occur when a salt bridge is placed between the two half cells. If the components of a redox reaction co-exist in solution without a phase interface, they are listed together with a comma separating oxidized and reduced forms. This is often the case when the current is introduced into a solution via an inert electrode (usually platinum metal); an example is the oxidation of iron(ii) to iron (III) for which the notation could be: Pt(s) Fe 2+ (1 M), Fe 3+ (1 M).

5 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page 110 The operation of a voltaic cell is summarized in Figure It represents a complete electrical circuit. In the wires external to the cell, the current is carried exclusively by a moving stream of electrons. However, within the solution, the current must be carried by migrating ionic species, such as in our example cell Zn 2+ and Cu 2+ ions. Within the salt bridge, if one is used, inert ions such as K + and Cl carry the charges and provide for current flow. We all know from daily experience that the voltage and power output of a voltaic cell (an automobile battery or a dry cell) is adversely affected. We can now see at least one reason by low temperatures might have a deleterious effect: Standard Half-Cell Reduction Potentials From the construction of our electrochemical demonstration cell, and from the cell notation, it should now be obvious to you that electrochemical halfreactions, which we arbitrarily introduced as a tool to balance redox reactions, have some degree of physical reality they can after all be built into electrochemical halfcells. It should, for example, be possible to uncouple the copper/zinc cell and construct other voltaic cells from them, for example a copper/lithium or a silver/zinc cell, etc. This is indeed possible. It would also be nice to be able to assign a voltage to each half-cell, but this turns out to be impossible, since the voltage depends on a complete electrical circuit being established. Nonetheless, so attractive is this idea of a half-cell potential, that chemists have figured out a way to do so. We do it by arbitrarily setting one redox half-reaction to zero, and measuring all other cells with respect to this arbitrary zero reference point. The universal reference standard for electrochemistry is the standard hydrogen electrode, or SHE. For example, if the zinc half-cell is combined with the SHE, as shown in Figure 21.7, a voltage is

6 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page 111 measured as V. We assign this cell voltage entirely to the zinc half cell, since SHE is zero. Alternatively, we combine the SHE with the copper half cell, and get V. Then we can establish the potential of the copper zinc cell as E = V = 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 as reductions. This leads to the compilation of the standard reduction potentials in aqueous solution which are compiled in Table 21.1, and in greater detail in Appendix J. An extract of this table is provided in the notes below. We can use this table in numerous ways. First of all, we can now construct an electrochemical cell out of any combination of half cell reactions. Thus we can go and obtain potentials for the examples cited above: Table of Standard Reduction Potentials in Aqueous Solution at 25 C Reduction Half-Reaction E (V) F 2 (g) + 2 e 2 F (aq) H 2 O 2 (aq) + 2 H 3 O + (aq) + 2 e 4 H 2 O(l) PbO 2 (s) + SO 2 4 (aq) + 4 H 3 O + (aq) + 2 e PbSO4(s) + 6 H2O(l) MnO 4 (aq) + 8 H 3 O + (aq) + 5 e Mn 2+ (aq) + 12 H 2 O(l) Au 3+ (aq) + 3 e Au(s) Cl 2 (g) + 2 e 2 Cl (aq) Cr 2 O 2 7 (aq) + 14 H 3 O + (aq) + 6 e 2 Cr 3+ (aq) + 21 H 2 O(l) O 2 (g) + 4 H 3 O + (aq) + 4 e 6 H 2 O(l) Br 2 (l) + 2 e 2 Br (aq) NO 3 (aq) + 4 H 3 O + (aq) + 3 e NO(g) + 6 H 2 O(l) OCl (aq) + H 2 O(l) + 2 e Cl (aq) + 2 OH (aq) Hg 2+ (aq) + 2 e Hg(l) Ag + (aq) + e Ag(s) Hg 2+ 2 (aq) + 2 e 2 Hg(l) Fe 3+ (aq) + e Fe 2+ (aq) I 2 (s) + 2 e 2 I (aq) O 2 (g) + 2 H 2 O(l) + 4 e 4 OH (aq) Cu 2+ (aq) + 2 e Cu(s) Sn 4+ (aq) + 2 e Sn 2+ (aq) H 3 O + (aq) + 2 e fi H 2 (g) + 2 H 2 O(l) 0.00 Sn 2+ (aq) + 2 e Sn(s) 0.14 Ni 2+ (aq) + 2 e Ni(s) 0.25 V 3+ (aq) + e V 2+ (aq) PbSO 4 (s) + 2 e Pb(s) + SO 2 4 (aq) Cd 2+ (aq) + 2 e Cd(s) 0.40 Fe 2+ (aq) + 2 e Fe(s) Zn 2+ (aq) + 2 e Zn(s) H 2 O(l) + 2 e H 2 (g) + 2 OH (aq) Al 3+ (aq) + 3 e Al(s) 1.66 Mg 2+ (aq) + 2 e Mg(s) 2.37 Na + (aq) + e Na(s) K + (aq) + e K(s) Li + (aq) + e Li(s) In volts (V) versus the standard hydrogen electrode Constructing cell potentials from standard half-cell potentials copper/lithium cell From the table we get the half-reactions: Cu 2+ (aq) + 2 e Cu(s) E = V Li + (aq) + e Li(s) E = V We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the lithium equation, making it the anode (where the oxidation will take place.)

7 Chemistry 2810 Lecture Notes Dr. R. T. Boeré Page 112 Li(s) Li + (aq) + e E = V now we have the cell potential for the overall reaction Cu 2+ (aq) + 2 Li(s) Cu(s) + 2 Li + (aq) E cell = V Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in the value of n.