Oxidation-Reduction Reactions

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1 Chapter 18 Oxidation-Reduction Reactions Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act on Your Strategy, and Check Your. Where a shorter solution is appropriate, the Sample Problem contains only two steps: Problem and. Where relevant, a Check Your step is also included in the shorter Sample Problems. s for Practice Problems Student Textbook page Problem Write a balanced net ionic equation for the reaction of zinc with aqueous iron(ii) chloride. Include the physical states of the reactants and products. Write a balanced chemical equation for the reaction. Include the physical states. Zn (s) + FeCl 2(aq) Fe (s) + ZnCl 2(aq) Write a balanced total ionic equation for the reaction. Zn (s) + Fe 2+ (aq) + 2Cl (aq) Fe (s) + Zn 2+ (aq) + 2Cl (aq) Eliminate the spectator ions (the chloride ions) to write the net ionic equation. Zn (s) + Fe 2+ (aq) Fe (s) + Zn 2+ (aq) 2. Problem Write a balanced net ionic equation for each reaction, including physical states. (a) magnesium and aqueous aluminum sulfate (b) a solution of silver nitrate with metallic cadmium (a) Write a balanced chemical equation for the reaction. Include the physical states. Mg (s) + Al 2 (SO 4 ) 3(aq) Al (s) + MgSO 4(aq) (unbalanced) 3Mg (s) + Al 2 (SO 4 ) 3(aq) 2Al (s) + 3MgSO 4(aq) (balanced) Write a balanced total ionic equation for the reaction. 3Mg (s) + 2Al 3+ 2 (aq) + 3SO 4 (aq) 2Al (s) + 3Mg 2+ 2 (aq) + 3SO 4 (aq) Eliminate the spectator ions (the sulfate ions) to write the net ionic equation. 3Mg (s) + 2Al 3+ (aq) 2Al (s) + 3Mg 2+ (aq) (b) Write a balanced chemical equation for the reaction. Include the physical states. AgNO 3(aq) + Cd (s) Ag (s) + Cd(NO 3 ) 2(aq) (unbalanced) 2AgNO 3(aq) + Cd (s) 2Ag (s) + Cd(NO 3 ) 2(aq) (balanced) Write a balanced total ionic equation for the reaction. 2Ag + (aq) + 2NO 3 (aq) + Cd (s) 2Ag (s) + Cd 2+ (aq) + 2NO 3 (aq) Eliminate the spectator ions (the nitrate ions) to write the net ionic equation. 2Ag + (aq) + Cd (s) 2Ag (s) + Cd 2+ (aq) 318

2 3. Problem Identify the reactant oxidized and the reactant reduced in each reaction in question 2. (a) The reactant oxidized is the reactant that loses electrons. The reactant reduced is the reactant that gains electrons. The net ionic equation shows that each magnesium atom loses two electrons to form a magnesium ion, and that each aluminum ion gains three electrons to form an aluminum atom. 3Mg (s) + 2Al 3+ (aq) 2Al (s) + 3Mg 2+ (aq) Therefore, magnesium is oxidized, and aluminum ions are reduced. (b) The reactant oxidized is the reactant that loses electrons. The reactant reduced is the reactant that gains electrons. The net ionic equation shows that each cadmium atom loses two electrons to form a cadmium ion, and that each silver ion gains one electron to form a silver atom. 2Ag + (aq) + Cd (s) 2Ag (s) + Cd 2+ (aq) Therefore, cadmium is oxidized, and silver ions are reduced. 4. Problem Identify the oxidizing agent and the reducing agent in each reaction in question 2. (a) The oxidizing agent is the reactant that accepts electrons. The reducing agent is the reactant that donates electrons. The net ionic equation shows that each aluminum ion accepts three electrons to form an aluminum atom, and that each magnesium atom donates two electrons to form a magnesium ion. 3Mg (s) + 2Al 3+ (aq) 2Al (s) + 3Mg 2+ (aq) Therefore, aluminum ions are the oxidizing agent, and magnesium is the reducing agent. (b) The oxidizing agent is the reactant that accepts electrons. The reducing agent is the reactant that donates electrons. The net ionic equation shows that each silver ion accepts one electron to form a silver atom, and that each cadmium atom donates two electrons to form a cadmium ion. 2Ag + (aq) + Cd (s) 2Ag (s) + Cd 2+ (aq) Therefore, silver ions are the oxidizing agent, and cadmium is the reducing agent. s for Practice Problems Student Textbook page Problem Write balanced half-reactions from the net ionic equation for the reaction between solid aluminum and aqueous iron(iii) sulfate. The sulfate ions are spectator ions and are not included. Al (s) + Fe 3+ (aq) Al 3+ (aq) + Fe (s) The net ionic equation shows that each aluminum atom loses three electrons to form an aluminum ion, and that each iron(iii) ion gains three electrons to form an iron atom. Therefore, the two half-reactions are as follows. 319

3 Oxidation: Al (s) Al 3+ (aq) + 3e Reduction: Fe 3+ (aq) + 3e Fe (s) 6. Problem Write balanced half-reactions from the following net ionic equations. (a) Fe (s) + Cu 2+ (aq) Fe 2+ (aq) + Cu (s) (b) Cd (s) + 2Ag + (aq) Cd 2+ (aq) + 2Ag (s) (a) The net ionic equation shows that each iron atom loses two electrons to form an iron(ii) ion, and that each copper(ii) ion gains two electrons to form a copper atom. Therefore, the two half-reactions are as follows. Oxidation: Fe (s) Fe 2+ (aq) + 2e Reduction: Cu 2+ (aq) + 2e Cu (s) (b) The net ionic equation shows that each cadmium atom loses two electrons to form a cadmium ion, and that each silver ion gains one electron to form a silver atom. Therefore, the two half-reactions are as follows. Oxidation: Cd (s) Cd 2+ (aq) + 2e Reduction: Ag + (aq) + 1e Ag (s) 7. Problem Write balanced half-reactions for each of the following reactions. (a) Sn (s) + PbCl 2(aq) SnCl 2(aq) + Pb (s) (b) Au(NO 3 ) 3(aq) + 3Ag (s) 3AgNO 3(aq) + Au (s) (c) 3Zn (s) + Fe 2 (SO 4 ) 3(aq) 3ZnSO 4(aq) + 2Fe (s) (a) Write a balanced total ionic equation for the reaction. Sn (s) + Pb 2+ (aq) + 2Cl (aq) Sn 2+ (aq) + 2Cl (aq) + Pb (s) Eliminate the spectator ions (the chloride ions) to write the net ionic equation. Sn (s) + Pb 2+ (aq) Sn 2+ (aq) + Pb (s) The net ionic equation shows that each tin atom loses two electrons to form a tin(ii) ion, and that each lead(ii) ion gains two electrons to form a lead atom. Therefore, the two half-reactions are as follows. Oxidation: Sn (s) Sn 2+ (aq) + 2e Reduction: Pb 2+ (aq) + 2e Pb (s) (b) Write a balanced total ionic equation for the reaction. Au 3+ (aq) + 3NO 3 (aq) + 3Ag (s) 3Ag + (aq) + 3NO 3 (aq) + Au (s) Eliminate the spectator ions (the nitrate ions) to write the net ionic equation. Au 3+ (aq) + 3Ag (s) 3Ag + (aq) + Au (s) The net ionic equation shows that each gold(iii) ion gains three electrons to form a gold atom, and that each silver atom loses one electron to form a silver ion. Therefore, the two half-reactions are as follows. Oxidation: Ag (s) Ag + (aq) + e Reduction: Au 3+ (aq) + 3e Au (s) (c) Write a balanced total ionic equation for the reaction. 3Zn (s) + 2Fe 3+ (aq) + 3SO 4 2 (aq) 3Zn 2+ (aq) + 3SO 4 2 (aq) + 2Fe (s) Eliminate the spectator ions (the sulfate ions) to write the net ionic equation. 3Zn (s) + 2Fe 3+ (aq) 3Zn 2+ (aq) + 2Fe (s) The net ionic equation shows that each zinc atom loses two electrons to form a zinc ion, and that each iron(iii) ion gains three electrons to form an iron atom. Therefore, the two half-reactions are as follows. 320

