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Chapter 8

Chapter 8. Reduction and oxidation. TOPICS Redox reactions and oxidation states (an overview) Reduction potentials and Gibbs energy Disproportionation Potential diagrams Frost--Ebsworth diagrams The effect of complex formation or precipitation on M z+ /M reduction potentials

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Chapter 8

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  1. Chapter 8 Reduction and oxidation • TOPICS • Redox reactions and oxidation states (an overview) • Reduction potentials and Gibbs energy • Disproportionation • Potential diagrams • Frost--Ebsworth diagrams • The effect of complex formation or precipitation on • Mz+/M reduction potentials • Applications of redox reactions to industrial processes

  2. Oxidation and reduction Oxidation refers to gaining oxygen, losing hydrogen or losing one or more electrons. Reduction refers to losing oxygen, gaining hydrogen or gaining one or more electrons Magnesium acts as the reducing agent or reductant , while O2 acts as the oxidizing agent or oxidant. Redoxis an abbreviation for reduction–oxidation.

  3. In an electrolytic cell, the passage of an electrical current initiates a redox reaction In a galvanic cell, a spontaneous redox reaction occurs and generates an electrical current Oxidation states An oxidation processis accompanied by an increase in the oxidation state of the element involved; conversely, a decrease in the oxidation state corresponds to a reduction step.

  4. 8.2 Standard reduction potentials, Eo, and relationships between Eo, DGo and K Half-cells and galvanic cells The reaction is written as an equilibrium The Daniell cell is an example of a galvanic cell.

  5. Eocell is the standard cell potential.

  6. The standard cell potential (at 298 K) for the Daniell cell is 1.10 V. Calculate the corresponding values of DGo and K and comment on the thermodynamic viability of the cell reaction:

  7. Defining and using standard reduction potentials, Eo For example, if Zn metal is placed into dilute acid, H2 is evolved. Thus, when the standard hydrogen electrode is connected in a galvanic cell with a Zn2+/Zn electrode, The following reaction is the spontaneous cell process.

  8. The following two half-reactions correspond to two half-cells that are combined to form an electrochemical cell: (a) What is the spontaneous cell reaction? (b) Calculate Eocell. (a) First, look up values of Eo for the half-reactions. (b) The cell potential difference is the difference between the standard reduction potentials of the two half-cells:

  9. For example, to investigate the reaction between Fe and aqueous Cl2, we consider redox couples 8.16–8.18. For reaction 8.19 (where z = 2), DGo = 347 kJ per mole of reaction, while for reaction 8.20 (z = 6), DGo =  810 kJ per mole of reaction. Per mole of Fe, the values of DGo =  347 and  810 kJ , revealing that reaction 8.20 is thermodynamically favoured over reaction 8.19.

  10. Dependence of reduction potentials on cell conditions

  11. Application of the Nernst equation to the Zn2+/Zn half-cell ( Eo =  0.76 V) gives E =  0.79 V for [Zn2+] = 0.10 mol dm3 The more negative value of E , corresponding to a more positive value of DG, signifies that it is more difficult to reduce Zn2+ at the lower concentration Now consider the effect of pH (pH=  log[H+]) on the oxidizing ability of [MnO4]in aqueous solution at 298 K. E = Eo when [H+] =1 mol dm3, and [Mn2+] = [MnO4] =1 mol dm3. As [H+] increases (i.e. the pH of the solution is lowered), the value of E becomes more positive. The fact that the oxidizing power of [MnO4] is lower in dilute acid than in concentrated acid explains why, for example, [MnO4] will not oxidize Cl in neutral solution, but liberates Cl2 from concentrated HCl.

  12. Worked example 8.4 Oxidation of Cr2+ ions in O2-free, acidic, aqueous solution Explain why an acidic, aqueous solution of Cr2+ ions liberates H2 from solution (assume standard conditions). What will be the effect of raising the pH of the solution?

  13. Raising the pH of the solution lowers the concentration of H+ ions. Let us consider a value of pH 3.0 with the ratio [Cr3+] : [Cr2+] remaining equal to 1. The 2H+ = H2 electrode now has a new reduction potential. Now we must consider the following combination of half-cells, taking Cr3+ =Cr2+ still to be under standard conditions Thus, although the reaction still has a negative value of DG , the increase in pH has made the oxidation of Cr2+ less thermodynamically favourable.

  14. 8.3 The effect of complex formation or precipitation on Mz+/M reduction potentials Half-cells involving silver halides Under standard conditions, Ag+ ions are reduced to Ag(equation 8.29), but if the concentration of Ag+ is lowered, application of the Nernst equation shows that the reduction potential becomes less positive (i.e. DG is less negative). In practice, a lower concentration of Ag+ ions can be achieved by dilution of the aqueous solution, but it can also be brought about by removal of Ag+ ions from solution by the formation of a stable complex or by precipitation of a sparingly soluble salt.

