Inorganic Chemistry Textbook PDF - Weller/Armstrong 7th Edition

Our textbook: Weller/Armstrong 7th edition Printed-Book Inorganic Chemistry by Mark Weller, Tina Overton, Jonathon Rourke & Fraser Armstrong. Oxford University Press, Publishers, 7th edition, 2018, ISBN-978-0-19- 876812-8....

Our textbook: Weller/Armstrong 7th edition Printed-Book Inorganic Chemistry by Mark Weller, Tina Overton, Jonathon Rourke & Fraser Armstrong. Oxford University Press, Publishers, 7th edition, 2018, ISBN-978-0-19- 876812-8. Dr. Yasser Hassan Assistant Professor of Inorganic Chemistry Chemistry Department, Qatar University Email: [email protected] 1  COLLEGE OF ARTS AND SCIENCES DEPARTMENT OF CHEMISTRY AND EARTH SCIENCES CHEMISTRY 221: Inorganic Chemistry (I) Chapter 6 Oxidation and Reduction Dr. Yasser Hassan 2 Oxidation and reduction: Chapter Outlines  Redox half reactions;  Standard potentials & spontaneity;  The electrochemical series;  Reaction in water;  Oxidation by atmospheric oxygen;  The influence of complexation;  Frost diagrams Oxidation and Reduction Reactions  Reaction that involves transferring electrons from one atom to another  These are called oxidation-reduction reactions  Also known as redox reactions  The element that loses electrons in the reaction is oxidized.  The substance that gains electrons in the reaction is reduced  You cannot have one without the other 5 Oxidation and Reduction Reactions Oxidation-Reduction Reactions: It is important to note that many redox reactions occur without the reactants being dissolved in water. 6 Oxidation and Reduction Reactions OIL RIG “OIL” = Oxidation is Loss of Electrons “RIG” Reduction is Gain of Electrons 7  Oxidation-Reduction Reactions oxygen removed Old definition reduction lead carbon oxide + carbon  lead + monoxide oxygen added oxidation Which substances are oxidized and reduced in this reaction? Reduction and oxidation always take place together. redox = reduction and oxidation magnesium + oxygen  magnesium oxide 2Mg(s) + O2(g)  2MgO(s) The term reduction comes from the Latin stem meaning "to lead back." Anything that that leads back to magnesium metal therefore involves reduction. 8 Recognition of Redox Reactions 1. Any reaction that has an element that is uncombined on one side and combined on the other is a redox reaction – Uncombined = Free element 2CO + O2 → 2 CO2 2N2O5 → 4 NO2 + O2 3C + Fe2O3 → 3 CO + 2 Fe Mg + Cl2 → MgCl2 2. Any reaction where a cation changes charge is redox CuCl + FeCl3 → FeCl2 + CuCl2 SnCl2 + F2 → SnCl2F2 Redox reactions describe all chemical reactions in which there is a net change in atomic charge 9 Oxidizing and Reducing Agents A species is oxidized when it loses electrons. – Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion. What is reduced is the oxidizing agent. A species is reduced when it gains H+ oxidizes Zn by taking electrons from it. electrons. – Here, each of the H+ gains an electron, What is oxidized is the reducing agent. and they combine to form H2. Zn reduces H+ by giving it electrons. 10 Oxidation-Reduction Reactions (Electron Transfer Reactions) 1 Ca(s) is oxidized (loses electrons) to become Ca2+ O2(g) is reduced (gains electrons) to become O2- 11 Oxidation-Reduction Reactions (Electron Transfer Reactions) 2 Zn 𝑠𝑠 + CuSO4 (𝑎𝑎𝑎𝑎) → ZnSO4 𝑎𝑎𝑎𝑎 + Cu (𝑠𝑠) Zn → Zn2+ + 2e− Zn is oxidized Zn is the reducing agent Cu2+ + 2e− → Cu Cu2+ is reduced Cu2+ is the oxidizing agent Reducing agent: Oxidizing agent: – Causes reduction – Causes oxidation – Looses electrons – Gains electrons 12 Oxidizing and Reducing Agents Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s) Zn Zn2+ + 2e- Zn is oxidized Zn is the reducing agent Cu2+ + 2e- Cu Cu2+ is reduced Cu2+ is the oxidizing agent Copper wire reacts with silver nitrate to form silver metal. What is the oxidizing agent in the reaction? Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s) Cu Cu2+ + 2e- Cu is the ?? agent Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent 13 Many redox reactions occur between reactants in the same physical state. Some examples are In gases:  2NO(g) + O2(g) → 2NO2(g)  2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g) In solution:  Fe3+(aq) + Cr2+(aq) → Fe2+(aq) + Cr3+(aq)  3CH3CH2OH(aq) + 2CrO42−(aq) + 10H+(aq) → 3CH3CHO(aq) + 2Cr3+(aq) + 8H2O(l) In biological systems:  ‘Mn4’(V,IV,IV,IV) + 2 H2O(l) → ‘Mn4’(IV,III,III,III) + 4 H+(aq) + O2(g) the production of O2 from water by a MnCa4O5 cofactor contained in one of the photosynthetic complexes of plants In solids:  LiCoO2(s) + 6C(s) → LiC6(s) + CoO2(s) lithium-ion battery during charging and its reverse takes place during discharge  CeO2(s) heat CeO2−δ(s) + δ/2 O2(g) followed by CeO2−δ(s) + δH2O → CeO2(s) + δH2(g) Solar thermal water splitting! 14 Oxidation Numbers The definitions of oxidation and reduction (loss and gain of electrons) apply to the formation of ionic compounds. However, these definitions do not accurately characterize the formation of molecular compounds. H2 (g) + Cl2 (g) → 2 HCl (g) S (g) + O2 (g) → SO2 (g) To keep track of electrons in redox reactions – oxidation numbers (also called oxidation state) are assigned to reactants and products. oxidation state = # electrons required to reduce or oxidize a compound to its elemental form 15 The Oxidation Numbers of Elements in their Compounds Oxidation number (O.N.) is also known as oxidation state  It is defined as the charge the atom would have if electrons were not shared but were transferred completely  For a binary ionic compound, the O.N. is equivalent to the ionic charge.  For covalent compounds or polyatomic ions, the O.N. is less obvious and can be determined by a given set of rules 16 The Oxidation Numbers of Elements in their Compounds Non-metals Transition metals 17 Oxidation Numbers The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 3. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1. 18 Oxidation Numbers 4. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2-, is –½. 5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. 6. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 7. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 19 Oxidation Numbers Test yourself What are the oxidation numbers of all the elements in HCO3- ? HCO3- O = –2 H = +1 3x(–2) + 1 + ? = –1 C = +4 20 Oxidation Numbers Test yourself What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7 21 What are the oxidation numbers of IF7 all the elements in each of these compounds? F = -1 NaIO3 IF7 K2Cr2O7 7x(-1) + ? = 0 NaIO3 I = +7 Na = +1 O = -2 K2Cr2O7 3x(-2) + 1 + ? = 0 O = -2 K = +1 I = +5 7x(-2) + 2x(+1) + 2x(?) = 0 Cr = +6 22  Test yourself Find charge (oxidation number) on S in following species: H2S H2S (−2), S2− S2− (−2) HS− HS− (−2) S8 S8 (0) SO2 SO2 (+4) SO3 SO3 (+6) H2SO4 H2SO4 (+6) 23 Test yourself Find charge on all elements in following species: Cs2Cr2O7 HClO4 BaNaPO4 K2TaF7 24 6.1 Redox half-reactions Redox reactions involve the transfer of electrons. 2Mg + O2 2MgO Equations written to show what happens to the electrons during oxidation and reduction are called half-equations. 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-) 2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e- 2Mg + O2 2MgO The oxidized and reduced species in a half-reaction constitute a redox couple. A couple is written with the oxidized species before the reduced, as in O2-/O2 and Mg/Mg2+, and typically the phases are not shown. 25 6.1 Redox half-reactions Combining half-reactions Write a balanced equation for the oxidation of Fe2+ by permanganate ions (MnO4−) in acid solution.  Many reactions occur in either acidic or basic solutions The Ion-Electron Method in Acidic Solution: 1) Divide the equation into two half-reactions. 2) Balance atoms other than H and O. 3) Balance O by adding water. 4) Balance H by adding hydrogen ion. 5) Balance net charge by adding electrons. 6) Make electron gain and loss equal: add half-reactions. 7) Cancel anything that’s the same on both sides of the equation. 26 6.1 Redox half-reactions Combining half-reactions Write a balanced equation for the oxidation of Fe2+ by permanganate ions (MnO4−) in acid solution. 27 6.1 Redox half-reactions Combining half-reactions Write a balanced equation for the Cr2O72– (oxidizing agent) in acidic media. 