{"id":2179,"date":"2026-01-21T15:59:18","date_gmt":"2026-01-21T20:59:18","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/?post_type=chapter&p=2179"},"modified":"2026-03-23T14:34:33","modified_gmt":"2026-03-23T18:34:33","slug":"review-of-basic-electrochemistry","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/chapter\/review-of-basic-electrochemistry\/","title":{"raw":"2. Review of Basic Electrochemistry","rendered":"2. Review of Basic Electrochemistry"},"content":{"raw":"Before we can delve into Eh-pH diagrams, the associated electrochemistry has to be in place. If this material is familiar, it may be skipped.\r\n

2.1 Electron Transfer Reactions<\/h2>\r\nMany (though not all) reactions involve transfer of electrons between reagents. These may be called electron transfer reactions, redox reactions or electrochemical reactions (the latter refers specifically to processes occurring at electrodes). In normal chemical systems the electron does not exist outside of being attached to atoms, i.e. free electrons do not survive. The energy of a free electron would be so high that it would react with the first thing it encounters. The transfer of electrons from one chemical species to another is a means of transferring energy. Electrons moving from one species to another are essentially energy in transit, just as heat is energy in transit.\r\nThere are two reasons that electrons transfer between chemical species. One is that the system (the chemical reaction) and its surroundings together have higher net entropy than the reactants - a spontaneous chemical reaction.* The electrons in one reactant (the reducing agent) are sufficiently energetic to be spontaneously accepted by another reactant (the oxidizing agent). The second reason is that we force an unfavourable electron transfer reaction by imparting extra energy to the electrons, by means of an external applied voltage. This is the basis for electrolysis.\r\n\r\nFurther, it can be seen that we can consider an electron transfer reaction to be comprised of two parts - something getting reduced, and something getting oxidized. This is the only way that electron transfer reactions can occur. The electrons must be taken from one reagent and accepted by another. (Again, free electrons do not exist.) Thus we can separate an electron transfer reaction into two half reactions, one a reduction and one an oxidation. It must be borne in mind that half reactions never occur in isolation; the electrons must come from one reactant be delivered to another.\r\n

2.2 The Electrochemical Potential<\/h2>\r\nThis is actually a quite straightforward quantity. It is simply the tendency of a chemical species to accept electrons and be reduced under the specified conditions (like pressure, temperature, activities of reactants and products). The higher the potential, the stronger the driving force for this to occur. Now we are specifically focusing on a reduction half reaction, e.g.\r\n\r\n\\[\\ce{Cu^{2+}_{aq} + 2e^- = Cu_s} \\tag{1}\\]\r\n
\r\n\r\nRecall that what drives chemical reactions (makes them spontaneous) is that over the system and its surroundings there is a net increase in entropy. Metaphorically, entropy can be thought of as the concentration of energy. It is intuitively reasonable that concentrated energy tends to naturally disperse; degrades to a lower \"concentration.\" The Gibbs free energy equation reflects this link between entropy and spontaneity. Since\r\n\u0394G = \u0394H - T \u0394S at some specified temperature T,\r\n\r\n\\[\\ce{\\frac{-\\Delta G}{T} = \\frac{-\\Delta H}{T} + \\Delta S}\\]\r\n\r\n\u0394H is the enthalpy change for the system. Then -\u0394H is the heat flow to the surroundings. At constant pressure \u0394H is the heat flow into the system. Then\r\n-\u0394H\/T = -q\u03a1<\/sub>\/T (q\u03a1<\/sub> being heat flow at constant pressure) and this has the form of the entropy. It can be shown (see suitable thermodynamics textbooks) that in fact\r\n-\u0394H\/T is the entropy change of the surroundings. Thus,\r\n\r\n\\[\\ce{-\\Delta G\/T = \\Delta S_{surroundings} + \\Delta S_{system} = Total\\ \\Delta S}\\]\r\n\r\nThis is the basis upon which we say that a reaction is spontaneous; if \u0394G < 0. It applies to all chemical processes.\r\n\r\n<\/div>\r\nThe Cu+2<\/sup> ion in aqueous solution has some potential to accept an electron and be reduced to Cu. If we can measure that electrochemical potential for any conceivable half reaction, we can understand and combine those half reactions to perform chemical transformations. The problem though is that there is no way to measure the reduction potential for a half reaction in isolation; electron transfer reactions require a reactant source of electrons and a reactant to accept them. For instance,\r\n\r\n\\[\r\n\\begin{align*}\r\n&\\ce{Cu^{2+}_{aq}}+\\ce{2e-}=\\ce{Cu_s}\\tag{2} \\\\\r\n&\\ce{H2_g}=\\ce{2H+_{aq}}+\\ce{2e-}\\tag{3} \\\\\r\n\\hline\r\n\\text{net: }&\\ce{Cu^{2+}_{aq}}+\\ce{H2_g}=\\ce{Cu_s}+\\ce{2H+_{aq}}\\tag{4}\r\n\\end{align*}\r\n\\]\r\n\r\nHydrogen gas is the source of the electrons; Cu+2<\/sup> accepts them. (It is the same in electrolysis reactions, which are thermodynamically unfavourable and are forced to go by imposition of a voltage; the electrons are forcibly taken from one reactant and forced onto another.) But, now we have two half reactions that are combined into an overall reaction. We can measure the potential difference<\/em> (voltage) that is manifested between the two half cells as illustrated in the figure below. This is the measure of the driving force for the electrons to transfer as per the reaction. But, electron transfer occurs. It does not undergo chemical transformation. The [pb_glossary id=\"2719\"]salt bridge[\/pb_glossary] allows the electrical circuit to be completed by ionic conduction between compartments, otherwise no current would flow.<\/a>\r\n\r\n[caption id=\"attachment_2861\" align=\"aligncenter\" width=\"1375\"]\"Diagram Figure 2.1. Schematic illustration of an electrochemical cell comprised of an H+\/H2 half reaction and a Cu+2\/Cu half reaction. The voltage of the cell can be measured between the electrodes (anode and cathode). The Pt electrode simply acts as a surface where electron transfer occurs. It does not undergo chemical transformation. The salt bridge allows the electrical circuit to be completed by ionic conduction between compartments, otherwise no current would flow.[\/caption]\r\n\r\nThe potential difference is the net effect of two half reactions. There is no way to make a half reaction occur in isolation, and so no way to measure the absolute reduction potential of a half reaction.\r\n\r\nThe solution to this conundrum is to select one half reaction and assign it a voltage of 0 V<\/em>. Then all other half reactions are referenced to this one standard. For half reaction (4<\/span>) then, occurring in a cell as illustrated in Figure 2.1<\/a>, we measure the potential difference of the cell under some set of standard conditions and arbitrarily<\/em> say that the half cell voltage for the H+<\/sup>\/H2<\/sub> half reaction will be 0 V. Then the whole of the cell voltage<\/em> is ascribed to the Cu+2<\/sup>\/Cu half reaction. Now we have a measure of the tendency of Cu+2<\/sup> to undergo reduction relative to the H+<\/sup>\/H2<\/sub> half reaction<\/em>. Then any half reaction, under this same standard set of conditions (pressure, temperature and composition) can be combined with the H+<\/sup>\/H2<\/sub> half cell and the cell's voltage is then assigned wholly to the half reaction of interest. In other words, every half reaction voltage is precisely equal to the voltage for a cell where the reduction half cell is the half reaction of interest and the [pb_glossary id=\"2661\"]anode [\/pb_glossary]is the H+<\/sup>\/H2<\/sub> half cell under specified standard conditions.<\/em><\/strong>\r\n\r\nSo, when we set up a cell like that in Figure 2.1<\/a>, at 25\u00b0C and 1 atm pressure with unit activities of the solutes (on the molal scale), we should be able to measure, in principle, a voltage of +0.34 V; the potential difference measured between the [pb_glossary id=\"2663\"]cathode [\/pb_glossary](Cu+2<\/sup>\/Cu) and the anode (H+<\/sup>\/H2<\/sub>), i.e. cathode minus anode*. This is the standard reduction potential for the Cu+2<\/sup>\/Cu couple, i.e. half reaction (1). This tells us that Cu+2<\/sup> has a greater tendency to be reduced than H+<\/sup>by 0.34 V under these conditions. Alternatively, if we were to use the Cr+3<\/sup>\/Cu+2<\/sup> couple instead of Cu+2<\/sup>\/Cu in Figure 2.1<\/a>, we would find that the cell's potential difference would be\r\n-0.42 V (cathode voltage - H+<\/sup>\/H2<\/sub> anode voltage). This indicates that Cr+3 is a weaker oxidizing agent (has a lower potential to be reduced) than H+, or in other words, H+<\/sup> is 0.42 V more strongly oxidizing than Cr+3<\/sup>.\r\n

