{"id":388,"date":"2017-08-21T06:47:02","date_gmt":"2017-08-21T06:47:02","guid":{"rendered":"http:\/\/americanboard.org\/Subjects\/chemistry\/?page_id=388"},"modified":"2017-09-19T19:43:56","modified_gmt":"2017-09-19T19:43:56","slug":"oxidation-and-reduction","status":"publish","type":"page","link":"https:\/\/americanboard.org\/Subjects\/chemistry\/oxidation-and-reduction\/","title":{"rendered":"Oxidation and Reduction"},"content":{"rendered":"<div class=\"row\">\n<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\/kinetics\">\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\/cell-potential-electric-work-and-free-energy\">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\">Oxidation and Reduction<\/h1>\n<h4>Objective<\/h4>\n<p>In this lesson we will review how to balance redox reactions. We will also review the Daniell Cell and workings of\u00a0electrolysis.<\/p>\n<h4>Previously we covered&#8230;<\/h4>\n<ul>\n<li>The rates of chemical reactions can be determined by\u00a0measuring the concentrations of reactants or products during the time\u00a0the reaction occurs.<\/li>\n<li>Rate equations show how the rate is affected by changes in concentration of reactants.<\/li>\n<li>Orders of reactions can only be determined experimentally.<\/li>\n<li>Graphs can be used to determine the order of reaction, the value of the rate constant, and the half-life of the\u00a0reaction.<\/li>\n<li>Collision Theory, based on the kinetic molecular theory, accounts for the observed characteristics of reaction\u00a0rates.<\/li>\n<li>Reaction mechanisms are a proposed series of steps by which a reaction is thought to proceed.<\/li>\n<li>Five factors affect the rate of reaction: Concentration,\u00a0temperature, surface area, nature of reactants, and presence of a\u00a0catalyst.<\/li>\n<li>The Arrhenius equation can be used to determine activation energies, the minimum energy required for a reaction\u00a0to occur.<\/li>\n<\/ul>\n<section>\n<h3>Redox Reactions<\/h3>\n<p>Redox\u00a0(oxidation-reduction)\u00a0reactions are ones in which electron(s) are transferred. In a redox\u00a0reaction:<\/p>\n<ul>\n<li>The\u00a0substance being <abbr title=\"Involves the gain of electrons. The substance becomes more negative.\">reduced<\/abbr>,\u00a0meaning it\u00a0gains electrons, is the oxidizing agent.<\/li>\n<li>The\u00a0substance being <abbr title=\"Oxidation: Involves the loss of electrons. The substance becomes more positive.\">oxidized<\/abbr>,\u00a0meaning\u00a0it loses electrons, is the reducing agent.<\/li>\n<\/ul>\n<h4>Balancing Redox Reactions<\/h4>\n<p>There are several steps\u00a0we\u00a0must use to properly\u00a0balance\u00a0redox reactions.<\/p>\n<ol>\n<li>Separating\u00a0the equations into half reactions<\/li>\n<li>Balancing\u00a0elements that are NOT hydrogen or\u00a0oxygen<\/li>\n<li>Balancing\u00a0oxygen by adding the appropriate\u00a0amount of water molecules to the opposite side of the equation<\/li>\n<li>Balancing\u00a0hydrogen by adding hydrogen ions to\u00a0the opposite side.<\/li>\n<li>Balancing\u00a0the charge by adding the appropriate\u00a0amount of electrons to the more positive side of the equation<\/li>\n<li>If\u00a0the two half-reaction equations have different amounts of\u00a0electrons, we multiply each equation to obtain the lowest common\u00a0multiple of\u00a0the two electron amounts.<\/li>\n<li>If\u00a0in basic solution, we add hydroxides to both\u00a0sides of the equation (in the amount of hydrogen ions still present).\u00a0Then we\u00a0combine the hydroxide ions with hydrogen ions on one side of the\u00a0equation to\u00a0form water molecules.<\/li>\n<li>We\u00a0add the two half reactions.<\/li>\n<li>We\u00a0cancel substances that are on both sides and readjust the coefficients if necessary.<\/li>\n<li>The total\u00a0charge should be equal on both sides of the\u00a0equation as well as having the same amount of each element on each side\u00a0of the\u00a0balanced equation.