Chapter 18 Oxidation-Reduction Reactions - …hrsbstaff.ednet.ns.ca/benoitn/chem12/electrochemistry/NF...Write balanced half-reactions from the net ionic equation for the reaction

Chapter 18

Oxidation-ReductionReactions

Note to teacher: You will notice that there are two different formats for the SampleProblems in the student textbook. Where appropriate, the Sample Problem contains thefull set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act onYour Strategy, and Check Your Solution. Where a shorter solution is appropriate, theSample Problem contains only two steps: Problem and Solution. Where relevant, a CheckYour Solution step is also included in the shorter Sample Problems.

Solutions for Practice ProblemsStudent Textbook page 715

1. ProblemWrite a balanced net ionic equation for the reaction of zinc with aqueous iron(II)chloride. Include the physical states of the reactants and products.

SolutionWrite a balanced chemical equation for the reaction. Include the physical states.Zn(s) + FeCl2(aq) → Fe(s) + ZnCl2(aq)

Write a balanced total ionic equation for the reaction.Zn(s) + Fe2+

(aq) + 2Cl−(aq) → Fe(s) + Zn2+(aq) + 2Cl−(aq)

Eliminate the spectator ions (the chloride ions) to write the net ionic equation.Zn(s) + Fe2+

(aq) → Fe(s) + Zn2+(aq)

2. ProblemWrite 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

Solution(a) Write a balanced chemical equation for the reaction. Include the physical states.

Mg(s) + Al2(SO4)3(aq) → Al(s) + MgSO4(aq) (unbalanced)3Mg(s) + Al2(SO4)3(aq) → 2Al(s) + 3MgSO4(aq) (balanced)Write a balanced total ionic equation for the reaction.3Mg(s) + 2Al3+

(aq) + 3SO42−

(aq) → 2Al(s) + 3Mg2+(aq) + 3SO4

2−(aq)

Eliminate the spectator ions (the sulfate ions) to write the net ionic equation.3Mg(s) + 2Al3+

(aq) → 2Al(s) + 3Mg2+(aq)

(b) Write a balanced chemical equation for the reaction. Include the physical states.AgNO3(aq) + Cd(s) → Ag(s) + Cd(NO3)2(aq) (unbalanced)2AgNO3(aq) + Cd(s) → 2Ag(s) + Cd(NO3)2(aq) (balanced)Write a balanced total ionic equation for the reaction.2Ag+

(aq) + 2NO3−

(aq) + Cd(s) → 2Ag(s) + Cd2+(aq) + 2NO3

−(aq)

Eliminate the spectator ions (the nitrate ions) to write the net ionic equation.2Ag+

(aq) + Cd(s) → 2Ag(s) + Cd2+(aq)

318Chapter 18 Oxidation-Reduction Reactions • MHR

3. ProblemIdentify the reactant oxidized and the reactant reduced in each reaction in question 2.

Solution(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 toform a magnesium ion, and that each aluminum ion gains three electrons to forman aluminum atom.3Mg(s) + 2Al3+

(aq) → 2Al(s) + 3Mg2+(aq)

Therefore, magnesium is oxidized, and aluminum ions are reduced.

(b) The reactant oxidized is the reactant that loses electrons. The reactant reduced isthe 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) + Cd2+(aq)

Therefore, cadmium is oxidized, and silver ions are reduced.

4. ProblemIdentify the oxidizing agent and the reducing agent in each reaction in question 2.

Solution(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 toform an aluminum atom, and that each magnesium atom donates two electrons toform a magnesium ion. 3Mg(s) + 2Al3+

(aq) → 2Al(s) + 3Mg2+(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 isthe 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) + Cd2+(aq)

Therefore, silver ions are the oxidizing agent, and cadmium is the reducing agent.


5. ProblemWrite balanced half-reactions from the net ionic equation for the reaction betweensolid aluminum and aqueous iron(III) sulfate. The sulfate ions are spectator ions andare not included.Al(s) + Fe3+

(aq) → Al3+(aq) + Fe(s)

SolutionThe net ionic equation shows that each aluminum atom loses three electrons to forman aluminum ion, and that each iron(III) ion gains three electrons to form an ironatom. Therefore, the two half-reactions are as follows.


Oxidation: Al(s) → Al3+(aq) + 3e−

Reduction: Fe3+(aq) + 3e− → Fe(s)

6. ProblemWrite balanced half-reactions from the following net ionic equations.(a) Fe(s) + Cu2+

(aq) → Fe2+(aq) + Cu(s)

(b) Cd(s) + 2Ag+(aq) → Cd2+

(aq) + 2Ag(s)

Solution(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 copperatom. Therefore, the two half-reactions are as follows.Oxidation: Fe(s) → Fe2+

(aq) + 2e−

Reduction: Cu2+(aq) + 2e− → Cu(s)

(b) The net ionic equation shows that each cadmium atom loses two electrons toform a cadmium ion, and that each silver ion gains one electron to form a silveratom. Therefore, the two half-reactions are as follows.Oxidation: Cd(s) → Cd2+

(aq) + 2e−

Reduction: Ag+(aq) + 1e− → Ag(s)

7. ProblemWrite balanced half-reactions for each of the following reactions.(a) Sn(s) + PbCl2(aq) → SnCl2(aq) + Pb(s)

(b) Au(NO3)3(aq) + 3Ag(s) → 3AgNO3(aq) + Au(s)

(c) 3Zn(s) + Fe2(SO4)3(aq) → 3ZnSO4(aq) + 2Fe(s)

Solution(a) Write a balanced total ionic equation for the reaction.

