{"id":2177,"date":"2018-04-11T23:52:09","date_gmt":"2018-04-12T03:52:09","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/oxidation-reduction-reactions\/"},"modified":"2019-05-14T16:03:32","modified_gmt":"2019-05-14T20:03:32","slug":"oxidation-reduction-reactions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/oxidation-reduction-reactions\/","title":{"raw":"6.4 Oxidation-Reduction Reactions","rendered":"6.4 Oxidation-Reduction Reactions"},"content":{"raw":"<div class=\"section\" id=\"ball-ch04_s06\" lang=\"en\">\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define <em>oxidation<\/em> and <em>reduction<\/em>.<\/li>\r\n \t<li>Assign oxidation numbers to atoms in simple compounds.<\/li>\r\n \t<li>Recognize a reaction as an oxidation-reduction reaction.<\/li>\r\n \t<li>Recognize composition, decomposition, combustion and single replacement reactions.<\/li>\r\n \t<li>Predict the products of a combustion reaction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-idm48520656\">\r\n<h2>Redox Reactions<\/h2>\r\n<p id=\"fs-idp3801440\">Earth\u2019s atmosphere contains about 20% molecular oxygen, O<sub>2<\/sub>, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term <strong>oxidation<\/strong> was originally used to describe chemical reactions involving O<sub>2<\/sub>, but its meaning has evolved to refer to a broad and important reaction class known as <em>oxidation-reduction (redox) reactions<\/em>. A few examples of such reactions will be used to develop a clear picture of this classification.<\/p>\r\n<p id=\"fs-idm49954608\">Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm5657872\" style=\"text-align: center\">$latex 2\\text{Na}(s) + \\text{Cl}_2(g) \\longrightarrow 2\\text{NaCl}(s)$<\/div>\r\n<p id=\"fs-idm102441680\">It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a <strong>half-reaction<\/strong>:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp15924368\" style=\"text-align: center\">$latex 2\\text{Na}(s) \\longrightarrow 2\\text{Na}^{+}(s) + 2\\text{e}^{-}$\r\n$latex \\text{Cl}_2(g) + 2\\text{e}^{-} \\longrightarrow 2\\text{Cl}^{-}(s)$<\/div>\r\n<p id=\"fs-idp97564400\">These equations show that Na atoms <em>lose electrons<\/em> while Cl atoms (in the Cl<sub>2<\/sub> molecule) <em>gain electrons<\/em>, the \u201c<em>s<\/em>\u201d subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp61672384\" style=\"text-align: center\">$latex \\begin{array}{r @ {{}={}} l} \\pmb{\\text{oxidation}} &amp; \\text{loss of electrons} \\\\[1em] \\pmb{\\text{reduction}} &amp; \\text{gain of electrons} \\end{array} $<\/div>\r\n<p id=\"fs-idp6686448\">In this reaction, then, sodium is <em>oxidized<\/em> and chlorine undergoes <strong>reduction<\/strong>. Viewed from a more active perspective, sodium functions as a <strong>reducing agent (reductant)<\/strong>, since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an <strong>oxidizing agent (oxidant)<\/strong>, as it effectively removes electrons from (oxidizes) sodium.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm29833328\" style=\"text-align: center\">$latex \\begin{array}{r @ {{}={}} l} \\pmb{\\text{reducing agent}} &amp; \\text{species that is oxidized} \\\\[1em] \\pmb{\\text{oxidizing agent}} &amp; \\text{species that is reduced} \\end{array} $<\/div>\r\n<p id=\"fs-idp108466096\">Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">$latex \\text{H}_2(g) + \\text{Cl}_2(g) \\longrightarrow 2 \\text{HCl}(g) $<\/div>\r\n<div class=\"equation\"><\/div>\r\n<div class=\"equation\" id=\"fs-idp37282464\" style=\"text-align: left\">The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called <em>oxidation number<\/em> has been defined. The <strong>oxidation number<\/strong> (or <strong>oxidation state<\/strong>) of an element in a compound is the charge its atoms would possess <em>if the compound was ionic<\/em>.<\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\">The following guidelines are used to assign oxidation numbers to each element in a molecule or ion:<\/div>\r\n<ol id=\"fs-idp29396208\">\r\n \t<li>The oxidation number of an atom in an elemental substance is zero.<\/li>\r\n \t<li>The oxidation number of a monatomic ion is equal to the ion\u2019s charge.<\/li>\r\n \t<li>Oxidation numbers for common non-metals are usually assigned as follows:\r\n<ul id=\"fs-idm48186672\">\r\n \t<li>Hydrogen: +1 when combined with nonmetals, \u22121 when combined with metals<\/li>\r\n \t<li>Oxygen: \u22122 in most compounds, sometimes \u22121 (so-called peroxides, O<sub>2<\/sub><sup>2\u2212<\/sup>), very rarely $latex -\\frac{1}{2}$ (so-called superoxides, O<sub>2<\/sub><sup>\u2212<\/sup>), positive values when combined with F (values vary)<\/li>\r\n \t<li>Halogens: \u22121 for F always, \u22121 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.<\/li>\r\n<\/ol>\r\n<p id=\"fs-idm72440048\">Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.<\/p>\r\n\r\n<div class=\"textbox shaded\" id=\"fs-idm24634320\">\r\n<h3>Example 1<\/h3>\r\n<p id=\"fs-idm73523056\">Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:<\/p>\r\n<p id=\"fs-idp9372816\">a) H<sub>2<\/sub>S<\/p>\r\n<p id=\"fs-idp5671152\">b) SO<sub>3<\/sub><sup>2\u2212<\/sup><\/p>\r\n<p id=\"fs-idp109909120\">c) Na<sub>2<\/sub>SO<sub>4<\/sub><\/p>\r\n&nbsp;\r\n<p id=\"fs-idp203498912\"><strong>Solution<\/strong>\r\na) According to guideline 1, the oxidation number for H is +1.<\/p>\r\n<p id=\"fs-idm32134656\">Using this oxidation number and the compound\u2019s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">$latex \\text{charge on H}_2 \\text{S} = 0 = (2 \\times +1) + (1 \\times x)$<\/div>\r\n<div class=\"equation\" style=\"text-align: center\">$latex x = 0 = - (2 \\times +1) = -2$<\/div>\r\n<div class=\"equation\" id=\"fs-idm58489232\" style=\"text-align: center\"><\/div>\r\n<p id=\"fs-idm1965888\">b) Guideline 3 suggests the oxidation number for oxygen is \u22122.<\/p>\r\n<p id=\"fs-idp181502096\">Using this oxidation number and the ion\u2019s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">$latex {\\text{charge on SO}_3}^{2-} = -2 = (3 \\times -2) + (1 \\times x)$<\/div>\r\n<div class=\"equation\" style=\"text-align: center\">$latex x = -2 - (3 \\times -2) = +4$<\/div>\r\n<div class=\"equation\" id=\"fs-idm22135232\" style=\"text-align: center\"><\/div>\r\n<p id=\"fs-idp180932672\">c) For ionic compounds, it\u2019s convenient to assign oxidation numbers for the cation and anion separately.<\/p>\r\n<p id=\"fs-idp50986592\">According to guideline 2, the oxidation number for sodium is +1.<\/p>\r\n<p id=\"fs-idm60591184\">Assuming the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:<\/p>\r\n\r\n<div class=\"equation\" style=\"text-align: center\">$latex {\\text{charge on SO}_4}^{2-} = -2 = (4 \\times -2) + (1 \\times x)$<\/div>\r\n<div class=\"equation\" id=\"fs-idp6634688\" style=\"text-align: center\">$latex x = -2 -(4 \\times -2) = +6$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp108038048\"><em><strong>Test Yourself<\/strong><\/em>\r\nAssign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:<\/p>\r\n<p id=\"fs-idp34225952\">a) K<u>N<\/u>O<sub>3 \u00a0 \u00a0\u00a0<\/sub>b) <u>Al<\/u>H<sub>3 \u00a0 \u00a0<\/sub>c) <span style=\"text-decoration: underline\">N<\/span>H<sub>4<\/sub><sup>+ \u00a0 \u00a0<\/sup>d) H<sub>2<\/sub><span style=\"text-decoration: underline\">P<\/span>O<sub>4<\/sub><sup>\u2212<\/sup><\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answers<\/strong><\/em>\r\n\r\na) N, +5 \u00a0 \u00a0 b) Al, +3 \u00a0 \u00a0 c) N, \u22123 \u00a0 \u00a0d) P, +5\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 2<\/h3>\r\n<p id=\"ball-ch04_s06_p07\" class=\"para\">Assign oxidation numbers to the atoms in each substance.<\/p>\r\n<p class=\"para\">a) Br<sub class=\"subscript\">2<\/sub> \u00a0 \u00a0 \u00a0b)\u00a0SiO<sub class=\"subscript\">2<\/sub> \u00a0 \u00a0 \u00a0c)\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) Br<sub class=\"subscript\">2<\/sub> is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.<\/p>\r\n<p class=\"simpara\">b) By rule 3, oxygen is normally assigned an oxidation number of \u22122. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.<\/p>\r\n<p class=\"simpara\">c) The compound barium nitrate can be separated into two parts: the Ba<sup class=\"superscript\">2+<\/sup> ion and the nitrate ion. Considering these separately, the Ba<sup class=\"superscript\">2+<\/sup> ion has an oxidation number of +2 by rule 2. Now consider the NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> ion. Oxygen is assigned an oxidation number of \u22122, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">x<\/em> +\u00a03(\u22122) = \u22121<\/span><\/span>\r\n<p id=\"ball-ch04_s06_p08\" class=\"para\">where <em class=\"emphasis\">x<\/em> is the oxidation number of the nitrogen atom and \u22121 represents the charge on the species. Evaluating,<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">x<\/em> +\u00a0(\u22126) = \u22121<\/span><\/span>\r\n<span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">x<\/em> = +5<\/span><\/span>\r\n<p id=\"ball-ch04_s06_p09\" class=\"para\">Thus, the oxidation number on the N atom in the nitrate ion is +5.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s06_p10\" class=\"para\">Assign oxidation numbers to the atoms in H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s06_p11\" class=\"para\">H = +1, O = \u22122, P = +5<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idp45838960\">Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. <strong>Oxidation-reduction (redox) reactions<\/strong> are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist <a href=\"#fs-idp180799104\" class=\"autogenerated-content\">Example 5c)<\/a>. Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp231200304\" style=\"text-align: center\">$latex \\pmb{\\text{oxidation}} = \\text{increase in oxidation number}$<\/div>\r\n<div class=\"equation\" style=\"text-align: center\">$latex \\pmb{\\text{reduction}} = \\text{decrease in oxidation number}$<\/div>\r\n<p id=\"fs-idm1410784\">Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl<sub>2<\/sub> to \u22121 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H<sub>2<\/sub> to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl<sub>2<\/sub> to \u22121 in HCl).<\/p>\r\n\r\n<h2>Classification of Redox Reactions<\/h2>\r\nFour classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.