1. Balance all atoms (but not O or H yet) E.g. Cr2O72–  2Cr3+ 2. Balance O by adding H2O to the other side E.g. Cr2O72–  2Cr3+ + 7H2O 3. Balance H by adding H+ to the other side E.g. 14H+ + Cr2O72–  2Cr3+ + 7H2O 4. Balance charge by adding electrons (e-) E.g. 14H+ + 6e– + Cr2O72–  2Cr3+ + 7H2O 28 6.1 Redox half-reactions Combining half-reactions: another example There are three other chemical species available in an acidic solution: H2O H+ e¯ water is present because the reaction is taking place in solution the hydrogen ion is available because it is in acid solution electrons are available because that's what is transferred in redox reactions All three will be used in getting the final answer. 29 6.1 Redox half-reactions Combining half-reactions: What about in basic solutions?  The simplest way to balance reactions in basic solution is to first balance them as if they were in acidic solution, then “convert” to basic solution: Additional Steps for Basic Solutions 1. Add to both sides of the equation the same number of OH– ion as there are H+ 2. Combine H+ and OH– to form H2O 3. Cancel any H2O if it is on both sides. 30 6.1 Redox half-reactions Example: Balance the following in basic solution: MnO -4 + C 2 O 24- → MnO 2 + CO 32- ANALYSIS : Balance as if in acidic solution then " convert". SOLUTION : C 2 O 24- + 2H 2 O → 2CO 32- + 4H + + 2e - MnO -4 + 4H + + 3e - → MnO 2 + 2H 2 O Net ionic : 3C 2 O 24- + 2H 2 O + 2MnO -4 → 6CO 32− + 4H + + 2MnO 2 Add OH - 3C 2 O 24- + 2H 2 O + 4OH - + 2MnO -4 → 6CO 32− + 4(H + + OH − ) + 2MnO 2 Form H 2 O 3C 2 O 24- + 2H 2 O + 4OH - + 2MnO -4 → 6CO 32− + 4H 2 O + 2MnO 2 Simplify 3C 2 O 24- + 4OH - + 2MnO -4 → 6CO 32− + 2H 2 O + 2MnO 2 31 6.1 Redox half-reactions Combining half-reactions: another example There are three other chemical species available in a basic solution besides the ones shown above. They are: H2O OH¯ e¯ water is present because the reaction is taking place in solution the hydroxide ion is available because it is in basic solution electrons are available because that's what is transferred in redox reactions. All three will be used in getting the final answer. 32 6.2 Standard potentials and spontaneity  Thermodynamic arguments can be used to identify which reactions are spontaneous (i.e., have a natural tendency to occur).  The thermodynamic criterion of spontaneity is that, at constant temperature and pressure, the reaction Gibbs energy change, ΔrG, is negative.  The free energy change is a measure of the change in the total entropy of a system and its surroundings at constant pressure; spontaneous processes are accompanied by a decrease in free energy. 33  Thermodynamics Free Energy and ∆G  Free energy (G) values tell us if reactions will occur.  They do not tell us how fast reactions will occur.  During a reaction  Reactant particles must physically collide  They must collide with enough energy to break the bonds in the reactant.  Some reactions require the addition of heat energy. This gives the reactants the extra energy needed (more collisions, harder collisions) for this process to occur. © 2017 Pearson Education, Inc.  Thermodynamics Free Energy and ∆G  The energy exchanged during a chemical reaction is free energy (G), or Gibbs free energy.  This is the amount of energy present in molecules available to do work.  The free energy change, ∆G, is the energy difference between the free energy present in the product and that in the reactant molecules. ∆G = ∆H - T∆S H is the enthalpy ∆G < 0 The reaction is spontaneous. T is the temperature ∆G > 0 The reaction is nonspontaneous. S is the entropy ∆G = 0 The reaction mixture is at equilibrium. Entropy changes in the surrounding are primarily determined by heat flow. The enthalpy (H) concerns the system  Thermodynamics Because ∆G includes both heat and randomness, the terms exergonic (gives off energy) and endergonic (requires energy) are used. For an exergonic reaction, the free energy of the reactants is greater than the free energy of the products and ∆G is negative. Reactions with a negative ∆G value do not require energy input and occur spontaneously. Spontaneous processes will continue to occur once started and do not require energy from the surroundings. © 2017 Pearson Education, Inc.  