Sn(s) + Pb2+(aq) + 2Cl−(aq) → Sn2+

(aq) + 2Cl−(aq) + Pb(s)

Eliminate the spectator ions (the chloride ions) to write the net ionic equation.Sn(s) + Pb2+

(aq) → Sn2+(aq) + Pb(s)

The net ionic equation shows that each tin atom loses two electrons to form atin(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) → Sn2+

(aq) + 2e−

Reduction: Pb2+(aq) + 2e− → Pb(s)

(b) Write a balanced total ionic equation for the reaction.Au3+

(aq) + 3NO3−

(aq) + 3Ag(s) → 3Ag+(aq) + 3NO3

−(aq) + Au(s)

Eliminate the spectator ions (the nitrate ions) to write the net ionic equation.Au3+

(aq) + 3Ag(s) → 3Ag+(aq) + Au(s)

The net ionic equation shows that each gold(III) ion gains three electrons to forma 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: Au3+(aq) + 3e− → Au(s)

(c) Write a balanced total ionic equation for the reaction.3Zn(s) + 2Fe3+

(aq) + 3SO42−

(aq) → 3Zn2+(aq) + 3SO4

2−(aq) + 2Fe(s)

Eliminate the spectator ions (the sulfate ions) to write the net ionic equation.3Zn(s) + 2Fe3+

(aq) → 3Zn2+(aq) + 2Fe(s)

The net ionic equation shows that each zinc atom loses two electrons to form azinc ion, and that each iron(III) ion gains three electrons to form an iron atom.Therefore, the two half-reactions are as follows.


Oxidation: Zn(s) → Zn2+(aq) + 2e−

Reduction: Fe3+(aq) + 3e− → Fe(s)

8. ProblemWrite 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 Hg2

2+.

SolutionWrite a balanced net ionic equation for the reaction.Hg2

2+(aq) → Hg(�) + Hg2+

(aq)

The net ionic equation shows that the reaction is a disproportionation in which equalnumbers 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) ionsgain electrons to form mercury atoms. Because only one Hg2

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 2.2Hg2

2+(aq) → 2Hg(�) + 2Hg2+

(aq)

The two half-reactions are as follows.Reduction: Hg2

2+(aq) + 2e− → 2Hg(�)

Oxidation: Hg22+

(aq) → 2Hg2+(aq) + 2e−

Note: For simplicity, the formula of mercury(I) ions is sometimes written as Hg+,even though these ions exist in the form Hg2

2+. If the formula of mercury(I) ions is written as Hg+, the balanced net ionic equation and the half-reactions are writtenas follows. 2Hg+

(aq) → Hg(�) + Hg2+(aq)

Reduction: Hg+(aq) + e− → Hg(�)

Oxidation: Hg+(aq) → Hg2+

(aq) + e−


9. ProblemDetermine the oxidation number of the specified element in each of the following.(a) N in NF3 (c) Cr in CrO4

2− (e) C in C12H22O11

(b) S in S8 (d) P in P2O5 (f) C in CHCl3

Solution(a) • Because the compound NF3 does not contain hydrogen or oxygen, rule 5

applies. Because NF3 is a compound, rule 6 also applies.• Nitrogen has an electronegativity of 3.04. Fluorine has an electronegativity of

3.98. 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 = 0x = 3

The oxidation number of nitrogen is +3.

(b) • Because S8 is the formula of a pure element, rule 1 applies. • From rule 1, the oxidation number of S in S8 is 0.

(c) • Because the polyatomic ion CrO42− contains oxygen, rule 4 applies. Because

CrO42− is a polyatomic ion, rule 7 also applies.


• 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 = −2x = 6

The oxidation number of chromium is +6.

(d) • Because the compound P2O5 contains oxygen, rule 4 applies. Because P2O5 is acompound, 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 = 02x = 10x = 5

The oxidation number of phosphorus is +5.

(e) • Because the compound C12H22O11 contains hydrogen and oxygen, rules 3 and4 apply. Because C12H22O11 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 + 22 − 22 = 012x = 0

x = 0The oxidation number of carbon is 0.

(f) • Because the compound CHCl3 contains hydrogen, rule 3 applies. BecauseCHCl3 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 2.55. Chlorine has an electronegativity of 3.16.Therefore, using the same reasoning as in rule 5, assign chlorine an oxidationnumber of −1.

• The oxidation number of carbon is unknown, so let it be x. From rule 6, thesum of the oxidation numbers is 0. Then,x + 1(+1) + 3(−1) = 0

x + 1 − 3 = 0x − 2 = 0

x = 2The oxidation number of carbon is +2.

10. ProblemDetermine the oxidation number of each element in each of the following.(a) H2SO3 (b) OH− (c) HPO4

2−

Solution(a) • Because the compound H2SO3 contains hydrogen and oxygen, rules 3 and 4

apply. Because H2SO3 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 sulfur is unknown, so let it be x. From rule 6, thesum of the oxidation numbers is 0. Then,2(+1) + x + 3(−2) = 0

2 + x − 6 = 0x − 4 = 0

x = 4The 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 4apply. 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) = −1−2 + 1 = −1

The oxidation number of oxygen is −2. The oxidation number of hydrogen is +1.

(c) • Because the polyatomic ion HPO42− contains hydrogen and oxygen, rules 3

and 4 apply. Because HPO42− 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 = −2x − 7 = −2

x = 5The oxidation number of hydrogen is +1. The oxidation number of phosphorus is +5. The oxidation number of oxygen is −2.

11. ProblemAs 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, OF2

(b) the peroxide ion, O22−

Solution(a) • Oxygen has an electronegativity of 3.44. Fluorine has an electronegativity

of 3.98. Therefore, using the same reasoning as in rule 5, assign fluorine an oxidation number of −1.

• Because OF2 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 = 0x = 2

The oxidation number of oxygen is +2.

(b) • Because O22− is a polyatomic ion, rule 7 applies.

• The oxidation number of oxygen is unknown, so let it be x. From rule 7, thesum of the oxidation numbers is −2. Then,2x = −2x = −1

The oxidation number of oxygen is −1.


12. ProblemDetermine the oxidation number of each element in each of the following ionic com-pounds by considering the ions separately. Hint: One formula unit of the compoundin part (c) contains two identical monatomic ions and one polyatomic ion.(a) Al(HCO3)3 (b) (NH4)3PO4 (c) K2H3IO6

Solution(a) • The compound Al(HCO3)3 contains Al3+ ions and HCO3

− ions. • Because Al3+ is a monatomic ion, rule 2 applies. From rule 2, the oxidation

number of aluminum is +3.• Because the polyatomic ion HCO3

− contains hydrogen and oxygen, rules 3 and4 apply. Because HCO3

− 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 HCO3− is −1. Then,

1(+1) + x + 3(−2) = −11 + x − 6 = −1

x − 5 = −1x = 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 (NH4)3PO4 contains NH4+ ions and PO4

3− ions. • Because the polyatomic ion NH4

+ contains hydrogen, rule 3 applies. BecauseNH4

+ 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 NH4+ is +1. Then,

x + 4(+1) = +1x + 4 = +1

x = +1 − 4x = −3

• Because the polyatomic ion PO43− contains oxygen, rule 4 applies. Because

PO43− 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 PO43− is −3. Then,

y + 4(−2) = −3y − 8 = −3

y = −3 + 8y = 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 ofoxygen is −2.

