{"id":3641,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=3641"},"modified":"2021-05-17T17:38:59","modified_gmt":"2021-05-17T17:38:59","slug":"the-nernst-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/the-nernst-equation\/","title":{"raw":"Potential, Free Energy, and Equilibrium","rendered":"Potential, Free Energy, and Equilibrium"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain the relations between potential, free energy change, and equilibrium constants<\/li>\r\n \t<li>Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium<\/li>\r\n \t<li>Use the Nernst equation to determine cell potentials under nonstandard conditions<\/li>\r\n<\/ul>\r\n<\/div>\r\nSo far in this module, the relationship between the cell potential and reaction\u00a0<em>spontaneity<\/em>\u00a0has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant\u00a0<em>strength<\/em>\u00a0was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties [latex]\\Delta G[\/latex] and K.\r\n\r\n<section id=\"fs-idm248064976\">\r\n<h3>[latex]E^\\circ[\/latex] and [latex]\\Delta G^\\circ[\/latex]<\/h3>\r\n<p id=\"fs-idm215301376\">The standard free energy change of a process, [latex]\\Delta G^\\circ[\/latex], was defined in a previous chapter as the maximum work that could be performed by a system,\u00a0[latex]w_{max}[\/latex]. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant,\u00a0[latex]w_{elec}[\/latex]:<\/p>\r\n\r\n<center>[latex]\\Delta G^\\circ = w_{max} = w_{elec}[\/latex]<\/center>\r\n<p id=\"fs-idm233319952\">The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:<\/p>\r\n\r\n<center>[latex]\\Delta G^\\circ = w_{elec} = -nFE_{cell}^\\circ[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]\\Delta G^\\circ = -nFE_{cell}^\\circ[\/latex]<\/center>\r\n<p id=\"fs-idm648263408\">where\u00a0<em>n<\/em>\u00a0is the number of moles of electrons transferred,\u00a0<em>F<\/em>\u00a0is\u00a0<strong>Faraday\u2019s constant<\/strong>, and\u00a0[latex]E_{cell}^\\circ[\/latex]\u00a0is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.<\/p>\r\n\r\n<\/section><section id=\"fs-idm218866032\">\r\n<h3>[latex]E^\\circ[\/latex] and K<\/h3>\r\n<p id=\"fs-idm242158160\">Combining a previously derived relation between [latex]\\Delta G^\\circ[\/latex] and K (see the chapter on thermodynamics) and the equation above relating [latex]\\Delta G^\\circ[\/latex] and\u00a0[latex]E_{cell}^\\circ[\/latex]\u00a0yields the following:<\/p>\r\n\r\n<center>[latex]\\Delta G^\\circ = -RT\\text{ln}K = -nFE_{cell}^\\circ[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]E_{cell}^\\circ = (\\dfrac{RT}{nF})\\text{ln}K[\/latex]<\/center>\r\n<p id=\"fs-idm242471888\">This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between\u00a0[latex]E^\\circ[\/latex], [latex]\\Delta G^\\circ[\/latex] and\u00a0[latex]K[\/latex]\u00a0is depicted in Figure 1, and a table correlating reaction spontaneity to values of these properties is provided in Table 1.<\/p>\r\n\r\n<\/section>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214142\/CNX_Chem_17_04_Relation.jpg\" alt=\"A diagram is shown that involves three double headed arrows positioned in the shape of an equilateral triangle. The vertices are labeled in red. The top vertex is labeled \u201cK.\u201c The vertex at the lower left is labeled \u201cdelta G superscript degree symbol.\u201d The vertex at the lower right is labeled \u201cE superscript degree symbol subscript cell.\u201d The right side of the triangle is labeled \u201cE superscript degree symbol subscript cell equals ( R T divided by n F ) l n K.\u201d The lower side of the triangle is labeled \u201cdelta G superscript degree symbol equals negative n F E superscript degree symbol subscript cell.\u201d The left side of the triangle is labeled \u201cdelta G superscript degree symbol equals negative R T l n K.\u201d\" width=\"650\" height=\"373\" \/> Figure\u00a01.\u00a0Graphic depicting the relation between three important thermodynamic properties.[\/caption]\r\n\r\nTable 1\r\n<table summary=\"Table 17.2 \">\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td><em>K<\/em><\/td>\r\n<td>\u0394<em>G<\/em>\u00b0<\/td>\r\n<td><em>E<\/em>\u00b0<sub>cell<\/sub><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>&gt; 1<\/td>\r\n<td>&lt; 0<\/td>\r\n<td>&gt; 0<\/td>\r\n<td>\r\n<p id=\"fs-idm239811792\">Reaction is spontaneous under standard conditions<\/p>\r\n<p id=\"fs-idm218962512\">Products more abundant at equilibrium<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>&lt; 1<\/td>\r\n<td>&gt; 0<\/td>\r\n<td>&lt; 0<\/td>\r\n<td>\r\n<p id=\"fs-idm248001280\">Reaction is non-spontaneous under standard conditions<\/p>\r\n<p id=\"fs-idm197346624\">Reactants more abundant at equilibrium<\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>= 1<\/td>\r\n<td>= 0<\/td>\r\n<td>= 0<\/td>\r\n<td>\r\n<p id=\"fs-idm249738032\">Reaction is at equilibrium under standard conditions<\/p>\r\n<p id=\"fs-idm197184048\">Reactants and products equally abundant<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div><\/div>\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=5559620&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=cW7RpCYAOdI&amp;video_target=tpm-plugin-xbyknyp2-cW7RpCYAOdI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/PracticeProblemCellPotential_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Practice Problem: Cell Potential, Equilibrium Constants, and Free Energy Change\" here (opens in new window)<\/a>.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes<\/h3>\r\nWhat is the standard free energy change and equilibrium constant for the following reaction at 25 \u00b0C?