4 Oxidation: Zn (s) Zn 2+ (aq) + 2e Reduction: Fe 3+ (aq) + 3e Fe (s) 8. Problem Write the net ionic equation and the half-reactions for the disproportionation of mercury(i) ions in aqueous solution to give liquid mercury and aqueous mercury(ii) ions. Assume that mercury(i) ions exist in solution as Hg Write a balanced net ionic equation for the reaction. 2+ Hg 2 (aq) Hg (l) + Hg 2+ (aq) The net ionic equation shows that the reaction is a disproportionation in which equal numbers of mercury(i) ions undergo oxidation and reduction. Half the mercury(i) ions lose electrons to form mercury(ii) ions, and the other half of the mercury(i) ions gain electrons to form mercury atoms. Because only one Hg 2+ 2 ion is present in the balanced net ionic equation, the half-reactions are easier to see if the net ionic equation is multiplied by Hg 2 (aq) 2Hg (l) + 2Hg 2+ (aq) The two half-reactions are as follows. 2+ Reduction: Hg 2 (aq) + 2e 2Hg (l) 2+ Oxidation: Hg 2 (aq) 2Hg 2+ (aq) + 2e Note: For simplicity, the formula of mercury(i) ions is sometimes written as Hg +, even though these ions exist in the form Hg If the formula of mercury(i) ions is written as Hg +, the balanced net ionic equation and the half-reactions are written as follows. 2Hg + (aq) Hg (l) + Hg 2+ (aq) Reduction: Hg + (aq) + e Hg (l) Oxidation: Hg + (aq) Hg 2+ (aq) + e s for Practice Problems Student Textbook page Problem Determine the oxidation number of the specified element in each of the following. (a) N in NF 3 2 (c) Cr in CrO 4 (e) C in C 12 H 22 O 11 (b) S in S 8 (d) P in P 2 O 5 (f) C in CHCl 3 (a) Because the compound NF 3 does not contain hydrogen or oxygen, rule 5 applies. Because NF 3 is a compound, rule 6 also applies. Nitrogen has an electronegativity of Fluorine has an electronegativity of Therefore, from rule 5, assign fluorine an oxidation number of 1. The oxidation number of nitrogen is unknown, so let it be x. From rule 6, the sum of the oxidation numbers is 0. Then, x + 3(1) = 0 x 3 = 0 x = 3 The oxidation number of nitrogen is +3. (b) Because S 8 is the formula of a pure element, rule 1 applies. From rule 1, the oxidation number of S in S 8 is 0. (c) Because the polyatomic ion CrO 4 2 contains oxygen, rule 4 applies. Because CrO 4 2 is a polyatomic ion, rule 7 also applies. 321

5 From rule 4, oxygen has its usual oxidation number of 2. The oxidation number of chromium is unknown, so let it be x. From rule 7, the sum of the oxidation numbers is 2. Then, x + 4(2) =2 x 8 =2 x = 6 The oxidation number of chromium is +6. (d) Because the compound P 2 O 5 contains oxygen, rule 4 applies. Because P 2 O 5 is a compound, rule 6 also applies. From rule 4, oxygen has its usual oxidation number of 2. The oxidation number of phosphorus is unknown, so let it be x. From rule 6, the sum of the oxidation numbers is 0. Then, 2x + 5(2) = 0 2x 10 = 0 2x = 10 x = 5 The oxidation number of phosphorus is +5. (e) Because the compound C 12 H 22 O 11 contains hydrogen and oxygen, rules 3 and 4 apply. Because C 12 H 22 O 11 is a compound, rule 6 also applies. From rule 3, hydrogen has its usual oxidation number of +1. From rule 4, oxygen has its usual oxidation number of 2. The oxidation number of carbon is unknown, so let it be x. From rule 6, the sum of the oxidation numbers is 0. Then, 12x + 22(+1) + 11(2) = 0 12x = 0 12x = 0 x = 0 The oxidation number of carbon is 0. (f) Because the compound CHCl 3 contains hydrogen, rule 3 applies. Because CHCl 3 is a compound, rule 6 also applies. From rule 3, hydrogen has its usual oxidation number of +1. The oxidation numbers of carbon and chlorine are both unknown. Carbon has an electronegativity of Chlorine has an electronegativity of Therefore, using the same reasoning as in rule 5, assign chlorine an oxidation number of 1. The oxidation number of carbon is unknown, so let it be x. From rule 6, the sum of the oxidation numbers is 0. Then, x + 1(+1) + 3(1) = 0 x = 0 x 2 = 0 x = 2 The oxidation number of carbon is Problem Determine the oxidation number of each element in each of the following. (a) H 2 SO 3 (b) OH (c) HPO 4 2 (a) Because the compound H 2 SO 3 contains hydrogen and oxygen, rules 3 and 4 apply. Because H 2 SO 3 is a compound, rule 6 also applies. From rule 3, hydrogen has its usual oxidation number of +1. From rule 4, oxygen has its usual oxidation number of

6 The oxidation number of sulfur is unknown, so let it be x. From rule 6, the sum of the oxidation numbers is 0. Then, 2(+1) + x + 3(2) = x 6 = 0 x 4 = 0 x = 4 The oxidation number of hydrogen is +1. The oxidation number of sulfur is +4. The oxidation number of oxygen is 2. (b) Because the polyatomic ion OH contains hydrogen and oxygen, rules 3 and 4 apply. Because OH is a polyatomic ion, rule 7 also applies. From rule 3, hydrogen has its usual oxidation number of +1. From rule 4, oxygen has its usual oxidation number of 2. From rule 7, the sum of the oxidation numbers is 1. 1(2) + 1(+1) = =1 The oxidation number of oxygen is 2. The oxidation number of hydrogen is +1. (c) Because the polyatomic ion HPO 4 2 contains hydrogen and oxygen, rules 3 and 4 apply. Because HPO 4 2 is a polyatomic ion, rule 7 also applies. From rule 3, hydrogen has its usual oxidation number of +1. From rule 4, oxygen has its usual oxidation number of 2. The oxidation number of phosphorus is unknown, so let it be x. From rule 7, the sum of the oxidation numbers is 2. Then, 1(+1) + x + 4(2) =2 1 + x 8 =2 x 7 =2 x = 5 The oxidation number of hydrogen is +1. The oxidation number of phosphorus is +5. The oxidation number of oxygen is Problem As stated in rule 4, oxygen does not always have its usual oxidation number of 2. Determine the oxidation number of oxygen in each of the following. (a) the compound oxygen difluoride, OF 2 (b) the peroxide ion, O 2 2 (a) Oxygen has an electronegativity of Fluorine has an electronegativity of Therefore, using the same reasoning as in rule 5, assign fluorine an oxidation number of 1. Because OF 2 is a compound, rule 6 applies. The oxidation number of oxygen is unknown, so let it be x. From rule 6, the sum of the oxidation numbers is 0. Then, x + 2(1) = 0 x 2 = 0 x = 2 The oxidation number of oxygen is +2. (b) Because O 2 2 is a polyatomic ion, rule 7 applies. The oxidation number of oxygen is unknown, so let it be x. From rule 7, the sum of the oxidation numbers is 2. Then, 2x =2 x =1 The oxidation number of oxygen is