(c) • The compound K2H3IO6 contains K+ ions and H3IO62− ions. Because K+

is a monatomic ion, rule 2 applies. From rule 2, the oxidation number of potassium is +1.

• Because the polyatomic ion H3IO62− contains hydrogen and oxygen, rules 3

and 4 apply. Because H3IO62− 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 −2.


• The oxidation number of iodine is unknown, so let it be x. From rule 7, thesum of the oxidation numbers in H3IO6

2− is −2. Then,3(+1) + x + 6(−2) = −2

3 + x − 12 = −2x − 9 = −2

x = 7The oxidation number of potassium is +1. The oxidation number of hydrogenis +1. The oxidation number of iodine is +7. The oxidation number of oxygenis −2.


13. ProblemDetermine whether each reaction is a redox reaction.(a) H2O2 + 2Fe(OH)2 → 2Fe(OH)3

(b) PCl3 + 3H2O → H3PO3 + 3HCl

SolutionFind the oxidation number of each element in the reactants and products. Identifyany elements that undergo an increase or a decrease in oxidation number during the reaction.(a) H2O2 + 2Fe(OH)2 → 2Fe(OH)3

+1 −1 +2 −2 +1 +3 −2 +1

• The oxidation number of hydrogen is +1 on both sides of the equation, sohydrogen is neither oxidized nor reduced.

• The oxidation number of the oxygen atoms that originate in Fe(OH)2 is −2 onboth sides of the equation.

• The oxidation number of the oxygen atoms that originate in H2O2 decreasesfrom −1 to −2.

• The oxidation number of iron increases from +2 to +3.Because both oxygen and iron undergo changes in oxidation number, the reactionis a redox reaction.

(b) PCl3 + 3H2O → H3PO3 + 3HCl+3 −1 +1 −2 +1 +3 −2 +1 −1

No elements undergo changes in oxidation number, so the reaction is not a redox reaction.

14. ProblemIdentify the oxidizing agent and the reducing agent for the redox reaction(s) in theprevious question.

SolutionIn part (a) of the previous question,• The oxidation number of the oxygen atoms that originate in H2O2 decreases from

−1 to −2, so H2O2 is reduced. Therefore, H2O2 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 Fe2+, is the reducing agent.


15. ProblemFor the following balanced net ionic equation, identify the reactant that undergoesoxidation and the reactant that undergoes reduction.Br2 + 2ClO2

− → 2Br− + 2ClO2

SolutionFind the oxidation number of each element in the reactants and products. Identifyany elements that undergo an increase or a decrease in oxidation number during the reaction.Br2 + 2ClO2

− → 2Br− + 2ClO20 +3 −2 −1 +4 −2

• 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 ClO2− ions. Therefore, ClO2

− ions undergo oxidation.

• The oxidation number of bromine decreases from 0 to −1. The bromine atoms onthe reactant side are found in elemental bromine, Br2. Therefore, elementalbromine undergoes reduction.

16. ProblemNickel and copper are two metals that have played a role in the economy ofNewfoundland and Labrador. Nickel and copper ores usually contain the metals as sulfides, such as NiS and Cu2S. Do the extractions of these pure elemental metalsfrom their ores involve redox reactions? Explain your reasoning.

Solution• In NiS, nickel has an oxidation number of +2. Metallic nickel, Ni, is an element

with an oxidation number of 0.• In Cu2S, 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 reductionalways occur together, so the extraction processes must be redox reactions. (We donot need to know the other reactants to know that this conclusion is true.)


17. ProblemWrite a balanced half-reaction for the reduction of cerium(IV) ions to cerium(III) ions.

Solution• Represent the given reactant and product with the correct formulas.

Ce4+ → Ce3+

• Balance the atoms, if necessary. The atoms are already balanced.• Add an electron to the left side to balance the charges.

Ce4+ + e− → Ce3+


18. ProblemWrite a balanced half-reaction for the oxidation of bromide ions to bromine.


Br− → Br2

• Balance the bromine atoms.2Br− → Br2

• Add two electrons to the right side to balance the charges.2Br− → Br2 + 2e−

19. ProblemBalance each of the following half-reactions under acidic conditions.(a) O2 → H2O2 (b) H2O → O2 (c) NO3

− → N2

Solution(a) Step 1 The unbalanced half-reaction, including the correct formulas, is given.

O2 → H2O2

Step 2 There are no atoms to balance other than oxygen and hydrogen.Step 3 Add water molecules to balance the oxygen atoms, if necessary.

The oxygen atoms are already balanced.Step 4 The reaction occurs in acidic solution, so add hydrogen ions to balance

the hydrogen atoms.O2 + 2H+ → H2O2

Step 5 Add two electrons to the left side to balance the charges.O2 + 2H+ + 2e− → H2O2

(b) Step 1 The unbalanced half-reaction, including the correct formulas, is given.H2O → O2

Step 2 There are no atoms to balance other than oxygen and hydrogen.Step 3 The reaction occurs in aqueous solution, so add another water molecule

to balance the oxygen atoms.2H2O → O2

Step 4 The reaction occurs in acidic solution, so add hydrogen ions to balancethe hydrogen atoms.2H2O → O2 + 4H+

Step 5 Add four electrons to the right side to balance the charges.2H2O → O2 + 4H+ + 4e−

(c) Step 1 The unbalanced half-reaction, including the correct formulas, is given.NO3

− → N2

Step 2 Balance the nitrogen atoms.2NO3

− → N2

Step 3 The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms.2NO3

− → N2 + 6H2OStep 4 The reaction occurs in acidic solution, so add hydrogen ions to balance

the hydrogen atoms.2NO3

− + 12H+ → N2 + 6H2OStep 5 Add ten electrons to the left side to balance the charges.

