{"id":389,"date":"2017-08-21T06:47:12","date_gmt":"2017-08-21T06:47:12","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=389"},"modified":"2017-09-19T19:52:37","modified_gmt":"2017-09-19T19:52:37","slug":"cell-potential-electric-work-and-free-energy","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/cell-potential-electric-work-and-free-energy\/","title":{"rendered":"Cell Potential, Electric Work, and Free Energy"},"content":{"rendered":"<div class=\"twelve columns\" style=\"margin-top: 10%;\">\n<div class=\"advance\">\n<p><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/oxidation-and-reduction\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/energy-work-and-heat-flow\">Next Lesson\u00a0\u27a1<\/a><\/p>\n<\/div>\n<p><!-- UPDATE NEXT\/PREVIOUS ABOVE --><\/p>\n<p><!-- CONTENT STARTS HERE --><\/p>\n<h1 id=\"title\">Cell Potential, Electric Work, and Free Energy<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review cell potential and how to calculate emf. We will also review the relationships between\u00a0cell potential, work, and free energy.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>Balancing both acidic and basic redox reactions<\/li>\n<li>Understanding the workings of the Daniell Cell<\/li>\n<li>Determining the direction of electron flow between the anode and cathode<\/li>\n<li>Understanding electrolysis and its practical applications<\/li>\n<\/ul>\n<section>\n<h3>Cell Potential (<em>E<\/em>)<\/h3>\n<p>Cell potential is the driving force in\u00a0electrochemical\u00a0cells that moves\u00a0the electrons from\u00a0the\u00a0substance being oxidized to the substance being reduced. Cell potential is\u00a0measured\u00a0in volts (V) and is the difference between the electric potential of\u00a0two\u00a0electrodes.<\/p>\n<p>Cell potential also indicates whether a reaction is\u00a0spontaneous under\u00a0standard conditions (1 atm pressure, 298 K, 1.0 M solutions). A negative value\u00a0of cell potential means the reaction\u00a0is not spontaneous, while a\u00a0positive\u00a0value is spontaneous under standard conditions. Should the reaction in\u00a0an\u00a0equation need to be reversed, the sign of the value of the reduction\u00a0potential\u00a0will need to be reversed as well.<\/p>\n<h4>Tables of Standard Reduction Potential<\/h4>\n<p>Tables of <abbr title=\"The potential of the half-reaction under standard state conditions as measured against the potential of the standard hydrogen potential\">standard\u00a0reduction\u00a0potential<\/abbr> (<em>E<\/em>\u00b0) contain\u00a0given values for\u00a0reactions that undergo the reduction process in redox half-reactions,\u00a0thus\u00a0gaining electrons, or becoming more negative. Values of standard\u00a0reduction\u00a0potential are measured at 25\u00b0C, 101.325 kPa, and\u00a0the solutions\u2019\u00a0concentrations are 1 M. Tables of standard reduction potential are\u00a0determined\u00a0using hydrogen half cells. For a hydrogen half-cell to work it must be\u00a0connected to another half-cell. This will make a voltaic cell. By definition, the\u00a0potential of\u00a0the hydrogen half-cell is 0.00 Volts, so the voltage measured in the\u00a0voltaic\u00a0cell must be due to the other half-cell. Therefore, when the hydrogen\u00a0half-cell\u00a0is connected to another half-cell it is possible to determine the\u00a0potential of\u00a0the other half cell.