{"id":2320,"date":"2026-01-23T15:11:30","date_gmt":"2026-01-23T20:11:30","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/?post_type=chapter&p=2320"},"modified":"2026-03-23T15:43:57","modified_gmt":"2026-03-23T19:43:57","slug":"eh-ph-diagrams-theory","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/chapter\/eh-ph-diagrams-theory\/","title":{"raw":"3. Eh-pH diagrams - Theory","rendered":"3. Eh-pH diagrams – Theory"},"content":{"raw":"Leaching (any many other chemical processes) may involve acid-base reactions, electron transfer and complexation. All such processes can be represented on Eh-pH diagrams. This section will develop the equations needed to be able to draw the diagrams.\r\n

3.1 Reduction potential as a function of pH (the Eh Equation)<\/h2>\r\nConsider a generalized half reaction:\r\n\r\n\\[\\ce{aA + bB + mH^+ + ne^- = cC + dD} \\tag{32}\\]\r\n\r\nThe half reaction involves H+<\/sup>, reactants A and B, and produces products C and D. The factors a, b, m, c and d are stoichiommetric coefficients; n is the stoichiommetric coefficient for the moles of electrons per mole of half reaction. (Reactions of hydrometallurgical importance often involve acidic or basic pH.) We know that E\u00b0 for this half reaction is defined as \u0394E\u00b0 for the cell where the cathode is the reduction reaction under standard conditions and the anode is the standard H+<\/sup>\/H2<\/sub><\/em> half reaction. Further E for the half reaction under non-standard conditions <\/em>is defined as \u0394E for the cell where the cathode is the half reaction under any specified set of conditions and the anode is, again, the standard H+<\/sup>\/H2<\/sub> half cell<\/em>:\r\n\r\n[latex]\r\n\\begin{align*}\r\n\\ce{aA}&+\\ce{bB}+\\ce{mH+}+\\ce{ne-}=\\ce{cC}+\\ce{dD}\\tag{33} \\\\\r\n\\ce{nH+}&=\\ce{n\/2H2}+\\ce{ne-} \\\\\r\n\\hline\r\n\\text{net: }\\ce{aA}&+\\ce{bB}+\\ce{mH+(a_H+)}+\\ce{n\/2H2(1 atm)}\r\n=\\ce{cC}+\\ce{dD}+\\ce{nH+(a_H+ = 1)}*\\tag{34}\r\n\\end{align*}\r\n[\/latex]\r\n\r\nFor this cell,\r\n\r\n\\[\\Delta E = \\Delta E^{\\circ} - \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}(a_{H^+}=1)^n}{a_A^{a}a_B^{b}a_{H^+}^{n}(P_{H_2}=1)^{n\/2}} \\tag{35}\\]\r\n\r\n\\[\\Delta E = \\Delta E^{\\circ} - \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}a_{H^+}^{m}} \\tag{36}\\]\r\n\r\nThis is equivalent to writing,\r\n\r\n\\[\\;E = E^{\\circ} - \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}a_{H^+}^{m}} \\tag{37}\\]\r\n\r\nwhere E\u00b0 is the standard half reaction (32) voltage. This is what we would get if we used the half reaction alone to write the Nernst equation for E and ignored the electrons. This works because the effects of the standard H+\/H2 half reaction fall away in the expression for the cell voltage; E\u00b0H+\/H2 = 0 V and aH+ = 1 and PH2 = 1 in half reaction (34); all the factors that affect the cell voltage for reaction are ascribed to half reaction (23) alone. Equation (37) can now be developed into the form we will need.\r\n
\r\n\r\nSolute activities are on the molal scale. Properly gas pressures should be fugacities, although at 1 atm pressure fugacity is nearly equal to pressure in many cases. Some data compilations use pressures or fugacities on the bar scale instead; 1 atm = 1.01325 bar. In most cases the error caused by this difference is negligible. Ionic activities may differ greatly from the concentrations of the ions, particularly as concentration increases (which is of practical interest in hydrometallurgy), since ionic solutions are highly non-ideal.\r\n\r\n<\/div>\r\nFirst, from this point on we give E a new symbol, [pb_glossary id=\"2666\"]Eh[\/pb_glossary]; E = Eh. This only serves to indicate that the half reaction potential is referenced to the standard H+\/H2 half reaction.\r\n\r\nSubstituting \u2013DG\u00b0\/nF for DE\u00b0 and separating out the aH+ term,\r\n\r\n\\[\\;E_h = -\\frac{\\Delta G^{\\circ}}{nF} - \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}} - \\frac{2.303RT}{nF}\\log\\frac{1}{a_{H^+}^{m}} \\tag{38}\\]\r\n\r\n\\[\\;E_h = -\\frac{\\Delta G^{\\circ}}{nF} - \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}} - \\frac{2.303RT\\,m}{nF}\\,pH \\tag{39}\\]\r\n\r\nThis is the form we will use to calculate Eh as a function of pH, solute activities and gas pressures. The quantity,\r\n\r\n\\[\\frac{a_C^{c} a_D^{d}}{a_A^{a} a_B^{b}}\\tag{40}\\]\r\n\r\nwe will call Q-H. It is the reaction quotient for all species except H+; aH+ is taken care of by the pH term. Then,\r\n\r\n\\[\\;E_h = -\\frac{\\Delta G^{\\circ}}{nF} - \\frac{2.303RT}{nF}\\log Q_H - \\frac{2.303RT\\,m}{nF}\\,pH \\tag{41}\\]\r\n\r\nIn order to use this equation we must determine \u0394G\u00b0 data at the specified temperature, and specify the activities of the solutes and pressures of the gases. Then the terms,\r\n\r\n\\[-\\frac{\\Delta G^{\\circ}}{nF} \\quad \\text{and} \\quad \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}\\tag{42}\\]\r\n\r\nare constants and Eh is a linear function of pH with slope = -2.303RTm\/nF. These lines are plotted on Eh-pH diagrams to graphically show reduction potentials for couples as a function of pH.\r\n