2.3 Calculating \u0394E\u00b0 for Electron Transfer Reactions<\/h2>\r\nThe standard cell voltage is denoted \u0394E\u00b0, while the standard reduction potential is E\u00b0. Under non-standard conditions the superscript \u00b0 is dropped.\r\n
\r\n\r\nThis may be quite difficult in practice. There are many complicating factors that can make accurate potential measurements difficult or sometimes impossible. However, there are also plenty of indirect methods for obtaining such data. For instance, other thermodynamic measurements may allow us to obtain \u0394G\u00b0 for a reaction. Then through the relationships,\r\n\r\n\\[\\ce{\\Delta G^{\\circ} = -nF\\Delta E^{\\circ} = -RT\\ln K}\\]\r\n\r\nthe cell voltage can be obtained. Here n is the moles of electrons per mole of reaction and F is the Faraday constant (the charge of a mole of electrons, 96485 C\/mole of electrons). These quantities will be developed later.\r\n\r\n<\/div>\r\nMost generally we are interested in all electron transfer reactions, not just those that involve the H+<\/sup>\/H2<\/sub> couple. There is a very simple rule for determining the standard cell voltage for any electron transfer reaction:\r\n\r\n\\[\\ce{\\Delta E^{\\circ} = E^{\\circ}_{red} - E^{\\circ}_{ox}} \\tag{5}\\]\r\n\r\nwhere E\u00b0red<\/sub> is the standard reduction potential for the couple undergoing reduction, and E\u00b0ox<\/sub> is the standard reduction potential for the couple undergoing oxidation. DO NOT change the sign of E\u00b0ox. Simply calculate the appropriate difference in standard reduction potentials<\/em><\/strong>.\r\n\r\nConsider the reaction,\r\n\r\n\\[\r\n\\begin{align*}\r\n&\\ce{Cu^{2+}_{aq}}+\\ce{2e-}=\\ce{Cu_s}\\tag{6} \\\\\r\n&\\ce{Fe_s}=\\ce{Fe^{2+}_{aq}}+\\ce{2e-}\\tag{7} \\\\\r\n\\hline\r\n\\text{net: }&\\ce{Cu^{2+}_{aq}}+\\ce{Fe_s}=\\ce{Cu_s}+\\ce{Fe^{2+}_{aq}}\\tag{8}\r\n\\end{align*}\r\n\\]\r\n\r\n(This used to be the basis for recovery of copper metal from dilute copper leach solutions using scrap iron.) The standard reduction potential for Cu+2<\/sup>\/Cu is\r\n+0.34 V. That for Fe+2<\/sup>\/Fe, i.e. Fe+2<\/sup> + 2e- = Fe, is -0.44 V. In the reaction Cu+2<\/sup> is reduced and Fe is oxidized. By equation [1] the standard cell voltage is:\r\n\r\n\\[\\ce{\\Delta E^{\\circ} = 0.34 - (-0.44) = +0.78\\ V} \\tag{9}\\]\r\n\r\nThe cell is shown in Figure 2.2<\/a>. (The two half cells must be separated in order to be able to have the electrons pass through the external circuit and facilitate potential difference measurements, as explained in the figure caption.) So why don't we reverse the sign of E\u00b0Fe+2<\/sup>\/Fe in order to calculate \u0394E\u00b0? (Students often want to do this because it seems intuitively reasonable, since the half reaction has been reversed to become an oxidation half reaction.) BUT, the electrons can, in principle, be taken by Cu+2<\/sup> from Fe, or they can be taken by Fe+2<\/sup> from Cu. In other words, we have to compare the standard reduction potentials. Hence, do not flip the sign of E\u00b0ox.<\/em> What DE\u00b0 tells us is the net driving force for the electrons to be accepted by one couple, rather than the other, for the reaction as written. Note that E\u00b0Cu+2<\/sup>\/Cu > E\u00b0Fe+2<\/sup>\/Fe. This means that the Cu+2<\/sup>\/Cu electrode is positive with respect to the Fe+2\/Fe electrode, and then the Fe+2<\/sup>\/Fe electrode is negative with respect to the Cu+2<\/sup>\/Cu electrode.\r\n\r\nIf we wrote the reaction as,\r\n\r\n\\[\\ce{Cu_s + Fe^{2+}_{aq} = Cu^{2+}_{aq} + Fe_s} \\tag{10}\\]\r\n\r\nthen \u0394E\u00b0 would be,\r\n\r\n\\[\\ce{-0.44 - 0.34 = -0.78\\ V} \\tag{11}\\]\r\n\r\nIn this case we would be measuring the potential difference of the Fe+2<\/sup>\/Fe electrode with respect to the Cu+2<\/sup>\/Cu electrode. Now Fe+2<\/sup> + e- = Fe is the reduction and the oxidation is Cu = Cu+2<\/sup>+ 2e-<\/sup>. Clearly it is important which way round we write the reaction. *<\/a>\r\n\r\n[caption id=\"attachment_2862\" align=\"aligncenter\" width=\"1547\"]\"(a) Figure 2.2. (a) An electrochemical cell comprised of the Cu+2\/Cu and Fe+2\/Fe couples, where Cu+2 oxidizes Fe. The electrodes are both directly involved in the reaction, as well as being surfaces at which electron transfer occurs. (b) If Fe metal were placed in direct contact with a Cu+2-containing solution the electrons would transfer from Fe, through the plated Cu metal and to Cu+2 in solution. This arrangement could not be used to measure potential difference. The electrons cannot pass through an external circuit (even if we connected a wire to the Cu and the Fe), which is what is needed to do measurements, or use the energy released. Both arrangements involve the same reaction. In the former electrical energy can be extracted or input, and in the latter it cannot.[\/caption]\r\n\r\n \r\n\r\nThe second common error is to multiply \u0394E\u00b0 by the number of electrons involved (the value of n). The quantity n is the moles of electrons transferred per mole of reaction as written. For reaction (7<\/span>) n = 2 moles e-\/mole of reaction. One mole of reaction as per reaction (7)<\/span> means 1 mole Cu+2<\/sup> reacting with 1 mole Fe to form 1 mol Fe+2<\/sup> and 1 mole Cu. If we wrote the reaction instead as,\r\n\r\n\\[\r\n\\begin{align*}\r\n&\\tfrac{1}{2}\\ce{Cu^{2+}_{aq}}+\\ce{e-}=\\tfrac{1}{2}\\ce{Cu_s}\\tag{12}\\\\\r\n&\\tfrac{1}{2}\\ce{Fe_s}=\\tfrac{1}{2}\\ce{Fe^{2+}_{aq}}+\\ce{e-}\\tag{13}\\\\\r\n\\hline\r\n\\text{net: }&\\tfrac{1}{2}\\ce{Cu^{2+}_{aq}}+\\tfrac{1}{2}\\ce{Fe_s}=\\tfrac{1}{2}\\ce{Cu_s}+\\tfrac{1}{2}\\ce{Fe^{2+}_{aq}}\\tag{14}\r\n\\end{align*}\r\n\\]\r\n\r\nNow 1 mole of reaction means 0.5 mole Cu+2<\/sup> reacting with 0.5 mole Fe, etc., and n = 1. Multiplying \u0394E\u00b0 by n yields something that is no longer a voltage, but rather,\r\n\r\n\\[\\ce{n\\Delta E^{\\circ} = moles\\ e^- \\ volts\/mol\\ of\\ reaction} \\tag{15}\\]\r\n\r\nThis is actually more akin to energy. (i.e., converting moles of electrons to charge, we get charge\/mole x voltage = energy in J\/mole.) Further, the voltage is the potential for electrons to pass from one reactant to another under specified conditions, regardless of the number involved. It is the same if we pass 1 kmol of electrons or 1\/1000 mole of electrons. Finally, if we wrote,\r\n\r\n\\[\\ce{10{,}000Cu^{2+}_{aq} + 10{,}000Fe_s = 10{,}000Cu_s + 10{,}000Fe^{2+}_{aq}} \\tag{16}\\]\r\n\r\n(and there is no reason why we can't) would it seem reasonable to multiply 0.78 V by 10,000 and anticipate a cell voltage of 7800 V? Therefore, when calculating a standard potential difference DO NOT multiply <\/em><\/strong>\u0394E\u00b0 by n<\/em><\/strong>.\r\n