2NO3− + 12H+ + 10e− → N2 + 6H2O


20. ProblemBalance each of the following half-reactions under acidic conditions.(a) ClO3

− → Cl− (b) NO → NO−3 (c) Cr2O7

2− → Cr3+


ClO3− → Cl−

Step 2 Balance the chlorine atoms, if necessary. The chlorine atoms are already balanced.

Step 3 The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms.ClO3

− → Cl− + 3H2OStep 4 The reaction occurs in acidic solution, so add hydrogen ions to balance

the hydrogen atoms.ClO3

− + 6H+ → Cl− + 3H2OStep 5 Add six electrons to the left side to balance the charges.

ClO3− + 6H+ + 6e− → Cl− + 3H2O

(b) Step 1 The unbalanced half-reaction, including the correct formulas, is given.NO → NO3

−

Step 2 Balance the nitrogen atoms, if necessary. The nitrogen atoms are already balanced.

Step 3 The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms.NO + 2H2O → NO3

−

Step 4 The reaction occurs in acidic solution, so add hydrogen ions to balancethe hydrogen atoms.NO + 2H2O → NO3

− + 4H+

Step 5 Add three electrons to the right side to balance the charges.NO + 2H2O → NO3

− + 4H+ + 3e−

(c) Step 1 The unbalanced half-reaction, including the correct formulas, is given.Cr2O7

2− → Cr3+

Step 2 Balance the chromium atoms.Cr2O7

2− → 2Cr3+

Step 3 The reaction occurs in aqueous solution, so add water molecules to balance the oxygen atoms.Cr2O7

2− → 2Cr3+ + 7H2OStep 4 The reaction occurs in acidic solution, so add hydrogen ions to balance

the hydrogen atoms.Cr2O7

2− + 14H+ → 2Cr3+ + 7H2OStep 5 Add six electrons to the left side to balance the charges.

Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O



21. ProblemWrite a balanced half-reaction for the oxidation of chromium(II) ions tochromium(III) ions.


Cr2+ → Cr3+

• Balance the atoms, if necessary. The atoms are already balanced.• Add an electron to the right side to balance the charges.

Cr2+ → Cr3+ + e−

22. ProblemWrite a balanced half-reaction for the reduction of oxygen to oxide ions.


O2 → O2−

• Balance the oxygen atoms.O2 → 2O2−

• Add four electrons to the left side to balance the charges.O2 + 4e− → 2O2−

23. ProblemBalance each of the following half-reactions under basic conditions.(a) Al → Al(OH)4

− (d) CrO42− → Cr(OH)3

(b) CN− → CNO− (e) CO32− → C2O4

2−

(c) MnO4− → MnO2


Al → Al(OH)4−

Step 2 Balance the aluminum atoms, if necessary. The aluminum atoms arealready balanced.

Step 3 Balance the oxygen and hydrogen atoms as if the conditions are acidic.Add water molecules to balance the oxygen atoms.Al + 4H2O → Al(OH)4

−

Add hydrogen ions to balance the hydrogen atoms.Al + 4H2O → Al(OH)4

− + 4H+

Step 4 Adjust for basic conditions by adding four hydroxide ions to each side.Al + 4H2O + 4OH− → Al(OH)4

− + 4H+ + 4OH−

Step 5 Combine the hydrogen ions and hydroxide ions on the right side intowater molecules.Al + 4H2O + 4OH− → Al(OH)4

− + 4H2OStep 6 Remove four water molecules from each side.

Al + 4OH− → Al(OH)4−

Step 7 Add three electrons to the right side to balance the charges.Al + 4OH− → Al(OH)4

− + 3e−

(b) Step 1 The unbalanced half-reaction, including the correct formulas, is given.CN− → CNO−

Step 2 Balance the carbon and nitrogen atoms, if necessary. These atoms arealready balanced.


Step 3 Balance the oxygen and hydrogen atoms as if the conditions are acidic.Add water molecules to balance the oxygen atoms.CN− + H2O → CNO−

Add hydrogen ions to balance the hydrogen atoms.CN− + H2O → CNO− + 2H+

Step 4 Adjust for basic conditions by adding two hydroxide ions to each side.CN− + H2O + 2OH− → CNO− + 2H+ + 2OH−

Step 5 Combine the hydrogen ions and hydroxide ions on the right side intowater molecules.CN− + H2O + 2OH− → CNO− + 2H2O

Step 6 Remove a water molecule from each side.CN− + 2OH− → CNO− + H2O

Step 7 Add two electrons to the right side to balance the charges.CN− + 2OH− → CNO− + H2O + 2e−

(c) Step 1 The unbalanced half-reaction, including the correct formulas, is given.MnO4

− → MnO2

Step 2 Balance the manganese atoms, if necessary. The manganese atoms arealready balanced.

Step 3 Balance the oxygen and hydrogen atoms as if the conditions are acidic.Add water molecules to balance the oxygen atoms.MnO4

− → MnO2 + 2H2OAdd hydrogen ions to balance the hydrogen atoms.MnO4

− + 4H+ → MnO2 + 2H2OStep 4 Adjust for basic conditions by adding four hydroxide ions to each side.

MnO4− + 4H+ + 4OH− → MnO2 + 2H2O + 4OH−

Step 5 Combine the hydrogen ions and hydroxide ions on the left side intowater molecules.MnO4

− + 4H2O → MnO2 + 2H2O + 4OH−

Step 6 Remove two water molecules from each side.MnO4

− + 2H2O → MnO2 + 4OH−

Step 7 Add three electrons to the left side to balance the charges.MnO4

− + 2H2O + 3e− → MnO2 + 4OH−

(d) Step 1 The unbalanced half-reaction, including the correct formulas, is given.CrO4

2− → Cr(OH)3

Step 2 Balance the chromium atoms, if necessary. The chromium atoms arealready balanced.