Thermodynamics © 2017 Pearson Education, Inc.  Thermodynamics Reactions with a positive ∆G value require energy input from their surroundings and are nonspontaneous. Nonspontaneous processes do not occur naturally and require energy input. A canoe can go upstream only if the paddler inputs energy. © 2017 Pearson Education, Inc.  5.1 Thermodynamics Some reactions need a little “push” to get started. The energy necessary to align the reactant molecules and cause them to collide with enough force to react is the activation energy. © 2017 Pearson Education, Inc. 6.2 Standard potentials and spontaneity  It is usually sufficient to consider the standard reaction Gibbs energy, ΔrG⦵, which is related to the equilibrium constant in this equation: ⦵ =Plimsoll ΔrG = −RT ln K ⦵ ΔrG⦵ is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states  A negative value of ΔrG⦵ corresponds to K > 1  therefore, to a ‘favorable’ reaction in the sense that the products dominate the reactants at equilibrium. 40 Free Energy and Equilibrium ∆G = ∆Go + RT ln(Q) ∆Go = -RT ln(K) at equilibrium R = Universal gas constant = 8.314 J/K mol T = Temperature in Kelvin K = Equilibrium constant = [Pproducts]/[Preactants] System is at equilibrium (no net movement) when ∆Go = 0 (K = 1) If the system is NOT at equilibrium, use Q instead of K Galvanic Cell: Measuring Standard Reaction Gibbs Energies  Standard reaction Gibbs energies may be measured by setting up a galvanic cell, an electrochemical cell in which a chemical reaction is used to generate an electric current, in which the reaction driving the electric current through the external circuit is the reaction of interest The cathode is the electrode at which reduction occurs and the anode is the site of oxidation.  In practice, we must ensure that the cell is acting reversibly in a thermodynamic sense, which means that the potential difference must be measured with no current flowing. 42 Galvanic Cell: Measuring Standard Reaction Gibbs Energies The potential that corresponds to the ΔrG⦵ of a half-reaction is written E° , with ΔrG⦵ = -ν FE ⦵ where ν is the stoichiometric coefficient of the electrons transferred when the half-reactions are combined, and F is Faraday’s constant (F = 96.48 kC mol−1) The potential E⦵ is called the standard potential as shown in the table 6.2 for selected standard potentials at 298 K. Ecell = E high potential couple(OX) – E low potentials couple (red) ‘standard reduction potential’, to emphasize that by convention, the half-reaction is a reduction and written with the oxidized species and electrons on the left) 43 Galvanic Cell: Measuring Standard Reaction Gibbs Energies By convention, the specially chosen half-reaction is the reduction of hydrogen ions at pH = 0, 1 bar H2: H+(aq.) + e− ½ H2 (g) ΔrG⦵ = 0 (is arbitrarily set at zero) at all temperatures. Therefore: the standard potential of the H+/H2 couple is also zero at all temperatures. Note that the E⦵ values for couples (and their half-reactions) are called standard potentials and that their difference is denoted E⦵cell and called the standard cell potential. Thus: a reaction is favourable (in the sense K > 1) if the corresponding standard cell potential is positive 44 A short compilation of standard potentials values at 25 °c is given in table 6.2, this list is known as the electrochemical series. 45 The Standard Potential E˚ The tendency of an electron to migrate from one species to another is expressed in terms of the differences between their standard potentials.  The standard potential (E⦵) is a convenient metric for determining the relative strength of RedOx reactions at 1 M concentrations.  Think of E⦵ as, “How easily a reaction will happen.”  Very positive means very easy, very negative means very difficult.  You can change the sign of E˚ by reversing the reaction. Techniques such as cyclic voltammetry are used to measure reduction potentials under different conditions, and extract information about the reversibility and kinetics of electron-transfer reactions and influences of pH and complexation. 