Calculation of \u0394G\u00b0<\/h3>\r\nFrom thermodynamics it is known that for a chemical reaction involving reactants i and products j, with reactant stoichiommetric coefficients \u03bdi<\/sub> and product stoichiommetric coefficients \u03bdj<\/sub>,\r\n\r\n\\[\\Delta G^{\\circ} = \\sum_{j}\\nu_j \\Delta G_f^{\\circ\\,j} - \\sum_{i}\\nu_i \\Delta G_f^{\\circ\\,i} \\tag{43}\\]\r\n\r\nwhere \u0394G\u00b0fj and \u0394G\u00b0fi are the standard free energies of formation of the species involved in the reaction. These are defined for formation reactions, which for uncharged species involves only the elements as reactants to form only the specified product. Further, all species are at a specified temperature and in their standard states, i.e. pure and the most stable chemical form of the species and 1 atm pressure (or 1 bar, depending on the compilation). An example is,\r\n\r\n\\[\\ce{H2_g + \\tfrac{1}{8}S8_s = H2S_g,\\ P = 1\\ atm,\\ 25^{\\circ}C} \\tag{44}\\]\r\n
\r\n\r\nAn alternative rendering of the Eh equation can be developed. Free electrons do not exist in solutions, but they are free to move in metals and semiconductors (e.g. electrodes and wires). In this environment we can allow for the activity of electrons. This allows us to consider the half reaction alone, at least conceptually. We bear in mind that half reactions do not occur in isolation.<\/em> There must be two; the electrons have to come from one reagent and be transferred to another. That being said consider as an example the half reaction,\r\n\r\n\\[ \\ce{Tl^{3+}_{(aq)} + 2e^- = Tl^{+}_{(aq)}} \\]\r\n\r\nFrom thermodynamics,\r\n\r\n\\[\r\n\\Delta G^\\circ = -nF E^\\circ = -RT\\ln K = -2.303\\,RT \\log K= -2.303\\,RT \\log\\!\\left(\r\n\\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}} a_{e^-}^n}\r\n\\right) (n = 2) \\tag{A}\r\n\\]<\/small>\r\n\r\nRearranging gives:\r\n\r\n\\[\r\n\\frac{nF E^\\circ}{2.303\\,RT}= n \\log \\frac{1}{ae^-}\r\n+ \\log \\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}}}\r\n= n \\log a_{e^-}\r\n+ \\log \\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}}}\r\n\\]\r\n\r\n\\[\r\n\\frac{F E^\\circ}{2.303\\,RT}\r\n= -\\log a_{e^-}\r\n+ \\frac{1}{n} \\log\\!\\left(\r\n\\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}}}\r\n\\right)\r\n\\]\r\n\r\nAnalogously to pH let -log ae- = pe. Further, when we are dealing with E\u00b0 alone, this implies standard conditions, i.e. all reactants and products at unit activity. Then,\r\n\r\n\\[\r\n\\frac{F E^\\circ}{2.303\\,RT} = \\mathrm{pe}^\\circ\r\n\\qquad\\text{or}\\qquad\r\nE^\\circ = \\frac{2.303\\,RT}{F} \\,\\mathrm{pe}^\\circ\r\n\\]\r\n\r\nIf we then defined pe under non-standard conditions as,\r\n\r\n\\[\r\n\\frac{FE^\\circ}{2.303\\,RT} = \\mathrm{pe^\\circ}\r\n\\qquad\\text{or}\\qquad\r\nE^\\circ = \\frac{2.303\\,RT}{F} \\,\\mathrm{pe^\\circ}\r\n\\]\r\n\r\nand substitute into the Nernst equation, we obtain,\r\n\r\n\\[\r\n\\mathrm{pe}\r\n= \\mathrm{pe}^\\circ - \\frac{1}{n}\\log Q\r\n\\]\r\n\r\nThis is often used as an alternative way to express E\u00b0 and then Eh-pH diagrams become e-pH diagrams. However, we will not make use of it in this course.\r\n\r\n<\/div>\r\nSulfur may exist in a number of solid forms (allotropes). The most stable is the rhombohedral form (referring to the crystal structure) and this is what is implied here. By definition, \u0394G\u00b0f for the elements in their standard states = 0. Then all of the free energy change for the formation reaction is carried by the compound. These data allow us to calculate \u0394G\u00b0 for any chemical reaction as per equation (A).