<\/p>\n<table class=\"gas_law_table\">\n<thead>\n<tr>\n<th colspan=\"2\">Standard Reduction Potential<\/th>\n<\/tr>\n<tr>\n<th><strong>Reaction<\/strong><\/th>\n<th><strong><em>E<\/em>\u00b0 (V) <\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Be<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>&#8211;<\/sup> \u2194 Be(<em>s<\/em>)<\/td>\n<td valign=\"top\">-1.97<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Al<sup>3+<\/sup>(<em>aq<\/em>) + 3e<sup>&#8211;<\/sup> \u2194 Al(<em>s<\/em>)<\/td>\n<td valign=\"top\">-1.68<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Mn<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>&#8211;<\/sup> \u2194 Mn(<em>s<\/em>)<\/td>\n<td valign=\"top\">-1.18<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Fe<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>&#8211;<\/sup> \u2194 Fe(<em>s<\/em>)<\/td>\n<td valign=\"top\">-0.447<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Pb<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>&#8211;<\/sup> \u2194 Pb(<em>s<\/em>)<\/td>\n<td valign=\"top\">-0.125<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Cu<sup>2+<\/sup>(<em>aq<\/em>) +2e<sup>&#8211;<\/sup> \u2194 Cu(<em>s<\/em>)<\/td>\n<td valign=\"top\">+ 0.340<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Br<sub>2<\/sub> (<em>l<\/em>) + 2e<sup>\u2013<\/sup> \u2194 2Br<sup>&#8211;<\/sup> (<em>aq<\/em>)<\/td>\n<td valign=\"top\">+ 1.07<\/td>\n<\/tr>\n<tr>\n<td valign=\"top\">Ce<sup>4+<\/sup>(<em>aq<\/em>) + e<sup>&#8211;<\/sup> \u2194 Ce<sup>3+<\/sup>(<em>aq<\/em>)<\/td>\n<td valign=\"top\">+ 1.72<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3>Electromagnetic Force<\/h3>\n<p>The electromotive force (EMF) is the maximum potential difference between two electrodes of a galvanic or voltaic\u00a0cell. This quantity is related to the standard reduction potential of a\u00a0substance. Under standard conditions the emf is equal to the potential\u00a0difference of the cell. If the value is negative, the reaction is not\u00a0spontaneous and will not occur\u00a0as written\u00a0under standard conditions. The reverse reaction will be spontaneous.<\/p>\n<h4>Sample Problem<\/h4>\n<p>Determine the emf for the following reaction and\u00a0determine if it is\u00a0spontaneous under standard state conditions.<\/p>\n<p class=\"center\">Be<sup>2+<\/sup>(<em>aq<\/em>) + Pb(<em>s<\/em>) \u2192 Be(<em>s<\/em>)\u00a0+ Pb<sup>2+<\/sup>(<em>aq<\/em>)<\/p>\n<p>Divide the reaction into the\u00a0half-reactions and find\u00a0the potentials from a standard reduction potential table.<\/p>\n<p class=\"center\">Be<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>\u2013<\/sup> \u2192\u00a0Be(<em>s<\/em>); E\u00b0 =\u00a0-1.97 V<br \/>\nPb<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>&#8211;<\/sup> \u2192 Pb(<em>s<\/em>); E\u00b0 = -0.125 V<\/p>\n<p><strong>Note:<\/strong> This value given is the reverse of\u00a0the one\u00a0needed, so before using this value, we reverse the reaction.<\/p>\n<p>We know to do this because the Pb(<em>s<\/em>) is a reactant, not Pb<sup>2+<\/sup>.<\/p>\n<p class=\"center\">Pb(s) \u2192 Pb<sup>2+<\/sup>(<em>aq<\/em>)\u00a0+ 2e<sup>&#8211;<\/sup>;E\u00b0 = +0.125 V<\/p>\n<p>Now we add the two reactions together.<\/p>\n<p class=\"center\">Be<sup>2+<\/sup>(<em>aq<\/em>) + 2 e<sup>&#8211;<\/sup> \u2192 Be(<em>s<\/em>); E\u00b0 =\u00a0-1.97 V<br \/>\nPb(<em>s<\/em>) \u2192 Pb<sup>2+<\/sup>(<em>aq<\/em>) + 2 e<sup>\u2013<\/sup>;E\u00b0 =\u00a0+0.125 V<\/p>\n<p style=\"text-align: center;\">Be<sup>2+<\/sup>(<em>aq<\/em>) + Pb(<em>s<\/em>) \u2192 Be(<em>s<\/em>)\u00a0+ Pb<sup>2+<\/sup>(<em>aq<\/em>);E\u00b0 = -1.84 V<\/p>\n<p>Since the value is negative, the\u00a0reaction is NOT\u00a0spontaneous.<\/p>\n<h3>Nernst Equation<\/h3>\n<p>Another way to calculate the emf was determined by the\u00a0German chemist H.W.\u00a0Nernst who found a relationship between the conditions of the cell and\u00a0its\u00a0voltage. He determined that the emf can be calculated under\u00a0non-standard\u00a0conditions if the concentrations of the solutions are known.