46 5.4 The electrochemical series: KEY POINTS The oxidized member of a couple is a strong oxidizing agent if E⦵ is positive and large; the reduced member is a strong reducing agent if E⦵ is negative and large. A short compilation of standard potentials values at 25 °c is given in table 6.2, this list is known as the electrochemical series. Ox/Red couple with strongly +ve E⦵ (OX is strongly oxidizing) Ox/Red couple with strongly –ve E⦵ (Red is strongly reducing) Important feature of the electrochemical series: The reduced member of any couple in the series has thermodynamic tendency to reduce the oxidized member of any couple that is lies above it in the series. See example 6.3 (page 192) 47  Cu2+(aq) + 2 e- →Cu(s) Zn2+(aq) + 2 e- →Zn(s) Example: Calculate the E⦵ cell = E⦵ cathode –E⦵Eanode standard cell potential for the following half reactions: We note from the values of the potentials that Cu2+ (couple with high potential) is the more oxidizing and would be reduced by the species of low potential Zn. 48 6.5 The Nernst equation and Equilibrium ∆G = ∆G⦵ + RT ln(Q) ∆G⦵ = -RT ln(K) at equilibrium ∆rG = ∆rG⦵ + RT ln Q ; where Q is the reaction quotient. [RedA]a’ [OXB]b’ aOxA + bRedB a’RedA + b’OXB &Q= [OxA]a [RedB]b 𝑅𝑅𝑅𝑅 Ecell = E⦵cell - ln(Q) 𝛾𝛾𝐹𝐹 At equilibrium Ecell = 0 and Q = K ν FE⦵cell The Nernst equation ln K = RT Redox stability Redox stability:  When assessing the thermodynamic stability of a species in solution, consider all possible reactants: the solvent, other solute, the species itself, and dissolved oxygen. 6.6 The influence of pH Self study! KEY POINT Many redox reactions in aqueous solution involve transfer of H+ as well as electrons and the electrode potential therefore depends on the pH.  At lower pH values, the oxidized form is predominant  As pH increases, the reduced form becomes more dominant.  The equilibrium between the two forms is pH-dependent (0.059 V)ν E = E′ - pH νe 50 Redox stability Redox stability:  When assessing the thermodynamic stability of a species in solution, consider all possible reactants: the solvent, other solute, the species itself, and dissolved oxygen. 6.7 Reactions with water: Water may act as an oxidizing agent, when it is reduced to H2: H2O(l) + 1e- → ½ H2(g) + OH–(aq) For the equivalent reduction of hydronium ions in water at any pH (and partial pressure of H2 of 1 bar) we have seen that the Nernst equation gives H+ (aq) + e– → ½ H2 (g) E = –0.059 V x pH This is the reaction that chemists typically have in mind when they refer to ‘the reduction of water’. 51 Redox stability 6.7 Reactions with water: Water may act as an oxidizing agent, when it is reduced to H2: H2O(l) + 1e- → ½ H2(g) + OH–(aq) It can also act as reducing agent, when it is oxidized to O2: 2 H2O(l) →O2(g) + 4H+(aq)  When the partial pressure of O2 is 1 bar, the Nernst equation for the O2, 4H+/2 H2O half- reaction becomes E = 1.23 V –(0.059 V x pH)  Both H+ and O2 therefore have the same pH dependence for their reduction half-reactions.  The variation of these two potentials with pH is shown in Fig. 6.3. Oxidation by water: The conversion of water into oxygen gas and protons.  The reaction of a metal with water or aqueous acid is in fact the oxidation of the metal by water or hydrogen ions because the overall reaction is one of the following processes. H2O(l) + M(s) →M+(aq) + 1/2H2(g) + OH-(aq) H+(aq) + M(s) →M+(aq) + 1/2H2(g)  These reactions are thermodynamically favorable when M is an s-block metal, a 3d-series metal from Group 3 to at least Group 8 or 9 and beyond (Ti, V, Cr, Mn, Ni), or a lanthanoid. An example from Group 3 is: 6H+(aq) + 2Sc(s) →2Sc3+(aq) + 3H2(g)  When the standard potential for the reduction of a metal ion to the metal is negative, the metal should undergo oxidation in 1 M acid with the evolution of hydrogen. Production of H2 by electrolysis or photolysis of water is widely viewed as one of the renewable energy solutions for the future 53 Oxidation by water:  Although the reactions of magnesium and aluminum with moist air are spontaneous, both metals can be used for years in the presence of water and oxygen. They survive because they are passivated, or protected against reaction, by an impervious film of oxide.  