\r\n\r\nHowever, for ions in solution the problem is more complex. A particular ion, e.g. SO4<\/sub>2-<\/sup>, will not exist in solution without countercations. This is overcome by writing formation reactions for ions with H+<\/sup> and H2<\/sub> involved as well and<\/em> setting \u0394G\u00b0f<\/sub> for both H+<\/sup> and H2<\/sub> in their standard states to be 0. That \u0394G\u00b0f for H+ is 0 has already been implied by setting E\u00b0H+\/H2<\/sub> to 0:\r\n\r\n\\[\\ce{2H^+ + 2e^- = H2_g} \\tag{45}\\]\r\n\r\nSince,\r\n\r\n\\[\\;E^{\\circ} = 0 = -\\frac{\\Delta G^{\\circ}}{nF}\r\n= -\\frac{1}{nF}\\left(\\Delta G_f^{\\circ\\,H_2} - 2\\Delta G_f^{\\circ\\,H^+}\\right)\r\n\\tag{46}\\]\r\n\r\nand \u0394G\u00b0f<\/sub>H2<\/sup> = 0 by definition, it follows that \u0394G\u00b0f<\/sub>H+<\/sup> = 0 as well. (We can ignore the electron for the same reasons we did in developing the Eh equation. The standard H+<\/sup>\/H2<\/sub> half reaction is implicit as the anode.)\r\n\r\nTo illustrate, the formation reaction for SO4<\/sub>2-<\/sup> is then written as the sum of:\r\n\r\n[latex]\r\n\\begin{align*}\r\n\\ce{\\tfrac{1}{8}S8_s}&+\\ce{2O2_g}+\\ce{2e-}=\\ce{SO4^{2-}_{aq}}\\tag{45} \\\\\r\n\\ce{H2_g}&=\\ce{2H+_{aq}}+\\ce{2e-}\\tag{46} \\\\\r\n\\hline\r\n\\text{net: }\\ce{\\tfrac{1}{8}S8_s}&+\\ce{2O2_g}+\\ce{H2_g}\r\n=\\ce{SO4^{2-}_{aq}}+\\ce{2H+_{aq}}\\tag{47}\r\n\\end{align*}\r\n[\/latex]\r\n\r\nThen,\r\n\r\n[latex]\r\n\\begin{align*}\r\n\\Delta G_f^{\\circ}\r\n&= 2\\Delta G_f^{\\circ}(H^+) + \\Delta G_f^{\\circ}(SO_4^{2-})\r\n- \\tfrac{1}{8}\\Delta G_f^{\\circ}(S_8)\r\n- 2\\Delta G_f^{\\circ}(O_2)\r\n- \\Delta G_f^{\\circ}(H_2)\r\n\\tag{48} \\\\\r\n&= \\Delta G_f^{\\circ}(SO_4^{2-})\r\n\\tag{49}\r\n\\end{align*}\r\n[\/latex]\r\n\r\n(All other terms are 0 by definition.) Similarly, for an example of the formation reaction of a cation,\r\n\r\n[latex]\r\n\\begin{align*}\r\n\\ce{Cu_s}&=\\ce{Cu^{2+}_{aq}}+\\ce{2e-}\\tag{50} \\\\\r\n\\ce{2H+_{aq}}&+\\ce{2e-}=\\ce{H2_g}\\tag{51} \\\\\r\n\\hline\r\n\\text{Net: }\\ce{Cu_s}&+\\ce{2H+_{aq}}=\\ce{Cu^{2+}_{aq}}+\\ce{H2_g}\\tag{52}\r\n\\end{align*}\r\n[\/latex]\r\n\r\nTabulations of \u0394G\u00b0f data are readily available. Finally, we can now demonstrate the calculation of \u0394G\u00b0 for a half reaction. For example,\r\n\r\n\\[\\ce{SO4^{2-}_{aq} + 8H^{+}_{aq} + 6e^- = S_s + 4H2O_l} \\tag{53}\\]\r\n\r\nFor convenience we take 1\/8S8<\/sub> to be equivalently stated as S. The standard H+\/H2 anode half reaction is, as always, implicit. This can be explicitly added as,\r\n\r\n\\[\\ce{3H2_g\\ (P_{H2}=1\\ atm) = 6H^{+}_{aq}\\ (a_{H^+}=1) + 6e^-} \\tag{54}\\]\r\n\r\nrecalling that half reaction (53) is equivalent to a cell where the anode is the standard H+<\/sup>\/H2<\/sub> half reaction (54). And again the effects of standard H+<\/sup> and H2<\/sub> disappear. For half reaction (53),\r\n\r\n[latex]\r\n\\begin{align*}\r\n\\Delta G^{\\circ}\r\n&= 4\\Delta G_f^{\\circ}(H_2O) + \\Delta G_f^{\\circ}(S)\r\n- \\Delta G_f^{\\circ}(SO_4^{2-}) - 8\\Delta G_f^{\\circ}(H^+)\r\n\\tag{55} \\\\\r\n&= 4\\Delta G_f^{\\circ}(H_2O) - \\Delta G_f^{\\circ}(SO_4^{2-})\r\n\\tag{56}\r\n\\end{align*}\r\n[\/latex]\r\n\r\nUsually \u0394G\u00b0f<\/sub> data are expressed in kJ\/mol. As we shall see, when using the data in the equation for Eh, \u0394G\u00b0 must be converted to J\/mol.\r\n