<\/p>\n<p>His\u00a0equation, known as the Nernst equation, is:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s7_007.gif\" width=\"120\" height=\"41\" \/><\/center><em>R<\/em> is the Ideal Gas Constant, <em>T<\/em> is the Kelvin temperature, <em>n<\/em> is the number of moles of\u00a0electrons transferred and <em>Q<\/em> is the reaction quotient. If the reaction occurs at standard temperature of 298\u00a0K and the base-10 logarithm is used the equation can also be expressed as:<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s7_001.gif\" width=\"353\" height=\"45\" \/><\/center>The concentrations of the\u00a0reactants and\u00a0products are given as molarity of the solution. The concentrations of the products and reactants must be raised to\u00a0the exponent which is the coefficient of that substance in the balanced equation.<\/p>\n<h4>Sample Problem<\/h4>\n<p>Determine the voltage of a cell if the concentration of the manganese(II) ion is 1.00 M and the concentration of the\u00a0aluminum ion is 1.50 M.<\/p>\n<p class=\"center\">2Al(<em>s<\/em>) + 3Mn<sup>2+<\/sup>(<em>aq<\/em>) \u2192 2Al<sup>3+<\/sup>(<em>aq<\/em>) +\u00a03Mn(<em>s<\/em>)<\/p>\n<p>Split into the two half reactions<\/p>\n<p class=\"center\">2Al(<em>s<\/em>) \u2192 2Al<sup>3+<\/sup>(<em>aq<\/em>) + 6e<sup>&#8211;<\/sup> E\u00b0 = +1.68 V<br \/>\n(value reversed from table to reflect this equation)<br \/>\n3Mn<sup>2+<\/sup>(<em>aq<\/em>) + 6e<sup>&#8211;<\/sup> \u2192 3Mn(<em>s<\/em>); E\u00b0 = -1.18 V<\/p>\n<p>Now add the two reactions to obtain E\u00b0.<\/p>\n<p class=\"center\">E\u00b0 = 0.50 V<\/p>\n<p>Substitute\u00a0the values into the Nernst equation. Remember to\u00a0raise the concentrations to the power of their respective coefficients\u00a0in the\u00a0balanced equation.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s7_002.gif\" width=\"265\" height=\"50\" \/><\/center><\/p>\n<p class=\"center\" style=\"text-align: center;\">E = 0.49 V<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine\u00a0the emf of the following reaction under standard\u00a0conditions.<\/p>\n<p class=\"MsoNormal\" align=\"center\">Be + Fe<sup>2+<\/sup>(<em>aq<\/em>) \u2192 Fe + Be<sup>2+<\/sup>(<em>aq<\/em>)<br \/>\nFe<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>\u2013<\/sup> \u2192 Fe(<em>s<\/em>);<br \/>\nE\u00b0 = -0.447 V<br \/>\nBe<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>\u2013<\/sup> \u2192 Be (<em>s<\/em>);<br \/>\nE\u00b0 = -1.97 V<\/p>\n<ol>\n<li>-2.42 V<\/li>\n<li>-1.52 V<\/li>\n<li>1.52 V<\/li>\n<li>2.42 V<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The\u00a0correct answer is C. The Be half-reaction proceeds as an oxidation as shown in the complete equation. When\u00a0written as an oxidation the cell potential is reversed in sign and becomes +1.97 V. Adding +1.97 V and -0.447 V\u00a0gives a cell potential of + 1.52 V.<\/p>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Determine the voltage\u00a0of the following cell:<\/p>\n<p class=\"center\">Pb(<em>s<\/em>) | Pb<sup>2+<\/sup>(<em>aq<\/em>) (1.00M) || Cu<sup>2+<\/sup>(<em>aq<\/em>) (1.50M) |\u00a0Cu(<em>s<\/em>)<\/p>\n<p class=\"center\">Pb<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>\u2013<\/sup> <span class=\"MsoNormal\"> \u2192 <\/span>Pb(<em>s<\/em>); E<sup>o<\/sup> = &#8211; 0.125 V<br \/>\nCu<sup>2+<\/sup>(<em>aq<\/em>) +2e<sup>&#8211;<\/sup> <span class=\"MsoNormal\"> \u2192 <\/span>Cu(<em>s<\/em>);<br \/>\nE<sup>o<\/sup> = + 0.340 V<\/p>\n<ol>\n<li>-0.470 V<\/li>\n<li>-0.460 V<\/li>\n<li>0.460 V<\/li>\n<li>0.470 V<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<p class=\"q-reveal\">The correct answer is D.