  15. Reduction of Ag(I) when it is in the form of solid AgCl occurs according to reaction 8.31, and the relationship between equilibria 8.29–8.31 allows us to find, by difference, DGo for reaction 8.31. This leads to a value of Eo = + 0.22 V for this half-cell The difference in values of Eo for half-reactions 8.29 and 8.31 indicates that it is less easy to reduce Ag(I) in the form of solid AgCl than as hydrated Ag+. Silver iodide ( Ksp = 8.51x10 17) is less soluble than AgCl in aqueous solution, and so reduction of Ag(I) in the form of solid AgI is thermodynamically less favourable than reduction of AgCl. However, AgI is much more soluble in aqueous KI than AgCl is in aqueous KCl solution. The species present in the iodide solution is the complex [AgI3]2,

  16. Modifying the relative stabilities of different oxidation states of a metal consider the Mn3+/Mn2+ couple, for which equation 8.34 is appropriate for aqua species. In alkaline solution, both metal ions are precipitated, but Mn(III) much more completely than Mn(II) since values of Ksp for Mn(OH)3 and Mn(OH)2 are =1036 and = 2x10 13, respectively. Precipitation has the effect of significantly changing the half-cell potential for the reduction of Mn(III). In solutions in which [OH]= 1M, Mn(III) is stabilized with respect to reduction to Mn(II) as the value of Eo[OH]=1 for equation 8.35 illustrates. Compare this with equation 8.34.

  17. First, find the half-equations that are relevant to the question; note that pH 0 corresponds to standard conditions in which = [H+] 1 mol dm3.

  18. Most d- block metals resemble Mn in that higher oxidation states are more stable (with respect to reduction) in alkaline rather than acidic solutions. This follows from the fact that the hydroxide of the metal in its higher oxidation state is much less soluble than the hydroxide of the metal in its lower oxidation state. The following equations show the reduction of Co3+ The overall formation constant for [Co(NH3)6]3+ =1030

  19. A similar comparison can be made for the reduction of the hexaaqua ion of Fe3+ and the cyano complex (equations 8.38 and 8.39), and leads to the conclusion that the overall formation constant for [Fe(CN)6]3 is =107 times greater

  20. 8.4 Disproportionation reactions

  21. Worked example 8.6 Disproportionation of copper(I) Using appropriate data from Table 8.1, determine K (at 298 K) for the equilibrium: Three redox couples in Table 8.1 involve Cu(I), Cu(II) and Cu metal: The disproportionation of Cu(I) is the result of combining half-reactions (1) and (3). Thus The value indicates that disproportionation is thermodynamically favourable.

  22. 8.5 Potential diagrams Consider manganese as an example. Aqueous solution species may contain manganese in oxidation states ranging from Mn(II) to Mn(VII), and equations 8.42–8.46 give half-reactions for which standard reduction potentials can be determined experimentally. These potentials may be used to derive values of Eo for other half-reactions such as 8.47.

  23. Potential Diagrams – Latimer Diagrams

  24. The diagram corresponding to the two half-reactions Combining these two half-cells give reaction 8.48 for which Eocell = 1.20 V and DGo (298K) =  231 kJmol1 This indicates that reaction 8.48 is spontaneous. Similarly, at pH 0, Mn3+ is unstable with respect to disproportionation to MnO2 and Mn2+

  25. Worked example 8.7 Potential diagrams The following potential diagram summarizes some of the redox chemistry of iron in aqueous solution. Calculate the value of Eo for the reduction of Fe3+(aq) to iron metal. Although there are short cuts to this problem, the most rigorous method is to determine DGo (298K) for each step. Fe3+ to Fe2+is a one-electron reduction.

  26. 8.6 Frost-Ebsworth diagrams Frost--Ebsworth diagrams and their relationship to potential diagrams In a Frost-Ebsworth diagram, values of Go or, more commonly, Go/F for formation M(N) from M(0), where N is the oxidation state, are plotted againest increasing N, From the relationship:

  27. Fig. 8.4 Frost–Ebsworth diagrams in aqueous solution at pH 0, i.e. [H+] =1 mol dm3, for (a) chromium

  28. Fig. 8.4 Frost–Ebsworth diagrams in aqueous solution at pH 0, i.e. [H+] =1 mol dm3, for (b) phosphorus

  29. Fig. 8.4 Frost–Ebsworth diagrams in aqueous solution at pH 0, i.e. [H+] =1 mol dm3, for (c) nitrogen

  30. 8.7 The relationships between standard reduction potentials and some other quantities

  31. Worked example 8.9 Determination of DsolGo for an ionic salt

  32. Ellingham Diagram

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