{"id":3631,"date":"2015-05-06T03:51:00","date_gmt":"2015-05-06T03:51:00","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&p=3631"},"modified":"2020-12-30T23:56:10","modified_gmt":"2020-12-30T23:56:10","slug":"balancing-oxidation-reduction-reactions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/chemistryformajors\/chapter\/balancing-oxidation-reduction-reactions\/","title":{"raw":"Review of Redox Chemistry","rendered":"Review of Redox Chemistry"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Describe defining traits of redox chemistry<\/li>\r\n \t
  • Identify the oxidant and reductant of a redox reaction<\/li>\r\n \t
  • Balance chemical equations for redox reactions using the half-reaction method<\/li>\r\n<\/ul>\r\n<\/div>\r\n
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    Since reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier section of text (reaction stoichiometry).<\/p>\r\n\r\n

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    Oxidation Numbers<\/h3>\r\n

    By definition, a redox reaction is one that entails changes in\u00a0oxidation number<\/em>\u00a0(or\u00a0oxidation state<\/em>) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description,\u00a0the oxidation number of an atom in an element is equal to zero<\/em>. For an atom in a compound,\u00a0the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic<\/em>. Consequential to these rules,\u00a0the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule<\/em>. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered.<\/p>\r\n

    Simple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements\u2019 oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl, is comprised of Na+<\/sup>\u00a0cations and Cl\u2212<\/sup>\u00a0anions, and so oxidation numbers for sodium and chlorine are, +1 and \u22121, respectively. Calcium fluoride, CaF2<\/sub>, is comprised of Ca2+<\/sup>\u00a0cations and F\u2212<\/sup>\u00a0anions, and so oxidation numbers for calcium and fluorine are, +2 and \u22121, respectively.<\/p>\r\n

    Covalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent O\u2212H bonds. The shared electrons comprising an O\u2212H bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral.<\/p>\r\n

    Imagine that the polarization of shared electrons within the O\u2212H bonds of water were 100% complete\u2014the result would be\u00a0transfer<\/em>\u00a0of electrons from H to O, and water would be an ionic compound comprised of O2\u2212<\/sup>\u00a0anions and H+<\/sup>\u00a0cations. And so, the oxidations numbers for oxygen and hydrogen in water are \u22122 and +1, respectively. Applying this same logic to carbon tetrachloride, CCl4<\/sub>, yields oxidation numbers of +4 for carbon and \u22121 for chlorine. In the nitrate ion, [latex]\\text{NO}_3^-[\/latex], the oxidation number for nitrogen is +5 and that for oxygen is \u22122, summing to equal the 1\u2212 charge on the molecule:<\/p>\r\n\r\n

    [latex]\\text{(1 N atom)} \\left(\\dfrac{+5}{\\text{N atom}}\\right)+\\text{(3 O atoms)}\\left(\\dfrac{-2}{\\text{O atom}}\\right)=+5+-6=-1[\/latex]<\/center><\/section>
    \r\n

    Balancing Redox Equations<\/h3>\r\n

    The unbalanced equation below describes the decomposition of molten sodium chloride:<\/p>\r\n\r\n

    [latex]\\text{NaCl}(l) \\longrightarrow \\text{Na}(l) +\\text{Cl}_2(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{unbalanced}[\/latex]<\/center>\r\n

    This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes\u00a0reduction<\/em>) and that for Cl is increased from \u22121 to 0 (it undergoes\u00a0oxidation<\/em>). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na:<\/p>\r\n\r\n

    [latex]2\\text{NaCl}(l) \\longrightarrow 2\\text{Na}(l) +\\text{Cl}_2(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{balanced}[\/latex]<\/center>\r\n

    Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, H+<\/sup>(aq<\/em>) and OH\u2212<\/sup>(aq<\/em>), as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the\u00a0half-reaction method<\/em>\u00a0is helpful. This approach involves the following steps:<\/p>\r\n\r\n

      \r\n \t
    1. Write skeletal equations for the oxidation and reduction half-reactions.<\/li>\r\n \t
    2. Balance each half-reaction for all elements except H and O.<\/li>\r\n \t
    3. Balance each half-reaction for O by adding H2<\/sub>O.<\/li>\r\n \t
    4. Balance each half-reaction for H by adding H+<\/sup>.<\/li>\r\n \t
    5. Balance each half-reaction for charge by adding electrons.<\/li>\r\n \t
    6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.<\/li>\r\n \t
    7. Add the two half-reactions and simplify.<\/li>\r\n \t
    8. If the reaction takes place in a basic medium, add OH\u2212<\/sup>\u00a0ions the equation obtained in step 7 to neutralize the H+<\/sup>\u00a0ions (add in equal numbers to both sides of the equation) and simplify.<\/li>\r\n<\/ol>\r\n

      The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.<\/p>\r\n\r\n

      \r\n
      \r\n

      EXAMPLE\u00a0<\/span>1:Balancing Equations for Redox Reactions in Acidic Solutions<\/span><\/h3>\r\n<\/header>
      \r\n

      Question 1<\/h4>\r\nWrite the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.\r\n

      Solution<\/strong><\/h4>\r\n[reveal-answer q=\"576838\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"576838\"]\r\nFollowing the steps of the half-reaction method:\r\n
        \r\n \t
      1. Write skeletal equations for the oxidation and reduction half-reactions.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) [\/latex]<\/li>\r\n \t
      2. Balance each half-reaction for all elements except H and O.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) [\/latex]<\/li>\r\n \t
      3. Balance each half-reaction for O by adding H2<\/sub>O.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) + \\boldsymbol{2\\text{H}_2 \\text{O}} (l) [\/latex]<\/li>\r\n \t
      4. Balance each half-reaction for H by adding H<\/em>+<\/sup>.\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3\\text{H}^+} (aq) + \\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) + 2\\text{H}_2 \\text{O} (l) [\/latex]<\/li>\r\n \t
      5. Balance each half-reaction for charge by adding electrons.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq) + \\boldsymbol{2\\text{e}^-} [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3\\text{e}^-} + 3\\text{H}^+ (aq) + \\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) + 2\\text{H}_2 \\text{O} (l) [\/latex]<\/li>\r\n \t
      6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.<\/em>\r\n[latex]\\text{oxidation(x} 3): [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3}\\text{Cu} (s) \\rightarrow \\boldsymbol{3}\\text{Cu}^{2+} (aq) + \\boldsymbol{6}\\cancel{2}\\text{e}^- [\/latex]\r\n[latex]\\text{reduction(x} 2): [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{6}\\cancel{3}\\text{e}^- + \\boldsymbol{6}\\cancel{3}\\text{H}^+ (aq) + \\boldsymbol{2}\\text{HNO}_3 (aq) \\rightarrow \\boldsymbol{2}\\text{NO} (g) + \\boldsymbol{4}\\cancel{2}\\text{H}_2 \\text{O} (l) [\/latex]<\/li>\r\n \t
      7. Add the two half-reactions and simplify.<\/em>\r\n[latex]\\boldsymbol{3}\\text{Cu} (s) + \\boldsymbol{\\cancel{6}}\\text{e}^- + \\boldsymbol{6}\\text{H}^+ (aq) + \\boldsymbol{2}\\text{HNO}_3 (aq) \\rightarrow \\boldsymbol{3}\\text{Cu}^{2+} (aq) + \\boldsymbol{\\cancel{6}}\\text{e}^- + \\boldsymbol{2}\\text{NO} (g) + \\boldsymbol{4}\\text{H}_2 \\text{O} (l) [\/latex]\r\n[latex]\\boldsymbol{3}\\text{Cu} (s) + \\boldsymbol{6}\\text{H}^+ (aq) + \\boldsymbol{2}\\text{HNO}_3 (aq) \\rightarrow \\boldsymbol{3}\\text{Cu}^{2+} (aq) + \\boldsymbol{2}\\text{NO} (g) + \\boldsymbol{4}\\text{H}_2 \\text{O} (l) [\/latex]<\/li>\r\n \t
      8. If the reaction takes place in a basic medium, add OH\u2212<\/sup>\u00a0ions the equation obtained in step 7 to neutralize the H+<\/sup>\u00a0ions (add in equal numbers to both sides of the equation) and simplify.<\/em>\r\nThis step is not necessary since the solution is stipulated to be acidic.<\/li>\r\n<\/ol>\r\n

