Chapter 5 – Oxidation and Reduction Consider: +6 0 -2 +4 SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O reduction oxidation A redox reaction is the outcome of a reduction and an oxidation half reaction SO422- + 8e + 10H+ → H2S + 4H2O CH2O + H2O → CO2 + 4e + 4H+ 1 By convention, redox reactions are written as REDUCTION reactions SO42- + 8e + 10H+ → H2S + 4H2O (2) CO2 + 4e + 4H+ → CH2O + H2O (3) SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O (1) The overall equation is the difference of the chemical equations for the two half reactions; but number of electrons gained and lost must be the same (1) = (2) + 2 x (-3) Gibbs energy for reaction = difference of the standard Gibbs energies for the half reactions Because half reactions must always occur in pairs, only differences in G have any physical meaning Chose one half reaction as a standard, and measure all Gibbs energies relative to this half reaction 2H+(aq) + 2e → H2(g) Go = 0 2 Measure G for each half reaction by coupling to a half reaction of known G (such as the H+|H2 couple) in a galvanic cell Fig. 5.1 From physics, we know that: Potential (E) = - Work (w) / Charge (q) therefore, w = -qE Lett us now define L d fi a quantity tit called ll d a ffaraday d (F) and d llett F equall th the charge in coulombs per mole of electrons (96,485 C). Then q = nF and w= -nFE. From thermodynamics we know that: ∆G = ∆U -T∆S + ∆(PV) and U = heat + w. Therefore, at constant T and P: ∆G = w. Therefore: ∆G = -nFE and at standard state: ∆Go = -nFEo 3 Since by definition 2H+(aq) + 2e → H2(g) Go = 0 it follows immediatelyy that the standard reduction p potential is 2H+(aq) + 2e → H2(g) Eo = 0.00 V Given SO42- + 8e + 10H+ → H2S + 4H2O Eo = -0.221 V CO2 + 4e + 4H+ → CH2O + H2O Eo = -0.072 V SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O SO42- + 8e + 10H+ → H2S + 4H2O Eo = -0.221 V 2CH2O + 2H2O → 2CO2 + 8e + 8H+ Eo = 0.072 V Remember that you d ’ double don’t d bl Eo! Because G = -nFEo, doubling n will have doubled G already 4 SO42- + 2CH2O + 2H+ → H2S + 2CO2 + 2H2O Eo = -0.221 V + 0.072 V = -0.149 V Go = -nFEo = -8 x 96 485 C mol-1 x -0.149 -0 149 J C-1 = 115 kJ mol-1 G > 0 reaction is not spontaneous If Eo < 0 0, the reaction is not spontaneous For a spontaneous reaction, Eo > 0 The electrochemical series lists half reactions in order of Eo. The more positive Eo the stronger the species is as an oxidising agent Strong oxidising agent See Table 5.2 Weak oxidising agent 5 Example Calculate Eocell and G for the oxidation of copper by nitric acid Example Calculate Eocell and G for the oxidation of copper by nitric acid 6 The Nernst Equation The dependence of the electrode potential on the concentration can be obtained from the dependence of ∆rG on concentration. Recall that under non-standard conditions: ∆rG = ∆rG° + RT ln Q Where Q is the reaction quotient. For the general reaction: aA + bB + ne- Ý cC + dD [C]c [D]d Q [[A]]a [B]]b For systems at equilibrium: Q = Keq and ∆rG = 0 The Nernst Equation ∆rG = ∆rG° + RT ln Q (and we know: ∆G = -nFE ∆Go = -nFEo) -nFE = -nFE° + RT ln Q E Eo RT ln Q nF Converting to base 10 log: E E o 2.303RT log Q nF Nernst Equation Or E Eo 0.0592 log Q n (at 298 K) 7 At equilibrium, Q = K, and E = 0 because G = 0 Therefore: 0 E0 RT ln K nF RT ln K E o nF l K ln nFE o RT Example The overall equation for a fuel cell can be written as follows: 2H2(g) + O2(g) Ý 2H2O(l) Eº = 1.23 V What cell potential would be produced in a fuel cell operating with O2 and H2 at 5.0 bar and 298 K? 8 OVERPOTENTIALS Since Co2+ + 2e → Co 2H+ + 2e → H2 Eo = -0.277 V Eo = 0.000 V cobalt metal in contact with acid should release hydrogen vigorously Co + 2H+ → Co2+ + H2 Eocell = 0.277 V The reaction does occur but is slow. Thermodynamics tells us nothing about the speed (kinetics) of a reaction. E i i l observation: Empirical b ti reduction d ti off one couple l b by another th occurs att a significant rate only when the difference in potentials of the two couples exceeds a characteristic value known as the overpotential. To reduce H+ to H2: need an overpotential of about -0.6 V To oxidise H2O to O2: need an overpotential of about +0.6 V The Nernst Equation and pH The Nernst equation for the following half reaction at 25 °C and 1 bar H2(g) pressure 2H+(aq) + 2e- Ý H2(g) can be written as: E° = 0.00V pH2 E = E° - 0.0592 log + 2 [H ] n pH2 E = 0 - 0.0592 log + 2 [H ] 2 pH2 E = - 0.0592 log [H+] 1 E = - 0.0592 log [H+] 9 The Nernst Equation and pH 1 E = - 0.0592 log [H+] + ( 0 0592 log [H ]) 0 0592 log(1) - (-0.0592 E = - 0.0592 + E = - (-0.0592 log [H ]) E = - 0.0592 x pH Example What minimum potential must a couple have to be able to oxidise water at pH 7? Xn+ + ne → X O2 + 4H+ + 4e → 2H2O Eo = x V Eo = 1.23 V The Nernst equation for O2 + 4H+ + 4e- Ý 2H2O can be written as: 1 E = E° - 0.0592 log + 4 [H ] n 10 Example + E = 1.23V - (-0.0592 log [H ]) (pH = - log [H+]) E = 1.23V - 0.0592 x pH E = 1.23V - 0.0592 x 7 E = 0.82 V Redox stability in aqueous solution Let X be a species that is dissolved in water X + e → X- EoX Water can act as both a reducing agent: O2 + 4H+ + 4e → 2H2O Eo1 = 1.23V and an oxidising agent: 2H3O+ + 2e → 2H2O + H2 Eo2 = 0.00 V what are the permissible values of EoX for X to survive at a given pH? 11 Eo1 = 1.23V o E = E Q g o l V 2 9 5 n 0 . 0 O2 + 4H+ + 4e → 2H2O 1 H p x V 2 9 5 0 . 0 V 3 2 . 1 = E as we have seen before Similarly for 2H3O+ + 2e → 2H2O + H2 o E = E 2 Q g o l V 2 9 5 n 0 . 0 Eo2 = 0.00 V H p x V 2 9 5 0 . 0 V 0 0 . 0 = E Let’s look specifically at pH 7 V 2 8 . 0 = 7 x 2 9 5 0 . 0 3 2 . 1 = E Eo1 = 1.23V H p x V 2 9 5 0 . 0 V 3 2 . 