7 12. Problem Determine the oxidation number of each element in each of the following ionic compounds by considering the ions separately. Hint: One formula unit of the compound in part (c) contains two identical monatomic ions and one polyatomic ion. (a) Al(HCO 3 ) 3 (b) (NH 4 ) 3 PO 4 (c) K 2 H 3 IO 6 (a) The compound Al(HCO 3 ) 3 contains Al 3+ ions and HCO 3 ions. Because Al 3 + is a monatomic ion, rule 2 applies. From rule 2, the oxidation number of aluminum is +3. Because the polyatomic ion HCO 3 contains hydrogen and oxygen, rules 3 and 4 apply. Because HCO 3 is a polyatomic ion, rule 7 also applies. From rule 3, hydrogen has its usual oxidation number of +1. From rule 4, oxygen has its usual oxidation number of 2. The oxidation number of carbon is unknown, so let it be x. From rule 7, the sum of the oxidation numbers in HCO 3 is 1. Then, 1(+1) + x + 3(2) =1 1 + x 6 =1 x 5 =1 x = 4 The oxidation number of aluminum is +3. The oxidation number of hydrogen is +1. The oxidation number of carbon is +4. The oxidation number of oxygen is 2. (b) The compound (NH 4 ) 3 PO 4 contains NH 4 + ions and PO 4 3 ions. Because the polyatomic ion NH 4 + contains hydrogen, rule 3 applies. Because NH 4 + is a polyatomic ion, rule 7 also applies. From rule 3, hydrogen has its usual oxidation number of +1. The oxidation number of nitrogen is unknown, so let it be x. From rule 7, the sum of the oxidation numbers in NH 4 + is +1. Then, x + 4(+1) =+1 x + 4 =+1 x =+1 4 x =3 Because the polyatomic ion PO 4 3 contains oxygen, rule 4 applies. Because PO 4 3 is a polyatomic ion, rule 7 also applies. From rule 4, oxygen has its usual oxidation number of 2. The oxidation number of phosphorus is unknown, so let it be y. From rule 7, the sum of the oxidation numbers in PO 4 3 is 3. Then, y + 4(2) =3 y 8 =3 y =3 + 8 y = 5 The oxidation number of nitrogen is 3. The oxidation number of hydrogen is +1. The oxidation number of phosphorus is +5. The oxidation number of oxygen is 2. (c) The compound K 2 H 3 IO 6 contains K + ions and H 3 IO 6 2 ions. Because K + is a monatomic ion, rule 2 applies. From rule 2, the oxidation number of potassium is +1. Because the polyatomic ion H 3 IO 6 2 contains hydrogen and oxygen, rules 3 and 4 apply. Because H 3 IO 6 2 is a polyatomic ion, rule 7 also applies. From rule 3, hydrogen has its usual oxidation number of +1. From rule 4, the oxidation number of oxygen is

8 The oxidation number of iodine is unknown, so let it be x. From rule 7, the sum of the oxidation numbers in H 3 IO 6 2 is 2. Then, 3(+1) + x + 6(2) =2 3 + x 12 =2 x 9 =2 x = 7 The oxidation number of potassium is +1. The oxidation number of hydrogen is +1. The oxidation number of iodine is +7. The oxidation number of oxygen is 2. s for Practice Problems Student Textbook page Problem Determine whether each reaction is a redox reaction. (a) H 2 O 2 + 2Fe(OH) 2 2Fe(OH) 3 (b) PCl 3 + 3H 2 O H 3 PO 3 + 3HCl Find the oxidation number of each element in the reactants and products. Identify any elements that undergo an increase or a decrease in oxidation number during the reaction. (a) H 2 O 2 + 2Fe(OH) 2 2Fe(OH) The oxidation number of hydrogen is +1 on both sides of the equation, so hydrogen is neither oxidized nor reduced. The oxidation number of the oxygen atoms that originate in Fe(OH) 2 is 2 on both sides of the equation. The oxidation number of the oxygen atoms that originate in H 2 O 2 decreases from 1 to 2. The oxidation number of iron increases from +2 to +3. Because both oxygen and iron undergo changes in oxidation number, the reaction is a redox reaction. (b) PCl 3 + 3H 2 O H 3 PO 3 + 3HCl No elements undergo changes in oxidation number, so the reaction is not a redox reaction. 14. Problem Identify the oxidizing agent and the reducing agent for the redox reaction(s) in the previous question. In part (a) of the previous question, The oxidation number of the oxygen atoms that originate in H 2 O 2 decreases from 1 to 2, so H 2 O 2 is reduced. Therefore, H 2 O 2 is the oxidizing agent. The oxidation number of iron increases from +2 to +3. The iron atoms on the reactant side of the equation exist in the compound Fe(OH) 2. Therefore, Fe(OH) 2 is oxidized, and Fe(OH) 2, or the ion Fe 2+, is the reducing agent. 325

9 15. Problem For the following balanced net ionic equation, identify the reactant that undergoes oxidation and the reactant that undergoes reduction. Br 2 + 2ClO 2 2Br + 2ClO 2 Find the oxidation number of each element in the reactants and products. Identify any elements that undergo an increase or a decrease in oxidation number during the reaction. Br 2 + 2ClO 2 2Br + 2ClO The oxidation number of oxygen is the same on both sides of the equation. The oxidation number of chlorine increases from +3 to +4. The chlorine atoms on the reactant side are found in ClO 2 ions. Therefore, ClO 2 ions undergo oxidation. The oxidation number of bromine decreases from 0 to 1. The bromine atoms on the reactant side are found in elemental bromine, Br 2. Therefore, elemental bromine undergoes reduction. 16. Problem Nickel and copper are two metals that have played a role in the economy of Newfoundland and Labrador. Nickel and copper ores usually contain the metals as sulfides, such as NiS and Cu 2 S. Do the extractions of these pure elemental metals from their ores involve redox reactions? Explain your reasoning. In NiS, nickel has an oxidation number of +2. Metallic nickel, Ni, is an element with an oxidation number of 0. In Cu 2 S, copper has an oxidation number of +1. Metallic copper, Cu, is an element with an oxidation number of 0. Because each of these metals undergoes a decrease in its oxidation number during extraction, the extraction process must involve reduction. Oxidation and reduction always occur together, so the extraction processes must be redox reactions. (We do not need to know the other reactants to know that this conclusion is true.) s for Practice Problems Student Textbook page Problem Write a balanced half-reaction for the reduction of cerium(iv) ions to cerium(iii) ions. Represent the given reactant and product with the correct formulas. Ce 4+ Ce 3+ Balance the atoms, if necessary. The atoms are already balanced. Add an electron to the left side to balance the charges. Ce 4+ + e Ce