2.4 More on E\u00b0 and \u0394E\u00b0<\/h2>\r\nWe noted earlier that E\u00b0, the standard reduction potential, is a measure of the tendency of a species to be reduced. The highest standard reduction potential is about +3 V (e.g. E\u00b0F2\/F<\/sub>- = 2.87 V). The lowest is about -3 V (e.g. E\u00b0Li+\/Li<\/sub> = -3.05 V). Thus E\u00b0H+\/H2 = 0 V is in the middle of the range, which is consistent with the electronegativity for H being about half way between the highest and lowest values. A table of standard reduction (Table 2.1 below) potentials at 25\u00b0C is provided below. (Conversions to other temperatures can be performed.) The magnitude of the reduction potential indicates how strongly oxidizing a species is. The more strongly a chemical species wants electrons, the more weakly the reduced form will want to give them up. That is, a strong oxidant is reduced to form a weak reductant. For example, F2 is a very powerful oxidizing agent. It has a high reduction potential (2.87 V). This means that F- has a very weak tendency to give up an electron; it is a very weak reducing agent. Conversely, Li+ is a very weak oxidizing agent; it has a very low reduction potential. Hence Li+ does not have much tendency to accept electrons. It follows then that Li metal is a very powerful reducing agent. The trends are illustrated in the figure below.\r\n\r\n[caption id=\"attachment_2863\" align=\"aligncenter\" width=\"924\"]\"Diagram Figure 2.3. Trends in oxidizing and reducing strength for redox couples.[\/caption]\r\n\r\n[table id=117 \/]\r\n\r\nStandard conditions provides a common basis for comparison of E\u00b0 values and hence \u0394E\u00b0. However, there is deeper thermodynamic significance as well. In thermodynamics \u0394G (under any conditions, not just standard conditions) is the quantity we use to assess if a reaction at constant pressure is spontaneous or not. It turns out that DE\u00b0 is directly proportional to \u0394G\u00b0:\r\n\r\n\\[\\ce{\\Delta G^{\\circ} = -nF\\Delta E^{\\circ}} \\tag{17}\\]\r\n\r\nwhere, as before, n = moles of electrons per mole of reaction and F = the charge of a mole of electrons (96,485 C\/mole e-<\/sup>). The units of \u0394G\u00b0 are J\/mol. It can be readily seen that on the basis of the units, the equation is reasonable:\r\n
\\[\\ce{n\\ moles\\ e^-\/mol \\times F\\ C\/mole\\ e^- \\times \\Delta E^{\\circ}\\ V = V C\/mol = J\/mol} \\tag{18}\\]<\/div>\r\n
<\/div>\r\n
\r\n\r\nIn fact, \u0394E\u00b0 provides us with the same information as \u0394G\u00b0, just on a different scale, and specifically for electron transfer reactions. Based on equation [5] \u0394E\u00b0 > 0 means that a reaction is favourable under standard conditions. Further, the larger the value of DE\u00b0 the stronger the driving force for the reaction to occur, and the more energy that will be released as a result. Similarly, if DE\u00b0 < 0 the reaction is not favourable under standard conditions. Making it go will require the input of electrical energy. This is an electrolysis.\r\n

2.5 \u00a0E and \u0394E Under Non-Standard Conditions: the Nernst Equation<\/h2>\r\nAs noted above, the most general condition for assessing if a reaction as written is spontaneous is if \u0394G < 0 under the specified conditions. Changes in conditions can make a reaction favourable or unfavourable, depending on factors such as temperature, pressure and composition. (Note that partial pressure of a gas is analogous to concentration or activity of a solute in solution, or of a metal in an alloy, etc.) Just as \u0394G\u00b0 = -nF\u0394E\u00b0, so,\r\n\r\n\\[\\ce{\\Delta G = -nF\\Delta E} \\tag{19}\\]\r\n\r\nFor a chemical reaction \u0394G is a function of pressure, temperature and composition:\r\n\r\n\\[\\ce{\\Delta G = \\Delta G^{\\circ} + RT\\ln Q} \\tag{20}\\]\r\n\r\nwhere R is the gas constant (8.31441 J\/mol K), T is the absolute temperature (Kelvin) and Q is the reaction quotient. The reaction quotient was presented in Chemistry Review Part II.\r\n
\r\n\r\nBy definition 1 VC = 1 J. When a charge of 1 C passes through a voltage of 1 V an energy of 1 J is consumed or released, depending on whether the charge is going from + to \u2013 or from \u2013 to +. Introductory textbooks on the subject typically show that,\r\n\r\n\\[\\ce{-w_e = \\Delta G^{\\circ} = -nF\\Delta E^{\\circ}}\\]\r\n\r\nwhere -we is the maximum possible electrical work (or more generally, non pressure-volume work) that can be extracted from an electrochemical cell.\r\n\r\n<\/div>\r\n \r\n\r\nBriefly, it has exactly the same form as the equilibrium constant, but where concentrations and\/or gas pressures (activities and fugacities, properly speaking) are not necessarily those corresponding to an equilibrium state. Q is an indicator of how far the system is from equilibrium. The reaction will naturally proceed with changes in composition until equilibrium is attained, i.e. Q \u2192 K. In order to force Q to diverge from K, energy must be supplied (e.g. an electrolysis). Another important relationship is,\r\n\r\n<\/div>\r\n\\[\\ce{\\Delta G^{\\circ} = -RT\\ln K = -nF\\Delta E^{\\circ}} \\tag{21}\\]\r\n\r\n(This is another important aspect of standard conditions.) Note that when \u0394G = 0, \u0394G\u00b0 = -RTln Q (equation [8]) and now Q = K. Thus all reactions naturally tend to drive towards \u0394G going to 0. Once this condition is reached, Q = K and the reaction has reached equilibrium. There will be no further net change in composition. Substituting \u0394G = -nF\u0394E into equation [8] yields,\r\n\r\n\\[\\ce{-nF\\Delta E = -nF\\Delta E^{\\circ} + RT\\ln Q} \\tag{22}\\]\r\n\r\n\\[\\ce{\\Delta E = \\Delta E^{\\circ} - \\frac{RT}{nF}\\ln Q} \\tag{23}\\]\r\n\r\nThis is the Nernst equation. For a spontaneous electron transfer reaction, \u0394E is a measure of how far the reaction is away from equilibrium. All natural processes tend toward decreasing \u0394E. Once equilibrium is attained \u0394E goes to 0 and the reaction ceases. For a galvanic cell (a battery), the voltage declines with time and eventually the battery dies. If \u0394E < 0 then the reaction as written is unfavourable. To make it go in this direction will require energy to overcome the negative \u0394E.\r\n\r\nBy this point it can be seen that if the reaction is reversed, so is the sign of \u0394G (or \u0394G\u00b0); if it is favourable in the direction written (\u0394G < 0), then it is unfavourable in the opposite direction (and \u0394G > 0). Likewise, reversing the direction of an electron transfer reaction requires that we reverse the sign of \u0394E\u00b0 (or \u0394E for non-standard conditions).\r\n