1 = E O2 + 4H+ + 4e → 2H2O 7 H p V 1 4 . 0 = 7 x 2 9 5 0 . 0 0 0 . 0 = E Eo2 = 0.00 V H p x V 2 9 5 0 . 0 V 0 0 . 0 = E 2H3O+ + 2e → 2H2O + H2 7 H p 12 Eo1 = 1.23V V 2 8 . 0 = E O2 + 4H+ + 4e → 2H2O 7 H p Eo2 = 0.00 V V 1 4 . 0 = E 2H3O+ + 2e → 2H2O + H2 7 H p X + e → X- EoX Suppose EoX < -0.41 V. For example, let EoX = -0.50 V 2X- → 2X + 2e 2H3O+ + 2e → H2 + 2H2O 2H3O+ + 2X- → 2X + 2H2O + H2 0.50 V -0.41 V 0.09 V The reaction is spontaneous and X will not survive. It will be oxidised by water. Eo1 = 1.23V V 2 8 . 0 = E O2 + 4H+ + 4e → 2H2O 7 H p Eo2 = 0.00 V V 1 4 . 0 = E 2H3O+ + 2e → 2H2O + H2 7 H p X + e → X- EoX Suppose EoX > 0.82 V. For example, let EoX = 0.90 V 4X + 4e → 4X2H2O → O2 + 4H+ + 4e 2H2O + 4X → 4X- + O2 + 4H+ 0.90 V -0.82 V 0.08 V The reaction is spontaneous and X will not survive. It will by reduced by water. 13 Finally suppose -0.41 V < EoX < 0.82 V. For example, let EoX = 0.10 V 4X + 4e → 4X2H2O → O2 + 4H+ + 4e 2H2O + 4X → 4X- + O2 + 4H+ 0.10 V -0.82 V -0.72 V The reaction is not spontaneous spontaneous. 2X- → 2X + 2e 2H3O+ + 2e → H2 + 2H2O 2H3O+ + 2X- → 2X + 2H2O + H2 -0.10 V -0.41 V -0.51 V The reaction is not spontaneous. Eo1 = 1.23V V 2 8 . 0 = E O2 + 4H+ + 4e → 2H2O 7 H p Eo2 = 0.00 V V 1 4 . 0 = E 2H3O+ + 2e → 2H2O + H2 7 H p 0.82 V For a species to be stable in water, its redox potential must lie between these values at pH 7. This is sometimes referred to as the Electrochemical Window -0.41 V 14 Recall, however, that the overpotential may make species with Eo values outside the electrochemical window kinetically stable – i.e., they can exist in solution because the kinetics of their reaction with water are slow By applying the Nernst equation, we can get the stability field of water at any pH. H2O oxidised above this line V 3 2 . 1 = E H p x V 2 9 5 0 . 0 V 0 0 . 0 = E H p x V 2 9 5 0 . 0 Figure 5.3 H2O reduced below this line 15 Example Is Co(III) (a) thermodynamically and (b) kinetically stable in water at pH 0 and at pH 7? O2|H2O Co(III) ( ) + e → Co(II) ( ) Eo = 1.92V 1.23 V pH 7 pH 0 0.00 V At pH = 0 0.82 V -0.41 V H+|H2 Example Is Co(III) (a) thermodynamically and (b) kinetically stable in water at pH 0 and at pH 7? O2|H2O Co(III) ( ) + e → Co(II) ( ) At pH = 7 Eo = 1.92V 1.23 V pH 7 pH 0 0.00 V 0.82 V -0.41 V H+|H2 16 O2|H2O Aluminium should react spontaneously with water at pH 7: Al3+ + 3e → Al Eo = -1.66 V H+|H2 2Al → 2Al3+ + 6e 6H+ + 6e → 3H2 Eo = 1.66 V EpH7 = -0.41 2Al + 6H+ → 2Al3+ + 3H2 Eo = 1.25 V pH 7 0.82 V -0.41 V It doesn’t because the surface is passivated by the formation of a tough coating of Al2O3 which sticks tightly to the bulk Al metal and protects it. This can be enhanced by anodising the metal. The metal is made the anode in an electrolytical cell and oxidised to form the protective film of the oxide. Nitric acid can be used to passivate some metals such as stainless steels. 