        The balanced equation for the reaction in an acidic solution is then<\/p>\r\n\r\n

        [latex]3\\text{Cu} (s) + 6\\text{H}^+ (aq) + 2\\text{HNO}_3 (aq) \\rightarrow 3\\text{Cu}^{2+} (aq) + 2\\text{NO} (g) + 4\\text{H}_2 \\text{O} (l) [\/latex]<\/center>[\/hidden-answer]\r\n

        Check Your Learning<\/strong><\/h4>\r\nThe reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.\r\n\r\nAnswer:\r\n\r\n
        [reveal-answer q=\"576839\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"576839\"]
        [latex]\\text{Cu} (s) + 2\\text{H}^+ (aq) + 2\\text{HNO}_3 (aq) \\rightarrow \\text{Cu}^{2+} (aq) + 2\\text{NO}_2 (g) + 2\\text{H}_2 \\text{O} (l) [\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/section><\/section><\/div>\r\n<\/header><\/div>\r\n
        \r\n
        \r\n

        EXAMPLE\u00a0<\/span>17.2:\u00a0<\/span>Balancing Equations for Redox Reactions in Basic Solutions<\/h3>\r\n<\/header>
        \r\n

        Question 1<\/h4>\r\nWrite the balanced equation representing reaction between aqueous permanganate ion,\u00a0[latex]\\text{MnO}_4^-[\/latex], and solid chromium(III) hydroxide, Cr(OH)3<\/sub>, to yield solid manganese(IV) oxide, MnO2<\/sub>, and aqueous chromate ion,\u00a0[latex]\\text{CrO}_4^{2-}[\/latex].\u00a0The reaction takes place in a basic solution.\r\n

        Solution<\/strong><\/h4>\r\n[reveal-answer q=\"576840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"576840\"]\r\nFollowing the steps of the half-reaction method:\r\n
          \r\n \t
        1. Write skeletal equations for the oxidation and reduction half-reactions.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) [\/latex]<\/li>\r\n \t
        2. Balance each half-reaction for all elements except H and O.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) [\/latex]<\/li>\r\n \t
        3. Balance each half-reaction for O by adding H2<\/sub>O.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{\\text{H}_2 \\text{O} (l)} + \\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) + \\boldsymbol{2 \\text{H}_2 \\text{O} (l)}[\/latex]<\/li>\r\n \t
        4. Balance each half-reaction for H by adding H+<\/sup>.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{H}_2\\text{O}\\ (l) + \\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) +\\boldsymbol{5 \\text{H}^+ (aq)} [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{4 \\text{H}^+ (aq)} +\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\r\n \t
        5. Balance each half-reaction for charge by adding electrons.<\/em>\r\n[latex]\\text{oxidation:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{H}_2\\text{O}\\ (l) + \\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) +5 \\text{H}^+ (aq) + \\boldsymbol{3e^-} [\/latex]\r\n[latex]\\text{reduction:} [\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3e^-} +4 \\text{H}^+ (aq) +\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\r\n \t
        6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.<\/em>\r\nThis step is not necessary since the number of electrons is already in balance.<\/li>\r\n \t
        7. Add the two half-reactions and simplify.<\/em>\r\n[latex] \\cancel{\\text{H}_2\\text{O} (l)} + \\text{Cr(OH)}_3 (s) +\\cancel{3e^-} + \\cancel{4\\text{H}^+ (aq)} + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\cancel{5} \\text{H}^+ (aq) + \\cancel{3e^-} + \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]\r\n[latex] \\text{Cr(OH)}_3 (s) + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\text{H}^+ (aq) + \\text{MnO}_2 (s) + \\text{H}_2 \\text{O} (l)[\/latex]<\/li>\r\n \t
        8. If the reaction takes place in a basic medium, add OH\u2212<\/sup>\u00a0ions the equation obtained in step 7 to neutralize the H+<\/sup>\u00a0ions (add in equal numbers to both sides of the equation) and simplify.<\/em>\r\n[latex] \\boldsymbol{\\text{OH}^- (aq)} + \\text{Cr(OH)}_3 (s) + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\text{H}^+ (aq) + \\boldsymbol{\\text{OH}^- (aq)} + \\text{MnO}_2 (s) + \\text{H}_2 \\text{O} (l)[\/latex]\r\n[latex] \\text{OH}^- (aq) + \\text{Cr(OH)}_3 (s) + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n

          Check Your Learning<\/strong><\/h4>\r\nAqueous permanganate ion may also be reduced using aqueous bromide ion, Br\u2212<\/sup>, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, BrO3<\/sub>\u2212<\/sup>. Write the balanced equation for this reaction occurring in a basic medium.\r\n\r\n
          \r\n

          Answer:<\/p>\r\n[reveal-answer q=\"576841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"576841\"]\r\n[latex] \\text{H}_2\\text{O} (l) + 2\\text{MnO}_4^- (aq) + \\text{Br}^- (aq) \\rightarrow 2\\text{MnO}_2 (s) + \\text{BrO}_3^- (aq) + 2\\text{OH}^- (aq)[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/section><\/section><\/div>\r\n<\/header><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

          \r\n

          Key Concepts and Summary<\/h3>\r\nRedox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, H+<\/sup>\u00a0and OH\u2212<\/sup>, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions.\r\n\r\n<\/div>\r\n
          \r\n