Magnesium oxide and aluminum oxide both form a protective skin on the parent metal beneath.  A similar passivation occurs with iron, copper, and zinc.  The process of ‘anodizing’ a metal, in which the metal is made an anode in an electrolytic cell, is one in which partial oxidation produces a smooth, hard passivating film on its surface. KEY POINT For metals with large, negative standard potentials, reaction with aqueous acids leads to the production of H2 unless a passivating oxide layer is formed. 54 Reduction by water: Reduction by water : The conversion of water into hydrogen gas and hydroxide ions. The strongly positive potential of the couple O2,4H+/2H2O shows that acidified water is a poor reducing agent except towards strong oxidizing agents. Example of a metal ion Co3+ that is reduced by water in the couple Co3+ /Co2+ E°=1.92 V. 2H2O(l) + 4Co3+(aq) →4Co2+(aq) + O2(g) + 4H+(aq) ; Ecell=+0.69  Because H+ ions are produced in the reaction, lower acidity (higher pH) favours the oxidation; lowering the concentration of H+ ions encourages the formation of the products.  Only a few oxidizing agents (such as Co3+(aq)) can oxidize water rapidly enough to give appreciable rates of O2 evolution.  Most important to have standard potentials greater than +1.23 V  Ce4+/Ce3+ (E⦵ = +1.76 V),  the acidified dichromate ion couple Cr2O72−/Cr3+ (E⦵ = + 1.38 V),  and the acidified permanganate couple MnO4−/Mn2+ (E⦵ = +1.51 V). it remains a challenge for inorganic chemists to find good catalysts for O2 evolution for the desire to generate H2 (a ‘green’ fuel) from water by electrolysis or photolysis 55 The stability field of water  A reducing agent that can reduce water to H2 rapidly, or an oxidizing agent that can oxidize water to O2 rapidly, cannot survive in aqueous solution. KEY POINT The stability field of water shows the region of pH and reduction potential where couples are neither oxidized by nor reduce hydrogen ions. Pourbaix diagram The stability field of water, is the range of values of potential and pH for which water is thermodynamically stable towards both oxidation and reduction. 56 The stability field of water  Remember the Nernst equation for H+ (aq) + e– → ½ H2 (g) E = –0.059 V x pH -----6.9 O2(g) + 4 H+(aq) + 4 e− → 2 H2O E =1.23V - (0.059V pH) -----6.10  Any species with a potential more negative than that given in eqn 6.9 can reduce water (specifically, can reduce H+) with the production of H2;  hence the lower line defines the low-potential boundary of the stability field.  Similarly, any species with a potential more positive than that given in eqn 6.10 can liberate O2 from water Pourbaix diagram  and the upper line gives the high-potential boundary. 57 6.8 Oxidation by atmospheric oxygen: The Statue of Liberty The oxidation of copper roofs to a green substance known as verdigris (typically ‘basic copper carbonate’) is an example of atmospheric oxidation in a damp environment. The familiar green surface is a passive layer of an almost impenetrable hydrated copper(II) carbonate, sulfate or, near the sea, chloride. 58 https://www.acs.org/pressroom/reactions/library/the-statue-of-libertys-true-colors.html 6.9 Disproportionation and comproportionation Disproportionation, a redox reaction in which the oxidation number of an element is simultaneously raised and lowered. 2 Cu+(aq) → Cu2+(aq) + Cu(s) Cu+(aq) + e− → Cu(s) E⦵ = +0.52 V Cu2+(aq) + e− → Cu+(aq) E⦵ = +0.16 V Because E⦵cell = 0.52 V − 0.16 V = +0.36 V for the disproportionation reaction, K = 1.3 × 106 at 298 K, so, the reaction is highly favourable. In comproportionation, the reverse of disproportionation, two species with the same element in different oxidation states form a product in which the element is in an intermediate oxidation state. Ag2+(aq) + Ag(s) → 2Ag+ (aq) E⦵cell=+1.18 V The large positive potential indicates that Ag(II) and Ag(0) are completely converted to Ag(I) in aqueous solution (K = 1 × 1020 at 298 K). 59 5.10 The influence of complexation  The formation of a more thermodynamically stable complex when the metal is in the higher oxidation state of a couple favors oxidation and makes the standard potential more negative;  the formation of a more stable complex when the metal is in the lower oxidation state of the couple favors reduction and the standard potential becomes more positive.  