\r\n<p id=\"ball-ch04_s04_p02\" class=\"para editable block\">1 - A <strong><span class=\"margin_term\"><a class=\"glossterm\">composition reaction<\/a><\/span><\/strong>\u00a0(sometimes also called a <em class=\"emphasis\">combination reaction<\/em> or a <em class=\"emphasis\">synthesis reaction<\/em>) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sub class=\"subscript\">2<\/sub>(g) +\u00a0O<sub class=\"subscript\">2<\/sub>(g)\u00a0<\/span><\/span>$latex \\longrightarrow$<span class=\"informalequation block\"><span class=\"mathphrase\"> 2 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p03\" class=\"para editable block\">water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance\u2014water\u2014as a product. So this is a composition reaction.<\/p>\r\n<p id=\"ball-ch04_s04_p04\" class=\"para editable block\">2 - A <strong><span class=\"margin_term\"><a class=\"glossterm\">decomposition reaction<\/a><\/span><\/strong>\u00a0starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 NaHCO<sub class=\"subscript\">3<\/sub>(s) $latex \\longrightarrow$\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a0CO<sub class=\"subscript\">2<\/sub>(g) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p05\" class=\"para editable block\">sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.<\/p>\r\n<p id=\"ball-ch04_s04_p06\" class=\"para editable block\">Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 3<\/h3>\r\n<p id=\"ball-ch04_s04_p07\" class=\"para\">Identify each equation as a composition reaction, a decomposition reaction, or neither.<\/p>\r\n<p class=\"para\">a) Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a03 SO<sub class=\"subscript\">3<\/sub>\u00a0$latex \\longrightarrow$ Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/p>\r\n<p class=\"para\">b) NaCl +\u00a0AgNO<sub class=\"subscript\">3<\/sub>\u00a0$latex \\longrightarrow$ AgCl +\u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\r\n<p class=\"para\">c) (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub>\u00a0$latex \\longrightarrow$ Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a04 H<sub class=\"subscript\">2<\/sub>O +\u00a0N<sub class=\"subscript\">2<\/sub><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) In this equation, two substances combine to make a single substance. This is a composition reaction.<\/p>\r\n<p class=\"simpara\">b) Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.<\/p>\r\n<p class=\"simpara\">c) A single substance reacts to make multiple substances. This is a decomposition reaction.<\/p>\r\n&nbsp;\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s04_p08\" class=\"para\">Identify the equation as a composition reaction, a decomposition reaction, or neither.<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub>\u00a0$latex \\longrightarrow$ C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub><\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s04_p09\" class=\"para\">decomposition<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idp112552240\">3 -<strong> Combustion reactions<\/strong> in which the reductant, also called a <em>fuel,<\/em>\u00a0and oxidant, molecular oxygen, react vigorously and produce significant amounts of heat, and often light, in the form of a flame. \u00a0Combustion reactions produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N<sub class=\"subscript\">2<\/sub>. Many reactants, called <em class=\"emphasis\">fuels<\/em>, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O. For example, the balanced chemical equation for the combustion of methane, CH<sub class=\"subscript\">4<\/sub>, is as follows:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp91244624\" style=\"text-align: left\">\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">CH<sub class=\"subscript\">4<\/sub> +\u00a02 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ CO<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p11\" class=\"para editable block\">Kerosene can be approximated with the formula C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">26<\/sub>, and its combustion equation is<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">26<\/sub> +\u00a037 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 24 CO<sub class=\"subscript\">2<\/sub> +\u00a026 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p12\" class=\"para editable block\">Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH, whose combustion equation is<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH +\u00a03 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 CO<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s04_p13\" class=\"para editable block\">If nitrogen is present in the original fuel, it is converted to N<sub class=\"subscript\">2<\/sub>, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>, we have<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 4 CO<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O +\u00a02 N<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 4<\/h3>\r\n<p id=\"ball-ch04_s04_p14\" class=\"para\">Complete and balance each combustion equation.<\/p>\r\n<p class=\"para\">a) the combustion of propane, C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub><\/p>\r\n<p class=\"para\">b) the combustion of ammonia, NH<sub class=\"subscript\">3<\/sub><\/p>\r\n\r\n\r\n[caption id=\"attachment_2172\" align=\"aligncenter\" width=\"269\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-294x300.jpg\" alt=\"\" width=\"269\" height=\"274\" class=\"wp-image-2172\" \/><\/a> <strong>Figure 1.<\/strong> Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. \u00a0Source: \u201cflowers and propane\u201d by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic[\/caption]\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) The products of the reaction are CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O, so our unbalanced equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ CO<sub class=\"subscript\">2<\/sub> +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n<p id=\"ball-ch04_s04_p15\" class=\"para\">Balancing (and you may have to go back and forth a few times to balance this), we get<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 3 CO<sub class=\"subscript\">2<\/sub> +\u00a04 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n\r\nb) The nitrogen atoms in ammonia will react to make N<sub class=\"subscript\">2<\/sub>, while the hydrogen atoms will react with O<sub class=\"subscript\">2<\/sub> to make H<sub class=\"subscript\">2<\/sub>O:\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">NH<sub class=\"subscript\">3<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ N<sub class=\"subscript\">2<\/sub> +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n<p id=\"ball-ch04_s04_p16\" class=\"para\">To balance this equation without fractions (which is the convention), we get<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">4 NH<sub class=\"subscript\">3<\/sub> +\u00a03 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 N<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s04_p17\" class=\"para\">Complete and balance the combustion equation for cyclopropanol, C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">6<\/sub>O.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s04_p18\" class=\"para\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">6<\/sub>O +\u00a04 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 3 CO<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp4633776\" class=\"textbox shaded\">\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-4-3.png\" alt=\"\u00a0\" width=\"135\" height=\"84\" class=\"alignleft\" \/>\r\n<p id=\"fs-idm5712736\">Watch a brief <a href=\"http:\/\/openstaxcollege.org\/l\/16hybridrocket\">video<\/a> showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at\u00a03 s (green flame) use a liquid fuel\/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-idm580304\">4 - <strong>Single-displacement (replacement) reactions<\/strong> are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp4619296\" style=\"text-align: center\">$latex \\text{Zn}(s) + 2\\text{HCl}(aq) \\longrightarrow \\text{ZnCl}_2(aq) + \\text{H}_2(g)$<\/div>\r\n<p id=\"fs-idm50858768\">Metallic elements may also be oxidized by solutions of other metal salts; for example:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm10324592\" style=\"text-align: center\">$latex \\text{Cu}(s) + 2 \\text{AgNO}_3(aq) \\longrightarrow \\text{Cu(NO}_3)_2(aq) + 2 \\text{Ag}(s)$<\/div>\r\n<p id=\"fs-idm10678768\">This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu<sup>2+<\/sup> ions dissolve in the solution to yield a characteristic blue color (<a href=\"#CNX_Chem_04_02_CuAgNO3\" class=\"autogenerated-content\">Figure 2<\/a>).<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_02_CuAgNO3\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_CuAgNO3.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_CuAgNO3-3.jpg\" alt=\"This figure contains three photographs. In a, a coiled copper wire is shown beside a test tube filled with a clear, colorless liquid. In b, the wire has been inserted into the test tube with the clear, colorless liquid. In c, the test tube contains a light blue liquid and the coiled wire appears to have a fuzzy silver gray coating.\" width=\"975\" height=\"232\" \/><\/a> <strong>Figure 2.<\/strong> (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<div class=\"textbox shaded\" id=\"fs-idp180799104\">\r\n<h3>Example 5<\/h3>\r\n<p id=\"fs-idm59303872\">Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.<\/p>\r\n<p id=\"fs-idm23437408\">a) $latex \\text{ZnCO}_3(s) \\longrightarrow \\text{ZnO}(s) + \\text{CO}_2(g)$<\/p>\r\n<p id=\"fs-idm32376704\">b) $latex 2\\text{Ga}(l) + 3\\text{Br}_2(l) \\longrightarrow 2\\text{GaBr}_3(s)$<\/p>\r\nc) $latex 2\\text{H}_2 \\text{O}_2(aq) \\longrightarrow 2\\text{H}_2 \\text{O}(l) + \\text{O}_2(g)$\r\n\r\nd) $latex \\text{BaCl}_2(aq) + \\text{K}_2 \\text{SO}_4(aq) \\longrightarrow \\text{BaSO}_4(s) + 2\\text{KCl}(aq)$\r\n<p id=\"fs-idp64660848\">e) $latex \\text{C}_2 \\text{H}_4(g) + 3\\text{O}_2(g) \\longrightarrow 2\\text{CO}_2(g) + 2\\text{H}_2 \\text{O}(l)$<\/p>\r\n&nbsp;\r\n\r\n<strong>Solution<\/strong>\r\nRedox reactions are identified per definition if one or more elements undergo a change in oxidation number.