Step 3 Balance the oxygen and hydrogen atoms as if the conditions are acidic.Add water molecules to balance the oxygen atoms.CrO4

2− → Cr(OH)3 + H2OAdd hydrogen ions to balance the hydrogen atoms.CrO4

2− + 5H+ → Cr(OH)3 + H2OStep 4 Adjust for basic conditions by adding five hydroxide ions to each side.

CrO42− + 5H+ + 5OH− → Cr(OH)3 + H2O + 5OH−

Step 5 Combine the hydrogen ions and hydroxide ions on the left side intowater molecules.CrO4

2− + 5H2O → Cr(OH)3 + H2O + 5OH−

Step 6 Remove one water molecule from each side.CrO4

2− + 4H2O → Cr(OH)3 + 5OH−

Step 7 Add three electrons to the left side to balance the charges.CrO4

2− + 4H2O + 3e− → Cr(OH)3 + 5OH−


(e) Step 1 The unbalanced half-reaction, including the correct formulas, is given.CO3

2− → C2O42−

Step 2 Balance the carbon atoms.2CO3

2− → C2O42−

Step 3 Balance the oxygen and hydrogen atoms as if the conditions are acidic.Add water molecules to balance the oxygen atoms.2CO3

2− → C2O42− + 2H2O

Add hydrogen ions to balance the hydrogen atoms.2CO3

2− + 4H+ → C2O42− + 2H2O

Step 4 Adjust for basic conditions by adding four hydroxide ions to each side.2CO3

2− + 4H+ + 4OH− → C2O42− + 2H2O + 4OH−

Step 5 Combine the hydrogen ions and hydroxide ions on the left side intowater molecules.2CO3

2− + 4H2O → C2O42− + 2H2O + 4OH−

Step 6 Remove two water molecules from each side.2CO3

2− + 2H2O → C2O42− + 4OH−

Step 7 Add two electrons to the left side to balance the charges.2CO3

2− + 2H2O + 2e− → C2O42− + 4OH−

24. ProblemBalance each of the following half-reactions.(a) FeO4

2− → Fe3+ (acidic conditions)(b) ClO2

− → Cl− (basic conditions)


FeO42− → Fe3+

Step 2 Balance the iron atoms, if necessary. The iron atoms are balanced.Step 3 Add water molecules to balance the oxygen atoms.

FeO42− → Fe3+ + 4H2O

Step 4 Add hydrogen ions to balance the hydrogen atoms.FeO4

2− + 8H+ → Fe3+ + 4H2OStep 5 Add three electrons to the left side to balance the charges.

FeO42− + 8H+ + 3e− → Fe3+ + 4H2O

(b) Step 1 The unbalanced half-reaction, including the correct formulas, is given.ClO2

− → Cl−

Step 2 Balance the chlorine atoms, if necessary. The chlorine atoms are balanced.ClO2

− → Cl−

Step 3 Balance the oxygen and hydrogen atoms as if the conditions are acidic.Add water molecules to balance the oxygen atoms.ClO2

− → Cl− + 2H2OAdd hydrogen ions to balance the hydrogen atoms.ClO2

− + 4H+ → Cl− + 2H2OStep 4 Adjust for basic conditions by adding four hydroxide ions to each side.

ClO2− + 4H+ + 4OH− → Cl− + 2H2O + 4OH−

Step 5 Combine the hydrogen ions and hydroxide ions on the left side intowater molecules.ClO2

− + 4H2O → Cl− + 2H2O + 4OH−

Step 6 Remove two water molecules from each side.ClO2

− + 2H2O → Cl− + 4OH−

Step 7 Add four electrons to the left side to balance the charges.ClO2

− + 2H2O + 4e− → Cl− + 4OH−


Solutions for Practice ProblemsStudent Textbook pages 738–739

25. ProblemBalance each of the following redox equations by inspection. Write the balanced half-reactions in each case.(a) Na + F2 → NaF (b) Mg + N2 → Mg3N2 (c) HgO → Hg + O2

Solution(a) • Balance the fluorine atoms.

Na + F2 → 2NaF• Balance the sodium atoms.

2Na + F2 → 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 toform 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: F2 → F−

F2 → 2F−

F2 + 2e− → 2F− (balanced)

(b) • Balance the magnesium atoms.3Mg + N2 → Mg3N2

• Mg3N2 contains Mg2+ ions and N3− ions. Therefore, each magnesium atomloses two electrons to form a magnesium ion, and each nitrogen atom gainsthree 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 → Mg2+

Mg → Mg2+ + 2e− (balanced)Reduction half-reaction: N2 → N3−

N2 → 2N3−

N2 + 6e− → 2N3− (balanced)

(c) • Balance the oxygen atoms.2HgO → Hg + O2

• Balance the mercury atoms.2HgO → 2Hg + O2 (balanced)

• HgO contains Hg2+ ions and O2− ions. Therefore, each mercury(II) ion gainstwo electrons to form a mercury atom, and each oxide ion loses two electronsto 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.


Reduction half-reaction:Hg2+ → HgHg2+ + 2e− → Hg (balanced)Oxidation half-reaction:O2− → O2

2O2− → O2

2O2− → O2 + 4e− (balanced)

26. ProblemBalance the following equation by the half-reaction method.Cu2+ + I− → CuI + I3

−

SolutionStep 1 The unbalanced net ionic equation is given.

Cu2+ + I− → CuI + I3−

(Note: CuI is not written in ionic form because the electronegativity difference between copper and iodine is quite small (0.76). This compoundis covalent.)

Step 2 Divide the unbalanced net ionic equation into two half-reactions. Assign oxidation numbers to all the elements, if necessary. If you notice that theoxidation number of copper decreases from +2 to +1, there is no need toassign other oxidation numbers. Copper must be involved in the reductionhalf-reaction.Reduction half-reaction: Cu2− + I− → CuIOxidation half-reaction: I− → I3

−

Step 3 Balance the half-reactions independently.Reduction: Cu2+ + I− + e− → CuI (balanced)Oxidation: 3I− → I3

−

3I− → I3− + 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.2Cu2+ + 2I− + 2e− → 2CuI

Step 6 Add the half-reactions.2Cu2+ + 2I− + 2e− → 2CuI

3I− → I3− + 2e−

2Cu2+ + 5I− + 2e− → 2CuI + I3− + 2e−

Step 7 Simplify by removing two electrons from both sides.2Cu2+ + 5I− → 2CuI + I3

− (balanced)

27. ProblemBalance each of the following ionic equations for acidic conditions. Identify the oxidizing agent and the reducing agent in each case.(a) MnO4

− + Ag → Mn2+ + Ag+

(b) Hg + NO3− + Cl− → HgCl42− + NO2

(c) AsH3 + Zn2+ → H3AsO4 + Zn(d) I3

− → I− + IO3−

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 takesplace in acidic solution.