2.6 The Standard Oxidation Potential<\/h2>\r\nSometimes people characterize half reactions by standard oxidation potentials, rather than standard reduction potentials. This can be confusing. As always in thermodynamics, keeping the conventions straight is crucially important. A generic oxidation half reaction is,\r\n\r\n\\[\\ce{M_s = M^{n+} + ne^-} \\tag{24}\\]\r\n\r\nConsider again the Cu+2<\/sup>\/Cu reduction half reaction, Cu+2<\/sup> + 2e- = Cu. We defined the standard reduction potential as the cell voltage (cathode - anode) where the cathode was the Cu+2<\/sup>\/Cu couple and the anode was the H+\/H2 couple (set to 0 V). It tells us that Cu+2<\/sup> is 0.34 V more prone to accept electrons than is H+. And if you set up the cell, as in Figure 2.1<\/a>, and measure the potential of the Cu+2<\/sup>\/Cu electrode with respect to the H+<\/sup>\/H2<\/sub> electrode, the voltage will be +0.34 V. This means that Cu is more prone to hold onto its electrons than is H2<\/sub>; Cu is harder to oxidize than H2<\/sub>. Quantitatively, H2 is 0.34 V more prone to be oxidized than Cu, or, since E\u00b0H+<\/sup>\/H2<\/sub> = 0 V (by convention, for either its reduction or oxidation potential), the oxidation potential for the Cu+2<\/sup>\/Cu couple is -0.34 V with respect to standard H+<\/sup>\/H2<\/sub>. In other words, the standard oxidation potential is just the negative of the standard reduction potential.\r\n\r\nRegardless, the cell voltages work out the same. For the moment designate the standard reduction potential as E\u00b0red<\/strong><\/sup> and the standard oxidation potential as E\u00b0ox<\/strong><\/sup>. Consider again the reaction,\r\n\r\n\\[\\ce{Cu^{2+}_{aq} + Fe_s = Cu_s + Fe^{2+}_{aq}} \\tag{25}\\]\r\n\r\nFrom the perspective of reduction potentials,\r\n\r\n\\[\\ce{Cu^{2+}_{aq} + 2e^- = Cu_s}\\qquad E^{\\circ}_{red} = +0.34\\ V \\tag{26}\\]\r\n\r\n\\[\\ce{Fe^{2+}_{aq} + 2e^- = Fe_s}\\qquad E^{\\circ}_{red} = -0.44\\ V \\tag{27}\\]\r\n\r\n\\[\\ce{\\Delta E^{\\circ} = E^{\\circ}_{red} - E^{\\circ}_{ox}\r\n= E^{\\circ}_{red,cathode} - E^{\\circ}_{red,anode}\r\n= E^{\\circ}_{red,Cu^{2+}\/Cu} - E^{\\circ}_{red,Fe^{2+}\/Fe}}\\]\r\n\r\n\\[\\ce{= 0.34 - (-0.44) = +0.78\\ V} \\tag{28}\\]\r\n\r\nFrom the point of view of oxidation potentials,\r\n\r\n\\[\\ce{Cu_s = Cu^{2+}_{aq} + 2e^-}\\qquad E^{\\circ}_{ox} = -0.34\\ V \\tag{29}\\]\r\n\r\n\\[\\ce{Fe_s = Fe^{2+}_{aq} + 2e^-}\\qquad E^{\\circ}_{ox} = +0.44\\ V \\tag{30}\\]\r\n\r\nAnalogously to the preceding case,\r\n\r\n\\[\\ce{\\Delta E^{\\circ}\r\n= E^{\\circ}_{ox} - E^{\\circ}_{red}\r\n= E^{\\circ}_{ox,anode} - E^{\\circ}_{ox,cathode}\r\n= E^{\\circ}_{Fe\/Fe^{2+}} - E^{\\circ}_{Cu\/Cu^{2+}}\r\n= 0.44 - (-0.34) = +0.78\\ V} \\tag{31}\\]\r\n\r\nIn this text we will only use reduction potentials.<\/em>","rendered":"

Before we can delve into Eh-pH diagrams, the associated electrochemistry has to be in place. If this material is familiar, it may be skipped.<\/p>\n

2.1 Electron Transfer Reactions<\/h2>\n

Many (though not all) reactions involve transfer of electrons between reagents. These may be called electron transfer reactions, redox reactions or electrochemical reactions (the latter refers specifically to processes occurring at electrodes). In normal chemical systems the electron does not exist outside of being attached to atoms, i.e. free electrons do not survive. The energy of a free electron would be so high that it would react with the first thing it encounters. The transfer of electrons from one chemical species to another is a means of transferring energy. Electrons moving from one species to another are essentially energy in transit, just as heat is energy in transit.
\nThere are two reasons that electrons transfer between chemical species. One is that the system (the chemical reaction) and its surroundings together have higher net entropy than the reactants – a spontaneous chemical reaction.* The electrons in one reactant (the reducing agent) are sufficiently energetic to be spontaneously accepted by another reactant (the oxidizing agent). The second reason is that we force an unfavourable electron transfer reaction by imparting extra energy to the electrons, by means of an external applied voltage. This is the basis for electrolysis.<\/p>\n

Further, it can be seen that we can consider an electron transfer reaction to be comprised of two parts – something getting reduced, and something getting oxidized. This is the only way that electron transfer reactions can occur. The electrons must be taken from one reagent and accepted by another. (Again, free electrons do not exist.) Thus we can separate an electron transfer reaction into two half reactions, one a reduction and one an oxidation. It must be borne in mind that half reactions never occur in isolation; the electrons must come from one reactant be delivered to another.<\/p>\n

2.2 The Electrochemical Potential<\/h2>\n

This is actually a quite straightforward quantity. It is simply the tendency of a chemical species to accept electrons and be reduced under the specified conditions (like pressure, temperature, activities of reactants and products). The higher the potential, the stronger the driving force for this to occur. Now we are specifically focusing on a reduction half reaction, e.g.<\/p>\n

\\[\\ce{Cu^{2+}_{aq} + 2e^- = Cu_s} \\tag{1}\\]<\/p>\n

\n

Recall that what drives chemical reactions (makes them spontaneous) is that over the system and its surroundings there is a net increase in entropy. Metaphorically, entropy can be thought of as the concentration of energy. It is intuitively reasonable that concentrated energy tends to naturally disperse; degrades to a lower “concentration.” The Gibbs free energy equation reflects this link between entropy and spontaneity. Since
\n\u0394G = \u0394H – T \u0394S at some specified temperature T,<\/p>\n

\\[\\ce{\\frac{-\\Delta G}{T} = \\frac{-\\Delta H}{T} + \\Delta S}\\]<\/p>\n

\u0394H is the enthalpy change for the system. Then -\u0394H is the heat flow to the surroundings. At constant pressure \u0394H is the heat flow into the system. Then
\n-\u0394H\/T = -q\u03a1<\/sub>\/T (q\u03a1<\/sub> being heat flow at constant pressure) and this has the form of the entropy. It can be shown (see suitable thermodynamics textbooks) that in fact
\n-\u0394H\/T is the entropy change of the surroundings. Thus,<\/p>\n

\\[\\ce{-\\Delta G\/T = \\Delta S_{surroundings} + \\Delta S_{system} = Total\\ \\Delta S}\\]<\/p>\n

This is the basis upon which we say that a reaction is spontaneous; if \u0394G < 0. It applies to all chemical processes.<\/p>\n<\/div>\n

The Cu+2<\/sup> ion in aqueous solution has some potential to accept an electron and be reduced to Cu. If we can measure that electrochemical potential for any conceivable half reaction, we can understand and combine those half reactions to perform chemical transformations. The problem though is that there is no way to measure the reduction potential for a half reaction in isolation; electron transfer reactions require a reactant source of electrons and a reactant to accept them. For instance,<\/p>\n

\\[
\n\\begin{align*}
\n&\\ce{Cu^{2+}_{aq}}+\\ce{2e-}=\\ce{Cu_s}\\tag{2} \\\\
\n&\\ce{H2_g}=\\ce{2H+_{aq}}+\\ce{2e-}\\tag{3} \\\\
\n\\hline
\n\\text{net: }&\\ce{Cu^{2+}_{aq}}+\\ce{H2_g}=\\ce{Cu_s}+\\ce{2H+_{aq}}\\tag{4}
\n\\end{align*}
\n\\]<\/p>\n

Hydrogen gas is the source of the electrons; Cu+2<\/sup> accepts them. (It is the same in electrolysis reactions, which are thermodynamically unfavourable and are forced to go by imposition of a voltage; the electrons are forcibly taken from one reactant and forced onto another.) But, now we have two half reactions that are combined into an overall reaction. We can measure the potential difference<\/em> (voltage) that is manifested between the two half cells as illustrated in the figure below. This is the measure of the driving force for the electrons to transfer as per the reaction. But, electron transfer occurs. It does not undergo chemical transformation. The salt bridge<\/a> allows the electrical circuit to be completed by ionic conduction between compartments, otherwise no current would flow.<\/a><\/p>\n

\"Diagram
Figure 2.1. Schematic illustration of an electrochemical cell comprised of an H+\/H2 half reaction and a Cu+2\/Cu half reaction. The voltage of the cell can be measured between the electrodes (anode and cathode). The Pt electrode simply acts as a surface where electron transfer occurs. It does not undergo chemical transformation. The salt bridge allows the electrical circuit to be completed by ionic conduction between compartments, otherwise no current would flow.<\/figcaption><\/figure>\n