\r\n<p style=\"text-align: center;\">[latex]2{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+\\text{Fe}\\left(s\\right)\\rightleftharpoons \\text{2Ag}\\left(s\\right)+{\\text{Fe}}^{2+}\\left(aq\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"578755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"578755\"]\r\n\r\nThe reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in <a class=\"target-chapter\" href=\".\/chapter\/standard-electrode-half-cell-potentials\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\text{anode (oxidation):}&amp;\\text{Fe}\\left(s\\right)\\longrightarrow {\\text{Fe}}^{2+}\\left(aq\\right)+{\\text{2e}}^{-}&amp;{E}_{{\\text{Fe}}^{2+}\\text{\/Fe}}^{\\circ }=-\\text{0.447 V}\\\\ \\text{cathode (reduction):}&amp;2\\times \\left({\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{e}}^{-}\\longrightarrow \\text{Ag}\\left(s\\right)\\right)&amp;{E}_{{\\text{Ag}}^{\\text{+}}\\text{\/Ag}}^{\\circ }=\\text{0.7996 V}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }=\\qquad{E}_{{\\text{Ag}}^{\\text{+}}\\text{\/Ag}}^{\\circ }-{E}_{{\\text{Fe}}^{2+}\\text{\/Fe}}^{\\circ }=\\text{+1.247 V}[\/latex]<\/p>\r\nRemember that the cell potential for the cathode is not multiplied by two when determining the standard cell potential. With <em>n<\/em> = 2, the equilibrium constant is then\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{E}_{\\text{cell}}^{\\circ }&amp;=&amp;\\frac{\\text{0.0592 V}}{n}\\log{K}\\\\K&amp;=&amp;{10}^{n\\times {E}_{\\text{cell}}^{\\circ }\\text{\/}\\text{0.0592 V}}\\\\{}&amp;=&amp;{10}^{2\\times \\text{1.247 V\/0.0592 V}}\\\\{}&amp;=&amp;{10}^{42.128}\\\\{}&amp;=&amp;1.3\\times {10}^{42}\\end{array}[\/latex]<\/p>\r\nThe two equilibrium constants differ slightly due to rounding in the constants 0.0257 V and 0.0592 V. The standard free energy is then\r\n<p style=\"text-align: center;\">[latex]\\Delta {G}^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Delta G^{\\circ }=-2\\times\\text{96,485}\\frac{\\text{J}}{\\text{V}\\cdot \\text{mol}}\\times \\text{1.247 V}=-240.6\\frac{\\text{kJ}}{\\text{mol}}[\/latex]<\/p>\r\nThe reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The\u00a0<em data-effect=\"italics\">K<\/em>\u00a0value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.\r\n\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp110872768\">Check Your Learning<\/h4>\r\nWhat is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?\r\n<p style=\"text-align: center;\">[latex]\\text{Sn}\\left(s\\right)+2{\\text{Cu}}^{2+}\\left(aq\\right)\\rightleftharpoons {\\text{Sn}}^{2+}\\left(aq\\right)+2{\\text{Cu}}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"279511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"279511\"]\r\n\r\nSpontaneous; <em>n<\/em> = 2; [latex]{E}_{\\text{cell}}^{\\circ }=\\text{+0.291 V}[\/latex]; [latex]\\Delta G^{\\circ }=-56.2\\frac{\\text{kJ}}{\\text{mol}}[\/latex]; <em>K<\/em> = 6.8 [latex]\\times [\/latex] 10<sup>9<\/sup>.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Potentials at Nonstandard Conditions: The Nernst Equation<\/h3>\r\n<p id=\"fs-idp76396192\">Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.<\/p>\r\n\r\n<center>[latex]\\Delta G = \\Delta G^\\circ + RT\\text{ln}Q[\/latex]<\/center>\r\n<p id=\"fs-idm983728\">Notice the reaction quotient,\u00a0[latex]Q[\/latex], appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the\u00a0<strong>Nernst equation<\/strong>:<\/p>\r\n\r\n<center>[latex]-nFE_{cell} = -nFE_{cell}^\\circ + RT\\text{ln}Q[\/latex]<\/center>&nbsp;\r\n\r\n<center>[latex]E_{cell} = E_{cell}^\\circ - \\dfrac{RT}{nF}\\text{ln}Q[\/latex]<\/center>\r\n<p id=\"fs-idp24059424\">This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred,\u00a0[latex]n[\/latex] the temperature,\u00a0[latex]T[\/latex], and the reaction mixture composition as reflected in\u00a0[latex]Q[\/latex]. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and a factor converting from natural to base-10 logarithms have been included.<\/p>\r\n\r\n<center>[latex]E_{cell} = E_{cell}^\\circ - \\dfrac{0.0592 \\text{V}}{n}\\text{log}Q[\/latex]<\/center>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2: Predicting Redox Spontaneity Under Nonstandard Conditions<\/h3>\r\nUse the Nernst equation to predict the spontaneity of the redox reaction shown below.\r\n<p style=\"text-align: center;\">[latex]\\text{Co}\\left(s\\right)+{\\text{Fe}}^{2+}\\left(aq,1.94M\\right)\\longrightarrow {\\text{Co}}^{2+}\\left(aq\\text{, 0.15}M\\right)+\\text{Fe}\\left(s\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"323406\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"323406\"]<\/p>\r\nThere are two ways to solve the problem. If the thermodynamic information in <a class=\"target-chapter\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\" rel=\"noopener\">Standard Thermodynamic Properties for Selected Substances<\/a> were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/standard-electrode-half-cell-potentials\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>. Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from\u00a0<a class=\"target-chapter\" href=\".\/chemistryformajorsxmaster\/chapter\/standard-electrode-half-cell-potentials-missing-formulas\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>\u00a0and the problem,\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\text{Anode (oxidation):}&amp;\\text{Co}\\left(s\\right)\\longrightarrow {\\text{Co}}^{2+}\\left(aq\\right)+{\\text{2e}}^{-}&amp;{E}_{{\\text{Co}}^{2+}\\text{\/Co}}^{\\circ }=-\\text{0.28 V}\\\\ \\text{Cathode (reduction):}&amp;{\\text{Fe}}^{2+}\\left(aq\\right)+{\\text{2e}}^{-}\\longrightarrow \\text{Fe}\\left(s\\right)&amp;{E}_{{\\text{Fe}}^{2+}\\text{\/Fe}}^{\\circ }=-\\text{0.447 V}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }=\\qquad{-}\\text{0.447 V}-\\left(-\\text{0.28 V}\\right)=-\\text{0.17 V}[\/latex]<\/p>\r\nNotice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}Q &amp;=&amp;\\frac{\\left[\\text{Co}^{2+}\\right]}{\\left[\\text{Fe}^{2+}\\right]}=\\dfrac{0.15M}{1.94M}=0.077\\\\{E}_{\\text{cell}}&amp;=&amp;{E}_{\\text{cell}}^{\\circ }-\\frac{0.0592 V}{n}\\log{Q}\\\\{E}_{\\text{cell}}&amp;=&amp;-0.1\\text{7 V}-\\frac{0.0592 V}{2}\\log{0.077}\\\\{E}_{\\text{cell}}&amp;=&amp;-\\text{0.17 V}+0.033 V=-0.014 V\\end{array}[\/latex]<\/p>\r\n<p id=\"fs-idp160381408\">The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idp72968608\">Check Your Learning<\/h4>\r\nWhat is the cell potential for the following reaction at room temperature?