10 18. Problem Write a balanced half-reaction for the oxidation of bromide ions to bromine. Represent the given reactant and product with the correct formulas. Br Br 2 Balance the bromine atoms. 2Br Br 2 Add two electrons to the right side to balance the charges. 2Br Br 2 + 2e 19. Problem Balance each of the following half-reactions under acidic conditions. (a) O 2 H 2 O 2 (b) H 2 O O 2 (c) NO 3 N 2 (a) Step 1 Step 3 Step 4 Step 5 (b) Step 1 Step 3 Step 4 Step 5 (c) Step 1 Step 3 Step 4 Step 5 The unbalanced half-reaction, including the correct formulas, is given. O 2 H 2 O 2 There are no atoms to balance other than oxygen and hydrogen. Add water molecules to balance the oxygen atoms, if necessary. The oxygen atoms are already balanced. The reaction occurs in acidic solution, so add hydrogen ions to balance the hydrogen atoms. O 2 + 2H + H 2 O 2 Add two electrons to the left side to balance the charges. O 2 + 2H + + 2e H 2 O 2 The unbalanced half-reaction, including the correct formulas, is given. H 2 O O 2 There are no atoms to balance other than oxygen and hydrogen. The reaction occurs in aqueous solution, so add another water molecule to balance the oxygen atoms. 2H 2 O O 2 The reaction occurs in acidic solution, so add hydrogen ions to balance the hydrogen atoms. 2H 2 O O 2 + 4H + Add four electrons to the right side to balance the charges. 2H 2 O O 2 + 4H + + 4e The unbalanced half-reaction, including the correct formulas, is given. NO 3 N 2 Balance the nitrogen atoms. 2NO 3 N 2 The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms. 2NO 3 N 2 + 6H 2 O The reaction occurs in acidic solution, so add hydrogen ions to balance the hydrogen atoms. 2NO H + N 2 + 6H 2 O Add ten electrons to the left side to balance the charges. 2NO H e N 2 + 6H 2 O 327

11 20. Problem Balance each of the following half-reactions under acidic conditions. (a) ClO 3 Cl (b) NO NO 3 (c) Cr 2 O 7 2 Cr 3+ (a) Step 1 Step 3 Step 4 Step 5 (b) Step 1 Step 3 Step 4 Step 5 (c) Step 1 Step 3 Step 4 Step 5 The unbalanced half-reaction, including the correct formulas, is given. ClO 3 Cl Balance the chlorine atoms, if necessary. The chlorine atoms are already balanced. The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms. ClO 3 Cl + 3H 2 O The reaction occurs in acidic solution, so add hydrogen ions to balance the hydrogen atoms. ClO 3 + 6H + Cl + 3H 2 O Add six electrons to the left side to balance the charges. ClO 3 + 6H + + 6e Cl + 3H 2 O The unbalanced half-reaction, including the correct formulas, is given. NO NO 3 Balance the nitrogen atoms, if necessary. The nitrogen atoms are already balanced. The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms. NO + 2H 2 O NO 3 The reaction occurs in acidic solution, so add hydrogen ions to balance the hydrogen atoms. NO + 2H 2 O NO 3 + 4H + Add three electrons to the right side to balance the charges. NO + 2H 2 O NO 3 + 4H + + 3e The unbalanced half-reaction, including the correct formulas, is given. Cr 2 O 2 7 Cr 3+ Balance the chromium atoms. Cr 2 O 2 7 2Cr 3+ The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms. Cr 2 O 2 7 2Cr H 2 O The reaction occurs in acidic solution, so add hydrogen ions to balance the hydrogen atoms. Cr 2 O H + 2Cr H 2 O Add six electrons to the left side to balance the charges. Cr 2 O H + + 6e 2Cr H 2 O 328

12 s for Practice Problems Student Textbook page Problem Write a balanced half-reaction for the oxidation of chromium(ii) ions to chromium(iii) ions. Represent the given reactant and product with the correct formulas. Cr 2+ Cr 3+ Balance the atoms, if necessary. The atoms are already balanced. Add an electron to the right side to balance the charges. Cr 2+ Cr 3+ + e 22. Problem Write a balanced half-reaction for the reduction of oxygen to oxide ions. Represent the given reactant and product with the correct formulas. O 2 O 2 Balance the oxygen atoms. O 2 2O 2 Add four electrons to the left side to balance the charges. O 2 + 4e 2O Problem Balance each of the following half-reactions under basic conditions. (a) Al Al(OH) 4 (d) CrO 2 4 Cr(OH) 3 (b) CN CNO (e) CO C 2 O 4 (c) MnO 4 MnO 2 (a) Step 1 Step 3 Step 4 Step 5 Step 6 Step 7 (b) Step 1 The unbalanced half-reaction, including the correct formulas, is given. Al Al(OH) 4 Balance the aluminum atoms, if necessary. The aluminum atoms are already balanced. Balance the oxygen and hydrogen atoms as if the conditions are acidic. Add water molecules to balance the oxygen atoms. Al + 4H 2 O Al(OH) 4 Add hydrogen ions to balance the hydrogen atoms. Al + 4H 2 O Al(OH) 4 + 4H + Adjust for basic conditions by adding four hydroxide ions to each side. Al + 4H 2 O + 4OH Al(OH) 4 + 4H + + 4OH Combine the hydrogen ions and hydroxide ions on the right side into water molecules. Al + 4H 2 O + 4OH Al(OH) 4 + 4H 2 O Remove four water molecules from each side. Al + 4OH Al(OH) 4 Add three electrons to the right side to balance the charges. Al + 4OH Al(OH) 4 + 3e The unbalanced half-reaction, including the correct formulas, is given. CN CNO Balance the carbon and nitrogen atoms, if necessary. These atoms are already balanced. 329

13 Step 3 Step 4 Step 5 Step 6 Step 7 (c) Step 1 Step 3 Step 4 Step 5 Step 6 Step 7 (d) Step 1 Step 3 Step 4 Step 5 Step 6 Step 7 Balance the oxygen and hydrogen atoms as if the conditions are acidic. Add water molecules to balance the oxygen atoms. CN + H 2 O CNO Add hydrogen ions to balance the hydrogen atoms. CN + H 2 O CNO + 2H + Adjust for basic conditions by adding two hydroxide ions to each side. CN + H 2 O + 2OH CNO + 2H + + 2OH Combine the hydrogen ions and hydroxide ions on the right side into water molecules. CN + H 2 O + 2OH CNO + 2H 2 O Remove a water molecule from each side. CN + 2OH CNO + H 2 O Add two electrons to the right side to balance the charges. CN + 2OH CNO + H 2 O + 2e The unbalanced half-reaction, including the correct formulas, is given. MnO 4 MnO 2 Balance the manganese atoms, if necessary. The manganese atoms are already balanced. Balance the oxygen and hydrogen atoms as if the conditions are acidic. Add water molecules to balance the oxygen atoms. MnO 4 MnO 2 + 2H 2 O Add hydrogen ions to balance the hydrogen atoms. MnO 4 + 4H + MnO 2 + 2H 2 O Adjust for basic conditions by adding four hydroxide ions to each side. MnO 4 + 4H + + 4OH MnO 2 + 2H 2 O + 4OH Combine the hydrogen ions and hydroxide ions on the left side into water molecules. MnO 4 + 4H 2 O MnO 2 + 2H 2 O + 4OH Remove two water molecules from each side. MnO 4 + 2H 2 O MnO 2 + 4OH Add three electrons to the left side to balance the charges. MnO 4 + 2H 2 O + 3e MnO 2 + 4OH The unbalanced half-reaction, including the correct formulas, is given. CrO 2 4 Cr(OH) 3 Balance the chromium atoms, if necessary. The chromium atoms are already balanced. Balance the oxygen and hydrogen atoms as if the conditions are acidic. Add water molecules to balance the oxygen atoms. CrO 2 4 Cr(OH) 3 + H 2 O Add hydrogen ions to balance the hydrogen atoms. CrO H + Cr(OH) 3 + H 2 O Adjust for basic conditions by adding five hydroxide ions to each side. CrO H + + 5OH Cr(OH) 3 + H 2 O + 5OH Combine the hydrogen ions and hydroxide ions on the left side into water molecules. CrO H 2 O Cr(OH) 3 + H 2 O + 5OH Remove one water molecule from each side. CrO H 2 O Cr(OH) 3 + 5OH Add three electrons to the left side to balance the charges. CrO H 2 O + 3e Cr(OH) 3 + 5OH 330