\r\n<p id=\"fs-idp31047840\">a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.<\/p>\r\n<p id=\"fs-idp218627312\">b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(<em>l<\/em>) to +3 in GaBr<sub>3<\/sub>(<em>s<\/em>). The reducing agent is Ga(<em>l<\/em>). Bromine is reduced, its oxidation number decreasing from 0 in Br<sub>2<\/sub>(<em>l<\/em>) to \u22121 in GaBr<sub>3<\/sub>(<em>s<\/em>). The oxidizing agent is Br<sub>2<\/sub>(<em>l<\/em>).<\/p>\r\n<p id=\"fs-idp223712368\">c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called <em>disproportionation reaction)<\/em>. Oxygen is oxidized, its oxidation number increasing from \u22121 in H<sub>2<\/sub>O<sub>2<\/sub>(<em>aq<\/em>) to 0 in O<sub>2<\/sub>(<em>g<\/em>). Oxygen is also reduced, its oxidation number decreasing from \u22121 in H<sub>2<\/sub>O<sub>2<\/sub>(<em>aq<\/em>) to \u22122 in H<sub>2<\/sub>O(<em>l<\/em>). For disproportionation reactions, the same substance functions as an oxidant and a reductant.<\/p>\r\n<p id=\"fs-idm40215792\">d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.<\/p>\r\n<p id=\"fs-idm55300832\">e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from \u22122 in C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>) to +4 in CO<sub>2<\/sub>(<em>g<\/em>). The reducing agent (fuel) is C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>). Oxygen is reduced, its oxidation number decreasing from 0 in O<sub>2<\/sub>(<em>g<\/em>) to \u22122 in H<sub>2<\/sub>O(<em>l<\/em>). The oxidizing agent is O<sub>2<\/sub>(<em>g<\/em>).<\/p>\r\n&nbsp;\r\n<p id=\"fs-idm9371232\"><em><strong>Test Yourself<\/strong><\/em>\r\nThis equation describes the production of tin(II) chloride:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm73301872\" style=\"text-align: center\">$latex \\text{Sn}(s) + 2\\text{HCl}(g) \\longrightarrow \\text{SnCl}_2(s) + \\text{H}_2(g)$<\/div>\r\n<p id=\"fs-idp98112752\">Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\nYes, a single-replacement reaction. Sn(<em>s<\/em>)\u00a0is the reductant, HCl(<em>g<\/em>) is the oxidant.\r\n\r\n<\/div>\r\n<section id=\"fs-idp98840016\"><\/section><\/section><section id=\"fs-idm51820592\" class=\"summary\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idp62302320\">Chemical reactions are classified according to similar patterns of behavior. Redox reactions involve a change in oxidation number for one or more reactant elements.\u00a0There are four classifications of chemical reactions: composition, decomposition, combustion and single displacement.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3 itemprop=\"educationalUse\">Exercises<\/h3>\r\n<div class=\"qandaset block\" id=\"ball-ch04_s06_qs01\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s06_qs01_p1\" class=\"para\">1. Is the reaction<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 K(s) +\u00a0Br<sub class=\"subscript\">2<\/sub>(\u2113) $latex \\longrightarrow$\u00a02 KBr(s)<\/span><\/span>\r\n<p id=\"ball-ch04_s06_qs01_p2\" class=\"para\">an oxidation-reduction reaction? Explain your answer.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">2. In the reaction<\/span><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 Ca(s) +\u00a0O<sub class=\"subscript\">2<\/sub>(g) $latex \\longrightarrow$\u00a02 CaO<\/span><\/span>\r\n<p id=\"ball-ch04_s06_qs01_p8\" class=\"para\">indicate what has lost electrons and what has gained electrons.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">3. In the reaction<\/span><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 Li(s) +\u00a0O<sub class=\"subscript\">2<\/sub>(g) $latex \\longrightarrow$\u00a0Li<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span>\r\n<p id=\"ball-ch04_s06_qs01_p14\" class=\"para\">indicate what has been oxidized and what has been reduced.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">4. Assign oxidation numbers to each atom in each substance.<\/span><\/p>\r\n\r\n<\/div>\r\na) \u00a0P<sub class=\"subscript\">4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sub>b) \u00a0SO<sub class=\"subscript\">2<\/sub>\r\n\r\nc) \u00a0SO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2\u2212 \u00a0 \u00a0 \u00a0 <\/sup>d) \u00a0Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>\r\n\r\n<span style=\"font-size: 1em\">5. . \u00a0Assign oxidation numbers to each atom in each substance.<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0CO \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) \u00a0CO<sub class=\"subscript\">2<\/sub>\r\n\r\nc) \u00a0NiCl<sub class=\"subscript\">2 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>d) \u00a0NiCl<sub class=\"subscript\">3<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">6. \u00a0Assign oxidation numbers to each atom in each substance.<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0CH<sub class=\"subscript\">2<\/sub>O \u00a0 \u00a0 \u00a0b) \u00a0NH<sub class=\"subscript\">3<\/sub>\r\n\r\nc) \u00a0Rb<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4 \u00a0 \u00a0<\/sub>d) \u00a0Zn(C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">7. \u00a0Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.<\/span>\r\n<div class=\"question\">\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 NO +\u00a0Cl<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 NOCl<\/span><\/span>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">8. \u00a0Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.<\/span>\r\n<div class=\"question\">\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 KrF<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O $latex \\longrightarrow$\u00a02 Kr +\u00a04 HF +\u00a0O<sub class=\"subscript\">2<\/sub><\/span><\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s06_qs01_p37\" class=\"para\">9. \u00a0Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 K +\u00a0MgCl<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 KCl +\u00a0Mg<\/span><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n10. Indicate what type, or types, of reaction each of the following represents:\r\n<p id=\"fs-idp150338768\">a) $latex \\text{Ca}(s) + \\text{Br}_2(l) \\longrightarrow \\text{CaBr}_2(s)$<\/p>\r\n<p id=\"fs-idm72172912\">b) $latex \\text{Ca(OH)}_2 (aq) + 2\\text{HBr}(aq) \\longrightarrow \\text{CaBr}_2(aq) + 2\\text{H}_2 \\text{O}(l)$<\/p>\r\n<p id=\"fs-idm52206560\">c) $latex \\text{C}_6 \\text{H}_{12}(l) + 9\\text{O}_2(g) \\longrightarrow 6\\text{CO}_2(g) + 6\\text{H}_2 \\text{O}(g)$<\/p>\r\n11. Indicate what type, or types, of reaction each of the following represents:\r\n<p id=\"fs-idm54390176\">a) $latex \\text{H}_2 \\text{O}(g) + \\text{C}(s) \\longrightarrow \\text{CO}(g) + \\text{H}_2(g)$<\/p>\r\n<p id=\"fs-idp157485296\">b) $latex 2\\text{KClO}_3(s) \\longrightarrow 2\\text{KCl}(s) + 3\\text{O}_2(g)$<\/p>\r\n<p id=\"fs-idp24875792\">c) $latex \\text{Al(OH)}_3(aq) + 3\\text{HCl}(aq) \\longrightarrow \\text{AlCl}_3(aq) + 3\\text{H}_2 \\text{O}(l)$<\/p>\r\n<p id=\"fs-idp213918688\">d) $latex \\text{Pb(NO}_3)_2(aq) + \\text{H}_2 \\text{SO}_4(sq) \\longrightarrow \\text{PbSO}_4(s) + 2\\text{HNO}_3(aq)$<\/p>\r\n12. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.\r\n\r\n13. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.\r\n<p id=\"fs-idp69126480\">a) H<sub>3<\/sub>PO<sub>4 \u00a0 \u00a0 \u00a0<\/sub>b) Al(OH)<sub>3 \u00a0 \u00a0 \u00a0<\/sub>c) SeO<sub>2<\/sub><\/p>\r\n<p id=\"fs-idm29033440\">d) KNO<sub>2 \u00a0 \u00a0 \u00a0\u00a0<\/sub>e) In<sub>2<\/sub>S<sub>3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>f) P<sub>4<\/sub>O<sub>6<\/sub><\/p>\r\n14. Classify the following as acid-base reactions or oxidation-reduction reactions:\r\n<p id=\"fs-idm48082864\">a) $latex \\text{Na}_2 \\text{S}(aq) + 2 \\text{HCl}(aq) \\longrightarrow 2 \\text{NaCl}(aq) + \\text{H}_2 \\text{S}(g)$<\/p>\r\n<p id=\"fs-idp23628560\">b) $latex 2\\text{Na}(s) + 2\\text{HCl}(aq) \\longrightarrow 2\\text{NaCl}(aq) + \\text{H}_2(g)$<\/p>\r\n<p id=\"fs-idp66120016\">c) $latex \\text{Mg}(s) + \\text{Cl}_2(g) \\longrightarrow \\text{MgCl}_2(aq) $<\/p>\r\n<p id=\"fs-idm52069888\">d) $latex \\text{MgO}(s) + 2\\text{HCl}(aq) \\longrightarrow \\text{MgCl}_2(s) + \\text{H}_2 \\text{O}(l)$<\/p>\r\n<p id=\"fs-idp171620944\">e) $latex \\text{K}_3 \\text{P}(s) + 2\\text{O}_2(g) \\longrightarrow \\text{K}_3 \\text{PO}_4(s)$<\/p>\r\n<p id=\"fs-idp44938592\">f) $latex 3\\text{KOH}(aq) + \\text{H}_3 \\text{PO}_4(aq) \\longrightarrow \\text{K}_3\\text{PO}_4(aq) + 3 \\text{H}_2 \\text{O}(l)$<\/p>\r\n15. Complete and balance the following acid-base equations:\r\n<p id=\"fs-idp63939984\">a) HCl gas reacts with solid Ca(OH)<sub>2<\/sub>(<em>s<\/em>).<\/p>\r\n<p id=\"fs-idm54386864\">b) A solution of Sr(OH)<sub>2<\/sub> is added to a solution of HNO<sub>3<\/sub>.<\/p>\r\n16. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.\r\n<p id=\"fs-idp68989776\">a) $latex \\text{Al}(s) + \\text{F}_2(g) \\longrightarrow $<\/p>\r\n<p id=\"fs-idp20677664\">b) $latex \\text{Al}(s) + \\text{CuBr}_2(aq) \\longrightarrow \\;\\text{(single displacement)}$<\/p>\r\n<p id=\"fs-idp67828208\">c) $latex \\text{P}_4(s) + \\text{O}_2(g) \\longrightarrow $<\/p>\r\n<p id=\"fs-idp56247296\">d) $latex \\text{Ca}(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\;\\text{(products are a strong base and a diatomic gas)}$<\/p>\r\n17. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?\r\n\r\n18. Great Lakes Chemical Company produces bromine, Br<sub>2<\/sub>, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl<sub>2<\/sub>.\r\n\r\n19. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO<sub>2<\/sub>. (Hint: Water is one of the products.)\r\n\r\n20. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:\r\n<p id=\"fs-idp89463616\">a) $latex \\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow $<\/p>\r\n<p id=\"fs-idp2916112\">b) $latex \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow $<\/p>\r\n21. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.\r\n<p id=\"fs-idm49310192\">a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)<\/p>\r\n<p id=\"fs-idm49309616\">b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction<\/p>\r\n<p id=\"fs-idp231133680\">c) gaseous H<sub>2<\/sub>S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)<\/p>\r\n<span style=\"font-size: 1em\">22. Which is a composition reaction and which is not?