Plan Your StrategyFollow the steps for balancing a net ionic equation by the half-reaction method foracidic conditions.Identify the oxidizing agent and the reducing agent from the oxidation numbers orthe half-reactions.

Act on Your Strategy(a) Step 1 The unbalanced net ionic equation is given.

MnO4− + Ag → Mn2+ + Ag+

Step 2 There is no need to assign oxidation numbers, because silver atoms, Ag,are clearly being oxidized to silver ions, Ag+. Therefore, permangananteions, MnO4

−, are being reduced.Write the two unbalanced half-reactions.Oxidation: Ag → Ag+

Reduction: MnO4− → Mn2+

Step 3 Balance the two half-reactions for acidic conditions.Oxidation: Ag → Ag+ + e− (balanced)Reduction: MnO4

− → Mn2+

MnO4− → Mn2+ + 4H2O

MnO4− + 8H+ → Mn2+ + 4H2O

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (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 Add the half-reactions.5Ag → 5Ag+ + 5e−

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O

5Ag + MnO4− + 8H+ + 5e− → 5Ag+ + 5e− + Mn2+ + 4H2O

Step 7 Simplify by removing 5 electrons from both sides.5Ag + MnO4

− + 8H+ → 5Ag+ + Mn2+ + 4H2O (balanced)From the half-reactions, MnO4

− is reduced, so this is the oxidizing agent.From the half-reactions, Ag is oxidized, so this is the reducing agent.

(b) Step 1 The unbalanced net ionic equation is given.Hg + NO3

− + Cl− → HgCl42− + NO2

Step 2 To write the half-reactions, assign oxidation numbers to all the elements.Hg + NO3

− + Cl− → HgCl42− + NO2

0 +5 −2 −1 +2 −1 +4 −2

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− → HgCl42−

Reduction: NO3− → NO2

Step 3 Balance the two half-reactions for acidic conditions.Oxidation: Hg + Cl− → HgCl42−

Hg + 4Cl− → HgCl42−

Hg + 4Cl− → HgCl42− + 2e− (balanced)Reduction: NO3

− → NO2

NO3− → NO2 + H2O

NO3− + 2H+ → NO2 + H2O

NO3− + 2H+ + e− → NO2 + H2O (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.2NO3

− + 4H+ + 2e− → 2NO2 + 2H2OStep 6 Add the half-reactions.

Hg + 4Cl− → HgCl42− + 2e−

2NO3− + 4H+ + 2e− → 2NO2 + 2H2O

Hg + 4Cl− + 2NO3− + 4H+ + 2e− → HgCl4

2− + 2e− + 2NO2 + 2H2OStep 7 Simplify by removing 2 electrons from both sides.

Hg + 4Cl− + 2NO3− + 4H+ → HgCl42− + 2NO2 + 2H2O (balanced)

From the oxidation numbers or the half-reactions, NO3− is reduced, so

this is the oxidizing agent.From the oxidation numbers or the half-reactions, Hg is oxidized, so thisis the reducing agent.

(c) Step 1 The unbalanced net ionic equation is given.AsH3 + Zn2+ → H3AsO4 + Zn

Step 2 There is no need to assign oxidation numbers, because zinc ions, Zn2+,are clearly being reduced to zinc atoms, Zn. Therefore, AsH3 is being oxidized.Write the two unbalanced half-reactions.Reduction: Zn2+ → ZnOxidation: AsH3 → H3AsO4

Step 3 Balance the two half-reactions for acidic conditions.Reduction: Zn2+ + 2e− → Zn (balanced)Oxidation: AsH3 → H3AsO4

AsH3 + 4H2O → H3AsO4

AsH3 + 4H2O → H3AsO4 + 8H+

AsH3 + 4H2O → H3AsO4 + 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.4Zn2+ + 8e− → 4Zn

Step 6 Add the half-reactions.4Zn2+ + 8e− → 4Zn

AsH3 + 4H2O → H3AsO4 + 8H+ + 8e−

4Zn2+ + 8e− + AsH3 + 4H2O → 4Zn + H3AsO4 + 8H+ + 8e−

Step 7 Simplify by removing 8 electrons from both sides.4Zn2+ + AsH3 + 4H2O → 4Zn + H3AsO4 + 8H+ (balanced)From the half-reactions, Zn2+ is reduced, so this is the oxidizing agent.From the half-reactions, AsH3 is oxidized, so this is the reducing agent.

(d) Step 1 The unbalanced net ionic equation is given.I3

− → I− + IO3−

Step 2 To write the half-reactions, assign oxidation numbers to both elements.I3

− → I− + IO3−

−1/3 −1 +5 −2

Iodine undergoes both oxidation and reduction. The reaction is a disproportionation, and I3

− is a reactant in both half-reactions. Write the two unbalanced half-reactions.Reduction: I3

− → I−

Oxidation: I3− → IO3

−

Step 3 Balance the two half-reactions for acidic conditions.Reduction: I3

− → I−


I3− → 3I−

I3− + 2e− → 3I− (balanced)

Oxidation: I3− → IO3

−

I3− → 3IO3

−

I3− + 9H2O → 3IO3

−

I3− + 9H2O → 3IO3

− + 18H+

I3− + 9H2O → 3IO3

− + 18H+ + 16e− (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.8I3

− + 16e− → 24I−

Step 6 Add the half-reactions.8I3

− + 16e− → 24I−

I3− + 9H2O → 3IO3

− + 18H+ + 16e−

8I3− + 16e− + I3

− + 9H2O → 24I− + 3IO3− + 18H+ + 16e−

Step 7 Simplify by adding 8I3− and I3

− on the left side, and by removing 16 electrons from both sides.9I3

− + 16e− + 9H2O → 24I− + 3IO3− + 18H+ + 16e−

9I3− + 9H2O → 24I− + 3IO3

− + 18H+

Divide by 3.3I3

− + 3H2O → 8I− + IO3− + 6H+

From the oxidation numbers or the half-reactions, I3− is both reduced

and oxidized, so it is both the oxidizing agent and the reducing agent.