14 (e) Step 1 Step 3 Step 4 Step 5 Step 6 Step 7 The unbalanced half-reaction, including the correct formulas, is given. CO C 2 O 4 Balance the carbon atoms. 2CO C 2 O 4 Balance the oxygen and hydrogen atoms as if the conditions are acidic. Add water molecules to balance the oxygen atoms. 2CO 2 3 C 2 O H 2 O Add hydrogen ions to balance the hydrogen atoms. 2CO H + C 2 O H 2 O Adjust for basic conditions by adding four hydroxide ions to each side. 2CO H + + 4OH C 2 O H 2 O + 4OH Combine the hydrogen ions and hydroxide ions on the left side into water molecules. 2CO H 2 O C 2 O H 2 O + 4OH Remove two water molecules from each side. 2CO H 2 O C 2 O OH Add two electrons to the left side to balance the charges. 2CO H 2 O + 2e C 2 O OH 24. Problem Balance each of the following half-reactions. (a) FeO 2 4 Fe 3+ (acidic conditions) (b) ClO 2 Cl (basic conditions) (a) Step 1 The unbalanced half-reaction, including the correct formulas, is given. FeO 2 4 Fe 3+ Balance the iron atoms, if necessary. The iron atoms are balanced. Step 3 Add water molecules to balance the oxygen atoms. FeO 2 4 Fe H 2 O Step 4 Add hydrogen ions to balance the hydrogen atoms. FeO H + Fe H 2 O Step 5 Add three electrons to the left side to balance the charges. FeO H + + 3e Fe H 2 O (b) Step 1 Step 3 Step 4 Step 5 Step 6 Step 7 The unbalanced half-reaction, including the correct formulas, is given. ClO 2 Cl Balance the chlorine atoms, if necessary. The chlorine atoms are balanced. ClO 2 Cl Balance the oxygen and hydrogen atoms as if the conditions are acidic. Add water molecules to balance the oxygen atoms. ClO 2 Cl + 2H 2 O Add hydrogen ions to balance the hydrogen atoms. ClO 2 + 4H + Cl + 2H 2 O Adjust for basic conditions by adding four hydroxide ions to each side. ClO 2 + 4H + + 4OH Cl + 2H 2 O + 4OH Combine the hydrogen ions and hydroxide ions on the left side into water molecules. ClO 2 + 4H 2 O Cl + 2H 2 O + 4OH Remove two water molecules from each side. ClO 2 + 2H 2 O Cl + 4OH Add four electrons to the left side to balance the charges. ClO 2 + 2H 2 O + 4e Cl + 4OH 331

15 s for Practice Problems Student Textbook pages Problem Balance each of the following redox equations by inspection. Write the balanced half-reactions in each case. (a) Na + F 2 NaF (b) Mg + N 2 Mg 3 N 2 (c) HgO Hg + O 2 (a) Balance the fluorine atoms. Na + F 2 2NaF Balance the sodium atoms. 2Na + F 2 2NaF (balanced) NaF contains Na + ions and F ions. Therefore, each sodium atom loses an electron to form a sodium ion, and each fluorine atom gains an electron to form a fluoride ion. To balance each half-reaction, write the symbol of the reactant and product, balance the atoms (if necessary), and add one or more electrons to balance the charges. Oxidation half-reaction: Na Na + Na Na + + e (balanced) Reduction half-reaction: F 2 F F 2 2F F 2 + 2e 2F (balanced) (b) Balance the magnesium atoms. 3Mg + N 2 Mg 3 N 2 Mg 3 N 2 contains Mg 2+ ions and N 3 ions. Therefore, each magnesium atom loses two electrons to form a magnesium ion, and each nitrogen atom gains three electrons to form a nitride ion. To balance each half-reaction, write the symbol of the reactant and product, balance the atoms (if necessary), and add one or more electrons to balance the charges. Oxidation half-reaction: Mg Mg 2+ Mg Mg e (balanced) Reduction half-reaction: N 2 N 3 N 2 2N 3 N 2 + 6e 2N 3 (balanced) (c) Balance the oxygen atoms. 2HgO Hg + O 2 Balance the mercury atoms. 2HgO 2Hg + O 2 (balanced) HgO contains Hg 2+ ions and O 2 ions. Therefore, each mercury(ii) ion gains two electrons to form a mercury atom, and each oxide ion loses two electrons to form an oxygen atom. To balance each half-reaction, write the symbol of the reactant and product, balance the atoms (if necessary), and add one or more electrons to balance the charges. 332

16 Reduction half-reaction: Hg 2+ Hg Hg e Hg (balanced) Oxidation half-reaction: O 2 O 2 2O 2 O 2 2O 2 O 2 + 4e (balanced) 26. Problem Balance the following equation by the half-reaction method. Cu 2+ + I CuI + I 3 Step 1 The unbalanced net ionic equation is given. Cu 2+ + I CuI + I 3 (Note: CuI is not written in ionic form because the electronegativity difference between copper and iodine is quite small (0.76). This compound is covalent.) Divide the unbalanced net ionic equation into two half-reactions. Assign oxidation numbers to all the elements, if necessary. If you notice that the oxidation number of copper decreases from +2 to +1, there is no need to assign other oxidation numbers. Copper must be involved in the reduction half-reaction. Reduction half-reaction: Cu 2 + I CuI Step 3 Oxidation half-reaction: I I 3 Balance the half-reactions independently. Reduction: Cu 2+ + I + e CuI (balanced) Oxidation: 3I I 3 3I I 3 + 2e (balanced) Step 4 The LCM of 1 and 2 is 2. Step 5 Multiply the reduction half-reaction by 2, so that equal numbers of electrons are lost and gained. 2Cu I + 2e 2CuI Step 6 Add the half-reactions. 2Cu I + 2e 2CuI 3I I 3 + 2e 2Cu I + 2e 2CuI + I 3 + 2e Step 7 Simplify by removing two electrons from both sides. 2Cu I 2CuI + I 3 (balanced) 27. Problem Balance each of the following ionic equations for acidic conditions. Identify the oxidizing agent and the reducing agent in each case. (a) MnO 4 + Ag Mn 2+ + Ag + (b) Hg + NO 3 + Cl HgCl NO 2 (c) AsH 3 + Zn 2+ H 3 AsO 4 + Zn (d) I 3 I + IO 3 What Is Required? You need to write a balanced net ionic equation for the given reaction for acidic conditions, and to identify the oxidizing agent and the reducing agent. What Is Given? You know the formulas of some reactants and products, and that the reaction takes place in acidic solution. 333