<\/span>\r\n\r\na) \u00a0NaCl +\u00a0AgNO<sub class=\"subscript\">3<\/sub>\u00a0$latex \\longrightarrow$ AgCl +\u00a0NaNO<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a0CaO +\u00a0CO<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ CaCO<sub class=\"subscript\">3<\/sub>\r\n\r\n<span style=\"font-size: 1em\">23. \u00a0Which is a composition reaction and which is not?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a02 SO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 SO<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a06 C +\u00a03 H<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">6<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">24. \u00a0Which is a decomposition reaction and which is not?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0HCl +\u00a0NaOH $latex \\longrightarrow$\u00a0NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>O\r\n\r\nb) \u00a0CaCO<sub class=\"subscript\">3<\/sub>\u00a0$latex \\longrightarrow$ CaO +\u00a0CO<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">25. \u00a0Which is a decomposition reaction and which is not?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0Na<sub class=\"subscript\">2<\/sub>O +\u00a0CO<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub>\u00a0$latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O +\u00a0SO<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">26. \u00a0Which is a combustion reaction and which is not?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">12<\/sub>O<sub class=\"subscript\">6<\/sub> +\u00a06 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 6 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O\r\n\r\nb) \u00a02 Fe<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">27. \u00a0Which is a combustion reaction and which is not?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0P<sub class=\"subscript\">4<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 P<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">5<\/sub>\r\n\r\nb) \u00a02 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">28. \u00a0Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.<\/span>\r\n\r\n<span style=\"font-size: 1em\">29. \u00a0Complete and balance each combustion equation.<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ ?\r\n\r\nb) \u00a0CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ ?\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<b>Answers<\/b>\r\n\r\n1. Yes; both K and Br are changing oxidation numbers.\r\n\r\n2. Ca has lost electrons, and O has gained electrons.\r\n\r\n3. Li has been oxidized, and O has been reduced.\r\n\r\n4. a) \u00a0P: 0\r\n\r\nb) \u00a0S: +4; O: \u22122\r\n\r\nc) \u00a0S: +2; O: \u22122\r\n\r\nd) \u00a0Ca: 2+; N: +5; O: \u22122\r\n\r\n5. a) \u00a0C: +2; O: \u22122\r\n\r\nb) \u00a0C: +4; O: \u22122\r\n\r\nc) \u00a0Ni: +2; Cl: \u22121\r\n\r\nd) \u00a0Ni: +3; Cl: \u22121\r\n\r\n6. a) \u00a0C: 0; H: +1; O: \u22122\r\n\r\nb) \u00a0N: \u22123; H: +1\r\n\r\nc) \u00a0Rb: +1; S: +6; O: \u22122\r\n\r\nd) \u00a0Zn: +2; C: 0; H: +1; O: \u22122\r\n\r\n7. N is being oxidized, and Cl is being reduced.\r\n\r\n8. O is being oxidized, and Kr is being reduced.\r\n\r\n9. K is being oxidized, and Mg is being reduced.\r\n<p id=\"fs-idm3578016\">10. a) oxidation-reduction (addition); b) acid-base (neutralization); c) oxidation-reduction (combustion)<\/p>\r\n<p id=\"fs-idp97519472\">11. a) single replacement; \u00a0 b) decomposition; \u00a0 c) acid-base; \u00a0 d) precipitation<\/p>\r\n12. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.\r\n<p id=\"fs-idm9292752\">13. a) H +1, P +5, O \u22122; \u00a0 \u00a0b) Al +3, H +1, O \u22122; \u00a0 \u00a0c) Se +4, O \u22122;<\/p>\r\nd) K +1, N +3, O \u22122; \u00a0 \u00a0e) In +3, S \u22122; \u00a0 \u00a0f) P +3, O \u22122\r\n<p id=\"fs-idm20956672\">14. a) acid-base; \u00a0 \u00a0b) oxidation-reduction: Na is oxidized, H<sup>+<\/sup> is reduced;<\/p>\r\nc) oxidation-reduction: Mg is oxidized, Cl<sub>2<\/sub> is reduced; \u00a0 \u00a0 d) acid-base;\r\n\r\ne) oxidation-reduction: P<sup>3\u2212<\/sup> is oxidized, O<sub>2<\/sub> is reduced; \u00a0 \u00a0 f) acid-base\r\n<p id=\"fs-idm141927648\">15.\u00a0a) $latex 2\\text{HCl}(g) + \\text{Ca(OH)}_2(s) \\longrightarrow \\text{CaCl}_2(s) + 2\\text{H}_2 \\text{O}(l)$;\r\nb) $latex \\text{Sr(OH)}_2(aq) + 2\\text{HNO}_3(aq) \\longrightarrow \\text{Sr(NO}_3)_2(aq) + 2\\text{H}_2 \\text{O}(l)$;<\/p>\r\n<p id=\"fs-idp102506480\">16.\u00a0a) $latex 2\\text{Al}(s) + 3\\text{F}_2 \\longrightarrow 2\\text{AlF}_3(s)$;\r\nb) $latex 2\\text{Al}(s) + 3\\text{CuBr}_2(aq) \\longrightarrow 3\\text{Cu}(s) + 2\\text{AlBr}_3(aq)$;\r\nc) $latex \\text{P}_4(s) + 5\\text{O}_2(g) \\longrightarrow \\text{P}_4 \\text{O}_{10}(s)$;\r\nd) $latex \\text{Ca}(s) + 2\\text{H}_2 \\text{O}(l) \\longrightarrow \\text{Ca(OH)}_2(aq) + \\text{H}_2(g)$;<\/p>\r\n<p id=\"fs-idp98817536\">17. $latex \\text{H}_2(g) + \\text{F}_2(g) \\longrightarrow 2\\text{HF}(g) $<\/p>\r\n<p id=\"fs-idm120462976\">18. $latex 2\\text{NaBr}(aq) + \\text{Cl}_2(g) \\longrightarrow 2\\text{NaCl}(aq) + \\text{Br}_2(l) $<\/p>\r\n<p id=\"fs-idm1428848\">19. $latex 2\\text{LiOH}(aq) + \\text{CO}_2(g) \\longrightarrow \\text{Li}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{O}(l) $<\/p>\r\n<p id=\"fs-idp19195056\">20.\u00a0a) $latex \\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{CaS}(s) + 2\\text{H}_2\\text{O}(l);$\r\nb) $latex \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{Na}_2 \\text{S}(aq) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l)$<\/p>\r\n<p id=\"fs-idp231134992\">21.\u00a0a) step 1: $latex \\text{N}_2(g) + 3\\text{H}_2(g) \\longrightarrow 2\\text{NH}_3(g)$,<\/p>\r\nstep 2: $latex \\text{NH}_3(g) + \\text{HNO}_3(aq) \\longrightarrow \\text{NH}_4 \\text{NO}_3(aq) \\longrightarrow \\text{NH}_4 \\text{NO}_3\\;\\text{(s) (after drying)}$\r\n\r\nb) $latex \\text{H}_2(g) + \\text{Br}_2(l) \\longrightarrow 2\\text{HBr}(g) $\r\nc) $latex \\text{Zn}(s) + \\text{S}(s) \\longrightarrow \\text{ZnS}(s) \\;\\text{and} \\;\\text{ZnS}(s) + 2\\text{HCl}(aq) \\longrightarrow \\text{ZnCl}_2(aq) + \\text{H}_2 \\text{S}(g) $\r\n\r\n22. a) \u00a0not composition \u00a0 \u00a0b) \u00a0composition\r\n\r\n23. a) \u00a0composition \u00a0 \u00a0<span style=\"font-size: 1em\">b) \u00a0composition<\/span>\r\n<div class=\"answer\">\r\n\r\n24. a) \u00a0not decomposition \u00a0 \u00a0<span style=\"font-size: 1em\">b) \u00a0decomposition<\/span>\r\n\r\n<span style=\"font-size: 1em\">25. a) \u00a0not decomposition \u00a0 \u00a0<\/span><span style=\"font-size: 1em\">b) \u00a0decomposition<\/span>\r\n<div class=\"answer\">\r\n<div class=\"answer\">\r\n\r\n26. a) \u00a0combustion \u00a0\u00a0<span style=\"font-size: 1em\">b) \u00a0combustion<\/span>\r\n<div class=\"answer\">\r\n\r\n27. a) \u00a0combustion \u00a0 \u00a0<span style=\"font-size: 1em\">b) \u00a0combustion<\/span>\r\n<div class=\"answer\">\r\n\r\n28. Yes; 2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O (answers will vary)\r\n\r\n29. a) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a06 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 4 CO<sub class=\"subscript\">2<\/sub> +\u00a05 H<sub class=\"subscript\">2<\/sub>O\r\n<div class=\"answer\">b) \u00a04 CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a03 O<sub class=\"subscript\">2<\/sub>\u00a0$latex \\longrightarrow$ 4 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O +\u00a02 N<sub class=\"subscript\">2<\/sub><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>\u00a0Glossary<\/h2>\r\n<strong>combustion reaction:\u00a0<\/strong>vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light\r\n\r\n<strong>half-reaction:\u00a0<\/strong>an equation that shows whether each reactant loses or gains electrons in a reaction.\r\n\r\n<strong>oxidation:\u00a0<\/strong>process in which an element\u2019s oxidation number is increased by loss of electrons\r\n\r\n<strong>oxidation-reduction reaction:\u00a0<\/strong>(also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements\r\n\r\n<strong>oxidation number:\u00a0<\/strong>(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic\r\n\r\n<strong>oxidizing agent:\u00a0<\/strong>(also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced\r\n\r\n<strong>reduction:\u00a0<\/strong>process in which an element\u2019s oxidation number is decreased by gain of electrons\r\n\r\n<strong>reducing agent:\u00a0<\/strong>(also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized\r\n\r\n<strong>single-displacement reaction:\u00a0<\/strong>(also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species\r\n\r\n<\/section>\r\n<div>\r\n<dl id=\"fs-idp64228624\" class=\"definition\"><\/dl>\r\n<dl id=\"fs-idm4237408\" class=\"definition\">\r\n \t<dd id=\"fs-idm4236768\"><\/dd>\r\n<\/dl>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"section\" id=\"ball-ch04_s06\" lang=\"en\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define <em>oxidation<\/em> and <em>reduction<\/em>.<\/li>\n<li>Assign oxidation numbers to atoms in simple compounds.<\/li>\n<li>Recognize a reaction as an oxidation-reduction reaction.<\/li>\n<li>Recognize composition, decomposition, combustion and single replacement reactions.<\/li>\n<li>Predict the products of a combustion reaction.<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-idm48520656\">\n<h2>Redox Reactions<\/h2>\n<p id=\"fs-idp3801440\">Earth\u2019s atmosphere contains about 20% molecular oxygen, O<sub>2<\/sub>, a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term <strong>oxidation<\/strong> was originally used to describe chemical reactions involving O<sub>2<\/sub>, but its meaning has evolved to refer to a broad and important reaction class known as <em>oxidation-reduction (redox) reactions<\/em>. A few examples of such reactions will be used to develop a clear picture of this classification.