Check Your SolutionIn each equation, the atoms are balanced and the charges are balanced.

28. ProblemBalance each of the following ionic equations for basic conditions. Identify the oxidizing agent and the reducing agent in each case.(a) CN− + MnO4

− → CNO− + MnO2

(b) H2O2 + ClO2 → ClO2− + O2

(c) ClO− + CrO2− → CrO4

2− + Cl2(d) Al + NO2

− → NH3 + AlO2−

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 takesplace in basic solution.

Plan Your StrategyFollow the steps for balancing a net ionic equation by the half-reaction method forbasic conditions.Identify the oxidizing agent and the reducing agent from the oxidation numbers orthe half-reactions.

Act on Your Strategy(a) Step 1 The unbalanced net ionic equation is given.

CN− + MnO4− → CNO− + MnO2

Step 2 To write the half-reactions, assign oxidation numbers to all the elements.CN− + MnO4

− → CNO− + MnO2

+2 −3 +7 −2 +4 −3 −2 +4 −2


The oxidation number of carbon increases, so cyanide ions, CN−, are oxidized.The oxidation number of manganese decreases, so permanganate ions, MnO4

−, are reduced.Write the two unbalanced half-reactions.Oxidation: CN− → CNO−

Reduction: MnO−4 → MnO2

Step 3 Balance the two half-reactions as if the conditions are acidic.Oxidation: CN− → CNO−

CN− + H2O → CNO−

CN− + H2O → CNO− + 2H+

CN− + H2O → CNO− + 2H+ + 2e− (balanced)Reduction: MnO4

− → MnO2

MnO4− → MnO2 + 2H2O

MnO4− + 4H+ → MnO2 + 2H2O

MnO4− + 4H+ + 3e− → MnO2 + 2H2O (balanced)

Step 4 Adjust for basic conditions.Oxidation: CN− + H2O + 2OH− → CNO− + 2H+ + 2e− + 2OH−

CN− + H2O + 2OH− → CNO− + 2H2O + 2e−

CN− + 2OH− → CNO− + H2O + 2e−

Reduction: MnO4− + 4H+ + 3e− + 4OH− → MnO2 + 2H2O + 4OH−

MnO4− + 4H2O + 3e− → MnO2 + 2H2O + 4OH−

MnO4− + 2H2O + 3e− → MnO2 + 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− + 3H2O + 6e−

2MnO4− + 4H2O + 6e− → 2MnO2 + 8OH−

Step 7 Add the half-reactions.3CN− + 6OH− → 3CNO− + 3H2O + 6e−

2MnO4− + 4H2O + 6e− → 2MnO2 + 8OH−

3CN− + 6OH− + 2MnO4− + 4H2O + 6e− → 3CNO− + 3H2O + 6e− + 2MnO2 + 8OH−

Step 8 Simplify by removing 6 electrons from both sides.3CN− + 6OH− + 2MnO4

− + 4H2O → 3CNO− + 3H2O + 2MnO2 + 8OH−

Step 9 Simplify by removing 3 water molecules and 6 hydroxide ions from both sides.3CN− + 2MnO4

− + H2O → 3CNO− + 2MnO2 + 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− + 3H2O + 2MnO4− + 8H+ + 6e− → 3CNO− + 6H+ + 6e− + 2MnO2 + 4H2O

Simplify: 3CN− + 2MnO4− + 2H+ → 3CNO− + 2MnO2 + H2O

Adjust for basic conditions:3CN− + 2MnO4

− + 2H+ + 2OH− → 3CNO− + 2MnO2 + H2O + 2OH−

3CN− + 2MnO4− + 2H2O → 3CNO− + 2MnO2 + H2O + 2OH−

3CN− + 2MnO4− + H2O → 3CNO− + 2MnO2 + 2OH− (balanced)


From the oxidation numbers or the half-reactions, MnO4− 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.H2O2 + ClO2 → ClO2

− + O2

Step 2 To write the half-reactions, assign oxidation numbers to all the elements.H2O2 + ClO2 → ClO2

− + O2+1 −1 +4 −2 +3 −2 0

The oxidation number of oxygen (from H2O2) increases, so H2O2

is oxidized.The oxidation number of chlorine decreases, so ClO2 is reduced.Write the two unbalanced half-reactions.Oxidation: H2O2 → O2

Reduction: ClO2 → ClO2−

Step 3 Balance the two half-reactions as if the conditions are acidic.Oxidation: H2O2 → O2

H2O2 → O2 + 2H+

H2O2 → O2 + 2H+ + 2e− (balanced)Reduction: ClO2 → ClO2

−

ClO2 + e− → ClO2− (balanced)

Step 4 Adjust for basic conditions.Oxidation: H2O2 → O2 + 2H+ + 2e−

H2O2 + 2OH− → O2 + 2H+ + 2e− + 2OH−

H2O2 + 2OH− → O2 + 2H2O + 2e−

Reduction: ClO2 + e− → ClO2−

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.2ClO2 + 2e− → 2ClO2

−

Step 7 Add the half-reactions.H2O2 + 2OH− → O2 + 2H2O + 2e−

2ClO2 + 2e− → 2ClO2−

H2O2 + 2OH− + 2ClO2 + 2e− → O2 + 2H2O + 2e− + 2ClO2−

Step 8 Simplify by removing 2 electrons from both sides.H2O2 + 2OH− + 2ClO2 → O2 + 2H2O + 2ClO2

− (balanced)From the oxidation numbers or the half-reactions, ClO2 is reduced, sothis is the oxidizing agent.From the oxidation numbers or the half-reactions, H2O2 is oxidized, sothis is the reducing agent.