<\/li>\n<\/ol>\n<h3>Example of a Redox Reaction in a Basic Solution<\/h3>\n<p>Balance the following redox reaction that occurs in a basic solution:<\/p>\n<p class=\"center\" style=\"text-align: center;\">NO<sub>2<\/sub><sup>-1<\/sup>(<em> aq<\/em>) + Al (<em>s<\/em>) \u2192 NH<sub>3<\/sub>(<em>g<\/em>) +\u00a0AlO<sub>2<\/sub><sup>1-<\/sup>(<em> aq<\/em>)<\/p>\n<p style=\"text-align: center;\">NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 NH<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p style=\"text-align: center;\">NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 NH<sub>3<\/sub>(<em>g<\/em>) +\u00a02H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p style=\"text-align: center;\">7H<sup>1+<\/sup>(<em>aq<\/em>) + NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 NH<sub>3<\/sub>(<em>g<\/em>) + 2H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p style=\"text-align: center;\">6e<sup>1-<\/sup> + 7H<sup>1+<\/sup>(<em> aq<\/em>)\u00a0+ NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 NH<sub>3<\/sub>(<em>g<\/em>) +\u00a02H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p class=\"center\" style=\"text-align: center;\">Al(<em>s<\/em>) \u2192 AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>)<\/p>\n<p style=\"text-align: center;\">2H<sub>2<\/sub>O(<em>l<\/em>) + Al(<em>s<\/em>) \u2192 AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>)<\/p>\n<p style=\"text-align: center;\">2H<sub>2<\/sub>O(<em>l<\/em>) + Al(<em>s<\/em>) \u2192 AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>) +\u00a04H<sup>1+<\/sup>(<em> aq<\/em>)<\/p>\n<p style=\"text-align: center;\">2H<sub>2<\/sub>O(<em>l<\/em>) + Al(<em>s<\/em>) \u2192 AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>) +\u00a04H<sup>1+<\/sup>(<em> aq<\/em>) + 3e<sup>1-<\/sup><\/p>\n<p>This equation will need to be multiplied by two to ensure the electrons\u00a0will be\u00a0equal in both reactions and will cancel.<\/p>\n<p>Then\u00a0add the two half reactions.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s6_001.gif\" width=\"654\" height=\"74\" \/><\/center>Cancel\u00a0like terms.<\/p>\n<p class=\"center\">2H<sub>2<\/sub>O(<em>l<\/em>) + 2Al(<em>s<\/em>)\u00a0+ NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 2AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>)\u00a0+ H<sup>1+<\/sup>(<em>aq<\/em>) + NH<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p>If\u00a0the equation were in acidic solution, this would be the final point,\u00a0but since\u00a0in basic solution, we need to add enough hydroxide ions to both sides of the\u00a0reaction to effectively cancel the hydrogen ions present and create an equal number of moles of water.<\/p>\n<p class=\"center\">OH<sup>1-<\/sup>(<em>aq<\/em>) +\u00a02H<sub>2<\/sub>O(<em>l<\/em>) + 2Al(<em>s<\/em>)\u00a0+ NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 2AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>)\u00a0+ H<sup>1+<\/sup>(<em>aq<\/em>) + NH<sub>3<\/sub>(<em>g<\/em>) + OH<sup>1-<\/sup>(<em>aq<\/em>)<\/p>\n<p>Combine\u00a0hydroxide and hydrogen to form water.<\/p>\n<p class=\"center\">OH<sup>1-<\/sup>(<em>aq<\/em>) +\u00a02H<sub>2<\/sub>O(<em>l<\/em>) + 2Al(<em>s<\/em>) + NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 2AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>)\u00a0+ H<sub>2<\/sub>O(<em>l<\/em>) + NH<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p>Cancel\u00a0one water from each side.<\/p>\n<p class=\"center\">OH<sup>1-<\/sup>(<em>aq<\/em>) +\u00a0H<sub>2<\/sub>O(<em>l<\/em>) + 2Al(<em>s<\/em>)\u00a0+ NO<sub>2<\/sub><sup>-1<\/sup>(<em>aq<\/em>) \u2192 2AlO<sub>2<\/sub><sup>1-<\/sup>(<em>aq<\/em>)\u00a0+ NH<sub>3<\/sub>(<em>g<\/em>)<\/p>\n<p>Then\u00a0check to ensure charges are equal.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s6_002.gif\" width=\"266\" height=\"65\" \/><\/center>Note that 6 moles of electrons were transferred.