<\/p>\n<p id=\"fs-idm49954608\">Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride:<\/p>\n<div class=\"equation\" id=\"fs-idm5657872\" style=\"text-align: center\">[latex]2\\text{Na}(s) + \\text{Cl}_2(g) \\longrightarrow 2\\text{NaCl}(s)[\/latex]<\/div>\n<p id=\"fs-idm102441680\">It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a <strong>half-reaction<\/strong>:<\/p>\n<div class=\"equation\" id=\"fs-idp15924368\" style=\"text-align: center\">[latex]2\\text{Na}(s) \\longrightarrow 2\\text{Na}^{+}(s) + 2\\text{e}^{-}[\/latex]<br \/>\n[latex]\\text{Cl}_2(g) + 2\\text{e}^{-} \\longrightarrow 2\\text{Cl}^{-}(s)[\/latex]<\/div>\n<p id=\"fs-idp97564400\">These equations show that Na atoms <em>lose electrons<\/em> while Cl atoms (in the Cl<sub>2<\/sub> molecule) <em>gain electrons<\/em>, the \u201c<em>s<\/em>\u201d subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur:<\/p>\n<div class=\"equation\" id=\"fs-idp61672384\" style=\"text-align: center\">[latex]\\begin{array}{r @ {{}={}} l} \\pmb{\\text{oxidation}} & \\text{loss of electrons} \\\\[1em] \\pmb{\\text{reduction}} & \\text{gain of electrons} \\end{array}[\/latex]<\/div>\n<p id=\"fs-idp6686448\">In this reaction, then, sodium is <em>oxidized<\/em> and chlorine undergoes <strong>reduction<\/strong>. Viewed from a more active perspective, sodium functions as a <strong>reducing agent (reductant)<\/strong>, since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an <strong>oxidizing agent (oxidant)<\/strong>, as it effectively removes electrons from (oxidizes) sodium.<\/p>\n<div class=\"equation\" id=\"fs-idm29833328\" style=\"text-align: center\">[latex]\\begin{array}{r @ {{}={}} l} \\pmb{\\text{reducing agent}} & \\text{species that is oxidized} \\\\[1em] \\pmb{\\text{oxidizing agent}} & \\text{species that is reduced} \\end{array}[\/latex]<\/div>\n<p id=\"fs-idp108466096\">Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding NaCl:<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\text{H}_2(g) + \\text{Cl}_2(g) \\longrightarrow 2 \\text{HCl}(g)[\/latex]<\/div>\n<div class=\"equation\"><\/div>\n<div class=\"equation\" id=\"fs-idp37282464\" style=\"text-align: left\">The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called <em>oxidation number<\/em> has been defined. The <strong>oxidation number<\/strong> (or <strong>oxidation state<\/strong>) of an element in a compound is the charge its atoms would possess <em>if the compound was ionic<\/em>.<\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\">The following guidelines are used to assign oxidation numbers to each element in a molecule or ion:<\/div>\n<ol id=\"fs-idp29396208\">\n<li>The oxidation number of an atom in an elemental substance is zero.<\/li>\n<li>The oxidation number of a monatomic ion is equal to the ion\u2019s charge.<\/li>\n<li>Oxidation numbers for common non-metals are usually assigned as follows:\n<ul id=\"fs-idm48186672\">\n<li>Hydrogen: +1 when combined with nonmetals, \u22121 when combined with metals<\/li>\n<li>Oxygen: \u22122 in most compounds, sometimes \u22121 (so-called peroxides, O<sub>2<\/sub><sup>2\u2212<\/sup>), very rarely [latex]-\\frac{1}{2}[\/latex] (so-called superoxides, O<sub>2<\/sub><sup>\u2212<\/sup>), positive values when combined with F (values vary)<\/li>\n<li>Halogens: \u22121 for F always, \u22121 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values)<\/li>\n<\/ul>\n<\/li>\n<li>The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion.<\/li>\n<\/ol>\n<p id=\"fs-idm72440048\">Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties.<\/p>\n<div class=\"textbox shaded\" id=\"fs-idm24634320\">\n<h3>Example 1<\/h3>\n<p id=\"fs-idm73523056\">Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species:<\/p>\n<p id=\"fs-idp9372816\">a) H<sub>2<\/sub>S<\/p>\n<p id=\"fs-idp5671152\">b) SO<sub>3<\/sub><sup>2\u2212<\/sup><\/p>\n<p id=\"fs-idp109909120\">c) Na<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp203498912\"><strong>Solution<\/strong><br \/>\na) According to guideline 1, the oxidation number for H is +1.<\/p>\n<p id=\"fs-idm32134656\">Using this oxidation number and the compound\u2019s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\text{charge on H}_2 \\text{S} = 0 = (2 \\times +1) + (1 \\times x)[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center\">[latex]x = 0 = - (2 \\times +1) = -2[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idm58489232\" style=\"text-align: center\"><\/div>\n<p id=\"fs-idm1965888\">b) Guideline 3 suggests the oxidation number for oxygen is \u22122.<\/p>\n<p id=\"fs-idp181502096\">Using this oxidation number and the ion\u2019s formula, guideline 4 may then be used to calculate the oxidation number for sulfur:<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]{\\text{charge on SO}_3}^{2-} = -2 = (3 \\times -2) + (1 \\times x)[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center\">[latex]x = -2 - (3 \\times -2) = +4[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idm22135232\" style=\"text-align: center\"><\/div>\n<p id=\"fs-idp180932672\">c) For ionic compounds, it\u2019s convenient to assign oxidation numbers for the cation and anion separately.<\/p>\n<p id=\"fs-idp50986592\">According to guideline 2, the oxidation number for sodium is +1.<\/p>\n<p id=\"fs-idm60591184\">Assuming the usual oxidation number for oxygen (-2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4:<\/p>\n<div class=\"equation\" style=\"text-align: center\">[latex]{\\text{charge on SO}_4}^{2-} = -2 = (4 \\times -2) + (1 \\times x)[\/latex]<\/div>\n<div class=\"equation\" id=\"fs-idp6634688\" style=\"text-align: center\">[latex]x = -2 -(4 \\times -2) = +6[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp108038048\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nAssign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions:<\/p>\n<p id=\"fs-idp34225952\">a) K<u>N<\/u>O<sub>3 \u00a0 \u00a0\u00a0<\/sub>b) <u>Al<\/u>H<sub>3 \u00a0 \u00a0<\/sub>c) <span style=\"text-decoration: underline\">N<\/span>H<sub>4<\/sub><sup>+ \u00a0 \u00a0<\/sup>d) H<sub>2<\/sub><span style=\"text-decoration: underline\">P<\/span>O<sub>4<\/sub><sup>\u2212<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answers<\/strong><\/em><\/p>\n<p>a) N, +5 \u00a0 \u00a0 b) Al, +3 \u00a0 \u00a0 c) N, \u22123 \u00a0 \u00a0d) P, +5<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 2<\/h3>\n<p id=\"ball-ch04_s06_p07\" class=\"para\">Assign oxidation numbers to the atoms in each substance.<\/p>\n<p class=\"para\">a) Br<sub class=\"subscript\">2<\/sub> \u00a0 \u00a0 \u00a0b)\u00a0SiO<sub class=\"subscript\">2<\/sub> \u00a0 \u00a0 \u00a0c)\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) Br<sub class=\"subscript\">2<\/sub> is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.<\/p>\n<p class=\"simpara\">b) By rule 3, oxygen is normally assigned an oxidation number of \u22122. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.<\/p>\n<p class=\"simpara\">c) The compound barium nitrate can be separated into two parts: the Ba<sup class=\"superscript\">2+<\/sup> ion and the nitrate ion. Considering these separately, the Ba<sup class=\"superscript\">2+<\/sup> ion has an oxidation number of +2 by rule 2. Now consider the NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup> ion. Oxygen is assigned an oxidation number of \u22122, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">x<\/em> +\u00a03(\u22122) = \u22121<\/span><\/span><\/p>\n<p id=\"ball-ch04_s06_p08\" class=\"para\">where <em class=\"emphasis\">x<\/em> is the oxidation number of the nitrogen atom and \u22121 represents the charge on the species. Evaluating,<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">x<\/em> +\u00a0(\u22126) = \u22121<\/span><\/span><br \/>\n<span class=\"informalequation\"><span class=\"mathphrase\"><em class=\"emphasis\">x<\/em> = +5<\/span><\/span><\/p>\n<p id=\"ball-ch04_s06_p09\" class=\"para\">Thus, the oxidation number on the N atom in the nitrate ion is +5.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s06_p10\" class=\"para\">Assign oxidation numbers to the atoms in H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s06_p11\" class=\"para\">H = +1, O = \u22122, P = +5<\/p>\n<\/div>\n<p id=\"fs-idp45838960\">Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. <strong>Oxidation-reduction (redox) reactions<\/strong> are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist <a href=\"#fs-idp180799104\" class=\"autogenerated-content\">Example 5c)<\/a>. Definitions for the complementary processes of this reaction class are correspondingly revised as shown here:<\/p>\n<div class=\"equation\" id=\"fs-idp231200304\" style=\"text-align: center\">[latex]\\pmb{\\text{oxidation}} = \\text{increase in oxidation number}[\/latex]<\/div>\n<div class=\"equation\" style=\"text-align: center\">[latex]\\pmb{\\text{reduction}} = \\text{decrease in oxidation number}[\/latex]<\/div>\n<p id=\"fs-idm1410784\">Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl<sub>2<\/sub> to \u22121 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H<sub>2<\/sub> to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl<sub>2<\/sub> to \u22121 in HCl).<\/p>\n<h2>Classification of Redox Reactions<\/h2>\n<p>Four classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.<\/p>\n<p id=\"ball-ch04_s04_p02\" class=\"para editable block\">1 &#8211; A <strong><span class=\"margin_term\"><a class=\"glossterm\">composition reaction<\/a><\/span><\/strong>\u00a0(sometimes also called a <em class=\"emphasis\">combination reaction<\/em> or a <em class=\"emphasis\">synthesis reaction<\/em>) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sub class=\"subscript\">2<\/sub>(g) +\u00a0O<sub class=\"subscript\">2<\/sub>(g)\u00a0<\/span><\/span>[latex]\\longrightarrow[\/latex]<span class=\"informalequation block\"><span class=\"mathphrase\"> 2 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p03\" class=\"para editable block\">water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance\u2014water\u2014as a product. So this is a composition reaction.<\/p>\n<p id=\"ball-ch04_s04_p04\" class=\"para editable block\">2 &#8211; A <strong><span class=\"margin_term\"><a class=\"glossterm\">decomposition reaction<\/a><\/span><\/strong>\u00a0starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 NaHCO<sub class=\"subscript\">3<\/sub>(s) [latex]\\longrightarrow[\/latex]\u00a0Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a0CO<sub class=\"subscript\">2<\/sub>(g) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p05\" class=\"para editable block\">sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.<\/p>\n<p id=\"ball-ch04_s04_p06\" class=\"para editable block\">Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch04_s04_p07\" class=\"para\">Identify each equation as a composition reaction, a decomposition reaction, or neither.