17 Some couples have very positive potentials but the species can still exist in water. Examples include: O2|H2O pH 7 Cr2O72-|Cr3+ Eo = 1.38 V 0.82 V MnO4-|Mn2+ Eo = 1.51 V Hence at pH 7 the Ecell for oxidation of water to produce O2 is 0.56 V and 0.69 V, respectively. H+|H2 -0.41 V Reason why Cr2O72- and MnO4- can still exist: Requires a 3e and a 5e transfer, respectively. Multiple electron transfer reactions are usually very slow. These species are under kinetic control Natural waters pH controlled by CO2/H2CO3/HCO3-/CO32- equilibrium potential controlled by amount of dissolved O2 and by the presence of reducing couples from organic matter Fig.5.12 18 Disproportionation reactions Since Cu+|Cu Cu2+|Cu+ Eo = 0.52V Eo = 0.16 V Both potentials lie within the stability field of water Cu+ ions neither oxidise nor reduce water But Cu+ ions in solution are unstable because they can undergo disproportionation 2Cu+ → Cu2+ + Cu(s) A disproportionation reaction is one where the oxidation state of an element is simultaneously increased and decreased Disproportionation occurs because: Cu+ + e → Cu Cu+ → Cu2+ + e Eo = 0.52 Eo = -0.16 2Cu+ → Cu + Cu2+ Eo = 0.36 V 19 Example Fe2+|Fe Fe3+|Fe2+ E° = -0.44 V E° = 0.77 V Do we expect Fe2+ to disproportionate in aqueous solution? Disproportionation reaction: 3Fe2+ → 2Fe(s) + Fe3+ In a comproportionation reaction two species of an element in two different oxidation states form a product in which the element is in an intermediate oxidation state Ag2+ + Ag → 2Ag+ Eo = 1.18 V 20 Example Show that the equilibrium constant lies far to the right for the following comproportionation reaction at 25 °C: Ag2+ + Ag → 2Ag+ Eo = 1.18 V Ag(II) and Ag(0) completely converted to Ag(I) in aqueous soln. Representing potential data Latimer diagrams summarise the standard potential (in V) between species of an element. ClO4+7 +1.20 ClO3+5 +1.18 HClO2 +3 +1.65 HClO +1.67 +1 Cl2 0 +1.36 Cl- in acid -1 So we can immediately write down the half reaction and the potential connecting any two species. 21 For example, ClO3- and HClO2 ClO3- HClO2 - Balance O by adding H2O ClO3- HClO2 + H2O - In acid, balance H by adding H+. (In base, add H2O to the side of the equation requiring H, and OH- to the other side.) ClO3- + 3H+ HClO2 + H2O - Balance charges by adding e ClO3- + 3H+ + 2e- ClO4+7 +1.20 ClO3- +1.18 HClO2 + H2O HClO2 +5 +1.65 HClO +3 +1.67 Eo = +1.18V Cl2 +1 +1.36 Cl- 0 in acid -1 Suppose we are interested in the potential between two non-adjacent couples. Cannot just add their standard potentials! ∆rG°overall = ∆rG°individual steps (First Law of Thermodynamics) i.e. ∆rG°(a+b) = ∆rG°(a) + ∆rG°(b) -noverall FEooverall = -n1FEo1 + -n2FEo2 + -n3FEo3 22 ClO4- +1.20 +7 ClO3- +1.18 HClO2 +5 Eo +1.65 +3 +1.67 HClO +1 Cl2 +1.36 Cl- 0 in acid -1 o + -n FEo o FE + -n FE -n 1 2 3 1 2 3 overall = -noverallF + n3Eo3 = n1Eo1 + n2Eo2 n1 + n2 + n3 = 2 x 1.65 + 1 x 1.67 + 1 x 1.36 2+1+1 = 1.65 V ClO4+7 +1.20 ClO3- +1.18 +5 HClO2 +1.65 +3 HClO +1.67 +1 Cl2 +1.36 Cl- 0 in acid -1 and this refers to the reaction HClO2 + 3H+ + 4e Cl- + 2H2O 23 Frost Diagrams A plot of NEº for the couple X(N)/X(0) vs. oxidation number (N) of the element: X(N) + Ne- → X(0) Eº NEº = -∆rGº/F Frost Diagrams What information can one obtain from the Frost diagram below? Lower = more thermodynamically stable Convex = disproportionation Concave = comproportionation Upper left = strong oxidising agent in this diagram Upper right= reducing agent in this diagram Frost diagram for Mn 24 Frost Diagrams Can be constructed from Latimer diagrams Example Construct a Frost diagram from the following Latimer diagram (in aq acidic solution) Tl3+ +1.25 Tl+ -0.34 Tl +0.72 Tl+ + e = Tl0 Eo = -0.34 V n=1 nEo = -0.34 V Tl3+ + 3e = Tl0 Eo = +0.72 V n=3 nEo = +2.16 V Tl+ + e = Tl0 Eo = -0.34 V n=1 nEo = -0.34 V Tl3+ + 3e = Tl0 Eo = +0.72 V n=3 nEo = +2.16 V 3 nEo 2 +2.16 V 1 0 -1 -0.34 V 0 1 2 3 4 N 25 Note that in general n2Eo2 Eo nEo n2Eo2 – n1Eo1 = n2 – n1 I.e. slope = Eº n1Eo1 n2 n1 N Tl+ is the most stable oxidation state Tl3+ is a strong oxidising agent because slopes of the lines to the other two states are large and positive 26 Example Consider the following half reactions for copper: Cu3+ + e- → Cu2+ Eo = 1.80 V Cu+ + e- → Cu Eo = 0.52 V Cu2+ + 2e- → Cu Eo = 0.34 V (1) Construct a Latimer diagram for copper (2) Construct a Frost diagram for copper Example Cu3+ + e- → Cu2+ Eo = 1.80 V Cu+ + e- → Cu Eo = 0.52 V Cu2+ + 2e- → Cu Eo = 0.34 V -noverall FEooverall = -n1FEo1 + -n2FEo2 noverall FE0overall - n2FE02 Eo1= n1F Eo1= (2 0.34 V) - (1 0.52 V) 1 Eo1 = 0.16 V 27 Example 3 2.49 V 2.5 Remember: y = n E0 nE /V 2 i.e. for Cu2+: y = 2 0.34 V = 0.68 V 1.5 1 0.52 V 0.5 0.68 V 0V 0 -0.5 -1 0 1 2 3 Oxidation number Example 3 Is Cu3+ a strong oxidizing agent? 2.49 V 2.5 nE /V 2 What is the oxidation number of Cu in the product when Cu3+ is used as an oxidizing agent? 1.5 1 0.5 0.68 V 0.52 V 0 0V -0.5 -1 0 1 2 3 Comment on the stability of Cu+ in acidic aqueous solution. Oxidation number 28 See SA pp. 164-168 Example From the Frost diagram for N below, estimate Eº for the NO3- | NO2- couple in acidic media (red line): 29 Example Pourbaix diagrams Plots of E vs pH We will, as an example, derive the Pourbaix diagram for iron Two Latimer diagrams pertain In acid ([H+] = 1 M): Fe3+ 0.77 V Fe(OH)2 -0.44 V Fe In alkali ([OH-] = 1 M) -0.56 V Fe3+ -0.887 V Fe(OH)2 Fe Pourbaix diagrams: •correlate Latimer diagrams at pH 0 and pH 14 •take into account speciation or oxidation state of the element 30 Fe3+ 0.77 V Fe(OH)2 -0.44 