Special Case 1: m = 0<\/h3>\r\nWhen m = 0 no protons are involved in the half reaction and the -2.303RTmpH\/nF term goes to zero. Then the Eh is a constant, independent of pH, under the specified conditions. \u00a0It is a flat line on the Eh-pH diagram.\r\n

Special Case 2: When H+ Appears as a Product in the Half Reaction<\/h3>\r\nNote than m is a positive number when H+<\/sup> appears as a reactant in the half reaction. If H+<\/sup> appears as a product (which it does in some cases), then,\r\n\r\n\\[\\;E_h\r\n= -\\frac{\\Delta G^{\\circ}}{nF}\r\n- \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}\r\n- \\frac{2.303RT}{nF}\\log\\!\\left(a_{H^+}^{\\,m}\\right)\r\n\\tag{57}\\]\r\n\r\nThen the pH term becomes,\r\n\r\n\\[-\\frac{2.303RT}{nF}\\log\\!\\left(a_{H^+}^{\\,m}\\right)\r\n= +\\frac{2.303RT\\,m}{nF}\\,pH \\tag{58}\\]\r\n\r\nIf we preserve the same form as the Eh equation [19], with a minus sign in front of the pH term, we need to make m be a negative number. Thus when H+<\/sup> appears as a product, m is negative.<\/em>\r\n

Special Case 3: Reactions that do not Involve Electrons (n = 0)<\/h3>\r\nAs mentioned previously, many reactions of interest involve H+ (or OH-) and do not involve electrons. The simplest case is the dissociation of an acid in aqueous solution. We can obtain an equation for the pH as a function of solute activities and gas pressures. Consider the general reaction:\r\n\r\n\\[\\ce{aA + bB + mH^+ <=> cC + dD} \\tag{59}\\]\r\n\r\n\\[\\Delta G^{\\circ} = -2.303RT\\log K \\tag{60}\\]\r\n\r\n\\[\\Delta G^{\\circ}\r\n= -2.303RT\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}a_{H^+}^{m}}\r\n\\tag{61}\\]\r\n\r\n\\[\\Delta G^{\\circ}\r\n= -2.303RT\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}\r\n- 2.303RT\\log\\frac{1}{a_{H^+}^{m}}\r\n\\tag{62}\\]\r\n\r\n\\[\\Delta G^{\\circ}\r\n= -2.303RT\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}\r\n- 2.303RT\\,m\\,pH\r\n\\tag{63}\\]\r\n\r\nRearranging yields,\r\n\r\n\\[\\;pH\r\n= -\\frac{\\Delta G^{\\circ}}{2.303RT\\,m}\r\n- \\frac{1}{m}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}\r\n= -\\frac{\\Delta G^{\\circ}}{2.303RT\\,m}\r\n- \\frac{1}{m}\\log Q_H\r\n\\tag{64}\\]\r\n\r\nTo use this equation we again need \u0394G\u00b0 for the reaction and the activities for the solutes and pressures for the gases need to be specified. Again, if H+<\/sup> appears as a product, then the sign of m is negative.<\/em> Once the quantities have been specified the equation yields a number for pH, i.e. pH is a constant, independent of Eh. If we plot this on a diagram of Eh vs. pH it appears as a vertical line; constant at all Eh. What it means is that on this line the reactants and products have the activities\/pressures specified. At higher pH the activities of the reactants become larger relative to those of the products, as seen by rearranging equation (64):\r\n\r\n\\[\\;pH = -\\frac{\\Delta G^{\\circ}}{2.303RT\\,m} + \\frac{1}{m}\\log\\frac{a_A^{a}a_B^{b}}{a_C^{c}a_D^{d}} \\tag{65}\\]\r\n\r\nSimilarly, and to alter the idea a little, if we increase the activities of the products relative to the reactants, the pH will decrease.\r\n

Special Case 4: n = 0 and m = 0<\/h3>\r\nAn example of such a reaction is:\r\n\r\n\\[\\ce{2FeO(OH)_s = Fe2O3_s + H2O_l} \\tag{66}\\]\r\n\r\nThis is a perfectly valid reaction. (In leaching of nickel laterite ores, high pressures and temperature are used to effect just this reaction. Nickel may be bound with Fe(O)OH and is released as Ni+2<\/sup>aq<\/sub> when an acidic solution with the ore is heated in an autoclave at well above 100\u00b0C. Acid is not consumed in this reaction, but it is needed as a catalyst.) The reaction does not involve H+<\/sup> or e-<\/sup>. (This can be confirmed by balancing the reaction starting from 2 Fe(O)OH s<\/sub> = Fe2<\/sub>O3 s<\/sub>.) Hence it cannot be depicted on an Eh-pH diagram. What this means is that one or the other of Fe(O)OH or Fe2<\/sub>O3<\/sub> must be chosen when considering acid-base and electron transfer reactions involving these compounds. There is usually good reason to select the pertinent species from the context of the problem. For instance, at about 60\u00b0C Fe(O)OH is what forms when Fe+3 is precipitated from aqueous solution, while Fe2<\/sub>O3<\/sub> is the more thermodynamically stable and what forms by Fe+3<\/sup> precipitation at above 100\u00b0C.","rendered":"