<\/p>\n<p class=\"para\">a) Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a03 SO<sub class=\"subscript\">3<\/sub>\u00a0[latex]\\longrightarrow[\/latex] Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/p>\n<p class=\"para\">b) NaCl +\u00a0AgNO<sub class=\"subscript\">3<\/sub>\u00a0[latex]\\longrightarrow[\/latex] AgCl +\u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\n<p class=\"para\">c) (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">7<\/sub>\u00a0[latex]\\longrightarrow[\/latex] Cr<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a04 H<sub class=\"subscript\">2<\/sub>O +\u00a0N<sub class=\"subscript\">2<\/sub><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) In this equation, two substances combine to make a single substance. This is a composition reaction.<\/p>\n<p class=\"simpara\">b) Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.<\/p>\n<p class=\"simpara\">c) A single substance reacts to make multiple substances. This is a decomposition reaction.<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s04_p08\" class=\"para\">Identify the equation as a composition reaction, a decomposition reaction, or neither.<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub>\u00a0[latex]\\longrightarrow[\/latex] C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s04_p09\" class=\"para\">decomposition<\/p>\n<\/div>\n<p id=\"fs-idp112552240\">3 &#8211;<strong> Combustion reactions<\/strong> in which the reductant, also called a <em>fuel,<\/em>\u00a0and oxidant, molecular oxygen, react vigorously and produce significant amounts of heat, and often light, in the form of a flame. \u00a0Combustion reactions produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N<sub class=\"subscript\">2<\/sub>. Many reactants, called <em class=\"emphasis\">fuels<\/em>, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O. For example, the balanced chemical equation for the combustion of methane, CH<sub class=\"subscript\">4<\/sub>, is as follows:<\/p>\n<div class=\"equation\" id=\"fs-idp91244624\" style=\"text-align: left\">\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">CH<sub class=\"subscript\">4<\/sub> +\u00a02 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] CO<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p11\" class=\"para editable block\">Kerosene can be approximated with the formula C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">26<\/sub>, and its combustion equation is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">12<\/sub>H<sub class=\"subscript\">26<\/sub> +\u00a037 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 24 CO<sub class=\"subscript\">2<\/sub> +\u00a026 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p12\" class=\"para editable block\">Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH, whose combustion equation is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">5<\/sub>OH +\u00a03 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 CO<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p13\" class=\"para editable block\">If nitrogen is present in the original fuel, it is converted to N<sub class=\"subscript\">2<\/sub>, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>, we have<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">2<\/sub>N<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 4 CO<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O +\u00a02 N<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"ball-ch04_s04_p14\" class=\"para\">Complete and balance each combustion equation.<\/p>\n<p class=\"para\">a) the combustion of propane, C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub><\/p>\n<p class=\"para\">b) the combustion of ammonia, NH<sub class=\"subscript\">3<\/sub><\/p>\n<figure id=\"attachment_2172\" aria-describedby=\"caption-attachment-2172\" style=\"width: 269px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-294x300.jpg\" alt=\"\" width=\"269\" height=\"274\" class=\"wp-image-2172\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-294x300.jpg 294w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-768x783.jpg 768w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1.jpg 1004w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-65x66.jpg 65w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-225x229.jpg 225w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/5345065044_0d15179564_b-1-350x357.jpg 350w\" sizes=\"auto, (max-width: 269px) 100vw, 269px\" \/><\/a><figcaption id=\"caption-attachment-2172\" class=\"wp-caption-text\"><strong>Figure 1.<\/strong> Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. \u00a0Source: \u201cflowers and propane\u201d by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) The products of the reaction are CO<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O, so our unbalanced equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] CO<sub class=\"subscript\">2<\/sub> +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p15\" class=\"para\">Balancing (and you may have to go back and forth a few times to balance this), we get<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">8<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 3 CO<sub class=\"subscript\">2<\/sub> +\u00a04 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p>b) The nitrogen atoms in ammonia will react to make N<sub class=\"subscript\">2<\/sub>, while the hydrogen atoms will react with O<sub class=\"subscript\">2<\/sub> to make H<sub class=\"subscript\">2<\/sub>O:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">NH<sub class=\"subscript\">3<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] N<sub class=\"subscript\">2<\/sub> +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch04_s04_p16\" class=\"para\">To balance this equation without fractions (which is the convention), we get<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">4 NH<sub class=\"subscript\">3<\/sub> +\u00a03 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 N<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s04_p17\" class=\"para\">Complete and balance the combustion equation for cyclopropanol, C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">6<\/sub>O.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s04_p18\" class=\"para\">C<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">6<\/sub>O +\u00a04 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 3 CO<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp4633776\" class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-4-3.png\" alt=\"\u00a0\" width=\"135\" height=\"84\" class=\"alignleft\" \/><\/p>\n<p id=\"fs-idm5712736\">Watch a brief <a href=\"http:\/\/openstaxcollege.org\/l\/16hybridrocket\">video<\/a> showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA. The first engines firing at\u00a03 s (green flame) use a liquid fuel\/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture.<\/p>\n<\/div>\n<p id=\"fs-idm580304\">4 &#8211; <strong>Single-displacement (replacement) reactions<\/strong> are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals:<\/p>\n<div class=\"equation\" id=\"fs-idp4619296\" style=\"text-align: center\">[latex]\\text{Zn}(s) + 2\\text{HCl}(aq) \\longrightarrow \\text{ZnCl}_2(aq) + \\text{H}_2(g)[\/latex]<\/div>\n<p id=\"fs-idm50858768\">Metallic elements may also be oxidized by solutions of other metal salts; for example:<\/p>\n<div class=\"equation\" id=\"fs-idm10324592\" style=\"text-align: center\">[latex]\\text{Cu}(s) + 2 \\text{AgNO}_3(aq) \\longrightarrow \\text{Cu(NO}_3)_2(aq) + 2 \\text{Ag}(s)[\/latex]<\/div>\n<p id=\"fs-idm10678768\">This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu<sup>2+<\/sup> ions dissolve in the solution to yield a characteristic blue color (<a href=\"#CNX_Chem_04_02_CuAgNO3\" class=\"autogenerated-content\">Figure 2<\/a>).<\/p>\n<figure id=\"CNX_Chem_04_02_CuAgNO3\"><figcaption>\n<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_04_CuAgNO3.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_04_CuAgNO3-3.jpg\" alt=\"This figure contains three photographs. In a, a coiled copper wire is shown beside a test tube filled with a clear, colorless liquid. In b, the wire has been inserted into the test tube with the clear, colorless liquid. In c, the test tube contains a light blue liquid and the coiled wire appears to have a fuzzy silver gray coating.\" width=\"975\" height=\"232\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong> (a) A copper wire is shown next to a solution containing silver(I) ions. (b) Displacement of dissolved silver ions by copper ions results in (c) accumulation of gray-colored silver metal on the wire and development of a blue color in the solution, due to dissolved copper ions. (credit: modification of work by Mark Ott)<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<div class=\"textbox shaded\" id=\"fs-idp180799104\">\n<h3>Example 5<\/h3>\n<p id=\"fs-idm59303872\">Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant.<\/p>\n<p id=\"fs-idm23437408\">a) [latex]\\text{ZnCO}_3(s) \\longrightarrow \\text{ZnO}(s) + \\text{CO}_2(g)[\/latex]<\/p>\n<p id=\"fs-idm32376704\">b) [latex]2\\text{Ga}(l) + 3\\text{Br}_2(l) \\longrightarrow 2\\text{GaBr}_3(s)[\/latex]<\/p>\n<p>c) [latex]2\\text{H}_2 \\text{O}_2(aq) \\longrightarrow 2\\text{H}_2 \\text{O}(l) + \\text{O}_2(g)[\/latex]<\/p>\n<p>d) [latex]\\text{BaCl}_2(aq) + \\text{K}_2 \\text{SO}_4(aq) \\longrightarrow \\text{BaSO}_4(s) + 2\\text{KCl}(aq)[\/latex]<\/p>\n<p id=\"fs-idp64660848\">e) [latex]\\text{C}_2 \\text{H}_4(g) + 3\\text{O}_2(g) \\longrightarrow 2\\text{CO}_2(g) + 2\\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Solution<\/strong><br \/>\nRedox reactions are identified per definition if one or more elements undergo a change in oxidation number.<\/p>\n<p id=\"fs-idp31047840\">a) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.<\/p>\n<p id=\"fs-idp218627312\">b) This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga(<em>l<\/em>) to +3 in GaBr<sub>3<\/sub>(<em>s<\/em>). The reducing agent is Ga(<em>l<\/em>). Bromine is reduced, its oxidation number decreasing from 0 in Br<sub>2<\/sub>(<em>l<\/em>) to \u22121 in GaBr<sub>3<\/sub>(<em>s<\/em>). The oxidizing agent is Br<sub>2<\/sub>(<em>l<\/em>).