The formation of metal complexes affects standard potentials Mν+(aq) + e− → M(ν−1)+(aq) MLν+(aq) + e− → ML(ν−1)+(aq) because the ability of a complex (ML) formed by coordination of a ligand (L) to accept or release an electron differs from that of the corresponding aqua ion (M). 60 5.10 The influence of complexation The change in standard potential for the ML redox couple relative to that of M reflects the degree to which the ligand L coordinates more strongly to the oxidized or reduced form of M. In certain cases, the standard potential associated with particular oxidation states may be varied over more than two volts depending on the choice of ligand. For instance, the standard potential for one-electron reduction of Fe(III) complexes ranges between E > +1 V for L = bpy to E < −1 V when L is enterobactin 2,2′-bipyridine (bpy) 61 naturally occurring ligand enterobactin 5.10 The influence of complexation A change in standard potential due to complexation is analyzed by considering a generic thermodynamic cycle  Because the sum of reaction Gibbs energies round the cycle is zero, we can write – FE⦵ (M) – RT ln Kred + FE ⦵ (ML) + RT ln Kox = 0 where Kox and Kred are equilibrium constants for L binding to Mν+ and M(ν−1)+, respectively (of the form K = [ML]/[M][L]), and we have used ΔrG⦵ = −RT ln K in each case. RT Kox E⦵ (M) –E ⦵ (ML) = ln F Kred At 25°C, and with ln x = ln 10 log x Kox E⦵ (M) –E ⦵ (ML) = (0.059 V) log Kred Thus, every ten-fold increase in the equilibrium constant for ligand binding to Mν+ 62 compared to M(ν−1)+ decreases the reduction potential by 0.059 V. Diagrammatic presentation of potential data 6.13 Frost diagrams A Frost diagram (also known as an oxidation state diagram) of an element X is a plot of the relative free energy of a species νE⦵ for the couple X(N)/X(0) against the oxidation number, N, of the element.  (ν is the net number of electrons that is transferred to form each oxidation state, starting from N = 0)  Frost diagrams depict whether a particular species X(N) is a good oxidizing agent or reducing agent.  They also provide an important guide for identifying the oxidation states of an element that are inherently stable or unstable.  Graphically illustrate the stability of different The general form of a Frost diagram oxidation states relative to its elemental form (i.e., relative to oxidation state= 0) 63 6.13 Frost diagrams (a) Gibbs energies of formation for different oxidation states so, the position of a nitrogen compound  A Frost diagram shows how the on this graph gives us its stability Gibbs energies of formation of relative to N2 different oxidation states of an element vary with oxidation number.  The most stable oxidation state of an element corresponds to the species that lies lowest in its Frost diagram.  Frost diagrams are conveniently constructed by using electrode potential data.  Only NH4+(aq) is exergonic (ΔfG⦵< 0); FIGURE 6.6 The Frost diagram for nitrogen: the steeper the slope of a line, the higher the standard potential for the  all other species are endergonic redox couple. The red line refers to standard (acid) (ΔfG⦵ > 0) conditions (pH = 0), the blue line refers to pH = 14. 64 6.13 Frost diagrams (b) Interpretation KEY POINTS Frost diagrams may be used to gauge the inherent stabilities of different oxidation states of an element and to decide whether particular species are good oxidizing or reducing agents. The slope of a line connecting two species having different oxidation numbers is the reduction potential for that redox couple. FIGURE 6.8 The general structure of a region of a Frost diagram used to establish the relationship between the slope of a line connecting species having different oxidation numbers and the standard potential of the corresponding redox couple. 