Leaching (any many other chemical processes) may involve acid-base reactions, electron transfer and complexation. All such processes can be represented on Eh-pH diagrams. This section will develop the equations needed to be able to draw the diagrams.<\/p>\n

3.1 Reduction potential as a function of pH (the Eh Equation)<\/h2>\n

Consider a generalized half reaction:<\/p>\n

\\[\\ce{aA + bB + mH^+ + ne^- = cC + dD} \\tag{32}\\]<\/p>\n

The half reaction involves H+<\/sup>, reactants A and B, and produces products C and D. The factors a, b, m, c and d are stoichiommetric coefficients; n is the stoichiommetric coefficient for the moles of electrons per mole of half reaction. (Reactions of hydrometallurgical importance often involve acidic or basic pH.) We know that E\u00b0 for this half reaction is defined as \u0394E\u00b0 for the cell where the cathode is the reduction reaction under standard conditions and the anode is the standard H+<\/sup>\/H2<\/sub><\/em> half reaction. Further E for the half reaction under non-standard conditions <\/em>is defined as \u0394E for the cell where the cathode is the half reaction under any specified set of conditions and the anode is, again, the standard H+<\/sup>\/H2<\/sub> half cell<\/em>:<\/p>\n

[latex]\\begin{align*} \\ce{aA}&+\\ce{bB}+\\ce{mH+}+\\ce{ne-}=\\ce{cC}+\\ce{dD}\\tag{33} \\\\ \\ce{nH+}&=\\ce{n\/2H2}+\\ce{ne-} \\\\ \\hline \\text{net: }\\ce{aA}&+\\ce{bB}+\\ce{mH+(a_H+)}+\\ce{n\/2H2(1 atm)} =\\ce{cC}+\\ce{dD}+\\ce{nH+(a_H+ = 1)}*\\tag{34} \\end{align*}[\/latex]<\/p>\n

For this cell,<\/p>\n

\\[\\Delta E = \\Delta E^{\\circ} – \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}(a_{H^+}=1)^n}{a_A^{a}a_B^{b}a_{H^+}^{n}(P_{H_2}=1)^{n\/2}} \\tag{35}\\]<\/p>\n

\\[\\Delta E = \\Delta E^{\\circ} – \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}a_{H^+}^{m}} \\tag{36}\\]<\/p>\n

This is equivalent to writing,<\/p>\n

\\[\\;E = E^{\\circ} – \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}a_{H^+}^{m}} \\tag{37}\\]<\/p>\n

where E\u00b0 is the standard half reaction (32) voltage. This is what we would get if we used the half reaction alone to write the Nernst equation for E and ignored the electrons. This works because the effects of the standard H+\/H2 half reaction fall away in the expression for the cell voltage; E\u00b0H+\/H2 = 0 V and aH+ = 1 and PH2 = 1 in half reaction (34); all the factors that affect the cell voltage for reaction are ascribed to half reaction (23) alone. Equation (37) can now be developed into the form we will need.<\/p>\n

\n

Solute activities are on the molal scale. Properly gas pressures should be fugacities, although at 1 atm pressure fugacity is nearly equal to pressure in many cases. Some data compilations use pressures or fugacities on the bar scale instead; 1 atm = 1.01325 bar. In most cases the error caused by this difference is negligible. Ionic activities may differ greatly from the concentrations of the ions, particularly as concentration increases (which is of practical interest in hydrometallurgy), since ionic solutions are highly non-ideal.<\/p>\n<\/div>\n

First, from this point on we give E a new symbol, Eh<\/a>; E = Eh. This only serves to indicate that the half reaction potential is referenced to the standard H+\/H2 half reaction.<\/p>\n

Substituting \u2013DG\u00b0\/nF for DE\u00b0 and separating out the aH+ term,<\/p>\n

\\[\\;E_h = -\\frac{\\Delta G^{\\circ}}{nF} – \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}} – \\frac{2.303RT}{nF}\\log\\frac{1}{a_{H^+}^{m}} \\tag{38}\\]<\/p>\n

\\[\\;E_h = -\\frac{\\Delta G^{\\circ}}{nF} – \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}} – \\frac{2.303RT\\,m}{nF}\\,pH \\tag{39}\\]<\/p>\n

This is the form we will use to calculate Eh as a function of pH, solute activities and gas pressures. The quantity,<\/p>\n

\\[\\frac{a_C^{c} a_D^{d}}{a_A^{a} a_B^{b}}\\tag{40}\\]<\/p>\n

we will call Q-H. It is the reaction quotient for all species except H+; aH+ is taken care of by the pH term. Then,<\/p>\n