          Try It<\/h3>\r\n
            \r\n \t
          1. If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?<\/li>\r\n \t
          2. For the scenario in the previous question, how many electrons moved through the circuit?<\/li>\r\n \t
          3. For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.\r\n
              \r\n \t
            1. [latex]{\\text{Fe}}^{3+}+{\\text{3e}}^{-}\\longrightarrow \\text{Fe}[\/latex]<\/li>\r\n \t
            2. [latex]\\text{Cr}\\longrightarrow {\\text{Cr}}^{3+}+{\\text{3e}}^{-}[\/latex]<\/li>\r\n \t
            3. [latex]{\\text{MnO}}_{4}{}^{2-}\\longrightarrow {\\text{MnO}}_{4}{}^{-}+{\\text{e}}^{-}[\/latex]<\/li>\r\n \t
            4. [latex]{\\text{Li}}^{\\text{+}}+{\\text{e}}^{-}\\longrightarrow \\text{Li}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            5. For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.\r\n
                \r\n \t
              1. [latex]{\\text{Cl}}^{-}\\longrightarrow {\\text{Cl}}_{2}[\/latex]<\/li>\r\n \t
              2. [latex]{\\text{Mn}}^{2+}\\longrightarrow {\\text{MnO}}_{2}[\/latex]<\/li>\r\n \t
              3. [latex]{\\text{H}}_{2}\\longrightarrow {\\text{H}}^{\\text{+}}[\/latex]<\/li>\r\n \t
              4. [latex]{\\text{NO}}_{3}{}^{-}\\longrightarrow \\text{NO}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              5. Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half-reactions in an acidic solution.\r\n
                  \r\n \t
                1. [latex]\\text{Ca}\\longrightarrow {\\text{Ca}}^{2+}+{\\text{2e}}^{-}[\/latex], [latex]{\\text{F}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2F}}^{-}[\/latex]<\/li>\r\n \t
                2. [latex]\\text{Li}\\longrightarrow {\\text{Li}}^{\\text{+}}+{\\text{e}}^{-}[\/latex], [latex]{\\text{Cl}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Cl}}^{-}[\/latex]<\/li>\r\n \t
                3. [latex]\\text{Fe}\\longrightarrow {\\text{Fe}}^{3+}+{\\text{3e}}^{-}[\/latex], [latex]{\\text{Br}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Br}}^{-}[\/latex]<\/li>\r\n \t
                4. [latex]\\text{Ag}\\longrightarrow {\\text{Ag}}^{\\text{+}}+{\\text{e}}^{-}[\/latex], [latex]{\\text{MnO}}_{4}{}^{-}+{\\text{4H}}^{\\text{+}}+{\\text{3e}}^{-}\\longrightarrow {\\text{MnO}}_{2}+{\\text{2H}}_{2}\\text{O}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"273830\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"273830\"]\r\n\r\n1.\u00a0Because A = C\/s, time must be converted to seconds. Round to two significant digits based on the initial values in the problem. [latex]\\text{35 min}\\times \\frac{\\text{60 s}}{\\text{1 min}}\\times \\frac{\\text{2.5 C}}{\\text{1 s}}=\\text{5250 C}=5.3\\times {10}^{3}\\text{C}[\/latex]\r\n\r\n3.\u00a0(a) reduction (b) oxidation (c) oxidation (d) reduction\r\n\r\n5.\u00a0The\u00a0balanced reactions are as follows:\r\n
                    \r\n \t
                  1. [latex]\\begin{array}{rl}{} \\text{oxidation:}&\\text{Ca}\\longrightarrow {\\text{Ca}}^{2+}+{\\text{2e}}^{-}\\\\ \\text{reduction:}&{\\text{F}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2F}}^{-}\\\\ \\\\ \\text{overall:}&{\\text{F}}_{2}+\\text{Ca}\\longrightarrow {\\text{2F}}^{-}+{\\text{Ca}}^{2+}\\\\ \\\\ \\text{acidic solution:}&{\\text{F}}_{2}+\\text{Ca}\\longrightarrow {\\text{2F}}^{-}+{\\text{Ca}}^{2+}\\end{array}[\/latex]<\/li>\r\n \t
                  2. [latex]\\begin{array}{rl}\\\\ \\text{oxidation:}&\\text{2}\\times \\left(\\text{Li}\\longrightarrow {\\text{Li}}^{\\text{+}}+{\\text{e}}^{-}\\right)\\\\ \\text{reduction:}&{\\text{Cl}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Cl}}^{-}\\\\ \\\\ \\text{overall:}&{\\text{Cl}}_{2}+\\text{2Li}\\longrightarrow {\\text{2Li}}^{\\text{+}}+{\\text{2Cl}}^{-}\\\\ \\\\ \\text{acidic solution:}&{\\text{Cl}}_{2}+\\text{2Li}\\longrightarrow {\\text{2Li}}^{\\text{+}}+{\\text{2Cl}}^{-}\\end{array}[\/latex]<\/li>\r\n \t
                  3. [latex]\\begin{array}{rl}\\\\ \\text{oxidation:}&\\text{2}\\times \\left(\\text{Fe}\\longrightarrow {\\text{Fe}}^{3+}+{\\text{3e}}^{-}\\right)\\\\ \\text{reduction:}&\\text{3}\\times \\left({\\text{Br}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Br}}^{-}\\right)\\\\ \\\\ \\text{overall:}&\\text{3Br}+\\text{2Fe}\\longrightarrow {\\text{2Fe}}^{3+}+{\\text{6Br}}^{-}\\\\ \\\\ \\text{acidic solution:}&{\\text{3Br}}_{2}+\\text{2Fe}\\longrightarrow {\\text{2Fe}}^{3+}+{\\text{6Br}}^{-}\\end{array}[\/latex]<\/li>\r\n \t
                  4. [latex]\\begin{array}{rl}\\\\ \\text{oxidation:}&\\text{3}\\times \\left(\\text{Ag}\\longrightarrow {\\text{Ag}}^{\\text{+}}+{\\text{e}}^{-}\\right)\\\\ \\text{reduction:}&{\\text{MnO}}_{4}{}^{-}+{\\text{4H}}^{\\text{+}}+3{\\text{e}}^{-}\\longrightarrow {\\text{MnO}}_{2}{}^{-}+{\\text{2H}}_{2}\\text{O}\\\\ \\\\ \\text{overall:}&{\\text{MnO}}_{4}+{\\text{4H}}^{\\text{+}}+\\text{3Ag}\\longrightarrow {\\text{3Ag}}^{\\text{+}}+{\\text{MnO}}_{2}+{\\text{2H}}_{2}\\text{O}\\\\ \\\\ \\text{acidic solution:}&{\\text{MnO}}_{4}+{\\text{4H}}^{\\text{+}}+\\text{3Ag}\\longrightarrow {\\text{3Ag}}^{\\text{+}}+{\\text{MnO}}_{2}+{\\text{2H}}_{2}\\text{O}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
                    \r\n