\r\n<p style=\"text-align: center;\">[latex]\\text{Al}\\left(s\\right)\\mid {\\text{Al}}^{3+}\\left(aq,0.15M\\right)\\parallel {\\text{Cu}}^{2+}\\left(aq,0.025M\\right)\\mid \\text{Cu}\\left(s\\right)[\/latex]<\/p>\r\nWhat are the values of <em>n<\/em> and Q for the overall reaction? Is the reaction spontaneous under these conditions?\r\n[reveal-answer q=\"303221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"303221\"]\r\n\r\n<em>n<\/em> = 6; <em>Q<\/em> = 1440; <em>E<\/em><sub>cell<\/sub> = +1.97 V, spontaneous[\/hidden-answer]\r\n\r\n<\/div>\r\nA\u00a0<strong><span id=\"term646\">concentration cell<\/span><\/strong>\u00a0is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.\r\n<div class=\"textbox examples\">\r\n<h3>Example 3:\u00a0Concentration Cells<\/h3>\r\nWhat is the cell potential of the concentration cell described by\r\n<p style=\"text-align: center;\">[latex]\\text{Zn}\\left(s\\right)\\mid {\\text{Zn}}^{2+}\\left(aq\\text{, 0.10}M\\right)\\parallel {\\text{Zn}}^{2+}\\left(aq\\text{, 0.50}M\\right)\\mid \\text{Zn}\\left(s\\right)[\/latex]<\/p>\r\n[reveal-answer q=\"13739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"13739\"]\r\n\r\nFrom the information given:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\text{Anode:}&amp;\\text{Zn}\\left(s\\right)\\longrightarrow {\\text{Zn}}^{2+}\\left(aq\\text{, 0.10}M\\right)+{\\text{2e}}^{-}&amp;{E}_{\\text{anode}}^{\\circ }=-\\text{0.7618 V}\\\\ \\text{Cathode:}&amp;{\\text{Zn}}^{2+}\\left(aq\\text{, 0.50}M\\right)+{\\text{2e}}^{-}\\longrightarrow \\text{Zn}\\left(s\\right)&amp;{E}_{\\text{cathode}}^{\\circ }=-\\text{0.7618 V}\\\\ \\\\ \\text{Overall:}&amp;{\\text{Zn}}^{2+}\\left(aq\\text{, 0.50}M\\right)\\longrightarrow {\\text{Zn}}^{2+}\\left(aq\\text{, 0.10}M\\right)&amp;{E}_{\\text{cell}}^{\\circ }=\\text{0.000 V}\\end{array}[\/latex]<\/p>\r\nSubstituting into the Nernst equation,\r\n<p style=\"text-align: center;\">[latex]{E}_{\\text{cell}}=\\text{0.000 V}-\\dfrac{\\text{0.0592 V}}{2}\\log\\dfrac{0.10}{0.50}=+0.021 V[\/latex]<\/p>\r\n<p id=\"fs-idp47526144\">The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (<em data-effect=\"italics\">E<\/em><sub>cathode<\/sub>\u00a0&gt;\u00a0<em data-effect=\"italics\">E<\/em><sub>anode<\/sub>).<\/p>\r\n[\/hidden-answer]\r\n<h4 id=\"fs-idm11024080\">Check Your Learning<\/h4>\r\nThe concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?\r\n<div id=\"fs-idm1444240\" class=\"ui-has-child-title\"><header>\r\n<h4 class=\"os-title\"><span id=\"96556\" class=\"os-title-label\">ANSWER:<\/span><\/h4>\r\n<\/header><section>[reveal-answer q=\"13740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"13740\"]\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp18466656\"><em>E<\/em><sub>cell<\/sub>\u00a0= 0.000 V; [Zn<sup>2+<\/sup>]<sub>cathode<\/sub>\u00a0= [Zn<sup>2+<\/sup>]<sub>anode<\/sub>\u00a0= 0.30\u00a0<em>M<\/em><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nPotential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction\u2019s cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of\u00a0<em>K<\/em>). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient,\u00a0<em>Q<\/em>, as described by the Nernst equation.\r\n<h4>Key Equations<\/h4>\r\n<ul>\r\n \t<li>[latex]{E}_{\\text{cell}}^{\\circ }=\\dfrac{RT}{nF}\\ln{K}[\/latex]<\/li>\r\n \t<li>[latex]{E}_{\\text{cell}}^{\\circ }=\\dfrac{0.02\\text{57 V}}{n}\\ln{K}=\\dfrac{0.0\\text{592 V}}{n}\\log{K}\\left(\\text{at 298.15}K\\right)[\/latex]<\/li>\r\n \t<li>[latex]{E}_{\\text{cell}}={E}_{\\text{cell}}^{\\circ }-\\dfrac{RT}{nF}\\ln{Q}\\text{(Nernst equation)}[\/latex]<\/li>\r\n \t<li>[latex]{E}_{\\text{cell}}={E}_{\\text{cell}}^{\\circ }-\\dfrac{0.02\\text{57 V}}{n}\\ln{Q}={E}_{\\text{cell}}^{\\circ }-\\dfrac{0.05\\text{92 V}}{n}\\log{Q}\\left(\\text{at 298.15}K\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\Delta{G}=-nF{E}_{\\text{cell}}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta {G}^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }[\/latex]<\/li>\r\n \t<li>[latex]{w}_{\\text{ele}}={w}_{\\text{max}}=-nF{E}_{\\text{cell}}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Try It<\/h3>\r\n<ol>\r\n \t<li id=\"fs-idp166722736\">For the standard cell potentials given here, determine the \u0394<em>G<\/em>\u00b0 for the cell in kJ.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>0.000 V, n = 2<\/li>\r\n \t<li>+0.434 V, n = 2<\/li>\r\n \t<li>\u22122.439 V, n = 1<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>For the \u0394<em>G<\/em>\u00b0 values given here, determine the standard cell potential for the cell.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>12 kJ\/mol, n = 3<\/li>\r\n \t<li>\u221245 kJ\/mol, n = 1<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15\u00a0K.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\text{Hg}\\left(l\\right)+{\\text{S}}^{2-}\\left(aq\\text{, 0.10}M\\right)+2{\\text{Ag}}^{\\text{+}}\\left(aq\\text{, 0.25}M\\right)\\longrightarrow 2\\text{Ag}\\left(s\\right)+\\text{HgS}\\left(s\\right)[\/latex]<\/li>\r\n \t<li>The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 <em>M<\/em> aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 <em>M<\/em> nickel(II) nitrate solution.