\\[\\;E_h = -\\frac{\\Delta G^{\\circ}}{nF} – \\frac{2.303RT}{nF}\\log Q_H – \\frac{2.303RT\\,m}{nF}\\,pH \\tag{41}\\]<\/p>\n

In order to use this equation we must determine \u0394G\u00b0 data at the specified temperature, and specify the activities of the solutes and pressures of the gases. Then the terms,<\/p>\n

\\[-\\frac{\\Delta G^{\\circ}}{nF} \\quad \\text{and} \\quad \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}\\tag{42}\\]<\/p>\n

are constants and Eh is a linear function of pH with slope = -2.303RTm\/nF. These lines are plotted on Eh-pH diagrams to graphically show reduction potentials for couples as a function of pH.<\/p>\n

Calculation of \u0394G\u00b0<\/h3>\n

From thermodynamics it is known that for a chemical reaction involving reactants i and products j, with reactant stoichiommetric coefficients \u03bdi<\/sub> and product stoichiommetric coefficients \u03bdj<\/sub>,<\/p>\n

\\[\\Delta G^{\\circ} = \\sum_{j}\\nu_j \\Delta G_f^{\\circ\\,j} – \\sum_{i}\\nu_i \\Delta G_f^{\\circ\\,i} \\tag{43}\\]<\/p>\n

where \u0394G\u00b0fj and \u0394G\u00b0fi are the standard free energies of formation of the species involved in the reaction. These are defined for formation reactions, which for uncharged species involves only the elements as reactants to form only the specified product. Further, all species are at a specified temperature and in their standard states, i.e. pure and the most stable chemical form of the species and 1 atm pressure (or 1 bar, depending on the compilation). An example is,<\/p>\n

\\[\\ce{H2_g + \\tfrac{1}{8}S8_s = H2S_g,\\ P = 1\\ atm,\\ 25^{\\circ}C} \\tag{44}\\]<\/p>\n

\n

An alternative rendering of the Eh equation can be developed. Free electrons do not exist in solutions, but they are free to move in metals and semiconductors (e.g. electrodes and wires). In this environment we can allow for the activity of electrons. This allows us to consider the half reaction alone, at least conceptually. We bear in mind that half reactions do not occur in isolation.<\/em> There must be two; the electrons have to come from one reagent and be transferred to another. That being said consider as an example the half reaction,<\/p>\n

\\[ \\ce{Tl^{3+}_{(aq)} + 2e^- = Tl^{+}_{(aq)}} \\]<\/p>\n

From thermodynamics,
\n
\n\\[
\n\\Delta G^\\circ = -nF E^\\circ = -RT\\ln K = -2.303\\,RT \\log K= -2.303\\,RT \\log\\!\\left(
\n\\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}} a_{e^-}^n}
\n\\right) (n = 2) \\tag{A}
\n\\]<\/small><\/p>\n

Rearranging gives:<\/p>\n

\\[
\n\\frac{nF E^\\circ}{2.303\\,RT}= n \\log \\frac{1}{ae^-}
\n+ \\log \\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}}}
\n= n \\log a_{e^-}
\n+ \\log \\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}}}
\n\\]<\/p>\n

\\[
\n\\frac{F E^\\circ}{2.303\\,RT}
\n= -\\log a_{e^-}
\n+ \\frac{1}{n} \\log\\!\\left(
\n\\frac{a_{\\ce{Tl+}}}{a_{\\ce{Tl^{3+}}}}
\n\\right)
\n\\]<\/p>\n

Analogously to pH let -log ae- = pe. Further, when we are dealing with E\u00b0 alone, this implies standard conditions, i.e. all reactants and products at unit activity. Then,<\/p>\n

\\[
\n\\frac{F E^\\circ}{2.303\\,RT} = \\mathrm{pe}^\\circ
\n\\qquad\\text{or}\\qquad
\nE^\\circ = \\frac{2.303\\,RT}{F} \\,\\mathrm{pe}^\\circ
\n\\]<\/p>\n

If we then defined pe under non-standard conditions as,<\/p>\n

\\[
\n\\frac{FE^\\circ}{2.303\\,RT} = \\mathrm{pe^\\circ}
\n\\qquad\\text{or}\\qquad
\nE^\\circ = \\frac{2.303\\,RT}{F} \\,\\mathrm{pe^\\circ}
\n\\]<\/p>\n

and substitute into the Nernst equation, we obtain,<\/p>\n

\\[
\n\\mathrm{pe}
\n= \\mathrm{pe}^\\circ – \\frac{1}{n}\\log Q
\n\\]<\/p>\n

This is often used as an alternative way to express E\u00b0 and then Eh-pH diagrams become e-pH diagrams. However, we will not make use of it in this course.<\/p>\n<\/div>\n

Sulfur may exist in a number of solid forms (allotropes). The most stable is the rhombohedral form (referring to the crystal structure) and this is what is implied here. By definition, \u0394G\u00b0f for the elements in their standard states = 0. Then all of the free energy change for the formation reaction is carried by the compound. These data allow us to calculate \u0394G\u00b0 for any chemical reaction as per equation (A).<\/p>\n