                    Try It<\/h3>\r\n
                      \r\n \t
                    1. Balance the following in acidic solution:\r\n
                        \r\n \t
                      1. [latex]{\\text{H}}_{2}{\\text{O}}_{2}+{\\text{Sn}}^{2+}\\longrightarrow {\\text{H}}_{2}\\text{O}+{\\text{Sn}}^{4+}[\/latex]<\/li>\r\n \t
                      2. [latex]{\\text{PbO}}_{2}+\\text{Hg}\\longrightarrow {\\text{Hg}}_{2}{}^{2+}+{\\text{Pb}}^{2+}[\/latex]<\/li>\r\n \t
                      3. [latex]\\text{Al}+{\\text{Cr}}_{2}{\\text{O}}_{7}{}^{2-}\\longrightarrow {\\text{Al}}^{3+}+{\\text{Cr}}^{3+}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                      4. Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.<\/li>\r\n \t
                      5. Balance the following in basic solution:\r\n
                          \r\n \t
                        1. [latex]{\\text{SO}}_{3}{}^{2-}\\left(aq\\right)+{\\text{Cu(OH)}}_{2}\\left(s\\right)\\longrightarrow {\\text{SO}}_{4}{}^{2-}\\left(aq\\right)+\\text{Cu(OH)}\\left(s\\right)[\/latex]<\/li>\r\n \t
                        2. [latex]{\\text{O}}_{2}\\left(g\\right)+{\\text{Mn(OH)}}_{2}\\left(s\\right)\\longrightarrow {\\text{MnO}}_{2}\\left(s\\right)[\/latex]<\/li>\r\n \t
                        3. [latex]{\\text{NO}}_{3}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\left(g\\right)\\longrightarrow \\text{NO}\\left(g\\right)[\/latex]<\/li>\r\n \t
                        4. [latex]\\text{Al}\\left(s\\right)+{\\text{CrO}}_{4}{}^{2-}\\left(aq\\right)\\longrightarrow {\\text{Al(OH)}}_{3}\\left(s\\right)+{\\text{Cr(OH)}}_{4}{}^{-}\\left(aq\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                        5. Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.<\/li>\r\n \t
                        6. Why is it not possible for hydroxide ion (OH\u2212<\/sup>) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?<\/li>\r\n \t
                        7. Why is it not possible for hydrogen ion (H+<\/sup>) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?<\/li>\r\n \t
                        8. Why must the charge balance in oxidation-reduction reactions?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"135772\"]Show Selected Solutions[\/reveal-answer]\r\n[hidden-answer a=\"135772\"]\r\n\r\n2.\u00a0Oxidized:\r\n
                            \r\n \t
                          1. Ag<\/li>\r\n \t
                          2. Sn2+<\/sup><\/li>\r\n \t
                          3. Hg<\/li>\r\n \t
                          4. Al<\/li>\r\n<\/ol>\r\nReduced:\r\n
                              \r\n \t
                            1. [latex]{\\text{Hg}}_{2}{}^{2+}[\/latex]<\/li>\r\n \t
                            2. H2<\/sub>O2<\/sub><\/li>\r\n \t
                            3. PbO2<\/sub><\/li>\r\n \t
                            4. [latex]{\\text{Cr}}_{2}{\\text{O}}_{7}{}^{2-}[\/latex]<\/li>\r\n<\/ol>\r\nOxidizing agent:\r\n
                                \r\n \t
                              1. [latex]{\\text{Hg}}_{2}{}^{2+}[\/latex]<\/li>\r\n \t
                              2. H2<\/sub>O2<\/sub><\/li>\r\n \t
                              3. PbO2<\/sub><\/li>\r\n \t
                              4. [latex]{\\text{Cr}}_{2}{\\text{O}}_{7}{}^{2-}[\/latex]<\/li>\r\n<\/ol>\r\nReducing agent:\r\n
                                  \r\n \t
                                1. Ag<\/li>\r\n \t
                                2. Sn2+<\/sup><\/li>\r\n \t
                                3. Hg<\/li>\r\n \t
                                4. Al<\/li>\r\n<\/ol>\r\n4.\u00a0Oxidized = reducing agent:\r\n
                                    \r\n \t
                                  1. [latex]{\\text{SO}}_{3}{}^{2-}[\/latex]<\/li>\r\n \t
                                  2. Mn(OH)2<\/sub><\/li>\r\n \t
                                  3. H2<\/sub><\/li>\r\n \t
                                  4. Al<\/li>\r\n<\/ol>\r\nreduced = oxidizing agent:\r\n
                                      \r\n \t
                                    1. Cu(OH)2<\/sub><\/li>\r\n \t
                                    2. O2<\/sub><\/li>\r\n \t
                                    3. [latex]{\\text{NO}}_{3}{}^{-}[\/latex]<\/li>\r\n \t
                                    4. [latex]{\\text{CrO}}_{4}{}^{2-}[\/latex]<\/li>\r\n<\/ol>\r\n6.\u00a0In basic solution, [OH\u2212<\/sup>] > 1 \u00d7\u00a010\u22127<\/sup>M<\/em> > [H+<\/sup>]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not<\/em> appear as a reactant or product in basic solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

                                      Glossary<\/h2>\r\ncircuit: <\/b>path taken by a current as it flows because of an electrical potential difference\r\n\r\ncurrent: <\/b>flow of electrical charge; the SI unit of charge is the coulomb (C) and current is measured in amperes [latex]\\left(\\text{1 A}=1\\frac{\\text{C}}{\\text{s}}\\right)[\/latex]\r\n\r\nelectrical potential: <\/b>energy per charge; in electrochemical systems, it depends on the way the charges are distributed within the system; the SI unit of electrical potential is the volt [latex]\\left(\\text{1 V}=1\\frac{\\text{J}}{\\text{C}}\\right)[\/latex]\r\n\r\nhalf-reaction method: <\/b>method that produces a balanced overall oxidation-reduction reaction by splitting the reaction into an oxidation \u201chalf\u201d and reduction \u201chalf,\u201d balancing the two half-reactions, and then combining the oxidation half-reaction and reduction half-reaction in such a way that the number of electrons generated by the oxidation is exactly canceled by the number of electrons required by the reduction\r\n\r\noxidation half-reaction: <\/b>the \u201chalf\u201d of an oxidation-reduction reaction involving oxidation; the half-reaction in which electrons appear as products; balanced when each atom type, as well as the charge, is balanced\r\n\r\nreduction half-reaction: <\/b>the \u201chalf\u201d of an oxidation-reduction reaction involving reduction; the half-reaction in which electrons appear as reactants; balanced when each atom type, as well as the charge, is balanced","rendered":"
                                      \n
                                      \n

                                      Learning Outcomes<\/h3>\n
                                        \n
                                      • Describe defining traits of redox chemistry<\/li>\n
                                      • Identify the oxidant and reductant of a redox reaction<\/li>\n
                                      • Balance chemical equations for redox reactions using the half-reaction method<\/li>\n<\/ul>\n<\/div>\n
                                        \n
                                        \n
                                        \n

                                        Since reactions involving electron transfer are essential to the topic of electrochemistry, a brief review of redox chemistry is provided here that summarizes and extends the content of an earlier section of text (reaction stoichiometry).<\/p>\n

                                        \n

                                        Oxidation Numbers<\/h3>\n

                                        By definition, a redox reaction is one that entails changes in\u00a0oxidation number<\/em>\u00a0(or\u00a0oxidation state<\/em>) for one or more of the elements involved. The oxidation number of an element in a compound is essentially an assessment of how the electronic environment of its atoms is different in comparison to atoms of the pure element. By this description,\u00a0the oxidation number of an atom in an element is equal to zero<\/em>. For an atom in a compound,\u00a0the oxidation number is equal to the charge the atom would have in the compound if the compound were ionic<\/em>. Consequential to these rules,\u00a0the sum of oxidation numbers for all atoms in a molecule is equal to the charge on the molecule<\/em>. To illustrate this formalism, examples from the two compound classes, ionic and covalent, will be considered.<\/p>\n

                                        Simple ionic compounds present the simplest examples to illustrate this formalism, since by definition the elements\u2019 oxidation numbers are numerically equivalent to ionic charges. Sodium chloride, NaCl, is comprised of Na+<\/sup>\u00a0cations and Cl\u2212<\/sup>\u00a0anions, and so oxidation numbers for sodium and chlorine are, +1 and \u22121, respectively. Calcium fluoride, CaF2<\/sub>, is comprised of Ca2+<\/sup>\u00a0cations and F\u2212<\/sup>\u00a0anions, and so oxidation numbers for calcium and fluorine are, +2 and \u22121, respectively.<\/p>\n