The potential difference is the net effect of two half reactions. There is no way to make a half reaction occur in isolation, and so no way to measure the absolute reduction potential of a half reaction.<\/p>\n

The solution to this conundrum is to select one half reaction and assign it a voltage of 0 V<\/em>. Then all other half reactions are referenced to this one standard. For half reaction (4<\/span>) then, occurring in a cell as illustrated in Figure 2.1<\/a>, we measure the potential difference of the cell under some set of standard conditions and arbitrarily<\/em> say that the half cell voltage for the H+<\/sup>\/H2<\/sub> half reaction will be 0 V. Then the whole of the cell voltage<\/em> is ascribed to the Cu+2<\/sup>\/Cu half reaction. Now we have a measure of the tendency of Cu+2<\/sup> to undergo reduction relative to the H+<\/sup>\/H2<\/sub> half reaction<\/em>. Then any half reaction, under this same standard set of conditions (pressure, temperature and composition) can be combined with the H+<\/sup>\/H2<\/sub> half cell and the cell’s voltage is then assigned wholly to the half reaction of interest. In other words, every half reaction voltage is precisely equal to the voltage for a cell where the reduction half cell is the half reaction of interest and the anode <\/a>is the H+<\/sup>\/H2<\/sub> half cell under specified standard conditions.<\/em><\/strong><\/p>\n

So, when we set up a cell like that in Figure 2.1<\/a>, at 25\u00b0C and 1 atm pressure with unit activities of the solutes (on the molal scale), we should be able to measure, in principle, a voltage of +0.34 V; the potential difference measured between the cathode <\/a>(Cu+2<\/sup>\/Cu) and the anode (H+<\/sup>\/H2<\/sub>), i.e. cathode minus anode*. This is the standard reduction potential for the Cu+2<\/sup>\/Cu couple, i.e. half reaction (1). This tells us that Cu+2<\/sup> has a greater tendency to be reduced than H+<\/sup>by 0.34 V under these conditions. Alternatively, if we were to use the Cr+3<\/sup>\/Cu+2<\/sup> couple instead of Cu+2<\/sup>\/Cu in Figure 2.1<\/a>, we would find that the cell’s potential difference would be
\n-0.42 V (cathode voltage – H+<\/sup>\/H2<\/sub> anode voltage). This indicates that Cr+3 is a weaker oxidizing agent (has a lower potential to be reduced) than H+, or in other words, H+<\/sup> is 0.42 V more strongly oxidizing than Cr+3<\/sup>.<\/p>\n

2.3 Calculating \u0394E\u00b0 for Electron Transfer Reactions<\/h2>\n

The standard cell voltage is denoted \u0394E\u00b0, while the standard reduction potential is E\u00b0. Under non-standard conditions the superscript \u00b0 is dropped.<\/p>\n

\n

This may be quite difficult in practice. There are many complicating factors that can make accurate potential measurements difficult or sometimes impossible. However, there are also plenty of indirect methods for obtaining such data. For instance, other thermodynamic measurements may allow us to obtain \u0394G\u00b0 for a reaction. Then through the relationships,<\/p>\n

\\[\\ce{\\Delta G^{\\circ} = -nF\\Delta E^{\\circ} = -RT\\ln K}\\]<\/p>\n

the cell voltage can be obtained. Here n is the moles of electrons per mole of reaction and F is the Faraday constant (the charge of a mole of electrons, 96485 C\/mole of electrons). These quantities will be developed later.<\/p>\n<\/div>\n

Most generally we are interested in all electron transfer reactions, not just those that involve the H+<\/sup>\/H2<\/sub> couple. There is a very simple rule for determining the standard cell voltage for any electron transfer reaction:<\/p>\n

\\[\\ce{\\Delta E^{\\circ} = E^{\\circ}_{red} – E^{\\circ}_{ox}} \\tag{5}\\]<\/p>\n

where E\u00b0red<\/sub> is the standard reduction potential for the couple undergoing reduction, and E\u00b0ox<\/sub> is the standard reduction potential for the couple undergoing oxidation. DO NOT change the sign of E\u00b0ox. Simply calculate the appropriate difference in standard reduction potentials<\/em><\/strong>.<\/p>\n

Consider the reaction,<\/p>\n

\\[
\n\\begin{align*}
\n&\\ce{Cu^{2+}_{aq}}+\\ce{2e-}=\\ce{Cu_s}\\tag{6} \\\\
\n&\\ce{Fe_s}=\\ce{Fe^{2+}_{aq}}+\\ce{2e-}\\tag{7} \\\\
\n\\hline
\n\\text{net: }&\\ce{Cu^{2+}_{aq}}+\\ce{Fe_s}=\\ce{Cu_s}+\\ce{Fe^{2+}_{aq}}\\tag{8}
\n\\end{align*}
\n\\]<\/p>\n

(This used to be the basis for recovery of copper metal from dilute copper leach solutions using scrap iron.) The standard reduction potential for Cu+2<\/sup>\/Cu is
\n+0.34 V. That for Fe+2<\/sup>\/Fe, i.e. Fe+2<\/sup> + 2e- = Fe, is -0.44 V. In the reaction Cu+2<\/sup> is reduced and Fe is oxidized. By equation [1] the standard cell voltage is:<\/p>\n

\\[\\ce{\\Delta E^{\\circ} = 0.34 – (-0.44) = +0.78\\ V} \\tag{9}\\]<\/p>\n

The cell is shown in Figure 2.2<\/a>. (The two half cells must be separated in order to be able to have the electrons pass through the external circuit and facilitate potential difference measurements, as explained in the figure caption.) So why don’t we reverse the sign of E\u00b0Fe+2<\/sup>\/Fe in order to calculate \u0394E\u00b0? (Students often want to do this because it seems intuitively reasonable, since the half reaction has been reversed to become an oxidation half reaction.) BUT, the electrons can, in principle, be taken by Cu+2<\/sup> from Fe, or they can be taken by Fe+2<\/sup> from Cu. In other words, we have to compare the standard reduction potentials. Hence, do not flip the sign of E\u00b0ox.<\/em> What DE\u00b0 tells us is the net driving force for the electrons to be accepted by one couple, rather than the other, for the reaction as written. Note that E\u00b0Cu+2<\/sup>\/Cu > E\u00b0Fe+2<\/sup>\/Fe. This means that the Cu+2<\/sup>\/Cu electrode is positive with respect to the Fe+2\/Fe electrode, and then the Fe+2<\/sup>\/Fe electrode is negative with respect to the Cu+2<\/sup>\/Cu electrode.<\/p>\n

If we wrote the reaction as,<\/p>\n

\\[\\ce{Cu_s + Fe^{2+}_{aq} = Cu^{2+}_{aq} + Fe_s} \\tag{10}\\]<\/p>\n

then \u0394E\u00b0 would be,<\/p>\n

\\[\\ce{-0.44 – 0.34 = -0.78\\ V} \\tag{11}\\]<\/p>\n

In this case we would be measuring the potential difference of the Fe+2<\/sup>\/Fe electrode with respect to the Cu+2<\/sup>\/Cu electrode. Now Fe+2<\/sup> + e- = Fe is the reduction and the oxidation is Cu = Cu+2<\/sup>+ 2e–<\/sup>. Clearly it is important which way round we write the reaction. *<\/a><\/p>\n

\"(a)
Figure 2.2. (a) An electrochemical cell comprised of the Cu+2\/Cu and Fe+2\/Fe couples, where Cu+2 oxidizes Fe. The electrodes are both directly involved in the reaction, as well as being surfaces at which electron transfer occurs. (b) If Fe metal were placed in direct contact with a Cu+2-containing solution the electrons would transfer from Fe, through the plated Cu metal and to Cu+2 in solution. This arrangement could not be used to measure potential difference. The electrons cannot pass through an external circuit (even if we connected a wire to the Cu and the Fe), which is what is needed to do measurements, or use the energy released. Both arrangements involve the same reaction. In the former electrical energy can be extracted or input, and in the latter it cannot.<\/figcaption><\/figure>\n

 <\/p>\n

The second common error is to multiply \u0394E\u00b0 by the number of electrons involved (the value of n). The quantity n is the moles of electrons transferred per mole of reaction as written. For reaction (7<\/span>) n = 2 moles e-\/mole of reaction. One mole of reaction as per reaction (7)<\/span> means 1 mole Cu+2<\/sup> reacting with 1 mole Fe to form 1 mol Fe+2<\/sup> and 1 mole Cu. If we wrote the reaction instead as,<\/p>\n