V Fe The half reaction Fe33+ + e → Fe22+ Eo = 0.77 0 77 V does not involve a proton so Eo is independent of pH 31 Fe3+ + e → Fe2+ Fe3+ will precipitate out of solution as pH is increased. We can calculate the pH at which this will occur from the KSP for Fe(OH)3. Fe(OH)3(s) Ý Fe3+ + 3OH– KSP = 4.11 x 10-37 M4 At what pH will [Fe3+] = 1.00 M? KSP = 4.11 x 10-37 M4 = [Fe3+][OH–]3 [OH–] = (4.11 x 10-37/1)0.333 = 7.43 x 10-13 M So [H+] = 10-14/7.43 x 10-13 = 1.35 x 10-2 M hence pH = 1.87 32 Fe(OH)3 Ý Fe3+ + 3OH- Vertical lines in a Pourbaix diagram indicate where two species of an element in the same oxidation state are in equilibrium To calculate the Fe(OH)3|Fe2+ line... 33 Fe3+ + e → Fe2+ Go = -74.3 kJ mol-1 Eo = 0.77 V 3OH- + 3H+ → 3H2O -239.7 kJ mol-1 Fe(OH)3 → Fe3+ + 3OHFe(OH)3 + 3H+ + e → Fe2+ + 3H2O 207.6 kJ mol-1 Go = -nFEo -106.4 kJ mol-1 = -1 x 96485 x 0.77 Go = -RT ln KSP = -8.315 x 298 x ln (4.11 x 10-37) Go = -nFEo Eo = -Go /nF = 106400/1 x 96485 = 1.10 V Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O Eo = 1.10 V E = Eo – RT/nF ln Q E = 1.10 – 3 x 0.0592 x pH This must cross the Fe3+/Fe(OH)3 line when 0.77 = 1.10 – 3(0.0592)pH or pH = 1.87 which confirms the result we got from the KSP calclation Fe3+ 0.77 V Fe(OH)2 -0.44 V Fe 34 1.1 Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O 1.1 Fe(OH)3 + 3H+ + e → Fe2+ + 3H2O 35 From the KSP for Fe(OH)2 Fe(OH)2 Ý Fe2+ + 2OH– KSP = 1.61 x 10-15 M3 At what hat pH will ill [Fe2+] = 1.00 1 00 M? KSP = 1.61 x 10-15 M3 = [Fe2+][OH–]2 [OH–] = (1.61 x 10-15/1)0.5 = 4.01 x 10-8 M So [H+] = 10-14/4.01 x 10-8 = 2.49 x 10-7 M hence pH = 6.61 1.1 Fe(OH)2 Ý Fe2+ + 2OH- 36 The half reaction Fe2+ + 2e → Fe Eo = -0.44 V does not involve a proton so Eo is independent of pH 1.1 Fe2+ + 2e → Fe An expression for the potential for the Fe(OH)3|Fe(OH)2 couple can be derived from the following data Fe(OH)3 + 3H+ + e → Fe2+ + 2H2O Eo 1.10 V Go -106.4 kJ mol-1 3H2O → 3H+ + 3OH- 239.7 kJ mol-1 Fe2+ + 2OH- → Fe(OH)2 -84.4 kJ mol-1 Fe(OH)3 + e → Fe(OH)2 + OH- Eo -0.51 V Go 48.9 kJ mol-1 E = Eo – RT/nF ln Q E = -0 0.51 51 + 0 0.0592 0592 x pOH E = -0.51 + 0.0592 x (14 – pH) E = 0.316 – 0.0592 x pH 37 1.1 0.316 Fe(OH)3 + e → Fe(OH)2 + OH- ...and finally the value of Fe(OH)2|Fe couple can be found by similar considerations, and the Nernst equation applied. E = -0.060 – 0.0592 x pH 38 Overlaying Pourbaix diagrams The feasibility of a reaction can be predicted by overlaying the relevant Pourbaix diagrams 39 stability field for As(V) stability field for As(III) At pH < 5.5 and at pH > 9, Fe3+ has the potential to oxidise As3+ to As5+ 5.5 9 40 For example 0.65 0.45 Fe(OH)3 + e + 3H+ → Fe2+ + 3H2O E = 0.65 2 As3+ → As5+ + 2e E = -0.45 As3+ + 2Fe(OH)3 + 6H+ → 2Fe2+ + 6H2O + As5+ E = 0.20 V For 5.5 < pH < 9 As5+ will oxidise Fe2+ to Fe3+ 41