<\/li>\r\n \t<li>The cell made of a half-cell in which 1.0 <em>M<\/em> aqueous bromine is oxidized to 0.11 <em>M<\/em> bromide ion and a half-cell in which aluminum ion at 0.023 <em>M<\/em> is reduced to aluminum metal. Assume the standard reduction potential for Br<sub>2<\/sub>(<em>l<\/em>) is the same as that of Br<sub>2<\/sub>(<em>aq<\/em>).<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Determine \u0394<em>G<\/em> and \u0394<em>G<\/em>\u00b0 for each of the reactions in the previous problem.<\/li>\r\n \t<li>Use the data in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/standard-electrode-half-cell-potentials\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>\u00a0to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/li>\r\n \t<li>[latex]\\text{CdS}\\left(s\\right)\\rightleftharpoons {\\text{Cd}}^{2+}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)\\text{at 377 K}[\/latex]<\/li>\r\n \t<li>[latex]{\\text{Hg}}^{2+}\\left(aq\\right)+4{\\text{Br}}^{-}\\left(aq\\right)\\rightleftharpoons {\\left[{\\text{HgBr}}_{4}\\right]}^{2-}\\left(aq\\right)[\/latex]<\/li>\r\n \t<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\text{at 25}^\\circ C[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"433103\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"433103\"]\r\n\r\n1.\u00a0[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }=-\\left(2\\right)\\left(96485\\frac{\\text{C}}{\\text{V mol}}\\right)\\left(\\text{0.000 V}\\right)=\\text{0 J\/mol}=\\text{0 kJ\/mol}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }=-\\left(2\\right)\\left(96485\\frac{\\text{C}}{\\text{V mol}}\\right)\\left(\\text{+0.434 V}\\right)=-\\text{83749 J\/mol}=-\\text{83.7 kJ\/mol}[\/latex]<\/li>\r\n \t<li>[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }=-\\left(1\\right)\\left(96485\\frac{\\text{C}}{\\text{V mol}}\\right)\\left(-\\text{2.439 V}\\right)=\\text{+235327 J\/mol}=\\text{+235.3 kJ\/mol}[\/latex]<\/li>\r\n<\/ol>\r\n3.\u00a0All reactions are at 298.15 K and use [latex]{E}_{\\text{cell}}={E}_{\\text{cell}}^{\\circ }-\\frac{RT}{nF}\\ln{Q}={E}_{\\text{cell}}^{\\circ }-\\frac{0.0592}{n}\\log{Q}:[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]\\begin{array}{rl}\\text{anode: Hg}&amp;\\left(l\\right)+\\text{S}^{2-}\\left(aq\\text{, 0.25}M\\right)\\longrightarrow \\text{HgS}\\left(s\\right)+{2\\text{e}}^{-}{E}_{\\text{anode}}^{\\circ}=-0.70\\text{ V}\\\\ \\text{cathode:}&amp;2\\times\\left(\\text{Ag}^{+}\\left(aq\\text{, 0.25}M\\right)+{\\text{e}}^{-}\\longrightarrow \\text{Ag}\\left(s\\right)\\right){E}_{\\text{cathode}}^{\\circ}=\\text{0.7996 V}\\\\ \\\\ \\text{overall: Hg}&amp;\\left(l\\right)+{\\text{S}}^{2-}\\left(aq\\right)+{\\text{2Ag}}^{\\text{+}}\\left(aq\\right)\\longrightarrow \\text{2Ag}\\left(s\\right)+\\text{HgS}\\left(s\\right)\\end{array}[\/latex]\r\n[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }=\\text{0.7996 V}-\\left(-\\text{0.70 V}\\right)=\\text{1.50 V}\\left(\\text{spontaneous}\\right)[\/latex]\r\n[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cell}}^{\\circ }-\\frac{\\text{0.0592 V}}{2}\\log\\frac{1}{0.10\\times {0.25}^{2}}=\\text{1.43 V (spontaneous)}[\/latex]<\/li>\r\n \t<li>standard cell potential: 1.405 V, spontaneous; cell potential under stated conditions: 1.423 V, spontaneous<\/li>\r\n \t<li>Oxidation occurs at the anode and reduction at the cathode.<\/li>\r\n<\/ol>\r\n5.\u00a0All use [latex]K={e}^{nF{E}_{\\text{cell}}^{\\circ }\\text{\/}RT}[\/latex] with [latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }\\text{:}[\/latex]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>1.7 \u00d7 10<sup>\u221210<\/sup><\/li>\r\n \t<li>2.6 \u00d7 10<sup>\u221221<\/sup><\/li>\r\n \t<li>8.9 \u00d7 10<sup>19<\/sup><\/li>\r\n \t<li>1.0 \u00d7 10<sup>\u221214<\/sup><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<b>concentration cell: <\/b>galvanic cell in which the two half-cells are the same except for the concentration of the solutes; spontaneous when the overall reaction is the dilution of the solute\r\n\r\n<b>electrical work (<em>w<\/em><sub>ele<\/sub>): <\/b>negative of total charge times the cell potential; equal to <em>w<\/em><sub>max<\/sub> for the system, and so equals the free energy change (\u0394<em>G<\/em>)\r\n\r\n<b>Faraday\u2019s constant (F): <\/b>charge on 1 mol of electrons; <em>F<\/em> = 96,485 C\/mol e<sup>\u2212<\/sup>\r\n\r\n<b>Nernst equation: <\/b>equation that relates the logarithm of the reaction quotient (<em>Q<\/em>) to nonstandard cell potentials; can be used to relate equilibrium constants to standard cell potentials","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Explain the relations between potential, free energy change, and equilibrium constants<\/li>\n<li>Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium<\/li>\n<li>Use the Nernst equation to determine cell potentials under nonstandard conditions<\/li>\n<\/ul>\n<\/div>\n<p>So far in this module, the relationship between the cell potential and reaction\u00a0<em>spontaneity<\/em>\u00a0has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant\u00a0<em>strength<\/em>\u00a0was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties [latex]\\Delta G[\/latex] and K.<\/p>\n<section id=\"fs-idm248064976\">\n<h3>[latex]E^\\circ[\/latex] and [latex]\\Delta G^\\circ[\/latex]<\/h3>\n<p id=\"fs-idm215301376\">The standard free energy change of a process, [latex]\\Delta G^\\circ[\/latex], was defined in a previous chapter as the maximum work that could be performed by a system,\u00a0[latex]w_{max}[\/latex]. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant,\u00a0[latex]w_{elec}[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\Delta G^\\circ = w_{max} = w_{elec}[\/latex]<\/div>\n<p id=\"fs-idm233319952\">The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential:<\/p>\n<div style=\"text-align: center;\">[latex]\\Delta G^\\circ = w_{elec} = -nFE_{cell}^\\circ[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]\\Delta G^\\circ = -nFE_{cell}^\\circ[\/latex]<\/div>\n<p id=\"fs-idm648263408\">where\u00a0<em>n<\/em>\u00a0is the number of moles of electrons transferred,\u00a0<em>F<\/em>\u00a0is\u00a0<strong>Faraday\u2019s constant<\/strong>, and\u00a0[latex]E_{cell}^\\circ[\/latex]\u00a0is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes.<\/p>\n<\/section>\n<section id=\"fs-idm218866032\">\n<h3>[latex]E^\\circ[\/latex] and K<\/h3>\n<p id=\"fs-idm242158160\">Combining a previously derived relation between [latex]\\Delta G^\\circ[\/latex] and K (see the chapter on thermodynamics) and the equation above relating [latex]\\Delta G^\\circ[\/latex] and\u00a0[latex]E_{cell}^\\circ[\/latex]\u00a0yields the following:<\/p>\n<div style=\"text-align: center;\">[latex]\\Delta G^\\circ = -RT\\text{ln}K = -nFE_{cell}^\\circ[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]E_{cell}^\\circ = (\\dfrac{RT}{nF})\\text{ln}K[\/latex]<\/div>\n<p id=\"fs-idm242471888\">This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between\u00a0[latex]E^\\circ[\/latex], [latex]\\Delta G^\\circ[\/latex] and\u00a0[latex]K[\/latex]\u00a0is depicted in Figure 1, and a table correlating reaction spontaneity to values of these properties is provided in Table 1.<\/p>\n<\/section>\n<div style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/05\/23214142\/CNX_Chem_17_04_Relation.jpg\" alt=\"A diagram is shown that involves three double headed arrows positioned in the shape of an equilateral triangle. The vertices are labeled in red. The top vertex is labeled \u201cK.\u201c The vertex at the lower left is labeled \u201cdelta G superscript degree symbol.\u201d The vertex at the lower right is labeled \u201cE superscript degree symbol subscript cell.\u201d The right side of the triangle is labeled \u201cE superscript degree symbol subscript cell equals ( R T divided by n F ) l n K.\u201d The lower side of the triangle is labeled \u201cdelta G superscript degree symbol equals negative n F E superscript degree symbol subscript cell.\u201d The left side of the triangle is labeled \u201cdelta G superscript degree symbol equals negative R T l n K.\u201d\" width=\"650\" height=\"373\" \/><\/p>\n<p class=\"wp-caption-text\">Figure\u00a01.\u00a0Graphic depicting the relation between three important thermodynamic properties.<\/p>\n<\/div>\n<p>Table 1<\/p>\n<table summary=\"Table 17.2\">\n<tbody>\n<tr valign=\"top\">\n<td><em>K<\/em><\/td>\n<td>\u0394<em>G<\/em>\u00b0<\/td>\n<td><em>E<\/em>\u00b0<sub>cell<\/sub><\/td>\n<td><\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>&gt; 1<\/td>\n<td>&lt; 0<\/td>\n<td>&gt; 0<\/td>\n<td>\n<p id=\"fs-idm239811792\">Reaction is spontaneous under standard conditions<\/p>\n<p id=\"fs-idm218962512\">Products more abundant at equilibrium<\/p>\n<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>&lt; 1<\/td>\n<td>&gt; 0<\/td>\n<td>&lt; 0<\/td>\n<td>\n<p id=\"fs-idm248001280\">Reaction is non-spontaneous under standard conditions<\/p>\n<p id=\"fs-idm197346624\">Reactants more abundant at equilibrium<\/p>\n<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>= 1<\/td>\n<td>= 0<\/td>\n<td>= 0<\/td>\n<td>\n<p id=\"fs-idm249738032\">Reaction is at equilibrium under standard conditions<\/p>\n<p id=\"fs-idm197184048\">Reactants and products equally abundant<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div><\/div>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=5559620&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=cW7RpCYAOdI&amp;video_target=tpm-plugin-xbyknyp2-cW7RpCYAOdI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3-us-west-2.amazonaws.com\/Chemistry\/transcripts\/PracticeProblemCellPotential_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Practice Problem: Cell Potential, Equilibrium Constants, and Free Energy Change&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes<\/h3>\n<p>What is the standard free energy change and equilibrium constant for the following reaction at 25 \u00b0C?<\/p>\n<p style=\"text-align: center;\">[latex]2{\\text{Ag}}^{\\text{+}}\\left(aq\\right)+\\text{Fe}\\left(s\\right)\\rightleftharpoons \\text{2Ag}\\left(s\\right)+{\\text{Fe}}^{2+}\\left(aq\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q578755\">Show Solution<\/span><\/p>\n<div id=\"q578755\" class=\"hidden-answer\" style=\"display: none\">\n<p>The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in <a class=\"target-chapter\" href=\".\/chapter\/standard-electrode-half-cell-potentials\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\text{anode (oxidation):}&\\text{Fe}\\left(s\\right)\\longrightarrow {\\text{Fe}}^{2+}\\left(aq\\right)+{\\text{2e}}^{-}&{E}_{{\\text{Fe}}^{2+}\\text{\/Fe}}^{\\circ }=-\\text{0.447 V}\\\\ \\text{cathode (reduction):}&2\\times \\left({\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{e}}^{-}\\longrightarrow \\text{Ag}\\left(s\\right)\\right)&{E}_{{\\text{Ag}}^{\\text{+}}\\text{\/Ag}}^{\\circ }=\\text{0.7996 V}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }=\\qquad{E}_{{\\text{Ag}}^{\\text{+}}\\text{\/Ag}}^{\\circ }-{E}_{{\\text{Fe}}^{2+}\\text{\/Fe}}^{\\circ }=\\text{+1.247 V}[\/latex]<\/p>\n<p>Remember that the cell potential for the cathode is not multiplied by two when determining the standard cell potential. With <em>n<\/em> = 2, the equilibrium constant is then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}{E}_{\\text{cell}}^{\\circ }&=&\\frac{\\text{0.0592 V}}{n}\\log{K}\\\\K&=&{10}^{n\\times {E}_{\\text{cell}}^{\\circ }\\text{\/}\\text{0.0592 V}}\\\\{}&=&{10}^{2\\times \\text{1.247 V\/0.0592 V}}\\\\{}&=&{10}^{42.128}\\\\{}&=&1.3\\times {10}^{42}\\end{array}[\/latex]<\/p>\n<p>The two equilibrium constants differ slightly due to rounding in the constants 0.0257 V and 0.0592 V. The standard free energy is then<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta {G}^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\Delta G^{\\circ }=-2\\times\\text{96,485}\\frac{\\text{J}}{\\text{V}\\cdot \\text{mol}}\\times \\text{1.247 V}=-240.6\\frac{\\text{kJ}}{\\text{mol}}[\/latex]<\/p>\n<p>The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The\u00a0<em data-effect=\"italics\">K<\/em>\u00a0value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp110872768\">Check Your Learning<\/h4>\n<p>What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Sn}\\left(s\\right)+2{\\text{Cu}}^{2+}\\left(aq\\right)\\rightleftharpoons {\\text{Sn}}^{2+}\\left(aq\\right)+2{\\text{Cu}}^{\\text{+}}\\left(aq\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q279511\">Show Solution<\/span><\/p>\n<div id=\"q279511\" class=\"hidden-answer\" style=\"display: none\">\n<p>Spontaneous; <em>n<\/em> = 2; [latex]{E}_{\\text{cell}}^{\\circ }=\\text{+0.291 V}[\/latex]; [latex]\\Delta G^{\\circ }=-56.2\\frac{\\text{kJ}}{\\text{mol}}[\/latex]; <em>K<\/em> = 6.8 [latex]\\times[\/latex] 10<sup>9<\/sup>.<\/div>\n<\/div>\n<\/div>\n<h3>Potentials at Nonstandard Conditions: The Nernst Equation<\/h3>\n<p id=\"fs-idp76396192\">Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose.<\/p>\n<div style=\"text-align: center;\">[latex]\\Delta G = \\Delta G^\\circ + RT\\text{ln}Q[\/latex]<\/div>\n<p id=\"fs-idm983728\">Notice the reaction quotient,\u00a0[latex]Q[\/latex], appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the\u00a0<strong>Nernst equation<\/strong>:<\/p>\n<div style=\"text-align: center;\">[latex]-nFE_{cell} = -nFE_{cell}^\\circ + RT\\text{ln}Q[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center;\">[latex]E_{cell} = E_{cell}^\\circ - \\dfrac{RT}{nF}\\text{ln}Q[\/latex]<\/div>\n<p id=\"fs-idp24059424\">This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred,\u00a0[latex]n[\/latex] the temperature,\u00a0[latex]T[\/latex], and the reaction mixture composition as reflected in\u00a0[latex]Q[\/latex]. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and a factor converting from natural to base-10 logarithms have been included.<\/p>\n<div style=\"text-align: center;\">[latex]E_{cell} = E_{cell}^\\circ - \\dfrac{0.0592 \\text{V}}{n}\\text{log}Q[\/latex]<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2: Predicting Redox Spontaneity Under Nonstandard Conditions<\/h3>\n<p>Use the Nernst equation to predict the spontaneity of the redox reaction shown below.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Co}\\left(s\\right)+{\\text{Fe}}^{2+}\\left(aq,1.94M\\right)\\longrightarrow {\\text{Co}}^{2+}\\left(aq\\text{, 0.15}M\\right)+\\text{Fe}\\left(s\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q323406\">Show Solution<\/span><\/p>\n<div id=\"q323406\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are two ways to solve the problem. If the thermodynamic information in <a class=\"target-chapter\" href=\".\/chapter\/standard-thermodynamic-properties-for-selected-substances-2\/\" target=\"_blank\" rel=\"noopener\">Standard Thermodynamic Properties for Selected Substances<\/a> were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/standard-electrode-half-cell-potentials\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>. Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from\u00a0<a class=\"target-chapter\" href=\".\/chemistryformajorsxmaster\/chapter\/standard-electrode-half-cell-potentials-missing-formulas\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>\u00a0and the problem,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\text{Anode (oxidation):}&\\text{Co}\\left(s\\right)\\longrightarrow {\\text{Co}}^{2+}\\left(aq\\right)+{\\text{2e}}^{-}&{E}_{{\\text{Co}}^{2+}\\text{\/Co}}^{\\circ }=-\\text{0.28 V}\\\\ \\text{Cathode (reduction):}&{\\text{Fe}}^{2+}\\left(aq\\right)+{\\text{2e}}^{-}\\longrightarrow \\text{Fe}\\left(s\\right)&{E}_{{\\text{Fe}}^{2+}\\text{\/Fe}}^{\\circ }=-\\text{0.447 V}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }=\\qquad{-}\\text{0.447 V}-\\left(-\\text{0.28 V}\\right)=-\\text{0.17 V}[\/latex]<\/p>\n<p>Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}Q &=&\\frac{\\left[\\text{Co}^{2+}\\right]}{\\left[\\text{Fe}^{2+}\\right]}=\\dfrac{0.15M}{1.94M}=0.077\\\\{E}_{\\text{cell}}&=&{E}_{\\text{cell}}^{\\circ }-\\frac{0.0592 V}{n}\\log{Q}\\\\{E}_{\\text{cell}}&=&-0.1\\text{7 V}-\\frac{0.0592 V}{2}\\log{0.077}\\\\{E}_{\\text{cell}}&=&-\\text{0.17 V}+0.033 V=-0.014 V\\end{array}[\/latex]<\/p>\n<p id=\"fs-idp160381408\">The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous.<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idp72968608\">Check Your Learning<\/h4>\n<p>What is the cell potential for the following reaction at room temperature?<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Al}\\left(s\\right)\\mid {\\text{Al}}^{3+}\\left(aq,0.15M\\right)\\parallel {\\text{Cu}}^{2+}\\left(aq,0.025M\\right)\\mid \\text{Cu}\\left(s\\right)[\/latex]<\/p>\n<p>What are the values of <em>n<\/em> and Q for the overall reaction? Is the reaction spontaneous under these conditions?