17 Plan Your Strategy Follow the steps for balancing a net ionic equation by the half-reaction method for acidic conditions. Identify the oxidizing agent and the reducing agent from the oxidation numbers or the half-reactions. Act on Your Strategy (a) Step 1 The unbalanced net ionic equation is given. MnO 4 + Ag Mn 2+ + Ag + There is no need to assign oxidation numbers, because silver atoms, Ag, are clearly being oxidized to silver ions, Ag +. Therefore, permanganante ions, MnO 4, are being reduced. Write the two unbalanced half-reactions. Oxidation: Ag Ag + Reduction: MnO 4 Mn 2+ Step 3 Balance the two half-reactions for acidic conditions. Oxidation: Ag Ag + + e (balanced) Reduction: MnO 4 Mn 2+ MnO 4 Mn H 2 O MnO 4 + 8H + Mn H 2 O MnO 4 + 8H + + 5e Mn H 2 O (balanced) Step 4 The LCM of 1 and 5 is 5. Step 5 Multiply the oxidation half-reaction by 5, so that equal numbers of electrons are lost and gained. 5Ag 5Ag + + 5e Step 6 Step 7 (b) Step 1 Add the half-reactions. 5Ag 5Ag + + 5e MnO 4 + 8H + + 5e Mn H 2 O 5Ag + MnO 4 + 8H + + 5e 5Ag + + 5e + Mn H 2 O Simplify by removing 5 electrons from both sides. 5Ag + MnO 4 + 8H + 5Ag + + Mn H 2 O (balanced) From the half-reactions, MnO 4 is reduced, so this is the oxidizing agent. From the half-reactions, Ag is oxidized, so this is the reducing agent. The unbalanced net ionic equation is given. Hg + NO 3 + Cl HgCl NO 2 To write the half-reactions, assign oxidation numbers to all the elements. Hg + NO 3 + Cl HgCl NO The oxidation number of mercury increases, so mercury is oxidized. The oxidation number of nitrogen decreases, so nitrate ions are reduced. Write the two unbalanced half-reactions. Oxidation: Hg + Cl 2 HgCl 4 Reduction: NO 3 NO 2 Step 3 Balance the two half-reactions for acidic conditions. Oxidation: Hg + Cl 2 HgCl 4 Hg + 4Cl 2 HgCl 4 Hg + 4Cl HgCl e (balanced) Reduction: NO 3 NO 2 NO 3 NO 2 + H 2 O NO 3 + 2H + NO 2 + H 2 O NO 3 + 2H + + e NO 2 + H 2 O (balanced) Step 4 The LCM of 1 and 2 is

18 Step 5 Step 6 Step 7 Multiply the reduction half-reaction by 2, so that equal numbers of electrons are lost and gained. 2NO 3 + 4H + + 2e 2NO 2 + 2H 2 O Add the half-reactions. Hg + 4Cl HgCl e 2NO 3 + 4H + + 2e 2NO 2 + 2H 2 O Hg + 4Cl + 2NO 3 + 4H + + 2e HgCl e + 2NO 2 + 2H 2 O Simplify by removing 2 electrons from both sides. Hg + 4Cl + 2NO 3 + 4H + HgCl NO 2 + 2H 2 O (balanced) From the oxidation numbers or the half-reactions, NO 3 is reduced, so this is the oxidizing agent. From the oxidation numbers or the half-reactions, Hg is oxidized, so this is the reducing agent. (c) Step 1 The unbalanced net ionic equation is given. AsH 3 + Zn 2+ H 3 AsO 4 + Zn There is no need to assign oxidation numbers, because zinc ions, Zn 2+, are clearly being reduced to zinc atoms, Zn. Therefore, AsH 3 is being oxidized. Write the two unbalanced half-reactions. Reduction: Zn 2+ Zn Oxidation: AsH 3 H 3 AsO 4 Step 3 Balance the two half-reactions for acidic conditions. Reduction: Zn e Zn (balanced) Oxidation: AsH 3 H 3 AsO 4 AsH 3 + 4H 2 O H 3 AsO 4 AsH 3 + 4H 2 O H 3 AsO 4 + 8H + AsH 3 + 4H 2 O H 3 AsO 4 + 8H + + 8e (balanced) Step 4 The LCM of 2 and 8 is 8. Step 5 Multiply the reduction half-reaction by 4, so that equal numbers of electrons are lost and gained. 4Zn e 4Zn Step 6 Add the half-reactions. 4Zn e 4Zn AsH 3 + 4H 2 O H 3 AsO 4 + 8H + + 8e 4Zn e + AsH 3 + 4H 2 O 4Zn + H 3 AsO 4 + 8H + + 8e Step 7 Simplify by removing 8 electrons from both sides. 4Zn 2+ + AsH 3 + 4H 2 O 4Zn + H 3 AsO 4 + 8H + (balanced) From the half-reactions, Zn 2+ is reduced, so this is the oxidizing agent. From the half-reactions, AsH 3 is oxidized, so this is the reducing agent. (d) Step 1 Step 3 The unbalanced net ionic equation is given. I 3 I + IO 3 To write the half-reactions, assign oxidation numbers to both elements. I 3 I + IO 3 1/ Iodine undergoes both oxidation and reduction. The reaction is a disproportionation, and I 3 is a reactant in both half-reactions. Write the two unbalanced half-reactions. Reduction: I 3 I Oxidation: I 3 IO 3 Balance the two half-reactions for acidic conditions. Reduction: I 3 I 335

19 I 3 3I I 3 + 2e 3I (balanced) Oxidation: I 3 IO 3 I 3 3IO 3 I 3 + 9H 2 O 3IO 3 I 3 + 9H 2 O 3IO H + I 3 + 9H 2 O 3IO H e (balanced) Step 4 The LCM of 2 and 16 is 16. Step 5 Multiply the reduction half-reaction by 8, so that equal numbers of electrons are lost and gained. 8I e 24I Step 6 Add the half-reactions. 8I e 24I I 3 + 9H 2 O 3IO H e 8I e + I 3 + 9H 2 O 24I + 3IO H e Step 7 Simplify by adding 8I 3 and I 3 on the left side, and by removing 16 electrons from both sides. 9I e + 9H 2 O 24I + 3IO H e 9I 3 + 9H 2 O 24I + 3IO H + Divide by 3. 3I 3 + 3H 2 O 8I + IO 3 + 6H + From the oxidation numbers or the half-reactions, I 3 is both reduced and oxidized, so it is both the oxidizing agent and the reducing agent. Check Your In each equation, the atoms are balanced and the charges are balanced. 28. Problem Balance each of the following ionic equations for basic conditions. Identify the oxidizing agent and the reducing agent in each case. (a) CN + MnO 4 CNO + MnO 2 (b) H 2 O 2 + ClO 2 ClO 2 + O 2 (c) ClO + CrO 2 CrO Cl 2 (d) Al + NO 2 NH 3 + AlO 2 What Is Required? You need to write a balanced net ionic equation for the given reaction for basic conditions, and to identify the oxidizing agent and the reducing agent. What Is Given? You know the formulas of some reactants and products, and that the reaction takes place in basic solution. Plan Your Strategy Follow the steps for balancing a net ionic equation by the half-reaction method for basic conditions. Identify the oxidizing agent and the reducing agent from the oxidation numbers or the half-reactions. Act on Your Strategy (a) Step 1 The unbalanced net ionic equation is given. CN + MnO 4 CNO + MnO 2 To write the half-reactions, assign oxidation numbers to all the elements. CN + MnO 4 CNO + MnO