<br \/>\nThe standard potential of the cell is +0.465 V. Using the Nernst equation we substitute the concentrations of the\u00a0products and the reactants.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s7_003.gif\" width=\"256\" height=\"45\" \/><br \/>\nE = 0.470 V<\/p>\n<\/section>\n<h3>Work (<em>w<\/em>)<\/h3>\n<p>The work done in the cell is equal to the amount of\u00a0charge in coulombs multiplied by the cell\u2019s potential difference:<\/p>\n<p class=\"center\"><em>w = -qE<\/em><\/p>\n<p><em>w<\/em> represents work, <em>q\u00a0<\/em>the charge in coulombs and <em>E<\/em> the cell\u2019s potential difference.<\/p>\n<p>If the cell is galvanic, the charge, <em>q<\/em>,\u00a0equals the number of moles, <em>n<\/em>,\u00a0multiplied by the Faraday constant, F, named for Michael Faraday. One faraday, <em>F<\/em>, is equal to 96,485 Coulombs\u00a0per mole\u00a0of electrons. This makes the equation become<\/p>\n<p class=\"center\"><em>w<\/em> = &#8211;<em>nFE<\/em>.<\/p>\n<h3>Free Energy (<em>G)<\/em><\/h3>\n<p>The maximum amount of work that can be done is equal to\u00a0the change in free\u00a0energy, so <em>w<\/em> = \u0394<em>G<\/em>. This makes the equation become<\/p>\n<p class=\"center\">\u0394<em>G<\/em> = <em>-nFE<\/em>.<\/p>\n<p>Under standard conditions, the equation becomes<\/p>\n<p class=\"center\">\u0394<em>G<\/em>\u00b0 = &#8211;<em>nFE<\/em>\u00b0.<\/p>\n<p>The units for <em>E<\/em>\u00b0 is Volts,\u00a0but a volt is equal to a J\/C. So the units for \u0394<em>G<\/em>\u00b0 will be either J or kJ.<\/p>\n<h3>Sample Problem<\/h3>\n<p>Determine the free energy change for the following reaction:<\/p>\n<p class=\"center\" style=\"text-align: center;\">Cu<sup>2+<\/sup>(<em>aq<\/em>) + Fe(<em>s<\/em>) \u2192 Cu(<em>s<\/em>)\u00a0+ Fe<sup>2+<\/sup>(<em>aq<\/em>)<br \/>\nCu<sup>2+<\/sup>(<em>aq<\/em>) + 2e<sup>\u2013<\/sup> \u2192 Cu(<em>s<\/em>); <em>E<\/em>\u00b0 = 0.34 V<br \/>\nFe(<em>s<\/em>) + 2e<sup>\u2013<\/sup> \u2192 Fe<sup>2+<\/sup>(<em>aq<\/em>); <em>E<\/em>\u00b0 =\u00a0-0.447 V<\/p>\n<p class=\"center\" style=\"text-align: center;\">\u0394<em>G<\/em><sup>o\u00a0<\/sup>= &#8211;<em>nFE<\/em>\u00b0<\/p>\n<p style=\"text-align: center;\">\u0394<em>G<\/em>\u00b0\u00a0= &#8211; (2 mol e<sup>&#8211;<\/sup>)(96,485\u00a0C\/mol e<sup>\u2013<\/sup>)(0.787 V)<sup><br \/>\n<\/sup><\/p>\n<p style=\"text-align: center;\">\u0394<em>G<\/em>\u00b0\u00a0= &#8211; 152 kJ<\/p>\n<p><!-- CONTENT ENDS HERE --><\/p>\n<p><!-- UPDATE NEXT\/PREVIOUS BELOW --><\/p>\n<div class=\"advance\"><a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/oxidation-and-reduction\">\u2b05 Previous Lesson<\/a>\u00a0<a class=\"button\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/reactions-and-reactivity\">Workshop Index<\/a>\u00a0<a class=\"button button-primary\" href=\"http:\/\/americanboard.org\/Subjects\/chemistry\/energy-work-and-heat-flow\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/section>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Cell Potential, Electric Work, and Free Energy Objective In this lesson we will review cell potential and how to calculate emf. We will also review the relationships between\u00a0cell potential, work, and free energy. Previously we covered&#8230; Balancing both acidic and basic redox reactions Understanding the workings of the Daniell Cell [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-389","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/389","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/comments?post=389"}],"version-history":[{"count":16,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/389\/revisions"}],"predecessor-version":[{"id":913,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/389\/revisions\/913"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}