<\/p>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Balance\u00a0the following reaction in acidic solution using the half-reaction\u00a0method, and\u00a0determine how many moles of electrons are transferred.<\/p>\n<p class=\"center\">Br(<em>aq<\/em>)<sup>1-<\/sup> + MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192\u00a0Br<sub>2<\/sub>(<em>l<\/em>) + Mn(<em>aq<\/em>)<sup>2+<\/sup><\/p>\n<ol>\n<li>1 mol of electrons<\/li>\n<li>2 mol of electrons<\/li>\n<li>5 mol of electrons<\/li>\n<li>10 mol of electrons<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The\u00a0correct answer is D.<\/p>\n<p class=\"center\" style=\"text-align: center;\">Br(<em>aq<\/em>)<sup>1-<\/sup> + MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Br<sub>2<\/sub>(<em>l<\/em>)\u00a0+ Mn(<em>aq<\/em>)<sup>2+<\/sup><br \/>\nBr(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Br<sub>2<\/sub>(<em>l<\/em>)<br \/>\n2Br(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Br<sub>2<\/sub>(<em>l<\/em>)<\/p>\n<p style=\"text-align: center;\">2Br(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Br<sub>2<\/sub>(<em>l<\/em>) +\u00a02e<sup>1-<\/sup><br \/>\nWhen balancing the half reaction with bromine, there are two electrons\u00a0in the equation.<\/p>\n<p class=\"center\" style=\"text-align: center;\">MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Mn(<em>aq<\/em>)<sup>2+<\/sup><\/p>\n<p style=\"text-align: center;\">MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Mn(<em>aq<\/em>)<sup>2+<\/sup> +\u00a04H<sub>2<\/sub>O(<em>l<\/em>)<br \/>\n8H(<em>aq<\/em>)<sup>1+<\/sup> +\u00a0MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Mn(<em>aq<\/em>)<sup>2+<\/sup> +\u00a04H<sub>2<\/sub>O(<em>l<\/em>)<br \/>\n5e<sup>1-<\/sup> + 8H(<em>aq<\/em>)<sup>1+<\/sup> +\u00a0MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192 Mn(<em>aq<\/em>)<sup>2+<\/sup> +\u00a04H<sub>2<\/sub>O(<em>l<\/em>)<br \/>\nThe half reaction with the manganese has five electrons.<\/p>\n<p style=\"text-align: center;\">The lowest common multiple of two and five is ten.<\/p>\n<p><center><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s6_003.gif\" width=\"654\" height=\"84\" \/><\/center><\/p>\n<p class=\"center\">16H(<em>aq<\/em>)<sup>1+<\/sup> + 10Br(<em>aq<\/em>)<sup>1-<\/sup> +<br \/>\n2MnO<sub>4<\/sub>(<em>aq<\/em>)<sup>1-<\/sup> \u2192 5Br<sub>2<\/sub>(<em>l<\/em>) + 2<br \/>\nMn(<em>aq<\/em>)<sup>2+<\/sup> + 8H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p align=\"left\">10 mol electrons are transferred.<\/p>\n<\/div>\n<\/section>\n<section class=\"question\">\n<h4>Question<\/h4>\n<p>Balance the following reaction that occurs in acidic solution. Determine the moles of electrons<br \/>\ntransferred.<\/p>\n<p class=\"center\">H<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) + Zn(<em>s<\/em>) \u2192 AsH<sub>3<\/sub>(<em>g<\/em>) + Zn(<em>aq<\/em>)<sup>2+<\/sup><\/p>\n<ol>\n<li>1 mol of electrons<\/li>\n<li>2 mol of electrons<\/li>\n<li>4 mol of electrons<\/li>\n<li>8 mol of electrons<\/li>\n<\/ol>\n<p><a class=\"button button-primary q-answer\"> Reveal Answer <\/a><\/p>\n<div class=\"q-reveal\">\n<p>The correct answer is D.<\/p>\n<p class=\"center\">H<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) + Zn(<em>s<\/em>) \u2192\u00a0AsH<sub>3<\/sub>(<em>g<\/em>) + Zn(<em>aq<\/em>)<sup>2+<\/sup><\/p>\n<p class=\"center\">Zn(<em>s<\/em>) \u2192 Zn(<em>aq<\/em>)<sup>2+<\/sup><br \/>\nZn(<em>s<\/em>) \u2192 Zn(<em>aq<\/em>)<sup>2+<\/sup> + 2e<sup>1-<\/sup><\/p>\n<p class=\"center\">H<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) \u2192 AsH<sub>3<\/sub>(<em>g<\/em>)<br \/>\nH<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) \u2192 AsH<sub>3<\/sub>(<em>g<\/em>) +\u00a04H<sub>2<\/sub>O(<em>l<\/em>)<br \/>\n8H(<em>aq)<\/em><sup>1+<\/sup> + H<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) \u2192\u00a0AsH<sub>3<\/sub>(<em>g<\/em>) + 4H<sub>2<\/sub>O(<em>l<\/em>)<br \/>\n8e<sup>1-<\/sup> + 8H(<em>aq<\/em>)<sup>1+<\/sup> + H<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) \u2192\u00a0AsH<sub>3<\/sub>(<em>g<\/em>) + 4H<sub>2<\/sub>O(<em>l<\/em>)<\/p>\n<p>One half-reaction shows 2 moles of electrons and the other eight. The least common multiple of two and eight is\u00a0eight.