20 The oxidation number of carbon increases, so cyanide ions, CN, are oxidized. The oxidation number of manganese decreases, so permanganate ions, MnO 4, are reduced. Write the two unbalanced half-reactions. Oxidation: CN CNO Reduction: MnO 4 MnO 2 Step 3 Balance the two half-reactions as if the conditions are acidic. Oxidation: CN CNO CN + H 2 O CNO CN + H 2 O CNO + 2H + CN + H 2 O CNO + 2H + + 2e (balanced) Reduction: MnO 4 MnO 2 MnO 4 MnO 2 + 2H 2 O MnO 4 + 4H + MnO 2 + 2H 2 O MnO 4 + 4H + + 3e MnO 2 + 2H 2 O (balanced) Step 4 Adjust for basic conditions. Oxidation: CN + H 2 O + 2OH CNO + 2H + + 2e + 2OH CN + H 2 O + 2OH CNO + 2H 2 O + 2e CN + 2OH CNO + H 2 O + 2e Reduction: MnO 4 + 4H + + 3e + 4OH MnO 2 + 2H 2 O + 4OH MnO 4 + 4H 2 O + 3e MnO 2 + 2H 2 O + 4OH MnO 4 + 2H 2 O + 3e MnO 2 + 4OH Step 5 The LCM of 2 and 3 is 6. Step 6 Multiply the oxidation half-reaction by 3, and multiply the reduction half-reaction by 2, so that equal numbers of electrons are lost and gained. 3CN + 6OH 3CNO + 3H 2 O + 6e 2MnO 4 + 4H 2 O + 6e 2MnO 2 + 8OH Step 7 Add the half-reactions. 3CN + 6OH 3CNO + 3H 2 O + 6e 2MnO 4 + 4H 2 O + 6e 2MnO 2 + 8OH 3CN + 6OH + 2MnO 4 + 4H 2 O + 6e 3CNO + 3H 2 O + 6e + 2MnO 2 + 8OH Step 8 Simplify by removing 6 electrons from both sides. 3CN + 6OH + 2MnO 4 + 4H 2 O 3CNO + 3H 2 O + 2MnO 2 + 8OH Step 9 Simplify by removing 3 water molecules and 6 hydroxide ions from both sides. 3CN + 2MnO 4 + H 2 O 3CNO + 2MnO 2 + 2OH (balanced) (Note: An alternative method is to use the LCM to combine the two half-reactions from step 3 to give a balanced net ionic equation for acidic conditions. The balanced net ionic equation is then adjusted for basic conditions. If this method is used, the following equations are obtained. Both methods give the same final result.) Balanced equation (acidic conditions): 3CN + 3H 2 O + 2MnO 4 + 8H + + 6e 3CNO + 6H + + 6e + 2MnO 2 + 4H 2 O Simplify: 3CN + 2MnO 4 + 2H+ 3CNO + 2MnO 2 + H 2 O Adjust for basic conditions: 3CN + 2MnO 4 + 2H + + 2OH 3CNO + 2MnO 2 + H 2 O + 2OH 3CN + 2MnO 4 + 2H 2 O 3CNO + 2MnO 2 + H 2 O + 2OH 3CN + 2MnO 4 + H 2 O 3CNO + 2MnO 2 + 2OH (balanced) 337

21 From the oxidation numbers or the half-reactions, MnO 4 is reduced, so this is the oxidizing agent. From the oxidation numbers or the half-reactions, CN is oxidized, so this is the reducing agent. (b) Step 1 The unbalanced net ionic equation is given. H 2 O 2 + ClO 2 ClO 2 + O 2 To write the half-reactions, assign oxidation numbers to all the elements. H 2 O 2 + ClO 2 ClO 2 + O The oxidation number of oxygen (from H 2 O 2 ) increases, so H 2 O 2 is oxidized. The oxidation number of chlorine decreases, so ClO 2 is reduced. Write the two unbalanced half-reactions. Oxidation: H 2 O 2 O 2 Reduction: ClO 2 ClO 2 Step 3 Balance the two half-reactions as if the conditions are acidic. Oxidation: H 2 O 2 O 2 H 2 O 2 O 2 + 2H + H 2 O 2 O 2 + 2H + + 2e (balanced) Reduction: ClO 2 ClO 2 ClO 2 + e ClO 2 (balanced) Step 4 Adjust for basic conditions. Oxidation: H 2 O 2 O 2 + 2H + + 2e H 2 O 2 + 2OH O 2 + 2H + + 2e + 2OH H 2 O 2 + 2OH O 2 + 2H 2 O + 2e Reduction: ClO 2 + e ClO 2 Step 5 The LCM of 1 and 2 is 2. Step 6 Multiply the reduction half-reaction by 2, so that equal numbers of electrons are lost and gained. 2ClO 2 + 2e 2ClO 2 Step 7 Add the half-reactions. H 2 O 2 + 2OH O 2 + 2H 2 O + 2e 2ClO 2 + 2e 2ClO 2 H 2 O 2 + 2OH + 2ClO 2 + 2e O 2 + 2H 2 O + 2e + 2ClO 2 Step 8 Simplify by removing 2 electrons from both sides. H 2 O 2 + 2OH + 2ClO 2 O 2 + 2H 2 O + 2ClO 2 (balanced) From the oxidation numbers or the half-reactions, ClO 2 is reduced, so this is the oxidizing agent. From the oxidation numbers or the half-reactions, H 2 O 2 is oxidized, so this is the reducing agent. (c) Step 1 The unbalanced net ionic equation is given. ClO + CrO 2 CrO Cl 2 To write the half-reactions, assign oxidation numbers to all the elements. ClO + CrO 2 CrO Cl The oxidation number of chromium increases, so CrO 2 is oxidized. The oxidation number of chlorine decreases, so ClO is reduced. Write the two unbalanced half-reactions. Oxidation: CrO 2 2 CrO 4 Reduction: ClO Cl 2 338

22 Step 3 Balance the two half-reactions as if the conditions are acidic. Oxidation: CrO 2 2 CrO 4 CrO H 2 O CrO 4 CrO 2 + 2H 2 O CrO H + CrO 2 + 2H 2 O CrO H + + 3e (balanced) Reduction: ClO Cl 2 2ClO Cl 2 2ClO Cl 2 + 2H 2 O 2ClO + 4H + Cl 2 + 2H 2 O 2ClO + 4H + + 2e Cl 2 + 2H 2 O (balanced) Step 4 Adjust for basic conditions. Oxidation: CrO 2 + 2H 2 O CrO H + + 3e CrO 2 + 2H 2 O + 4OH CrO H + + 3e + 4OH CrO 2 + 2H 2 O + 4OH CrO H 2 O + 3e CrO 2 + 4OH CrO H 2 O + 3e Reduction: 2ClO + 4H + + 2e Cl 2 + 2H 2 O 2ClO + 4H + + 2e + 4OH Cl 2 + 2H 2 O + 4OH 2ClO + 4H 2 O + 2e Cl 2 + 2H 2 O + 4OH 2ClO + 2H 2 O + 2e Cl 2 + 4OH Step 5 The LCM of 3 and 2 is 6. Step 6 Multiply the oxidation half-reaction by 2, and multiply the reduction halfreaction by 3, so that equal numbers of electrons are lost and gained. 2CrO 2 + 8OH 2CrO H 2 O + 6e 6ClO + 6H 2 O + 6e 3Cl OH Step 7 Add the half-reactions. 2CrO 2 + 8OH 2CrO H 2 O + 6e 6ClO + 6H 2 O + 6e 3Cl OH 2CrO 2 + 8OH + 6ClO + 6H 2 O + 6e 2CrO H 2 O + 6e + 3Cl OH Step 8 Simplify by removing 6 electrons from both sides. 2CrO 2 + 8OH + 6ClO + 6H 2 O 2CrO H 2 O + 3Cl OH Step 9 Simplify by removing 4 water molecules and 8 hydroxide ions from both sides. 2CrO 2 + 6ClO + 2H 2 O 2CrO Cl 2 + 4OH (balanced) From the oxidation numbers or the half-reactions, ClO is reduced, so this is the oxidizing agent. From the oxidation numbers or the half-reactions, CrO 2 is oxidized, so this is the reducing agent. (d) Step 1 The unbalanced net ionic equation is given. Al + NO 2 NH 3 + AlO 2 To write the half-reactions, assign oxidation numbers to all the elements. Al + NO 2 NH 3 + AlO The oxidation number of aluminum increases, so Al is oxidized. The oxidation number of nitrogen decreases, so NO 2 is reduced. Write the two unbalanced half-reactions. Oxidation: Al AlO 2 Reduction: NO 2 NH 3 339