\\[
\n\\begin{align*}
\n&\\tfrac{1}{2}\\ce{Cu^{2+}_{aq}}+\\ce{e-}=\\tfrac{1}{2}\\ce{Cu_s}\\tag{12}\\\\
\n&\\tfrac{1}{2}\\ce{Fe_s}=\\tfrac{1}{2}\\ce{Fe^{2+}_{aq}}+\\ce{e-}\\tag{13}\\\\
\n\\hline
\n\\text{net: }&\\tfrac{1}{2}\\ce{Cu^{2+}_{aq}}+\\tfrac{1}{2}\\ce{Fe_s}=\\tfrac{1}{2}\\ce{Cu_s}+\\tfrac{1}{2}\\ce{Fe^{2+}_{aq}}\\tag{14}
\n\\end{align*}
\n\\]<\/p>\n

Now 1 mole of reaction means 0.5 mole Cu+2<\/sup> reacting with 0.5 mole Fe, etc., and n = 1. Multiplying \u0394E\u00b0 by n yields something that is no longer a voltage, but rather,<\/p>\n

\\[\\ce{n\\Delta E^{\\circ} = moles\\ e^- \\ volts\/mol\\ of\\ reaction} \\tag{15}\\]<\/p>\n

This is actually more akin to energy. (i.e., converting moles of electrons to charge, we get charge\/mole x voltage = energy in J\/mole.) Further, the voltage is the potential for electrons to pass from one reactant to another under specified conditions, regardless of the number involved. It is the same if we pass 1 kmol of electrons or 1\/1000 mole of electrons. Finally, if we wrote,<\/p>\n

\\[\\ce{10{,}000Cu^{2+}_{aq} + 10{,}000Fe_s = 10{,}000Cu_s + 10{,}000Fe^{2+}_{aq}} \\tag{16}\\]<\/p>\n

(and there is no reason why we can’t) would it seem reasonable to multiply 0.78 V by 10,000 and anticipate a cell voltage of 7800 V? Therefore, when calculating a standard potential difference DO NOT multiply <\/em><\/strong>\u0394E\u00b0 by n<\/em><\/strong>.<\/p>\n

2.4 More on E\u00b0 and \u0394E\u00b0<\/h2>\n

We noted earlier that E\u00b0, the standard reduction potential, is a measure of the tendency of a species to be reduced. The highest standard reduction potential is about +3 V (e.g. E\u00b0F2\/F<\/sub>– = 2.87 V). The lowest is about -3 V (e.g. E\u00b0Li+\/Li<\/sub> = -3.05 V). Thus E\u00b0H+\/H2 = 0 V is in the middle of the range, which is consistent with the electronegativity for H being about half way between the highest and lowest values. A table of standard reduction (Table 2.1 below) potentials at 25\u00b0C is provided below. (Conversions to other temperatures can be performed.) The magnitude of the reduction potential indicates how strongly oxidizing a species is. The more strongly a chemical species wants electrons, the more weakly the reduced form will want to give them up. That is, a strong oxidant is reduced to form a weak reductant. For example, F2 is a very powerful oxidizing agent. It has a high reduction potential (2.87 V). This means that F- has a very weak tendency to give up an electron; it is a very weak reducing agent. Conversely, Li+ is a very weak oxidizing agent; it has a very low reduction potential. Hence Li+ does not have much tendency to accept electrons. It follows then that Li metal is a very powerful reducing agent. The trends are illustrated in the figure below.<\/p>\n