<\/p>\n<p class=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/s6_005.gif\" width=\"590\" height=\"80\" \/><\/p>\n<p>Eight moles of electrons are transferred. Now, cancel as needed.<\/p>\n<p class=\"center\">4Zn(<em>s<\/em>) + 8H(<em>aq<\/em>)<sup>1+<\/sup> + H<sub>3<\/sub>AsO<sub>4<\/sub>(<em>aq<\/em>) \u2192\u00a0AsH<sub>3<\/sub>(<em>g<\/em>) + 4H<sub>2<\/sub>O(<em>l<\/em>) + 4Zn(<em>aq<\/em>)<sup>2+<\/sup><\/p>\n<\/div>\n<\/section>\n<h3>Daniell Cell<\/h3>\n<p>The Daniell Cell is a device that uses the energy of a\u00a0spontaneous redox\u00a0reaction to produce electric current. It was devised in 1836 by British\u00a0chemist\u00a0and meteorologist John Frederick Daniell. The cell used a zinc anode, copper\u00a0cathode,\u00a0and solutions of zinc sulfate and copper sulfate. It is the predecessor\u00a0of the\u00a0current day galvanic cell. The galvanic cell uses a porous disk or a\u00a0salt\u00a0bridge to allow for the flow of ions between half-cells to maintain electrical neutrality.<\/p>\n<p>The Daniell Cell allows for the flow of electrons\u00a0from the anode to the cathode through an external circuit.\u00a0Useful work can be done by these electrons flowing through a wire as an electrical current.\u00a0The <abbr title=\"Electrode at which oxidation occurs in a galvanic cell.\">anode<\/abbr> is\u00a0the electrode\u00a0where oxidation occurs and the\u00a0<abbr title=\"Electrode at which reduction occurs in a galvanic cell\">cathode<\/abbr> is\u00a0the electrode\u00a0where reduction\u00a0occurs.<\/p>\n<p><center><img decoding=\"async\" src=\"http:\/\/americanboard.org\/Subjects\/Images\/chemistry\/img\/reactions\/Oxidation%20and%20Reduction6.electrocell.jpg\" alt=\"Image showing flow of electrons in Daniell Cell\" \/><\/center>The electron flow above can be written using the standard line notation:<\/p>\n<p class=\"center\">Zn(<em>s<\/em>)\u00a0| Zn(<em>aq<\/em>)<sup>2+ <\/sup>||\u00a0Cu(<em>aq<\/em>)<sup>2+<\/sup> | Cu(<em>s<\/em>)<\/p>\n<p>The double line (||) represents the salt bridge or\u00a0porous disk in the\u00a0drawing. By convention, the anode half-cell is written on the left, and cathode half-cell on the right. The\u00a0convention shows the anode material, the anode solution, the salt bridge, the cathode solution and then the\u00a0cathode material. You can tell from the line notation that the Zn is oxidized and the copper (II) ion is reduced\u00a0from their respective positions in the notation.<\/p>\n<p>The reaction, Zn(<em>s<\/em>)\u00a0+ Cu(<em>aq<\/em>)<sup>2+<\/sup>\u2192 Cu(<em>s<\/em>) +\u00a0Zn(<em>aq<\/em>)<sup>2+ <\/sup>,\u00a0can be split\u00a0into half reactions:<\/p>\n<p class=\"center\">Zn(<em>s<\/em>) \u2192 Zn(<em>aq<\/em>)<sup>2+ <\/sup>+\u00a02e<sup>&#8211;<\/sup><\/p>\n<p>Zinc is neutral and loses two electron<em>s<\/em>,\u00a0which means it is oxidized.<\/p>\n<p class=\"center\">Cu(<em>aq<\/em>)<sup>2+\u00a0<\/sup>+\u00a02e<sup>&#8211;<\/sup> \u2192\u00a0Cu(<em>s<\/em>)<\/p>\n<p>Copper has a charge of positive two, then gains two electrons to form\u00a0neutral\u00a0copper, hence it is reduced.<\/p>\n<\/section>\n<\/div>\n<h3>Electrolysis<\/h3>\n<p><abbr title=\"A process that forces a current through a cell to cause a reaction that is not spontaneous to occur\">Electrolysis<\/abbr> is the process of using electric current to decompose compounds into the separate elements and molecules and has many\u00a0practical applications.<\/p>\n<p>Electrolysis is used\u00a0to decompose solid table salt (sodium chloride) in\u00a0a Down\u2019s Cell. In\u00a0the Down\u2019s cell, the salt is heated until it is melted and\u00a0the sodium and\u00a0chlorine are separated, without being able to react with each other.<\/p>\n<p>Salt water may also be separated using electrolysis.