<\/p>\n<p id=\"fs-idp223712368\">c) This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called <em>disproportionation reaction)<\/em>. Oxygen is oxidized, its oxidation number increasing from \u22121 in H<sub>2<\/sub>O<sub>2<\/sub>(<em>aq<\/em>) to 0 in O<sub>2<\/sub>(<em>g<\/em>). Oxygen is also reduced, its oxidation number decreasing from \u22121 in H<sub>2<\/sub>O<sub>2<\/sub>(<em>aq<\/em>) to \u22122 in H<sub>2<\/sub>O(<em>l<\/em>). For disproportionation reactions, the same substance functions as an oxidant and a reductant.<\/p>\n<p id=\"fs-idm40215792\">d) This is not a redox reaction, since oxidation numbers remain unchanged for all elements.<\/p>\n<p id=\"fs-idm55300832\">e) This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from \u22122 in C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>) to +4 in CO<sub>2<\/sub>(<em>g<\/em>). The reducing agent (fuel) is C<sub>2<\/sub>H<sub>4<\/sub>(<em>g<\/em>). Oxygen is reduced, its oxidation number decreasing from 0 in O<sub>2<\/sub>(<em>g<\/em>) to \u22122 in H<sub>2<\/sub>O(<em>l<\/em>). The oxidizing agent is O<sub>2<\/sub>(<em>g<\/em>).<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm9371232\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nThis equation describes the production of tin(II) chloride:<\/p>\n<div class=\"equation\" id=\"fs-idm73301872\" style=\"text-align: center\">[latex]\\text{Sn}(s) + 2\\text{HCl}(g) \\longrightarrow \\text{SnCl}_2(s) + \\text{H}_2(g)[\/latex]<\/div>\n<p id=\"fs-idp98112752\">Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant.<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>Yes, a single-replacement reaction. Sn(<em>s<\/em>)\u00a0is the reductant, HCl(<em>g<\/em>) is the oxidant.<\/p>\n<\/div>\n<section id=\"fs-idp98840016\"><\/section>\n<\/section>\n<section id=\"fs-idm51820592\" class=\"summary\">\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idp62302320\">Chemical reactions are classified according to similar patterns of behavior. Redox reactions involve a change in oxidation number for one or more reactant elements.\u00a0There are four classifications of chemical reactions: composition, decomposition, combustion and single displacement.<\/p>\n<div class=\"textbox exercises\">\n<h3 itemprop=\"educationalUse\">Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch04_s06_qs01\">\n<div class=\"question\">\n<p id=\"ball-ch04_s06_qs01_p1\" class=\"para\">1. Is the reaction<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 K(s) +\u00a0Br<sub class=\"subscript\">2<\/sub>(\u2113) [latex]\\longrightarrow[\/latex]\u00a02 KBr(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s06_qs01_p2\" class=\"para\">an oxidation-reduction reaction? Explain your answer.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">2. In the reaction<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 Ca(s) +\u00a0O<sub class=\"subscript\">2<\/sub>(g) [latex]\\longrightarrow[\/latex]\u00a02 CaO<\/span><\/span><\/p>\n<p id=\"ball-ch04_s06_qs01_p8\" class=\"para\">indicate what has lost electrons and what has gained electrons.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">3. In the reaction<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 Li(s) +\u00a0O<sub class=\"subscript\">2<\/sub>(g) [latex]\\longrightarrow[\/latex]\u00a0Li<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s06_qs01_p14\" class=\"para\">indicate what has been oxidized and what has been reduced.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">4. Assign oxidation numbers to each atom in each substance.<\/span><\/p>\n<\/div>\n<p>a) \u00a0P<sub class=\"subscript\">4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/sub>b) \u00a0SO<sub class=\"subscript\">2<\/sub><\/p>\n<p>c) \u00a0SO<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2\u2212 \u00a0 \u00a0 \u00a0 <\/sup>d) \u00a0Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub><\/p>\n<p><span style=\"font-size: 1em\">5. . \u00a0Assign oxidation numbers to each atom in each substance.<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0CO \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) \u00a0CO<sub class=\"subscript\">2<\/sub><\/p>\n<p>c) \u00a0NiCl<sub class=\"subscript\">2 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>d) \u00a0NiCl<sub class=\"subscript\">3<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">6. \u00a0Assign oxidation numbers to each atom in each substance.<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0CH<sub class=\"subscript\">2<\/sub>O \u00a0 \u00a0 \u00a0b) \u00a0NH<sub class=\"subscript\">3<\/sub><\/p>\n<p>c) \u00a0Rb<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4 \u00a0 \u00a0<\/sub>d) \u00a0Zn(C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>)<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">7. \u00a0Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.<\/span><\/p>\n<div class=\"question\">\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 NO +\u00a0Cl<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 NOCl<\/span><\/span><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">8. \u00a0Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.<\/span><\/p>\n<div class=\"question\">\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 KrF<sub class=\"subscript\">2<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O [latex]\\longrightarrow[\/latex]\u00a02 Kr +\u00a04 HF +\u00a0O<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s06_qs01_p37\" class=\"para\">9. \u00a0Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 K +\u00a0MgCl<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 KCl +\u00a0Mg<\/span><\/span><\/p>\n<\/div>\n<\/div>\n<p>10. Indicate what type, or types, of reaction each of the following represents:<\/p>\n<p id=\"fs-idp150338768\">a) [latex]\\text{Ca}(s) + \\text{Br}_2(l) \\longrightarrow \\text{CaBr}_2(s)[\/latex]<\/p>\n<p id=\"fs-idm72172912\">b) [latex]\\text{Ca(OH)}_2 (aq) + 2\\text{HBr}(aq) \\longrightarrow \\text{CaBr}_2(aq) + 2\\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idm52206560\">c) [latex]\\text{C}_6 \\text{H}_{12}(l) + 9\\text{O}_2(g) \\longrightarrow 6\\text{CO}_2(g) + 6\\text{H}_2 \\text{O}(g)[\/latex]<\/p>\n<p>11. Indicate what type, or types, of reaction each of the following represents:<\/p>\n<p id=\"fs-idm54390176\">a) [latex]\\text{H}_2 \\text{O}(g) + \\text{C}(s) \\longrightarrow \\text{CO}(g) + \\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp157485296\">b) [latex]2\\text{KClO}_3(s) \\longrightarrow 2\\text{KCl}(s) + 3\\text{O}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp24875792\">c) [latex]\\text{Al(OH)}_3(aq) + 3\\text{HCl}(aq) \\longrightarrow \\text{AlCl}_3(aq) + 3\\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idp213918688\">d) [latex]\\text{Pb(NO}_3)_2(aq) + \\text{H}_2 \\text{SO}_4(sq) \\longrightarrow \\text{PbSO}_4(s) + 2\\text{HNO}_3(aq)[\/latex]<\/p>\n<p>12. Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer.<\/p>\n<p>13. Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides.<\/p>\n<p id=\"fs-idp69126480\">a) H<sub>3<\/sub>PO<sub>4 \u00a0 \u00a0 \u00a0<\/sub>b) Al(OH)<sub>3 \u00a0 \u00a0 \u00a0<\/sub>c) SeO<sub>2<\/sub><\/p>\n<p id=\"fs-idm29033440\">d) KNO<sub>2 \u00a0 \u00a0 \u00a0\u00a0<\/sub>e) In<sub>2<\/sub>S<sub>3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/sub>f) P<sub>4<\/sub>O<sub>6<\/sub><\/p>\n<p>14. Classify the following as acid-base reactions or oxidation-reduction reactions:<\/p>\n<p id=\"fs-idm48082864\">a) [latex]\\text{Na}_2 \\text{S}(aq) + 2 \\text{HCl}(aq) \\longrightarrow 2 \\text{NaCl}(aq) + \\text{H}_2 \\text{S}(g)[\/latex]<\/p>\n<p id=\"fs-idp23628560\">b) [latex]2\\text{Na}(s) + 2\\text{HCl}(aq) \\longrightarrow 2\\text{NaCl}(aq) + \\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp66120016\">c) [latex]\\text{Mg}(s) + \\text{Cl}_2(g) \\longrightarrow \\text{MgCl}_2(aq)[\/latex]<\/p>\n<p id=\"fs-idm52069888\">d) [latex]\\text{MgO}(s) + 2\\text{HCl}(aq) \\longrightarrow \\text{MgCl}_2(s) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idp171620944\">e) [latex]\\text{K}_3 \\text{P}(s) + 2\\text{O}_2(g) \\longrightarrow \\text{K}_3 \\text{PO}_4(s)[\/latex]<\/p>\n<p id=\"fs-idp44938592\">f) [latex]3\\text{KOH}(aq) + \\text{H}_3 \\text{PO}_4(aq) \\longrightarrow \\text{K}_3\\text{PO}_4(aq) + 3 \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p>15. Complete and balance the following acid-base equations:<\/p>\n<p id=\"fs-idp63939984\">a) HCl gas reacts with solid Ca(OH)<sub>2<\/sub>(<em>s<\/em>).<\/p>\n<p id=\"fs-idm54386864\">b) A solution of Sr(OH)<sub>2<\/sub> is added to a solution of HNO<sub>3<\/sub>.<\/p>\n<p>16. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms.<\/p>\n<p id=\"fs-idp68989776\">a) [latex]\\text{Al}(s) + \\text{F}_2(g) \\longrightarrow[\/latex]<\/p>\n<p id=\"fs-idp20677664\">b) [latex]\\text{Al}(s) + \\text{CuBr}_2(aq) \\longrightarrow \\;\\text{(single displacement)}[\/latex]<\/p>\n<p id=\"fs-idp67828208\">c) [latex]\\text{P}_4(s) + \\text{O}_2(g) \\longrightarrow[\/latex]<\/p>\n<p id=\"fs-idp56247296\">d) [latex]\\text{Ca}(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\;\\text{(products are a strong base and a diatomic gas)}[\/latex]<\/p>\n<p>17. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?<\/p>\n<p>18. Great Lakes Chemical Company produces bromine, Br<sub>2<\/sub>, from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl<sub>2<\/sub>.<\/p>\n<p>19. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO<sub>2<\/sub>. (Hint: Water is one of the products.)<\/p>\n<p>20. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:<\/p>\n<p id=\"fs-idp89463616\">a) [latex]\\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow[\/latex]<\/p>\n<p id=\"fs-idp2916112\">b) [latex]\\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow[\/latex]<\/p>\n<p>21. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required.<\/p>\n<p id=\"fs-idm49310192\">a) solid ammonium nitrate from gaseous molecular nitrogen via a two-step process (first reduce the nitrogen to ammonia, then neutralize the ammonia with an appropriate acid)<\/p>\n<p id=\"fs-idm49309616\">b) gaseous hydrogen bromide from liquid molecular bromine via a one-step redox reaction<\/p>\n<p id=\"fs-idp231133680\">c) gaseous H<sub>2<\/sub>S from solid Zn and S via a two-step process (first a redox reaction between the starting materials, then reaction of the product with a strong acid)<\/p>\n<p><span style=\"font-size: 1em\">22. Which is a composition reaction and which is not?