\"Diagram
Figure 2.3. Trends in oxidizing and reducing strength for redox couples.<\/figcaption><\/figure>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Table 2.1 - Selected standard reduction potentials in aqueous solutions. The number at the right of each half reaction is its standard reduction potential in volts at 25\u00b0C.<\/th>\n<\/tr>\n<\/thead>\n
Sm2+<\/sup> + 2e-<\/sup> \u2192 Sm<\/td>\n- 3.12<\/td>\nIn+<\/sup> + e-<\/sup> \u2192 In<\/td>\n-0.14<\/td>\n<\/tr>\n
Li+<\/sup> + e -<\/sup> \u2192 Li<\/td>\n-3.05<\/td>\nPb2+<\/sup> + 2e-<\/sup> \u2192 Pb<\/td>\n-0.13<\/td>\n<\/tr>\n
K+<\/sup> + e-<\/sup> \u2192 K<\/td>\n-2.93<\/td>\nO2<\/sub> + H2<\/sub>O + 2e-<\/sup><\/td>\n-0.08<\/td>\n<\/tr>\n
Rb+<\/sup> + e-<\/sup> \u2192 Rb<\/td>\n-2.92<\/td>\nHO2<\/sub>-<\/sup> + OH_<\/sup><\/td>\n<\/td>\n<\/tr>\n
Cs+<\/sup> + e-<\/sup> \u2192 Cs<\/td>\n-2.92<\/td>\nFe3+<\/sup> + 3e-<\/sup> \u2192 Fe<\/td>\n-4.40<\/td>\n<\/tr>\n
Ra2+<\/sup> + 2e-<\/sup> \u2192 Ra<\/td>\n-2.92<\/td>\nTi4-<\/sup> + e-<\/sup> \u2192 Ti3+<\/sup><\/td>\n0.00<\/td>\n<\/tr>\n
Ba2+<\/sup> + 2e-<\/sup> \u2192 Ba<\/td>\n-2.91<\/td>\n2H+<\/sup> + 2e-<\/sup> \u2192 H2<\/sub><\/td>\n0<\/td>\n<\/tr>\n
Sr2-<\/sup> + 2e-<\/sup> \u2192 Sr<\/td>\n-2.89<\/td>\nAgBr + e-<\/sup> \u2192 Ag + Br-<\/sup><\/td>\n0.07<\/td>\n<\/tr>\n
Ca2+<\/sup> + 2e \u2192 Ca<\/td>\n-2.87<\/td>\nSn4+<\/sup> + 2e-<\/sup> \u2192 Sn2+<\/sup><\/td>\n0.15<\/td>\n<\/tr>\n
Na+<\/sup> + e-<\/sup> \u2192 Na<\/td>\n-2.71<\/td>\nCu2-<\/sup> + e-<\/sup> \u2192 Cu+<\/sup><\/td>\n0.16<\/td>\n<\/tr>\n
La3+<\/sup> + 3e-<\/sup> \u2192 La<\/td>\n-2.52<\/td>\nBi3<\/sup> + 3e \u2192 Bi<\/td>\n0.20<\/td>\n<\/tr>\n
Ce3+<\/sup> + 3e-<\/sup> \u2192 Ce<\/td>\n-2.48<\/td>\nAgCl + e-<\/sup> \u2192 Ag + CI-<\/sup><\/td>\n0.2223<\/td>\n<\/tr>\n
Mf2+<\/sup> + 2e-<\/sup> \u2192 Mg<\/td>\n-2.36<\/td>\nHg2<\/sub>Cl2<\/sub> + 2e-<\/sup> \u2192 2Hg + 2Cl-<\/sup><\/td>\n0.27<\/td>\n<\/tr>\n
Be2+<\/sup> + 2e_<\/sup> \u2192 Be<\/td>\n-1.85<\/td>\nCu2+<\/sup> + 2e-<\/sup> \u2192 Cu<\/td>\n0.34<\/td>\n<\/tr>\n
U3+<\/sup> + 3e-<\/sup> \u2192 U<\/td>\n-1.79<\/td>\nO2<\/sub> + 2H2<\/sub>O + 4e-<\/sup> \u2192 4OH-<\/sup><\/td>\n0.40<\/td>\n<\/tr>\n
Al3+<\/sup> + 3e-<\/sup> \u2192 Al<\/td>\n-1.66<\/td>\nNiOOH + H2<\/sub>O + e-<\/sup> \u2192 Ni(OH)2<\/sub> + OH-<\/sup><\/td>\n0.49<\/td>\n<\/tr>\n
Ti2+<\/sup> + 2e-<\/sup> \u2192 Ti<\/td>\n-1.63<\/td>\nCu+<\/sup> + e-<\/sup> \u2192 Cu<\/td>\n0.52<\/td>\n<\/tr>\n
V2+<\/sup> + 2e-<\/sup> \u2192 V<\/td>\n-1.19<\/td>\nI3<\/sub>-<\/sup> + 2e \u2192 3I-<\/sup><\/td>\n0.53<\/td>\n<\/tr>\n
Mn2+<\/sup> + 2e-<\/sup> \u2192 Mn<\/td>\n-1.18<\/td>\nI2<\/sub> + 2e-<\/sup> \u2192 2I-<\/sup><\/td>\n0.54<\/td>\n<\/tr>\n
Cr2<\/sup> + 2e-<\/sup> \u2192 Cr<\/td>\n-0.91<\/td>\nHg2<\/sub>SO4<\/sub> + 2e-<\/sup> \u2192 2Hg + SO4<\/sub>2-<\/sup><\/td>\n0.62<\/td>\n<\/tr>\n
Fe(OH)2<\/sub> + 2e-<\/sup> \u2192 Fe + 2OH-<\/sup><\/td>\n-0.88<\/td>\nFe3+<\/sup> + e-<\/sup> \u2192 Fe2+<\/sup><\/td>\n0.77<\/td>\n<\/tr>\n
2H2<\/sub>O + 2e-<\/sup> \u2192 H2<\/sub> + 2OH-<\/sup><\/td>\n-0.83<\/td>\nAgF + e-<\/sup> \u2192 Ag + F-<\/sup><\/td>\n0.78<\/td>\n<\/tr>\n
Cd(OH)2<\/sub> + 2e-<\/sup> \u2192 Cd + 2OH-<\/sup><\/td>\n-0.81<\/td>\nHg2<\/sub>2+<\/sup> + 2e-<\/sup> \u2192 2Hg<\/td>\n0.79<\/td>\n<\/tr>\n
Zn2+<\/sup> + 2e-<\/sup> Ni2+<\/sup> + 2e-<\/sup> \u2192Ni Zn<\/td>\n-0.76<\/td>\nAg-<\/sup> + e-<\/sup> \u2192 Ag<\/td>\n0.80<\/td>\n<\/tr>\n
Cr3-<\/sup> + 3e-<\/sup> \u2192 Cr<\/td>\n-0.74<\/td>\n2Hg2+<\/sup> + 2e-<\/sup> Hgz<\/sub>2+<\/sup><\/td>\n0.92<\/td>\n<\/tr>\n
U4+<\/sup> + e-<\/sup> \u2192 U3+<\/sup><\/td>\n-0.61<\/td>\nPu4+<\/sup> + e-<\/sup> \u2192 Pu3+<\/sup><\/td>\n0.97<\/td>\n<\/tr>\n
O2<\/sub> - e-<\/sup> \u2192 O2<\/sub>-<\/sup><\/td>\n-0.56<\/td>\nBr2<\/sub> + 2e-<\/sup> \u2192 2Br-<\/sup><\/td>\n1.09<\/td>\n<\/tr>\n
In3+<\/sup> s-<\/sup> \u2192 In2+<\/sup><\/td>\n-0.49<\/td>\nPt2+<\/sup> + 2e-<\/sup> \u2192 Pt<\/td>\n1.20<\/td>\n<\/tr>\n
S + 2e-<\/sup> \u2192 S2<\/sup><\/td>\n-0.48<\/td>\nMnO2<\/sub> + 4H+<\/sup> + 2e-<\/sup> \u2192 Mn2+<\/sup> + 2H2<\/sub>O<\/td>\n1.23<\/td>\n<\/tr>\n
In3+<\/sup> + 2e-<\/sup> \u2192 In+<\/sup><\/td>\n-0.44<\/td>\nO2<\/sub> + 4H+<\/sup> 4e-<\/sup> \u2192 2H2<\/sub>O<\/td>\n1.23<\/td>\n<\/tr>\n
Fe2-<\/sup> + 2e \u2192 Fe<\/td>\n-0.44<\/td>\nCrO2<\/sub>O7<\/sub>2-<\/sup> + 14H+<\/sup> + 6e-<\/sup> \u2192 2Cr3+<\/sup> + 7H2<\/sup>O<\/td>\n1.33<\/td>\n<\/tr>\n
Cr3-<\/sup> + e-<\/sup> \u2192 Cr2+<\/sup><\/td>\n-0.41<\/td>\nCl2<\/sub> + 2e-<\/sup> \u2192 2C1-<\/sup><\/td>\n1.36<\/td>\n<\/tr>\n
Cd2+<\/sup> + 2e-<\/sup> \u2192 Cd<\/td>\n-0.40<\/td>\nAu3+<\/sup> + 3e-<\/sup> \u2192 Au<\/td>\n1.40<\/td>\n<\/tr>\n
In2+<\/sup> + e-<\/sup> \u2192 In+<\/sup><\/td>\n-0.40<\/td>\nMn3+<\/sup> + e-<\/sup> \u2192 Mn2+<\/sup><\/td>\n1.51<\/td>\n<\/tr>\n
Ti3+<\/sup> + e-<\/sup> \u2192 Ti2+<\/sup><\/td>\n-0.37<\/td>\nMnP4<\/sub>-<\/sup> + 8H+<\/sup> + 5e \u2192 Mn2+<\/sup> + 4H2<\/sub>O<\/td>\n1.51<\/td>\n<\/tr>\n
PcSO4<\/sub> + 2e-<\/sup> \u2192 Pb + SO4<\/sub>2-<\/sup><\/td>\n-0.36<\/td>\nCe4+<\/sup> + e-<\/sup> \u2192 Ce3+<\/sup><\/td>\n1.61<\/td>\n<\/tr>\n
In3+<\/sup> + 3e-<\/sup> \u2192 In<\/td>\n-0.34<\/td>\nPb4+<\/sub> + 2e-<\/sup> \u2192 Pb2+<\/sup><\/td>\n1.67<\/td>\n<\/tr>\n
Co2+<\/sup> + 2e-<\/sup> \u2192 CO<\/td>\n-0.28<\/td>\nAu+<\/sup> + e-<\/sup> \u2192 Au<\/td>\n1.69<\/td>\n<\/tr>\n
V3+<\/sup> + e-<\/sup> \u2192 V2+<\/sup><\/td>\n-0.26<\/td>\nCo3+<\/sup> + e-<\/sup> Ni2+<\/sup> + 2e-<\/sup> \u2192Ni Co2+<\/sup><\/td>\n1.81<\/td>\n<\/tr>\n
Ni2+<\/sup> + 2e-<\/sup> \u2192Ni<\/td>\n-0.23<\/td>\nAg2+<\/sup> + e-<\/sup> \u2192 Ag+<\/sup><\/td>\n1.98<\/td>\n<\/tr>\n
Agl + e \u2192 Ag + I-<\/sup><\/td>\n-0.15<\/td>\n52<\/sup>O<\/sub>8<\/sub>2-<\/sup> \u2192 2e- \u2192 28O4<\/sub>2-<\/sup><\/td>\n2.05<\/td>\n<\/tr>\n
Sn2<\/sup> + 2e-<\/sup> \u2192 Sn<\/td>\n-0.14<\/td>\nF2<\/sub> + 2e-<\/sup> \u2192 2F-<\/sup><\/td>\n2.87<\/td>\n<\/tr>\n<\/tbody>\n
Source: P.W. Atkins, Physical Chemistry, 3rd ed., 1986.<\/th>\n<\/tr>\n<\/tfoot>\n<\/table>\n

<\/p>\n

Standard conditions provides a common basis for comparison of E\u00b0 values and hence \u0394E\u00b0. However, there is deeper thermodynamic significance as well. In thermodynamics \u0394G (under any conditions, not just standard conditions) is the quantity we use to assess if a reaction at constant pressure is spontaneous or not. It turns out that DE\u00b0 is directly proportional to \u0394G\u00b0:<\/p>\n

\\[\\ce{\\Delta G^{\\circ} = -nF\\Delta E^{\\circ}} \\tag{17}\\]<\/p>\n

where, as before, n = moles of electrons per mole of reaction and F = the charge of a mole of electrons (96,485 C\/mole e–<\/sup>). The units of \u0394G\u00b0 are J\/mol. It can be readily seen that on the basis of the units, the equation is reasonable:<\/p>\n

\\[\\ce{n\\ moles\\ e^-\/mol \\times F\\ C\/mole\\ e^- \\times \\Delta E^{\\circ}\\ V = V C\/mol = J\/mol} \\tag{18}\\]<\/div>\n
<\/div>\n
\n