                                        Covalent compounds require a more challenging use of the formalism. Water is a covalent compound whose molecules consist of two H atoms bonded separately to a central O atom via polar covalent O\u2212H bonds. The shared electrons comprising an O\u2212H bond are more strongly attracted to the more electronegative O atom, and so it acquires a partial negative charge in the water molecule (relative to an O atom in elemental oxygen). Consequently, H atoms in a water molecule exhibit partial positive charges compared to H atoms in elemental hydrogen. The sum of the partial negative and partial positive charges for each water molecule is zero, and the water molecule is neutral.<\/p>\n

                                        Imagine that the polarization of shared electrons within the O\u2212H bonds of water were 100% complete\u2014the result would be\u00a0transfer<\/em>\u00a0of electrons from H to O, and water would be an ionic compound comprised of O2\u2212<\/sup>\u00a0anions and H+<\/sup>\u00a0cations. And so, the oxidations numbers for oxygen and hydrogen in water are \u22122 and +1, respectively. Applying this same logic to carbon tetrachloride, CCl4<\/sub>, yields oxidation numbers of +4 for carbon and \u22121 for chlorine. In the nitrate ion, [latex]\\text{NO}_3^-[\/latex], the oxidation number for nitrogen is +5 and that for oxygen is \u22122, summing to equal the 1\u2212 charge on the molecule:<\/p>\n

                                        [latex]\\text{(1 N atom)} \\left(\\dfrac{+5}{\\text{N atom}}\\right)+\\text{(3 O atoms)}\\left(\\dfrac{-2}{\\text{O atom}}\\right)=+5+-6=-1[\/latex]<\/div>\n<\/section>\n
                                        \n

                                        Balancing Redox Equations<\/h3>\n

                                        The unbalanced equation below describes the decomposition of molten sodium chloride:<\/p>\n

                                        [latex]\\text{NaCl}(l) \\longrightarrow \\text{Na}(l) +\\text{Cl}_2(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{unbalanced}[\/latex]<\/div>\n

                                        This reaction satisfies the criterion for redox classification, since the oxidation number for Na is decreased from +1 to 0 (it undergoes\u00a0reduction<\/em>) and that for Cl is increased from \u22121 to 0 (it undergoes\u00a0oxidation<\/em>). The equation in this case is easily balanced by inspection, requiring stoichiometric coefficients of 2 for the NaCl and Na:<\/p>\n

                                        [latex]2\\text{NaCl}(l) \\longrightarrow 2\\text{Na}(l) +\\text{Cl}_2(g)[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{balanced}[\/latex]<\/div>\n

                                        Redox reactions that take place in aqueous solutions are commonly encountered in electrochemistry, and many involve water or its characteristic ions, H+<\/sup>(aq<\/em>) and OH\u2212<\/sup>(aq<\/em>), as reactants or products. In these cases, equations representing the redox reaction can be very challenging to balance by inspection, and the use of a systematic approach called the\u00a0half-reaction method<\/em>\u00a0is helpful. This approach involves the following steps:<\/p>\n

                                          \n
                                        1. Write skeletal equations for the oxidation and reduction half-reactions.<\/li>\n
                                        2. Balance each half-reaction for all elements except H and O.<\/li>\n
                                        3. Balance each half-reaction for O by adding H2<\/sub>O.<\/li>\n
                                        4. Balance each half-reaction for H by adding H+<\/sup>.<\/li>\n
                                        5. Balance each half-reaction for charge by adding electrons.<\/li>\n
                                        6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.<\/li>\n
                                        7. Add the two half-reactions and simplify.<\/li>\n
                                        8. If the reaction takes place in a basic medium, add OH\u2212<\/sup>\u00a0ions the equation obtained in step 7 to neutralize the H+<\/sup>\u00a0ions (add in equal numbers to both sides of the equation) and simplify.<\/li>\n<\/ol>\n

                                          The examples below demonstrate the application of this method to balancing equations for aqueous redox reactions.<\/p>\n

                                          \n
                                          \n
                                          <\/div>\n<\/header>\n
                                          \n

                                          EXAMPLE\u00a0<\/span>1:Balancing Equations for Redox Reactions in Acidic Solutions<\/span><\/h3>\n<\/header>\n
                                          \n

                                          Question 1<\/h4>\n

                                          Write the balanced equation representing reaction between solid copper and nitric acid to yield aqueous copper(II) ions and nitrogen monoxide gas.<\/p>\n

                                          Solution<\/strong><\/h4>\n
                                          Show Solution<\/span><\/p>\n
                                          \nFollowing the steps of the half-reaction method:<\/p>\n
                                            \n
                                          1. Write skeletal equations for the oxidation and reduction half-reactions.<\/em>
                                            \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq)[\/latex]
                                            \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g)[\/latex]<\/li>\n
                                          2. Balance each half-reaction for all elements except H and O.<\/em>
                                            \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq)[\/latex]
                                            \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g)[\/latex]<\/li>\n
                                          3. Balance each half-reaction for O by adding H2<\/sub>O.<\/em>
                                            \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq)[\/latex]
                                            \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) + \\boldsymbol{2\\text{H}_2 \\text{O}} (l)[\/latex]<\/li>\n
                                          4. Balance each half-reaction for H by adding H<\/em>+<\/sup>.
                                            \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq)[\/latex]
                                            \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3\\text{H}^+} (aq) + \\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                          5. Balance each half-reaction for charge by adding electrons.<\/em>
                                            \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cu} (s) \\rightarrow \\text{Cu}^{2+} (aq) + \\boldsymbol{2\\text{e}^-}[\/latex]
                                            \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3\\text{e}^-} + 3\\text{H}^+ (aq) + \\text{HNO}_3 (aq) \\rightarrow \\text{NO} (g) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                          6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.<\/em>
                                            \n[latex]\\text{oxidation(x} 3):[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3}\\text{Cu} (s) \\rightarrow \\boldsymbol{3}\\text{Cu}^{2+} (aq) + \\boldsymbol{6}\\cancel{2}\\text{e}^-[\/latex]
                                            \n[latex]\\text{reduction(x} 2):[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{6}\\cancel{3}\\text{e}^- + \\boldsymbol{6}\\cancel{3}\\text{H}^+ (aq) + \\boldsymbol{2}\\text{HNO}_3 (aq) \\rightarrow \\boldsymbol{2}\\text{NO} (g) + \\boldsymbol{4}\\cancel{2}\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                          7. Add the two half-reactions and simplify.<\/em>
                                            \n[latex]\\boldsymbol{3}\\text{Cu} (s) + \\boldsymbol{\\cancel{6}}\\text{e}^- + \\boldsymbol{6}\\text{H}^+ (aq) + \\boldsymbol{2}\\text{HNO}_3 (aq) \\rightarrow \\boldsymbol{3}\\text{Cu}^{2+} (aq) + \\boldsymbol{\\cancel{6}}\\text{e}^- + \\boldsymbol{2}\\text{NO} (g) + \\boldsymbol{4}\\text{H}_2 \\text{O} (l)[\/latex]
                                            \n[latex]\\boldsymbol{3}\\text{Cu} (s) + \\boldsymbol{6}\\text{H}^+ (aq) + \\boldsymbol{2}\\text{HNO}_3 (aq) \\rightarrow \\boldsymbol{3}\\text{Cu}^{2+} (aq) + \\boldsymbol{2}\\text{NO} (g) + \\boldsymbol{4}\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                          8. If the reaction takes place in a basic medium, add OH\u2212<\/sup>\u00a0ions the equation obtained in step 7 to neutralize the H+<\/sup>\u00a0ions (add in equal numbers to both sides of the equation) and simplify.<\/em>
                                            \nThis step is not necessary since the solution is stipulated to be acidic.<\/li>\n<\/ol>\n