\u00a0Pure water is easier to\u00a0reduce\u00a0than sodium because\u00a0its reduction potential\u00a0(which will be explained in next lesson)\u00a0is much less than that of sodium.\u00a0The electrolysis of a sodium chloride \u00a0solution produces hydrogen gas and sodium hydroxide rather\u00a0than pure sodium\u00a0and pure chlorine. The sodium hydroxide can then be\u00a0purified using a mercury cell that allows the solid sodium from the\u00a0salt to\u00a0react with water to form hydrogen gas. The mercury cell process, also\u00a0called\u00a0the chlor-alkali process, has been replaced recently due to the health\u00a0hazards\u00a0associated with mercury. Now, an impermeable membrane is used in place\u00a0of the\u00a0mercury that allows only cations to flow through, and the result is\u00a0pure sodium\u00a0hydroxide.<\/p>\n<p>In addition, electrolysis is used to purify metals such\u00a0as copper. Huge\u00a0slabs of copper are placed in a tank containing aqueous copper sulfate\u00a0and very\u00a0thin sheets of pure copper are added to function as the cathode. The\u00a0copper\u00a0ions in the solution will form on the pure sheets of copper and the\u00a0impurities\u00a0will fall to the bottom of the solution producing 99.95% pure copper.<\/p>\n<p>Electrolysis may also be used to electroplate objects\u00a0such as tin cans and\u00a0chrome bumpers. The object to be plated is placed in a solution and serves as the\u00a0cathode. The anode will be the metal which is to be deposited on the object being plated. An important economic\u00a0advancement in history due to\u00a0electrolysis has been the production of aluminum. Aluminum oxide\u00a0(the main component of the ore\u00a0bauxite) was\u00a0too expensive to use since it has a very high melting point. Hall and\u00a0Heroult,\u00a0independently, but at about the same time, found that by using the\u00a0mixture of\u00a0aluminum oxide and sodium aluminum fluoride, a lower melting point\u00a0was\u00a0obtained. This causes the liquid aluminum to be deposited at the cathode as a precipitate. The carbon anode is\u00a0oxidized and bubbles away as carbon dioxide. The overall chemical reaction is:<\/p>\n<p class=\"center\">2Al<sub>2<\/sub>O<sub>3<\/sub> + 3C \u2192 4Al + 3CO<sub>2<\/sub><\/p>\n<p>The process produces 99.5% pure aluminum and uses about 5% of the entire\u00a0electricity output of the United States.\u00a0Several aluminum producing plants\u00a0have recently opened in Iceland,\u00a0using the natural geothermal power to operate the plants at a reduced\u00a0cost. Due to the work of Hall and Heroult, the price per pound of aluminum has\u00a0dropped\u00a0from $100,000 in the 1850\u2019s to about $1 today.<\/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\/kinetics\">\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\/cell-potential-electric-work-and-free-energy\">Next Lesson\u00a0\u27a1<\/a><\/div>\n<p><a class=\"backtotop\" href=\"#title\">Back to Top<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>\u2b05 Previous Lesson\u00a0Workshop Index\u00a0Next Lesson\u00a0\u27a1 Oxidation and Reduction Objective In this lesson we will review how to balance redox reactions. We will also review the Daniell Cell and workings of\u00a0electrolysis. Previously we covered&#8230; The rates of chemical reactions can be determined by\u00a0measuring the concentrations of reactants or products during the time\u00a0the reaction occurs. Rate equations [&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-388","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/388","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=388"}],"version-history":[{"count":14,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/388\/revisions"}],"predecessor-version":[{"id":911,"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/pages\/388\/revisions\/911"}],"wp:attachment":[{"href":"https:\/\/americanboard.org\/Subjects\/chemistry\/wp-json\/wp\/v2\/media?parent=388"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}