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q303221\">Show Solution<\/span><\/p>\n<div id=\"q303221\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>n<\/em> = 6; <em>Q<\/em> = 1440; <em>E<\/em><sub>cell<\/sub> = +1.97 V, spontaneous<\/div>\n<\/div>\n<\/div>\n<p>A\u00a0<strong><span id=\"term646\">concentration cell<\/span><\/strong>\u00a0is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3:\u00a0Concentration Cells<\/h3>\n<p>What is the cell potential of the concentration cell described by<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Zn}\\left(s\\right)\\mid {\\text{Zn}}^{2+}\\left(aq\\text{, 0.10}M\\right)\\parallel {\\text{Zn}}^{2+}\\left(aq\\text{, 0.50}M\\right)\\mid \\text{Zn}\\left(s\\right)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q13739\">Show Solution<\/span><\/p>\n<div id=\"q13739\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the information given:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rll}\\text{Anode:}&\\text{Zn}\\left(s\\right)\\longrightarrow {\\text{Zn}}^{2+}\\left(aq\\text{, 0.10}M\\right)+{\\text{2e}}^{-}&{E}_{\\text{anode}}^{\\circ }=-\\text{0.7618 V}\\\\ \\text{Cathode:}&{\\text{Zn}}^{2+}\\left(aq\\text{, 0.50}M\\right)+{\\text{2e}}^{-}\\longrightarrow \\text{Zn}\\left(s\\right)&{E}_{\\text{cathode}}^{\\circ }=-\\text{0.7618 V}\\\\ \\\\ \\text{Overall:}&{\\text{Zn}}^{2+}\\left(aq\\text{, 0.50}M\\right)\\longrightarrow {\\text{Zn}}^{2+}\\left(aq\\text{, 0.10}M\\right)&{E}_{\\text{cell}}^{\\circ }=\\text{0.000 V}\\end{array}[\/latex]<\/p>\n<p>Substituting into the Nernst equation,<\/p>\n<p style=\"text-align: center;\">[latex]{E}_{\\text{cell}}=\\text{0.000 V}-\\dfrac{\\text{0.0592 V}}{2}\\log\\dfrac{0.10}{0.50}=+0.021 V[\/latex]<\/p>\n<p id=\"fs-idp47526144\">The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (<em data-effect=\"italics\">E<\/em><sub>cathode<\/sub>\u00a0&gt;\u00a0<em data-effect=\"italics\">E<\/em><sub>anode<\/sub>).<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm11024080\">Check Your Learning<\/h4>\n<p>The concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?<\/p>\n<div id=\"fs-idm1444240\" class=\"ui-has-child-title\">\n<header>\n<h4 class=\"os-title\"><span id=\"96556\" class=\"os-title-label\">ANSWER:<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q13740\">Show Solution<\/span><\/p>\n<div id=\"q13740\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"os-note-body\">\n<p id=\"fs-idp18466656\"><em>E<\/em><sub>cell<\/sub>\u00a0= 0.000 V; [Zn<sup>2+<\/sup>]<sub>cathode<\/sub>\u00a0= [Zn<sup>2+<\/sup>]<sub>anode<\/sub>\u00a0= 0.30\u00a0<em>M<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Potential is a thermodynamic quantity reflecting the intrinsic driving force of a redox process, and it is directly related to the free energy change and equilibrium constant for the process. For redox processes taking place in electrochemical cells, the maximum (electrical) work done by the system is easily computed from the cell potential and the reaction stoichiometry and is equal to the free energy change for the process. The equilibrium constant for a redox reaction is logarithmically related to the reaction\u2019s cell potential, with larger (more positive) potentials indicating reactions with greater driving force that equilibrate when the reaction has proceeded far towards completion (large value of\u00a0<em>K<\/em>). Finally, the potential of a redox process varies with the composition of the reaction mixture, being related to the reactions standard potential and the value of its reaction quotient,\u00a0<em>Q<\/em>, as described by the Nernst equation.<\/p>\n<h4>Key Equations<\/h4>\n<ul>\n<li>[latex]{E}_{\\text{cell}}^{\\circ }=\\dfrac{RT}{nF}\\ln{K}[\/latex]<\/li>\n<li>[latex]{E}_{\\text{cell}}^{\\circ }=\\dfrac{0.02\\text{57 V}}{n}\\ln{K}=\\dfrac{0.0\\text{592 V}}{n}\\log{K}\\left(\\text{at 298.15}K\\right)[\/latex]<\/li>\n<li>[latex]{E}_{\\text{cell}}={E}_{\\text{cell}}^{\\circ }-\\dfrac{RT}{nF}\\ln{Q}\\text{(Nernst equation)}[\/latex]<\/li>\n<li>[latex]{E}_{\\text{cell}}={E}_{\\text{cell}}^{\\circ }-\\dfrac{0.02\\text{57 V}}{n}\\ln{Q}={E}_{\\text{cell}}^{\\circ }-\\dfrac{0.05\\text{92 V}}{n}\\log{Q}\\left(\\text{at 298.15}K\\right)[\/latex]<\/li>\n<li>[latex]\\Delta{G}=-nF{E}_{\\text{cell}}[\/latex]<\/li>\n<li>[latex]\\Delta {G}^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }[\/latex]<\/li>\n<li>[latex]{w}_{\\text{ele}}={w}_{\\text{max}}=-nF{E}_{\\text{cell}}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Try It<\/h3>\n<ol>\n<li id=\"fs-idp166722736\">For the standard cell potentials given here, determine the \u0394<em>G<\/em>\u00b0 for the cell in kJ.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>0.000 V, n = 2<\/li>\n<li>+0.434 V, n = 2<\/li>\n<li>\u22122.439 V, n = 1<\/li>\n<\/ol>\n<\/li>\n<li>For the \u0394<em>G<\/em>\u00b0 values given here, determine the standard cell potential for the cell.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>12 kJ\/mol, n = 3<\/li>\n<li>\u221245 kJ\/mol, n = 1<\/li>\n<\/ol>\n<\/li>\n<li>Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15\u00a0K.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\text{Hg}\\left(l\\right)+{\\text{S}}^{2-}\\left(aq\\text{, 0.10}M\\right)+2{\\text{Ag}}^{\\text{+}}\\left(aq\\text{, 0.25}M\\right)\\longrightarrow 2\\text{Ag}\\left(s\\right)+\\text{HgS}\\left(s\\right)[\/latex]<\/li>\n<li>The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 <em>M<\/em> aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 <em>M<\/em> nickel(II) nitrate solution.<\/li>\n<li>The cell made of a half-cell in which 1.0 <em>M<\/em> aqueous bromine is oxidized to 0.11 <em>M<\/em> bromide ion and a half-cell in which aluminum ion at 0.023 <em>M<\/em> is reduced to aluminum metal. Assume the standard reduction potential for Br<sub>2<\/sub>(<em>l<\/em>) is the same as that of Br<sub>2<\/sub>(<em>aq<\/em>).<\/li>\n<\/ol>\n<\/li>\n<li>Determine \u0394<em>G<\/em> and \u0394<em>G<\/em>\u00b0 for each of the reactions in the previous problem.<\/li>\n<li>Use the data in\u00a0<a class=\"target-chapter\" href=\".