                                            The balanced equation for the reaction in an acidic solution is then<\/p>\n

                                            [latex]3\\text{Cu} (s) + 6\\text{H}^+ (aq) + 2\\text{HNO}_3 (aq) \\rightarrow 3\\text{Cu}^{2+} (aq) + 2\\text{NO} (g) + 4\\text{H}_2 \\text{O} (l)[\/latex]<\/div>\n<\/div>\n<\/div>\n

                                            Check Your Learning<\/strong><\/h4>\n

                                            The reaction above results when using relatively diluted nitric acid. If concentrated nitric acid is used, nitrogen dioxide is produced instead of nitrogen monoxide. Write a balanced equation for this reaction.<\/p>\n

                                            Answer:<\/p>\n

                                            \n
                                            Show Solution<\/span><\/p>\n
                                            \n
                                            [latex]\\text{Cu} (s) + 2\\text{H}^+ (aq) + 2\\text{HNO}_3 (aq) \\rightarrow \\text{Cu}^{2+} (aq) + 2\\text{NO}_2 (g) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n
                                            \n
                                            \n
                                            <\/div>\n<\/header>\n
                                            \n

                                            EXAMPLE\u00a0<\/span>17.2:\u00a0<\/span>Balancing Equations for Redox Reactions in Basic Solutions<\/h3>\n<\/header>\n
                                            \n

                                            Question 1<\/h4>\n

                                            Write the balanced equation representing reaction between aqueous permanganate ion,\u00a0[latex]\\text{MnO}_4^-[\/latex], and solid chromium(III) hydroxide, Cr(OH)3<\/sub>, to yield solid manganese(IV) oxide, MnO2<\/sub>, and aqueous chromate ion,\u00a0[latex]\\text{CrO}_4^{2-}[\/latex].\u00a0The reaction takes place in a basic solution.<\/p>\n

                                            Solution<\/strong><\/h4>\n
                                            Show Solution<\/span><\/p>\n
                                            \nFollowing the steps of the half-reaction method:<\/p>\n
                                              \n
                                            1. Write skeletal equations for the oxidation and reduction half-reactions.<\/em>
                                              \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq)[\/latex]
                                              \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s)[\/latex]<\/li>\n
                                            2. Balance each half-reaction for all elements except H and O.<\/em>
                                              \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq)[\/latex]
                                              \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s)[\/latex]<\/li>\n
                                            3. Balance each half-reaction for O by adding H2<\/sub>O.<\/em>
                                              \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{\\text{H}_2 \\text{O} (l)} + \\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq)[\/latex]
                                              \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) + \\boldsymbol{2 \\text{H}_2 \\text{O} (l)}[\/latex]<\/li>\n
                                            4. Balance each half-reaction for H by adding H+<\/sup>.<\/em>
                                              \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{H}_2\\text{O}\\ (l) + \\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) +\\boldsymbol{5 \\text{H}^+ (aq)}[\/latex]
                                              \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{4 \\text{H}^+ (aq)} +\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                            5. Balance each half-reaction for charge by adding electrons.<\/em>
                                              \n[latex]\\text{oxidation:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\text{H}_2\\text{O}\\ (l) + \\text{Cr(OH)}_3 (s) \\rightarrow \\text{CrO}_4^{2-} (aq) +5 \\text{H}^+ (aq) + \\boldsymbol{3e^-}[\/latex]
                                              \n[latex]\\text{reduction:}[\/latex] \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]\\boldsymbol{3e^-} +4 \\text{H}^+ (aq) +\\text{MnO}_4^- (aq) \\rightarrow \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                            6. If necessary, multiply one or both half-reactions so that the number of electrons consumed in one is equal to the number produced in the other.<\/em>
                                              \nThis step is not necessary since the number of electrons is already in balance.<\/li>\n
                                            7. Add the two half-reactions and simplify.<\/em>
                                              \n[latex]\\cancel{\\text{H}_2\\text{O} (l)} + \\text{Cr(OH)}_3 (s) +\\cancel{3e^-} + \\cancel{4\\text{H}^+ (aq)} + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\cancel{5} \\text{H}^+ (aq) + \\cancel{3e^-} + \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]
                                              \n[latex]\\text{Cr(OH)}_3 (s) + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\text{H}^+ (aq) + \\text{MnO}_2 (s) + \\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n
                                            8. If the reaction takes place in a basic medium, add OH\u2212<\/sup>\u00a0ions the equation obtained in step 7 to neutralize the H+<\/sup>\u00a0ions (add in equal numbers to both sides of the equation) and simplify.<\/em>
                                              \n[latex]\\boldsymbol{\\text{OH}^- (aq)} + \\text{Cr(OH)}_3 (s) + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\text{H}^+ (aq) + \\boldsymbol{\\text{OH}^- (aq)} + \\text{MnO}_2 (s) + \\text{H}_2 \\text{O} (l)[\/latex]
                                              \n[latex]\\text{OH}^- (aq) + \\text{Cr(OH)}_3 (s) + \\text{MnO}_4^- (aq) \\rightarrow \\text{CrO}_4^{2-} (aq) + \\text{MnO}_2 (s) + 2\\text{H}_2 \\text{O} (l)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                                              Check Your Learning<\/strong><\/h4>\n

                                              Aqueous permanganate ion may also be reduced using aqueous bromide ion, Br\u2212<\/sup>, the products of this reaction being solid manganese(IV) oxide and aqueous bromate ion, BrO3<\/sub>\u2212<\/sup>. Write the balanced equation for this reaction occurring in a basic medium.<\/p>\n

                                              \n

                                              Answer:<\/p>\n

                                              Show Solution<\/span><\/p>\n
                                              \n[latex]\\text{H}_2\\text{O} (l) + 2\\text{MnO}_4^- (aq) + \\text{Br}^- (aq) \\rightarrow 2\\text{MnO}_2 (s) + \\text{BrO}_3^- (aq) + 2\\text{OH}^- (aq)[\/latex]\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
                                              \n

                                              Key Concepts and Summary<\/h3>\n

                                              Redox reactions are defined by changes in reactant oxidation numbers, and those most relevant to electrochemistry involve actual transfer of electrons. Aqueous phase redox processes often involve water or its characteristic ions, H+<\/sup>\u00a0and OH\u2212<\/sup>, as reactants in addition to the oxidant and reductant, and equations representing these reactions can be challenging to balance. The half-reaction method is a systematic approach to balancing such equations that involves separate treatment of the oxidation and reduction half-reactions.<\/p>\n<\/div>\n