In fact, \u0394E\u00b0 provides us with the same information as \u0394G\u00b0, just on a different scale, and specifically for electron transfer reactions. Based on equation [5] \u0394E\u00b0 > 0 means that a reaction is favourable under standard conditions. Further, the larger the value of DE\u00b0 the stronger the driving force for the reaction to occur, and the more energy that will be released as a result. Similarly, if DE\u00b0 < 0 the reaction is not favourable under standard conditions. Making it go will require the input of electrical energy. This is an electrolysis.<\/p>\n

2.5 \u00a0E and \u0394E Under Non-Standard Conditions: the Nernst Equation<\/h2>\n

As noted above, the most general condition for assessing if a reaction as written is spontaneous is if \u0394G < 0 under the specified conditions. Changes in conditions can make a reaction favourable or unfavourable, depending on factors such as temperature, pressure and composition. (Note that partial pressure of a gas is analogous to concentration or activity of a solute in solution, or of a metal in an alloy, etc.) Just as \u0394G\u00b0 = -nF\u0394E\u00b0, so,<\/p>\n

\\[\\ce{\\Delta G = -nF\\Delta E} \\tag{19}\\]<\/p>\n

For a chemical reaction \u0394G is a function of pressure, temperature and composition:<\/p>\n

\\[\\ce{\\Delta G = \\Delta G^{\\circ} + RT\\ln Q} \\tag{20}\\]<\/p>\n

where R is the gas constant (8.31441 J\/mol K), T is the absolute temperature (Kelvin) and Q is the reaction quotient. The reaction quotient was presented in Chemistry Review Part II.<\/p>\n

\n

By definition 1 VC = 1 J. When a charge of 1 C passes through a voltage of 1 V an energy of 1 J is consumed or released, depending on whether the charge is going from + to \u2013 or from \u2013 to +. Introductory textbooks on the subject typically show that,<\/p>\n

\\[\\ce{-w_e = \\Delta G^{\\circ} = -nF\\Delta E^{\\circ}}\\]<\/p>\n

where -we is the maximum possible electrical work (or more generally, non pressure-volume work) that can be extracted from an electrochemical cell.<\/p>\n<\/div>\n

 <\/p>\n

Briefly, it has exactly the same form as the equilibrium constant, but where concentrations and\/or gas pressures (activities and fugacities, properly speaking) are not necessarily those corresponding to an equilibrium state. Q is an indicator of how far the system is from equilibrium. The reaction will naturally proceed with changes in composition until equilibrium is attained, i.e. Q \u2192 K. In order to force Q to diverge from K, energy must be supplied (e.g. an electrolysis). Another important relationship is,<\/p>\n<\/div>\n

\\[\\ce{\\Delta G^{\\circ} = -RT\\ln K = -nF\\Delta E^{\\circ}} \\tag{21}\\]<\/p>\n

(This is another important aspect of standard conditions.) Note that when \u0394G = 0, \u0394G\u00b0 = -RTln Q (equation [8]) and now Q = K. Thus all reactions naturally tend to drive towards \u0394G going to 0. Once this condition is reached, Q = K and the reaction has reached equilibrium. There will be no further net change in composition. Substituting \u0394G = -nF\u0394E into equation [8] yields,<\/p>\n

\\[\\ce{-nF\\Delta E = -nF\\Delta E^{\\circ} + RT\\ln Q} \\tag{22}\\]<\/p>\n

\\[\\ce{\\Delta E = \\Delta E^{\\circ} – \\frac{RT}{nF}\\ln Q} \\tag{23}\\]<\/p>\n

This is the Nernst equation. For a spontaneous electron transfer reaction, \u0394E is a measure of how far the reaction is away from equilibrium. All natural processes tend toward decreasing \u0394E. Once equilibrium is attained \u0394E goes to 0 and the reaction ceases. For a galvanic cell (a battery), the voltage declines with time and eventually the battery dies. If \u0394E < 0 then the reaction as written is unfavourable. To make it go in this direction will require energy to overcome the negative \u0394E.<\/p>\n

By this point it can be seen that if the reaction is reversed, so is the sign of \u0394G (or \u0394G\u00b0); if it is favourable in the direction written (\u0394G < 0), then it is unfavourable in the opposite direction (and \u0394G > 0). Likewise, reversing the direction of an electron transfer reaction requires that we reverse the sign of \u0394E\u00b0 (or \u0394E for non-standard conditions).<\/p>\n

2.6 The Standard Oxidation Potential<\/h2>\n

Sometimes people characterize half reactions by standard oxidation potentials, rather than standard reduction potentials. This can be confusing. As always in thermodynamics, keeping the conventions straight is crucially important. A generic oxidation half reaction is,<\/p>\n

\\[\\ce{M_s = M^{n+} + ne^-} \\tag{24}\\]<\/p>\n

Consider again the Cu+2<\/sup>\/Cu reduction half reaction, Cu+2<\/sup> + 2e- = Cu. We defined the standard reduction potential as the cell voltage (cathode – anode) where the cathode was the Cu+2<\/sup>\/Cu couple and the anode was the H+\/H2 couple (set to 0 V). It tells us that Cu+2<\/sup> is 0.34 V more prone to accept electrons than is H+. And if you set up the cell, as in Figure 2.1<\/a>, and measure the potential of the Cu+2<\/sup>\/Cu electrode with respect to the H+<\/sup>\/H2<\/sub> electrode, the voltage will be +0.34 V. This means that Cu is more prone to hold onto its electrons than is H2<\/sub>; Cu is harder to oxidize than H2<\/sub>. Quantitatively, H2 is 0.34 V more prone to be oxidized than Cu, or, since E\u00b0H+<\/sup>\/H2<\/sub> = 0 V (by convention, for either its reduction or oxidation potential), the oxidation potential for the Cu+2<\/sup>\/Cu couple is -0.34 V with respect to standard H+<\/sup>\/H2<\/sub>. In other words, the standard oxidation potential is just the negative of the standard reduction potential.<\/p>\n

Regardless, the cell voltages work out the same. For the moment designate the standard reduction potential as E\u00b0red<\/strong><\/sup> and the standard oxidation potential as E\u00b0ox<\/strong><\/sup>. Consider again the reaction,<\/p>\n

\\[\\ce{Cu^{2+}_{aq} + Fe_s = Cu_s + Fe^{2+}_{aq}} \\tag{25}\\]<\/p>\n

From the perspective of reduction potentials,<\/p>\n

\\[\\ce{Cu^{2+}_{aq} + 2e^- = Cu_s}\\qquad E^{\\circ}_{red} = +0.34\\ V \\tag{26}\\]<\/p>\n

\\[\\ce{Fe^{2+}_{aq} + 2e^- = Fe_s}\\qquad E^{\\circ}_{red} = -0.44\\ V \\tag{27}\\]<\/p>\n

\\[\\ce{\\Delta E^{\\circ} = E^{\\circ}_{red} – E^{\\circ}_{ox}
\n= E^{\\circ}_{red,cathode} – E^{\\circ}_{red,anode}
\n= E^{\\circ}_{red,Cu^{2+}\/Cu} – E^{\\circ}_{red,Fe^{2+}\/Fe}}\\]<\/p>\n

\\[\\ce{= 0.34 – (-0.44) = +0.78\\ V} \\tag{28}\\]<\/p>\n

From the point of view of oxidation potentials,<\/p>\n

\\[\\ce{Cu_s = Cu^{2+}_{aq} + 2e^-}\\qquad E^{\\circ}_{ox} = -0.34\\ V \\tag{29}\\]<\/p>\n

\\[\\ce{Fe_s = Fe^{2+}_{aq} + 2e^-}\\qquad E^{\\circ}_{ox} = +0.44\\ V \\tag{30}\\]<\/p>\n

Analogously to the preceding case,<\/p>\n

\\[\\ce{\\Delta E^{\\circ}
\n= E^{\\circ}_{ox} – E^{\\circ}_{red}
\n= E^{\\circ}_{ox,anode} – E^{\\circ}_{ox,cathode}
\n= E^{\\circ}_{Fe\/Fe^{2+}} – E^{\\circ}_{Cu\/Cu^{2+}}
\n= 0.44 – (-0.34) = +0.78\\ V} \\tag{31}\\]<\/p>\n

In this text we will only use reduction potentials.<\/em><\/p>\n

Media Attributions<\/h2>