However, for ions in solution the problem is more complex. A particular ion, e.g. SO4<\/sub>2-<\/sup>, will not exist in solution without countercations. This is overcome by writing formation reactions for ions with H+<\/sup> and H2<\/sub> involved as well and<\/em> setting \u0394G\u00b0f<\/sub> for both H+<\/sup> and H2<\/sub> in their standard states to be 0. That \u0394G\u00b0f for H+ is 0 has already been implied by setting E\u00b0H+\/H2<\/sub> to 0:<\/p>\n

\\[\\ce{2H^+ + 2e^- = H2_g} \\tag{45}\\]<\/p>\n

Since,<\/p>\n

\\[\\;E^{\\circ} = 0 = -\\frac{\\Delta G^{\\circ}}{nF}
\n= -\\frac{1}{nF}\\left(\\Delta G_f^{\\circ\\,H_2} – 2\\Delta G_f^{\\circ\\,H^+}\\right)
\n\\tag{46}\\]<\/p>\n

and \u0394G\u00b0f<\/sub>H2<\/sup> = 0 by definition, it follows that \u0394G\u00b0f<\/sub>H+<\/sup> = 0 as well. (We can ignore the electron for the same reasons we did in developing the Eh equation. The standard H+<\/sup>\/H2<\/sub> half reaction is implicit as the anode.)<\/p>\n

To illustrate, the formation reaction for SO4<\/sub>2-<\/sup> is then written as the sum of:<\/p>\n

[latex]\\begin{align*} \\ce{\\tfrac{1}{8}S8_s}&+\\ce{2O2_g}+\\ce{2e-}=\\ce{SO4^{2-}_{aq}}\\tag{45} \\\\ \\ce{H2_g}&=\\ce{2H+_{aq}}+\\ce{2e-}\\tag{46} \\\\ \\hline \\text{net: }\\ce{\\tfrac{1}{8}S8_s}&+\\ce{2O2_g}+\\ce{H2_g} =\\ce{SO4^{2-}_{aq}}+\\ce{2H+_{aq}}\\tag{47} \\end{align*}[\/latex]<\/p>\n

Then,<\/p>\n

[latex]\\begin{align*} \\Delta G_f^{\\circ} &= 2\\Delta G_f^{\\circ}(H^+) + \\Delta G_f^{\\circ}(SO_4^{2-}) - \\tfrac{1}{8}\\Delta G_f^{\\circ}(S_8) - 2\\Delta G_f^{\\circ}(O_2) - \\Delta G_f^{\\circ}(H_2) \\tag{48} \\\\ &= \\Delta G_f^{\\circ}(SO_4^{2-}) \\tag{49} \\end{align*}[\/latex]<\/p>\n

(All other terms are 0 by definition.) Similarly, for an example of the formation reaction of a cation,<\/p>\n

[latex]\\begin{align*} \\ce{Cu_s}&=\\ce{Cu^{2+}_{aq}}+\\ce{2e-}\\tag{50} \\\\ \\ce{2H+_{aq}}&+\\ce{2e-}=\\ce{H2_g}\\tag{51} \\\\ \\hline \\text{Net: }\\ce{Cu_s}&+\\ce{2H+_{aq}}=\\ce{Cu^{2+}_{aq}}+\\ce{H2_g}\\tag{52} \\end{align*}[\/latex]<\/p>\n

Tabulations of \u0394G\u00b0f data are readily available. Finally, we can now demonstrate the calculation of \u0394G\u00b0 for a half reaction. For example,<\/p>\n

\\[\\ce{SO4^{2-}_{aq} + 8H^{+}_{aq} + 6e^- = S_s + 4H2O_l} \\tag{53}\\]<\/p>\n

For convenience we take 1\/8S8<\/sub> to be equivalently stated as S. The standard H+\/H2 anode half reaction is, as always, implicit. This can be explicitly added as,<\/p>\n

\\[\\ce{3H2_g\\ (P_{H2}=1\\ atm) = 6H^{+}_{aq}\\ (a_{H^+}=1) + 6e^-} \\tag{54}\\]<\/p>\n

recalling that half reaction (53) is equivalent to a cell where the anode is the standard H+<\/sup>\/H2<\/sub> half reaction (54). And again the effects of standard H+<\/sup> and H2<\/sub> disappear. For half reaction (53),<\/p>\n

[latex]\\begin{align*} \\Delta G^{\\circ} &= 4\\Delta G_f^{\\circ}(H_2O) + \\Delta G_f^{\\circ}(S) - \\Delta G_f^{\\circ}(SO_4^{2-}) - 8\\Delta G_f^{\\circ}(H^+) \\tag{55} \\\\ &= 4\\Delta G_f^{\\circ}(H_2O) - \\Delta G_f^{\\circ}(SO_4^{2-}) \\tag{56} \\end{align*}[\/latex]<\/p>\n

Usually \u0394G\u00b0f<\/sub> data are expressed in kJ\/mol. As we shall see, when using the data in the equation for Eh, \u0394G\u00b0 must be converted to J\/mol.<\/p>\n

Special Case 1: m = 0<\/h3>\n

When m = 0 no protons are involved in the half reaction and the -2.303RTmpH\/nF term goes to zero. Then the Eh is a constant, independent of pH, under the specified conditions. \u00a0It is a flat line on the Eh-pH diagram.<\/p>\n