                                              \n

                                              Try It<\/h3>\n
                                                \n
                                              1. If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?<\/li>\n
                                              2. For the scenario in the previous question, how many electrons moved through the circuit?<\/li>\n
                                              3. For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.\n
                                                  \n
                                                1. [latex]{\\text{Fe}}^{3+}+{\\text{3e}}^{-}\\longrightarrow \\text{Fe}[\/latex]<\/li>\n
                                                2. [latex]\\text{Cr}\\longrightarrow {\\text{Cr}}^{3+}+{\\text{3e}}^{-}[\/latex]<\/li>\n
                                                3. [latex]{\\text{MnO}}_{4}{}^{2-}\\longrightarrow {\\text{MnO}}_{4}{}^{-}+{\\text{e}}^{-}[\/latex]<\/li>\n
                                                4. [latex]{\\text{Li}}^{\\text{+}}+{\\text{e}}^{-}\\longrightarrow \\text{Li}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                                                5. For each of the following unbalanced half-reactions, determine whether an oxidation or reduction is occurring.\n
                                                    \n
                                                  1. [latex]{\\text{Cl}}^{-}\\longrightarrow {\\text{Cl}}_{2}[\/latex]<\/li>\n
                                                  2. [latex]{\\text{Mn}}^{2+}\\longrightarrow {\\text{MnO}}_{2}[\/latex]<\/li>\n
                                                  3. [latex]{\\text{H}}_{2}\\longrightarrow {\\text{H}}^{\\text{+}}[\/latex]<\/li>\n
                                                  4. [latex]{\\text{NO}}_{3}{}^{-}\\longrightarrow \\text{NO}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                                                  5. Given the following pairs of balanced half-reactions, determine the balanced reaction for each pair of half-reactions in an acidic solution.\n
                                                      \n
                                                    1. [latex]\\text{Ca}\\longrightarrow {\\text{Ca}}^{2+}+{\\text{2e}}^{-}[\/latex], [latex]{\\text{F}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2F}}^{-}[\/latex]<\/li>\n
                                                    2. [latex]\\text{Li}\\longrightarrow {\\text{Li}}^{\\text{+}}+{\\text{e}}^{-}[\/latex], [latex]{\\text{Cl}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Cl}}^{-}[\/latex]<\/li>\n
                                                    3. [latex]\\text{Fe}\\longrightarrow {\\text{Fe}}^{3+}+{\\text{3e}}^{-}[\/latex], [latex]{\\text{Br}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Br}}^{-}[\/latex]<\/li>\n
                                                    4. [latex]\\text{Ag}\\longrightarrow {\\text{Ag}}^{\\text{+}}+{\\text{e}}^{-}[\/latex], [latex]{\\text{MnO}}_{4}{}^{-}+{\\text{4H}}^{\\text{+}}+{\\text{3e}}^{-}\\longrightarrow {\\text{MnO}}_{2}+{\\text{2H}}_{2}\\text{O}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                      Show Selected Solutions<\/span><\/p>\n
                                                      \n

                                                      1.\u00a0Because A = C\/s, time must be converted to seconds. Round to two significant digits based on the initial values in the problem. [latex]\\text{35 min}\\times \\frac{\\text{60 s}}{\\text{1 min}}\\times \\frac{\\text{2.5 C}}{\\text{1 s}}=\\text{5250 C}=5.3\\times {10}^{3}\\text{C}[\/latex]<\/p>\n

                                                      3.\u00a0(a) reduction (b) oxidation (c) oxidation (d) reduction<\/p>\n

                                                      5.\u00a0The\u00a0balanced reactions are as follows:<\/p>\n

                                                        \n
                                                      1. [latex]\\begin{array}{rl}{} \\text{oxidation:}&\\text{Ca}\\longrightarrow {\\text{Ca}}^{2+}+{\\text{2e}}^{-}\\\\ \\text{reduction:}&{\\text{F}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2F}}^{-}\\\\ \\\\ \\text{overall:}&{\\text{F}}_{2}+\\text{Ca}\\longrightarrow {\\text{2F}}^{-}+{\\text{Ca}}^{2+}\\\\ \\\\ \\text{acidic solution:}&{\\text{F}}_{2}+\\text{Ca}\\longrightarrow {\\text{2F}}^{-}+{\\text{Ca}}^{2+}\\end{array}[\/latex]<\/li>\n
                                                      2. [latex]\\begin{array}{rl}\\\\ \\text{oxidation:}&\\text{2}\\times \\left(\\text{Li}\\longrightarrow {\\text{Li}}^{\\text{+}}+{\\text{e}}^{-}\\right)\\\\ \\text{reduction:}&{\\text{Cl}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Cl}}^{-}\\\\ \\\\ \\text{overall:}&{\\text{Cl}}_{2}+\\text{2Li}\\longrightarrow {\\text{2Li}}^{\\text{+}}+{\\text{2Cl}}^{-}\\\\ \\\\ \\text{acidic solution:}&{\\text{Cl}}_{2}+\\text{2Li}\\longrightarrow {\\text{2Li}}^{\\text{+}}+{\\text{2Cl}}^{-}\\end{array}[\/latex]<\/li>\n
                                                      3. [latex]\\begin{array}{rl}\\\\ \\text{oxidation:}&\\text{2}\\times \\left(\\text{Fe}\\longrightarrow {\\text{Fe}}^{3+}+{\\text{3e}}^{-}\\right)\\\\ \\text{reduction:}&\\text{3}\\times \\left({\\text{Br}}_{2}+{\\text{2e}}^{-}\\longrightarrow {\\text{2Br}}^{-}\\right)\\\\ \\\\ \\text{overall:}&\\text{3Br}+\\text{2Fe}\\longrightarrow {\\text{2Fe}}^{3+}+{\\text{6Br}}^{-}\\\\ \\\\ \\text{acidic solution:}&{\\text{3Br}}_{2}+\\text{2Fe}\\longrightarrow {\\text{2Fe}}^{3+}+{\\text{6Br}}^{-}\\end{array}[\/latex]<\/li>\n
                                                      4. [latex]\\begin{array}{rl}\\\\ \\text{oxidation:}&\\text{3}\\times \\left(\\text{Ag}\\longrightarrow {\\text{Ag}}^{\\text{+}}+{\\text{e}}^{-}\\right)\\\\ \\text{reduction:}&{\\text{MnO}}_{4}{}^{-}+{\\text{4H}}^{\\text{+}}+3{\\text{e}}^{-}\\longrightarrow {\\text{MnO}}_{2}{}^{-}+{\\text{2H}}_{2}\\text{O}\\\\ \\\\ \\text{overall:}&{\\text{MnO}}_{4}+{\\text{4H}}^{\\text{+}}+\\text{3Ag}\\longrightarrow {\\text{3Ag}}^{\\text{+}}+{\\text{MnO}}_{2}+{\\text{2H}}_{2}\\text{O}\\\\ \\\\ \\text{acidic solution:}&{\\text{MnO}}_{4}+{\\text{4H}}^{\\text{+}}+\\text{3Ag}\\longrightarrow {\\text{3Ag}}^{\\text{+}}+{\\text{MnO}}_{2}+{\\text{2H}}_{2}\\text{O}\\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
                                                        \n