23 Step 3 Balance the two half-reactions as if the conditions are acidic. Oxidation: Al AlO 2 Al + 2H 2 O AlO 2 Al + 2H 2 O AlO 2 + 4H + Al + 2H 2 O AlO 2 + 4H + + 3e (balanced) Reduction: NO 2 NH 3 NO 2 NH 3 + 2H 2 O NO 2 + 7H + NH 3 + 2H 2 O NO 2 + 7H + + 6e NH 3 + 2H 2 O (balanced) Step 4 Adjust for basic conditions. Oxidation: Al + 2H 2 O AlO 2 + 4H + + 3e Al + 2H 2 O + 4OH AlO 2 + 4H + + 3e + 4OH Al + 2H 2 O + 4OH AlO 2 + 4H 2 O + 3e Al + 4OH AlO 2 + 2H 2 O + 3e Reduction: NO 2 + 7H + + 6e NH 3 + 2H 2 O NO 2 + 7H + + 6e + 7OH NH 3 + 2H 2 O + 7OH NO 2 + 7H 2 O + 6e NH 3 + 2H 2 O + 7OH NO 2 + 5H 2 O + 6e NH 3 + 7OH Step 5 The LCM of 3 and 6 is 6. Step 6 Multiply the oxidation half-reaction by 2, so that equal numbers of electrons are lost and gained. 2Al + 8OH 2AlO 2 + 4H 2 O + 6e Step 7 Add the half-reactions. 2Al + 8OH 2AlO 2 + 4H 2 O + 6e NO 2 + 5H 2 O + 6e NH 3 + 7OH 2Al + 8OH + NO 2 + 5H 2 O + 6e 2AlO 2 + 4H 2 O + 6e + NH 3 + 7OH Step 8 Simplify by removing 6 electrons from both sides. 2Al + 8OH + NO 2 + 5H 2 O 2AlO 2 + 4H 2 O + NH 3 + 7OH Step 9 Simplify by removing 4 water molecules and 7 hydroxide ions from both sides. 2Al + OH + NO 2 + H 2 O 2AlO 2 + NH 3 (balanced) From the oxidation numbers or the half-reactions, NO 2 is reduced, so this is the oxidizing agent. From the oxidation numbers or the half-reactions, Al is oxidized, so this is the reducing agent. Check Your In each equation, the atoms are balanced and the charges are balanced. s for Practice Problems Student Textbook page Problem An analyst uses mol/l KMnO 4 to titrate a sample solution of H 2 O 2. The analyst knows that the sample solution is about 6% H 2 O 2 by mass. The analyst places g of H 2 O 2 solution in a flask, dilutes it with water, and adds a small amount of sulfuric acid to acidify it. It takes ml of KMnO 4 solution to reach the endpoint. What mass of pure H 2 O 2 was present. What is the mass percent of pure H 2 O 2 in the original solution? 340

24 What is Required? You need to determine the mass of H 2 O 2 in the sample and express the result as a mass percent. What is Given? You know that the concentration of KMnO 4 is mol/l and its volume is ml. The mass of H 2 O 2 solution is 1.284g. Plan Your Strategy Write the balance equation for the reaction. KMnO 4 can be taken as equivalent to MnO 4 Calculate the mol of MnO 4 added based on the concentration and volume of KMnO 4(aq). Use the mol ratio if the balanced equation to determine the mol of H 2 O 2 that reacts with the MnO 4. Use the equation m = n M to determine the mass of H 2 O 2. Use the mass of H 2 O 2 solution to express the concentration of the H 2 O 2 as mass percent. Act on Your Strategy The balanced equation for this reaction is 2MnO 4 + 5H 2 O 2 + 6H + 2Mn O 2 + 8H 2 O mol MnO 4 = C V = mol/l L = mol mol MnO 4 5 mol H2O2 = mol H 2 O 2 2 mol MnO4 mass H 2 O 2 = n M = mol g/ mol = g mass H2O g mass percent = 100% = 100% = % mass solution g Check Your The mass percent is close to the expected value of 6% and the answer has the correct unite and number of significant figures. 30. Problem A forensic chemist wants to determine the level of alcohol in a sample of blood plasma. The chemist titrates the plasma with a solution of potassium dichromate. The balanced equation is: 16H + + 2Cr 2 O C 2 H 5 OH 4Cr 3+ 2CO H 2 O If ml of mol/l Cr 2 OI 7 2 is required to titrate g of plasma, what is the mass percent alcohol in the plasma? What is Required? You must calculate the mass percent of alcohol in the plasma. What is Given? You know the volume and concentration of the 2Cr 2 O 7 2 solution and the mass of the plasma sample. Plan Your Strategy Use the formula n = C V to find the mol of 2Cr 2 O 7 2 and use the mol ratio from the balanced equation to determine the mol of C 2 H 5 OH that react. Use the formula m = n M to calculate the mass of C 2 H 5 OH. Use this value and the mass of the plasma sample to calculate the mass percent of C 2 H 5 OH. Act on Your Strategy n = C V = mol/l L = mol mol Cr 2 OI 1 mol C2H5OH 7 2 = mol C 2 H 5 OH 2 mol 2Cr O

25 mass of C 2 H 5 OH = mol g/mol = g mass C2H5OH g mass percent of C2H5OH = 100% = 100% = % mass of plasma g Check Your A small mass percent would be expected this answer seems reasonable. The units and number of significant figures are correct. 31. Problem An analyst titrates an acidified solution containing g of purified sodium oxalate, Na 2 C 2 O 4, with potassium permanganate solution, KMnO 4(aq). The purple endpoint is reached when the chemist has added ml of potassium permanganate solution. What is the molar concentration of the potassium permanganate solution? The balanced equation is 2MnO 4 + 5C 2 O H + 2Mn CO 2 + 8H 2 O What is Required? You must find the molar concentration of the titrating solution, potassium permanganate. What is Given? You know the mass of sodium oxalate and can find its molar mass using the periodic table. You also know the volume of potassium permanganate. Plan Your Strategy m Find the molar mass of Na 2 C 2 O 4 and determine the mol of using n =. Use the M mol ratio in the balanced equation to find the mol of KMnO 4 that has reacted. n Calculate the concentration of the KMnO 4 using the formula C =. V Act on Your Strategy M = g/mol g n = = mol Na 2C2O g/mol 2 mol Na 2 C 2 O 4 is the same as mol C 2 O 4 and mol KMnO 4 is the same as mol MnO 4 _ 2 mol MnO mol C2O4 = mol MnO4 2 5 mol C2O4 n mol C = = = mol/l V L Check Your The answer has the correct units and number of significant figures and seems to be reasonable. 32. Problem ml of a solution containing iron(ii) ions was titrated with a mol/l potassium dichromate solution. The endpoint was reached when ml of potassium dichromate solution had been added. What was the molar concentration of iron(ii) ions in the original, acidic solution? The unbalanced equation is: Cr 2 O Fe 2+ Cr 3+ + Fe 3 + What is Required? You must find the molar concentration of the Fe 2+ ions in the original solution. What is Given? You know the volume and concentration of the potassium dichromate solution and the volume of Fe 2+ ions. 342

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