Special Case 2: When H+ Appears as a Product in the Half Reaction<\/h3>\n

Note than m is a positive number when H+<\/sup> appears as a reactant in the half reaction. If H+<\/sup> appears as a product (which it does in some cases), then,<\/p>\n

\\[\\;E_h
\n= -\\frac{\\Delta G^{\\circ}}{nF}
\n– \\frac{2.303RT}{nF}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}
\n– \\frac{2.303RT}{nF}\\log\\!\\left(a_{H^+}^{\\,m}\\right)
\n\\tag{57}\\]<\/p>\n

Then the pH term becomes,<\/p>\n

\\[-\\frac{2.303RT}{nF}\\log\\!\\left(a_{H^+}^{\\,m}\\right)
\n= +\\frac{2.303RT\\,m}{nF}\\,pH \\tag{58}\\]<\/p>\n

If we preserve the same form as the Eh equation [19], with a minus sign in front of the pH term, we need to make m be a negative number. Thus when H+<\/sup> appears as a product, m is negative.<\/em><\/p>\n

Special Case 3: Reactions that do not Involve Electrons (n = 0)<\/h3>\n

As mentioned previously, many reactions of interest involve H+ (or OH-) and do not involve electrons. The simplest case is the dissociation of an acid in aqueous solution. We can obtain an equation for the pH as a function of solute activities and gas pressures. Consider the general reaction:<\/p>\n

\\[\\ce{aA + bB + mH^+ <=> cC + dD} \\tag{59}\\]<\/p>\n

\\[\\Delta G^{\\circ} = -2.303RT\\log K \\tag{60}\\]<\/p>\n

\\[\\Delta G^{\\circ}
\n= -2.303RT\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}a_{H^+}^{m}}
\n\\tag{61}\\]<\/p>\n

\\[\\Delta G^{\\circ}
\n= -2.303RT\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}
\n– 2.303RT\\log\\frac{1}{a_{H^+}^{m}}
\n\\tag{62}\\]<\/p>\n

\\[\\Delta G^{\\circ}
\n= -2.303RT\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}
\n– 2.303RT\\,m\\,pH
\n\\tag{63}\\]<\/p>\n

Rearranging yields,<\/p>\n

\\[\\;pH
\n= -\\frac{\\Delta G^{\\circ}}{2.303RT\\,m}
\n– \\frac{1}{m}\\log\\frac{a_C^{c}a_D^{d}}{a_A^{a}a_B^{b}}
\n= -\\frac{\\Delta G^{\\circ}}{2.303RT\\,m}
\n– \\frac{1}{m}\\log Q_H
\n\\tag{64}\\]<\/p>\n

To use this equation we again need \u0394G\u00b0 for the reaction and the activities for the solutes and pressures for the gases need to be specified. Again, if H+<\/sup> appears as a product, then the sign of m is negative.<\/em> Once the quantities have been specified the equation yields a number for pH, i.e. pH is a constant, independent of Eh. If we plot this on a diagram of Eh vs. pH it appears as a vertical line; constant at all Eh. What it means is that on this line the reactants and products have the activities\/pressures specified. At higher pH the activities of the reactants become larger relative to those of the products, as seen by rearranging equation (64):<\/p>\n

\\[\\;pH = -\\frac{\\Delta G^{\\circ}}{2.303RT\\,m} + \\frac{1}{m}\\log\\frac{a_A^{a}a_B^{b}}{a_C^{c}a_D^{d}} \\tag{65}\\]<\/p>\n

Similarly, and to alter the idea a little, if we increase the activities of the products relative to the reactants, the pH will decrease.<\/p>\n

Special Case 4: n = 0 and m = 0<\/h3>\n

An example of such a reaction is:<\/p>\n

\\[\\ce{2FeO(OH)_s = Fe2O3_s + H2O_l} \\tag{66}\\]<\/p>\n

This is a perfectly valid reaction. (In leaching of nickel laterite ores, high pressures and temperature are used to effect just this reaction. Nickel may be bound with Fe(O)OH and is released as Ni+2<\/sup>aq<\/sub> when an acidic solution with the ore is heated in an autoclave at well above 100\u00b0C. Acid is not consumed in this reaction, but it is needed as a catalyst.) The reaction does not involve H+<\/sup> or e–<\/sup>. (This can be confirmed by balancing the reaction starting from 2 Fe(O)OH s<\/sub> = Fe2<\/sub>O3 s<\/sub>.) Hence it cannot be depicted on an Eh-pH diagram. What this means is that one or the other of Fe(O)OH or Fe2<\/sub>O3<\/sub> must be chosen when considering acid-base and electron transfer reactions involving these compounds. There is usually good reason to select the pertinent species from the context of the problem. For instance, at about 60\u00b0C Fe(O)OH is what forms when Fe+3 is precipitated from aqueous solution, while Fe2<\/sub>O3<\/sub> is the more thermodynamically stable and what forms by Fe+3<\/sup> precipitation at above 100\u00b0C.<\/p>\n

definition<\/span>