                                                        Try It<\/h3>\n
                                                          \n
                                                        1. Balance the following in acidic solution:\n
                                                            \n
                                                          1. [latex]{\\text{H}}_{2}{\\text{O}}_{2}+{\\text{Sn}}^{2+}\\longrightarrow {\\text{H}}_{2}\\text{O}+{\\text{Sn}}^{4+}[\/latex]<\/li>\n
                                                          2. [latex]{\\text{PbO}}_{2}+\\text{Hg}\\longrightarrow {\\text{Hg}}_{2}{}^{2+}+{\\text{Pb}}^{2+}[\/latex]<\/li>\n
                                                          3. [latex]\\text{Al}+{\\text{Cr}}_{2}{\\text{O}}_{7}{}^{2-}\\longrightarrow {\\text{Al}}^{3+}+{\\text{Cr}}^{3+}[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                                                          4. Identify the species that undergoes oxidation, the species that undergoes reduction, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.<\/li>\n
                                                          5. Balance the following in basic solution:\n
                                                              \n
                                                            1. [latex]{\\text{SO}}_{3}{}^{2-}\\left(aq\\right)+{\\text{Cu(OH)}}_{2}\\left(s\\right)\\longrightarrow {\\text{SO}}_{4}{}^{2-}\\left(aq\\right)+\\text{Cu(OH)}\\left(s\\right)[\/latex]<\/li>\n
                                                            2. [latex]{\\text{O}}_{2}\\left(g\\right)+{\\text{Mn(OH)}}_{2}\\left(s\\right)\\longrightarrow {\\text{MnO}}_{2}\\left(s\\right)[\/latex]<\/li>\n
                                                            3. [latex]{\\text{NO}}_{3}{}^{-}\\left(aq\\right)+{\\text{H}}_{2}\\left(g\\right)\\longrightarrow \\text{NO}\\left(g\\right)[\/latex]<\/li>\n
                                                            4. [latex]\\text{Al}\\left(s\\right)+{\\text{CrO}}_{4}{}^{2-}\\left(aq\\right)\\longrightarrow {\\text{Al(OH)}}_{3}\\left(s\\right)+{\\text{Cr(OH)}}_{4}{}^{-}\\left(aq\\right)[\/latex]<\/li>\n<\/ol>\n<\/li>\n
                                                            5. Identify the species that was oxidized, the species that was reduced, the oxidizing agent, and the reducing agent in each of the reactions of the previous problem.<\/li>\n
                                                            6. Why is it not possible for hydroxide ion (OH\u2212<\/sup>) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in acidic solution?<\/li>\n
                                                            7. Why is it not possible for hydrogen ion (H+<\/sup>) to appear in either of the half-reactions or the overall equation when balancing oxidation-reduction reactions in basic solution?<\/li>\n
                                                            8. Why must the charge balance in oxidation-reduction reactions?<\/li>\n<\/ol>\n
                                                              Show Selected Solutions<\/span><\/p>\n
                                                              \n

                                                              2.\u00a0Oxidized:<\/p>\n

                                                                \n
                                                              1. Ag<\/li>\n
                                                              2. Sn2+<\/sup><\/li>\n
                                                              3. Hg<\/li>\n
                                                              4. Al<\/li>\n<\/ol>\n

                                                                Reduced:<\/p>\n

                                                                  \n
                                                                1. [latex]{\\text{Hg}}_{2}{}^{2+}[\/latex]<\/li>\n
                                                                2. H2<\/sub>O2<\/sub><\/li>\n
                                                                3. PbO2<\/sub><\/li>\n
                                                                4. [latex]{\\text{Cr}}_{2}{\\text{O}}_{7}{}^{2-}[\/latex]<\/li>\n<\/ol>\n

                                                                  Oxidizing agent:<\/p>\n

                                                                    \n
                                                                  1. [latex]{\\text{Hg}}_{2}{}^{2+}[\/latex]<\/li>\n
                                                                  2. H2<\/sub>O2<\/sub><\/li>\n
                                                                  3. PbO2<\/sub><\/li>\n
                                                                  4. [latex]{\\text{Cr}}_{2}{\\text{O}}_{7}{}^{2-}[\/latex]<\/li>\n<\/ol>\n

                                                                    Reducing agent:<\/p>\n

                                                                      \n
                                                                    1. Ag<\/li>\n
                                                                    2. Sn2+<\/sup><\/li>\n
                                                                    3. Hg<\/li>\n
                                                                    4. Al<\/li>\n<\/ol>\n

                                                                      4.\u00a0Oxidized = reducing agent:<\/p>\n

                                                                        \n
                                                                      1. [latex]{\\text{SO}}_{3}{}^{2-}[\/latex]<\/li>\n
                                                                      2. Mn(OH)2<\/sub><\/li>\n
                                                                      3. H2<\/sub><\/li>\n
                                                                      4. Al<\/li>\n<\/ol>\n

                                                                        reduced = oxidizing agent:<\/p>\n

                                                                          \n
                                                                        1. Cu(OH)2<\/sub><\/li>\n
                                                                        2. O2<\/sub><\/li>\n
                                                                        3. [latex]{\\text{NO}}_{3}{}^{-}[\/latex]<\/li>\n
                                                                        4. [latex]{\\text{CrO}}_{4}{}^{2-}[\/latex]<\/li>\n<\/ol>\n

                                                                          6.\u00a0In basic solution, [OH\u2212<\/sup>] > 1 \u00d7\u00a010\u22127<\/sup>M<\/em> > [H+<\/sup>]. Hydrogen ion cannot appear as a reactant because its concentration is essentially zero. If it were produced, it would instantly react with the excess hydroxide ion to produce water. Thus, hydrogen ion should not<\/em> appear as a reactant or product in basic solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                                                                          Glossary<\/h2>\n

                                                                          circuit: <\/b>path taken by a current as it flows because of an electrical potential difference<\/p>\n

                                                                          current: <\/b>flow of electrical charge; the SI unit of charge is the coulomb (C) and current is measured in amperes [latex]\\left(\\text{1 A}=1\\frac{\\text{C}}{\\text{s}}\\right)[\/latex]<\/p>\n

                                                                          electrical potential: <\/b>energy per charge; in electrochemical systems, it depends on the way the charges are distributed within the system; the SI unit of electrical potential is the volt [latex]\\left(\\text{1 V}=1\\frac{\\text{J}}{\\text{C}}\\right)[\/latex]<\/p>\n

                                                                          half-reaction method: <\/b>method that produces a balanced overall oxidation-reduction reaction by splitting the reaction into an oxidation \u201chalf\u201d and reduction \u201chalf,\u201d balancing the two half-reactions, and then combining the oxidation half-reaction and reduction half-reaction in such a way that the number of electrons generated by the oxidation is exactly canceled by the number of electrons required by the reduction<\/p>\n

                                                                          oxidation half-reaction: <\/b>the \u201chalf\u201d of an oxidation-reduction reaction involving oxidation; the half-reaction in which electrons appear as products; balanced when each atom type, as well as the charge, is balanced<\/p>\n

                                                                          reduction half-reaction: <\/b>the \u201chalf\u201d of an oxidation-reduction reaction involving reduction; the half-reaction in which electrons appear as reactants; balanced when each atom type, as well as the charge, is balanced<\/p>\n\n\t\t\t

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