\/chapter\/standard-electrode-half-cell-potentials\/\" target=\"_blank\" rel=\"noopener\">Standard Electrode (Half-Cell) Potentials<\/a>\u00a0to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\text{AgCl}\\left(s\\right)\\rightleftharpoons {\\text{Ag}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}}^{-}\\left(aq\\right)[\/latex]<\/li>\n<li>[latex]\\text{CdS}\\left(s\\right)\\rightleftharpoons {\\text{Cd}}^{2+}\\left(aq\\right)+{\\text{S}}^{2-}\\left(aq\\right)\\text{at 377 K}[\/latex]<\/li>\n<li>[latex]{\\text{Hg}}^{2+}\\left(aq\\right)+4{\\text{Br}}^{-}\\left(aq\\right)\\rightleftharpoons {\\left[{\\text{HgBr}}_{4}\\right]}^{2-}\\left(aq\\right)[\/latex]<\/li>\n<li>[latex]{\\text{H}}_{2}\\text{O}\\left(l\\right)\\rightleftharpoons {\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{-}\\left(aq\\right)\\text{at 25}^\\circ C[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q433103\">Show Selected Solutions<\/span><\/p>\n<div id=\"q433103\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }=-\\left(2\\right)\\left(96485\\frac{\\text{C}}{\\text{V mol}}\\right)\\left(\\text{0.000 V}\\right)=\\text{0 J\/mol}=\\text{0 kJ\/mol}[\/latex]<\/li>\n<li>[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }=-\\left(2\\right)\\left(96485\\frac{\\text{C}}{\\text{V mol}}\\right)\\left(\\text{+0.434 V}\\right)=-\\text{83749 J\/mol}=-\\text{83.7 kJ\/mol}[\/latex]<\/li>\n<li>[latex]\\Delta G^{\\circ }=-nF{E}_{\\text{cell}}^{\\circ }=-\\left(1\\right)\\left(96485\\frac{\\text{C}}{\\text{V mol}}\\right)\\left(-\\text{2.439 V}\\right)=\\text{+235327 J\/mol}=\\text{+235.3 kJ\/mol}[\/latex]<\/li>\n<\/ol>\n<p>3.\u00a0All reactions are at 298.15 K and use [latex]{E}_{\\text{cell}}={E}_{\\text{cell}}^{\\circ }-\\frac{RT}{nF}\\ln{Q}={E}_{\\text{cell}}^{\\circ }-\\frac{0.0592}{n}\\log{Q}:[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]\\begin{array}{rl}\\text{anode: Hg}&\\left(l\\right)+\\text{S}^{2-}\\left(aq\\text{, 0.25}M\\right)\\longrightarrow \\text{HgS}\\left(s\\right)+{2\\text{e}}^{-}{E}_{\\text{anode}}^{\\circ}=-0.70\\text{ V}\\\\ \\text{cathode:}&2\\times\\left(\\text{Ag}^{+}\\left(aq\\text{, 0.25}M\\right)+{\\text{e}}^{-}\\longrightarrow \\text{Ag}\\left(s\\right)\\right){E}_{\\text{cathode}}^{\\circ}=\\text{0.7996 V}\\\\ \\\\ \\text{overall: Hg}&\\left(l\\right)+{\\text{S}}^{2-}\\left(aq\\right)+{\\text{2Ag}}^{\\text{+}}\\left(aq\\right)\\longrightarrow \\text{2Ag}\\left(s\\right)+\\text{HgS}\\left(s\\right)\\end{array}[\/latex]<br \/>\n[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }=\\text{0.7996 V}-\\left(-\\text{0.70 V}\\right)=\\text{1.50 V}\\left(\\text{spontaneous}\\right)[\/latex]<br \/>\n[latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cell}}^{\\circ }-\\frac{\\text{0.0592 V}}{2}\\log\\frac{1}{0.10\\times {0.25}^{2}}=\\text{1.43 V (spontaneous)}[\/latex]<\/li>\n<li>standard cell potential: 1.405 V, spontaneous; cell potential under stated conditions: 1.423 V, spontaneous<\/li>\n<li>Oxidation occurs at the anode and reduction at the cathode.<\/li>\n<\/ol>\n<p>5.\u00a0All use [latex]K={e}^{nF{E}_{\\text{cell}}^{\\circ }\\text{\/}RT}[\/latex] with [latex]{E}_{\\text{cell}}^{\\circ }={E}_{\\text{cathode}}^{\\circ }-{E}_{\\text{anode}}^{\\circ }\\text{:}[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>1.7 \u00d7 10<sup>\u221210<\/sup><\/li>\n<li>2.6 \u00d7 10<sup>\u221221<\/sup><\/li>\n<li>8.9 \u00d7 10<sup>19<\/sup><\/li>\n<li>1.0 \u00d7 10<sup>\u221214<\/sup><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><b>concentration cell: <\/b>galvanic cell in which the two half-cells are the same except for the concentration of the solutes; spontaneous when the overall reaction is the dilution of the solute<\/p>\n<p><b>electrical work (<em>w<\/em><sub>ele<\/sub>): <\/b>negative of total charge times the cell potential; equal to <em>w<\/em><sub>max<\/sub> for the system, and so equals the free energy change (\u0394<em>G<\/em>)<\/p>\n<p><b>Faraday\u2019s constant (F): <\/b>charge on 1 mol of electrons; <em>F<\/em> = 96,485 C\/mol e<sup>\u2212<\/sup><\/p>\n<p><b>Nernst equation: <\/b>equation that relates the logarithm of the reaction quotient (<em>Q<\/em>) to nonstandard cell potentials; can be used to relate equilibrium constants to standard cell potentials<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3641\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry 2e. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/\">https:\/\/openstax.org\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Practice Problem: Cell Potential, Equilibrium Constants, and Free Energy Change. <strong>Authored by<\/strong>: Professor Dave Explains. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/cW7RpCYAOdI\">https:\/\/youtu.be\/cW7RpCYAOdI<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry 2e\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/chemistry-2e\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"Practice Problem: Cell Potential, Equilibrium Constants, and Free Energy Change\",\"author\":\"Professor Dave Explains\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/cW7RpCYAOdI\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3641","chapter","type-chapter","status-publish","hentry"],"part":2970,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3641","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3641\/revisions"}],"predecessor-version":[{"id":8157,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3641\/revisions\/8157"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/parts\/2970"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapters\/3641\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/media?parent=3641"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/pressbooks\/v2\/chapter-type?post=3641"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/contributor?post=3641"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/wp-json\/wp\/v2\/license?post=3641"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}