65 6.13 Frost diagrams (b) Interpretation  To interpret the qualitative information contained in a Frost diagram it is important to note (Fig. 6.8) that the slope of the line connecting two species having oxidation numbers N′′ and N′ is νE⦵/(N′ − N′′) = E⦵ (since ν = N′ − N′′). This simple rule leads to the following features 1. The steeper the line joining two points (left to right) in a Frost diagram, the more positive the standard potential of the corresponding couple (Fig. 6.9a). The slope of the lines connecting species on a Frost diagram is equal to the redox potential for The interpretation of a Frost diagram the rxn converting the species on the right to the to gauge (a) reduction potential species on the left 66 6.13 Frost diagrams (b) Interpretation the oxygen diagram  At the point corresponding to N = −1 (for H2O2), (−1) × E⦵ = −0.70 V,  and at N = −2 (for H2O), (−2) × −− E = −2.46 V.  The difference of the two values is −1.76 V.  The change in oxidation number of oxygen on going from H2O2 to H2O is −1.  Therefore, the slope of the line is (−1.76 V)/(−1) = +1.76V The most useful aspect of Frost diagrams is that The Frost diagram for oxygen they allow us to predict whether a RedOx in acidic solution (red line, pH = 0) and reaction will occur for a given pair of reagents alkaline solution (blue line, pH = 14). and what the outcome of the reaction will be. 67 6.13 Frost diagrams (b) Interpretation  To interpret the qualitative information contained in a Frost diagram it is important to note (Fig. 6.8) that the slope of the line connecting two species having oxidation numbers N′′ and N′ is νE⦵/(N′ − N′′) = E⦵ (since ν = N′ − N′′). Species liable This simple rule leads to the following features to undergo reduction Species liable to undergo oxidation 2. The oxidizing agent in the couple with the more positive slope (the more positive E⦵) is liable to undergo reduction (Fig. 6.9b). 3. The reducing agent of the couple with the less positive slope (the most negative E⦵) is liable to undergo oxidation (Fig. 6.9b). For instance, the steep slope connecting NO3− to (b) tendency towards oxidation and lower oxidation numbers in Fig. 6.6 shows that nitrate reduction is a good oxidizing agent under standard conditions. 68 6.13 Frost diagrams (b) Interpretation 4. A species in a Frost diagram is unstable with respect to disproportionation if its point lies above the line connecting the two adjacent species (on a convex curve).  When this criterion is satisfied, the standard potential for the couple to the left of the species is greater than that for the couple on the right.  the reaction Gibbs energy of a species with intermediate oxidation number lies above the average value for the two species on either side.  As a result, there is a tendency for the intermediate species to disproportionate into the two other species.  A specific example is NH2OH; as can be seen in Fig. 6.6, this compound is unstable with respect to disproportionation into NH3 and N2. 69 6.13 Frost diagrams (b) Interpretation 5. Two species will tend to comproportionate into an intermediate species that lies below the straight line joining the terminal species (on a concave curve).  A substance that lies below the line connecting its neighbours in a Frost diagram is inherently more stable than they are because their average molar Gibbs energy is higher (Fig. 6.9f) and hence comproportionation is thermodynamically favourable. Example: NH4NO3(s) → N2O(g)+2H2O(g)  thermodynamically spontaneous (∆rG⦵ = −168 kJ mol−1) and, once initiated by a detonation, explosively fast (blasting rocks)! 70 6.13 Frost diagrams FIGURE 6.9 The interpretation of a Frost diagram to gauge (a) reduction potential, (b) tendency towards oxidation and reduction, (c, d) disproportionation, and (e, f ) comproportionation. 71 6.13 Frost diagrams  Figure 6.10 shows the Frost diagram for manganese. Comment on the stability of Mn3+ in acidic aqueous solution.  What is the oxidation number of Mn in the product when MnO4− is used as an oxidizing agent in aqueous acid? 72 6.13 Frost diagrams Answer the following questions using the Frost diagram in Fig. 6.20. (a) What are the consequences of dissolving Cl2 in aqueous basic solution? (b) What are the consequences of dissolving Cl2 in aqueous acid? (c) Is the failure of HClO2 to disproportionate in aqueous solution a thermodynamic or a kinetic phenomenon? 73 End of Presentation