<\/span><\/p>\n<p>a) \u00a0NaCl +\u00a0AgNO<sub class=\"subscript\">3<\/sub>\u00a0[latex]\\longrightarrow[\/latex] AgCl +\u00a0NaNO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0CaO +\u00a0CO<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] CaCO<sub class=\"subscript\">3<\/sub><\/p>\n<p><span style=\"font-size: 1em\">23. \u00a0Which is a composition reaction and which is not?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a02 SO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 SO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a06 C +\u00a03 H<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">6<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">24. \u00a0Which is a decomposition reaction and which is not?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0HCl +\u00a0NaOH [latex]\\longrightarrow[\/latex]\u00a0NaCl +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>b) \u00a0CaCO<sub class=\"subscript\">3<\/sub>\u00a0[latex]\\longrightarrow[\/latex] CaO +\u00a0CO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">25. \u00a0Which is a decomposition reaction and which is not?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0Na<sub class=\"subscript\">2<\/sub>O +\u00a0CO<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] Na<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub><\/p>\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">3<\/sub>\u00a0[latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O +\u00a0SO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">26. \u00a0Which is a combustion reaction and which is not?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">12<\/sub>O<sub class=\"subscript\">6<\/sub> +\u00a06 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 6 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>b) \u00a02 Fe<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">27. \u00a0Which is a combustion reaction and which is not?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0P<sub class=\"subscript\">4<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 P<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">5<\/sub><\/p>\n<p>b) \u00a02 Al<sub class=\"subscript\">2<\/sub>S<sub class=\"subscript\">3<\/sub> +\u00a09 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 SO<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">28. \u00a0Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.<\/span><\/p>\n<p><span style=\"font-size: 1em\">29. \u00a0Complete and balance each combustion equation.<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] ?<\/p>\n<p>b) \u00a0CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] ?<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><b>Answers<\/b><\/p>\n<p>1. Yes; both K and Br are changing oxidation numbers.<\/p>\n<p>2. Ca has lost electrons, and O has gained electrons.<\/p>\n<p>3. Li has been oxidized, and O has been reduced.<\/p>\n<p>4. a) \u00a0P: 0<\/p>\n<p>b) \u00a0S: +4; O: \u22122<\/p>\n<p>c) \u00a0S: +2; O: \u22122<\/p>\n<p>d) \u00a0Ca: 2+; N: +5; O: \u22122<\/p>\n<p>5. a) \u00a0C: +2; O: \u22122<\/p>\n<p>b) \u00a0C: +4; O: \u22122<\/p>\n<p>c) \u00a0Ni: +2; Cl: \u22121<\/p>\n<p>d) \u00a0Ni: +3; Cl: \u22121<\/p>\n<p>6. a) \u00a0C: 0; H: +1; O: \u22122<\/p>\n<p>b) \u00a0N: \u22123; H: +1<\/p>\n<p>c) \u00a0Rb: +1; S: +6; O: \u22122<\/p>\n<p>d) \u00a0Zn: +2; C: 0; H: +1; O: \u22122<\/p>\n<p>7. N is being oxidized, and Cl is being reduced.<\/p>\n<p>8. O is being oxidized, and Kr is being reduced.<\/p>\n<p>9. K is being oxidized, and Mg is being reduced.<\/p>\n<p id=\"fs-idm3578016\">10. a) oxidation-reduction (addition); b) acid-base (neutralization); c) oxidation-reduction (combustion)<\/p>\n<p id=\"fs-idp97519472\">11. a) single replacement; \u00a0 b) decomposition; \u00a0 c) acid-base; \u00a0 d) precipitation<\/p>\n<p>12. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction.<\/p>\n<p id=\"fs-idm9292752\">13. a) H +1, P +5, O \u22122; \u00a0 \u00a0b) Al +3, H +1, O \u22122; \u00a0 \u00a0c) Se +4, O \u22122;<\/p>\n<p>d) K +1, N +3, O \u22122; \u00a0 \u00a0e) In +3, S \u22122; \u00a0 \u00a0f) P +3, O \u22122<\/p>\n<p id=\"fs-idm20956672\">14. a) acid-base; \u00a0 \u00a0b) oxidation-reduction: Na is oxidized, H<sup>+<\/sup> is reduced;<\/p>\n<p>c) oxidation-reduction: Mg is oxidized, Cl<sub>2<\/sub> is reduced; \u00a0 \u00a0 d) acid-base;<\/p>\n<p>e) oxidation-reduction: P<sup>3\u2212<\/sup> is oxidized, O<sub>2<\/sub> is reduced; \u00a0 \u00a0 f) acid-base<\/p>\n<p id=\"fs-idm141927648\">15.\u00a0a) [latex]2\\text{HCl}(g) + \\text{Ca(OH)}_2(s) \\longrightarrow \\text{CaCl}_2(s) + 2\\text{H}_2 \\text{O}(l)[\/latex];<br \/>\nb) [latex]\\text{Sr(OH)}_2(aq) + 2\\text{HNO}_3(aq) \\longrightarrow \\text{Sr(NO}_3)_2(aq) + 2\\text{H}_2 \\text{O}(l)[\/latex];<\/p>\n<p id=\"fs-idp102506480\">16.\u00a0a) [latex]2\\text{Al}(s) + 3\\text{F}_2 \\longrightarrow 2\\text{AlF}_3(s)[\/latex];<br \/>\nb) [latex]2\\text{Al}(s) + 3\\text{CuBr}_2(aq) \\longrightarrow 3\\text{Cu}(s) + 2\\text{AlBr}_3(aq)[\/latex];<br \/>\nc) [latex]\\text{P}_4(s) + 5\\text{O}_2(g) \\longrightarrow \\text{P}_4 \\text{O}_{10}(s)[\/latex];<br \/>\nd) [latex]\\text{Ca}(s) + 2\\text{H}_2 \\text{O}(l) \\longrightarrow \\text{Ca(OH)}_2(aq) + \\text{H}_2(g)[\/latex];<\/p>\n<p id=\"fs-idp98817536\">17. [latex]\\text{H}_2(g) + \\text{F}_2(g) \\longrightarrow 2\\text{HF}(g)[\/latex]<\/p>\n<p id=\"fs-idm120462976\">18. [latex]2\\text{NaBr}(aq) + \\text{Cl}_2(g) \\longrightarrow 2\\text{NaCl}(aq) + \\text{Br}_2(l)[\/latex]<\/p>\n<p id=\"fs-idm1428848\">19. [latex]2\\text{LiOH}(aq) + \\text{CO}_2(g) \\longrightarrow \\text{Li}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idp19195056\">20.\u00a0a) [latex]\\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{CaS}(s) + 2\\text{H}_2\\text{O}(l);[\/latex]<br \/>\nb) [latex]\\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{Na}_2 \\text{S}(aq) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idp231134992\">21.\u00a0a) step 1: [latex]\\text{N}_2(g) + 3\\text{H}_2(g) \\longrightarrow 2\\text{NH}_3(g)[\/latex],<\/p>\n<p>step 2: [latex]\\text{NH}_3(g) + \\text{HNO}_3(aq) \\longrightarrow \\text{NH}_4 \\text{NO}_3(aq) \\longrightarrow \\text{NH}_4 \\text{NO}_3\\;\\text{(s) (after drying)}[\/latex]<\/p>\n<p>b) [latex]\\text{H}_2(g) + \\text{Br}_2(l) \\longrightarrow 2\\text{HBr}(g)[\/latex]<br \/>\nc) [latex]\\text{Zn}(s) + \\text{S}(s) \\longrightarrow \\text{ZnS}(s) \\;\\text{and} \\;\\text{ZnS}(s) + 2\\text{HCl}(aq) \\longrightarrow \\text{ZnCl}_2(aq) + \\text{H}_2 \\text{S}(g)[\/latex]<\/p>\n<p>22. a) \u00a0not composition \u00a0 \u00a0b) \u00a0composition<\/p>\n<p>23. a) \u00a0composition \u00a0 \u00a0<span style=\"font-size: 1em\">b) \u00a0composition<\/span><\/p>\n<div class=\"answer\">\n<p>24. a) \u00a0not decomposition \u00a0 \u00a0<span style=\"font-size: 1em\">b) \u00a0decomposition<\/span><\/p>\n<p><span style=\"font-size: 1em\">25. a) \u00a0not decomposition \u00a0 \u00a0<\/span><span style=\"font-size: 1em\">b) \u00a0decomposition<\/span><\/p>\n<div class=\"answer\">\n<div class=\"answer\">\n<p>26. a) \u00a0combustion \u00a0\u00a0<span style=\"font-size: 1em\">b) \u00a0combustion<\/span><\/p>\n<div class=\"answer\">\n<p>27. a) \u00a0combustion \u00a0 \u00a0<span style=\"font-size: 1em\">b) \u00a0combustion<\/span><\/p>\n<div class=\"answer\">\n<p>28. Yes; 2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O (answers will vary)<\/p>\n<p>29. a) \u00a0C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">9<\/sub>OH +\u00a06 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 4 CO<sub class=\"subscript\">2<\/sub> +\u00a05 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<div class=\"answer\">b) \u00a04 CH<sub class=\"subscript\">3<\/sub>NO<sub class=\"subscript\">2<\/sub> +\u00a03 O<sub class=\"subscript\">2<\/sub>\u00a0[latex]\\longrightarrow[\/latex] 4 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O +\u00a02 N<sub class=\"subscript\">2<\/sub><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Glossary<\/h2>\n<p><strong>combustion reaction:\u00a0<\/strong>vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light<\/p>\n<p><strong>half-reaction:\u00a0<\/strong>an equation that shows whether each reactant loses or gains electrons in a reaction.<\/p>\n<p><strong>oxidation:\u00a0<\/strong>process in which an element\u2019s oxidation number is increased by loss of electrons<\/p>\n<p><strong>oxidation-reduction reaction:\u00a0<\/strong>(also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements<\/p>\n<p><strong>oxidation number:\u00a0<\/strong>(also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic<\/p>\n<p><strong>oxidizing agent:\u00a0<\/strong>(also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced<\/p>\n<p><strong>reduction:\u00a0<\/strong>process in which an element\u2019s oxidation number is decreased by gain of electrons<\/p>\n<p><strong>reducing agent:\u00a0<\/strong>(also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized<\/p>\n<p><strong>single-displacement reaction:\u00a0<\/strong>(also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species<\/p>\n<\/section>\n<div>\n<dl id=\"fs-idp64228624\" class=\"definition\"><\/dl>\n<dl id=\"fs-idm4237408\" class=\"definition\">\n<dd id=\"fs-idm4236768\"><\/dd>\n<\/dl>\n<\/div>\n<\/div>\n","protected":false},"author":330,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"6.4 Oxidation-Reduction Reactions","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[54],"class_list":["post-2177","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":2162,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2177","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/users\/330"}],"version-history":[{"count":18,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2177\/revisions"}],"predecessor-version":[{"id":4892,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2177\/revisions\/4892"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/2162"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2177\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=2177"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=2177"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=2177"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=2177"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}