(c) Step 1 The unbalanced net ionic equation is given.ClO− + CrO2

− → CrO42− + Cl2

Step 2 To write the half-reactions, assign oxidation numbers to all the elements.ClO− + CrO2

− → CrO42− + Cl2

+1 −2 +3 −2 +6 −2 0

The oxidation number of chromium increases, so CrO2− is oxidized.

The oxidation number of chlorine decreases, so ClO− is reduced.Write the two unbalanced half-reactions.Oxidation: CrO2

− → CrO42−

Reduction: ClO− → Cl2


Step 3 Balance the two half-reactions as if the conditions are acidic.Oxidation: CrO2

− → CrO42−

CrO2− + 2H2O → CrO4

2−


2− + 4H+


2− + 4H+ + 3e− (balanced)Reduction: ClO− → Cl22ClO− → Cl22ClO− → Cl2 + 2H2O2ClO− + 4H+ → Cl2 + 2H2O2ClO− + 4H+ + 2e− → Cl2 + 2H2O (balanced)

Step 4 Adjust for basic conditions.Oxidation: CrO2

− + 2H2O → CrO42− + 4H+ + 3e−

CrO2− + 2H2O + 4OH− → CrO4

2− + 4H+ + 3e− + 4OH−

CrO2− + 2H2O + 4OH− → CrO4

2− + 4H2O + 3e−

CrO2− + 4OH− → CrO4

2− + 2H2O + 3e−

Reduction: 2ClO− + 4H+ + 2e− → Cl2 + 2H2O2ClO− + 4H+ + 2e− + 4OH− → Cl2 + 2H2O + 4OH−

2ClO− + 4H2O + 2e− → Cl2 + 2H2O + 4OH−

2ClO− + 2H2O + 2e− → Cl2 + 4OH−

Step 5 The LCM of 3 and 2 is 6. Step 6 Multiply the oxidation half-reaction by 2, and multiply the reduction half-

reaction by 3, so that equal numbers of electrons are lost and gained.2CrO2

− + 8OH− → 2CrO42− + 4H2O + 6e−

6ClO− + 6H2O + 6e− → 3Cl2 + 12OH−

Step 7 Add the half-reactions.2CrO2

− + 8OH− → 2CrO42− + 4H2O + 6e−

6ClO− + 6H2O + 6e− → 3Cl2 + 12OH−

2CrO2− + 8OH− + 6ClO− + 6H2O + 6e− → 2CrO4

2− + 4H2O + 6e− + 3Cl2 + 12OH−

Step 8 Simplify by removing 6 electrons from both sides.2CrO2

− + 8OH− + 6ClO− + 6H2O → 2CrO42− + 4H2O + 3Cl2 + 12OH−

Step 9 Simplify by removing 4 water molecules and 8 hydroxide ions from both sides.2CrO2

− + 6ClO− + 2H2O → 2CrO42− + 3Cl2 + 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, CrO2

− is oxidized, so this is the reducing agent.

(d) Step 1 The unbalanced net ionic equation is given.Al + NO2

− → NH3 + AlO2−

Step 2 To write the half-reactions, assign oxidation numbers to all the elements.Al + NO2

− → NH3 + AlO2−

0 +3 −2 −3 +1 +3 −2

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 → AlO2

−

Reduction: NO2− → NH3


Step 3 Balance the two half-reactions as if the conditions are acidic.Oxidation: Al → AlO2

−

Al + 2H2O → AlO2−

Al + 2H2O → AlO2− + 4H+

Al + 2H2O → AlO2− + 4H+ + 3e− (balanced)

Reduction: NO2− → NH3

NO2− → NH3 + 2H2O

NO2− + 7H+ → NH3 + 2H2O

NO2− + 7H+ + 6e− → NH3 + 2H2O (balanced)

Step 4 Adjust for basic conditions.Oxidation: Al + 2H2O → AlO2

− + 4H+ + 3e−

Al + 2H2O + 4OH− → AlO2− + 4H+ + 3e− + 4OH−

Al + 2H2O + 4OH− → AlO2− + 4H2O + 3e−

Al + 4OH− → AlO2− + 2H2O + 3e−

Reduction: NO2− + 7H+ + 6e− → NH3 + 2H2O

NO2− + 7H+ + 6e− + 7OH− → NH3 + 2H2O + 7OH−

NO2− + 7H2O + 6e− → NH3 + 2H2O + 7OH−

NO2− + 5H2O + 6e− → NH3 + 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− → 2AlO2

− + 4H2O + 6e−

Step 7 Add the half-reactions.2Al + 8OH− → 2AlO2

− + 4H2O + 6e−

NO2− + 5H2O + 6e− → NH3 + 7OH−

2Al + 8OH− + NO2− + 5H2O + 6e− → 2AlO2

− + 4H2O + 6e− + NH3 + 7OH−

Step 8 Simplify by removing 6 electrons from both sides.2Al + 8OH− + NO2

− + 5H2O → 2AlO2− + 4H2O + NH3 + 7OH−

Step 9 Simplify by removing 4 water molecules and 7 hydroxide ions from both sides.2Al + OH− + NO2

− + H2O → 2AlO2− + NH3 (balanced)

From the oxidation numbers or the half-reactions, NO2− is reduced, so

this is the oxidizing agent.From the oxidation numbers or the half-reactions, Al is oxidized, so thisis the reducing agent.

Check Your SolutionIn each equation, the atoms are balanced and the charges are balanced.


29. ProblemAn analyst uses 0.02045 mol/L KMnO4 to titrate a sample solution of H2O2.The analyst knows that the sample solution is about 6% H2O2 by mass. The analystplaces 1.284 g of H2O2 solution in a flask, dilutes it with water, and adds a smallamount of sulfuric acid to acidify it. It takes 38.95 mL of KMnO4 solution to reachthe endpoint. What mass of pure H2O2 was present. What is the mass percent ofpure H2O2 in the original solution?


What is Required?You need to determine the mass of H2O2 in the sample and express the result as amass percent.

What is Given?You know that the concentration of KMnO4 is 0.02045 mol/L and its volume is38.95 mL. The mass of H2O2 solution is 1.284g.

Plan Your StrategyWrite the balance equation for the reaction. KMnO4 can be taken as equivalent toMnO4

–

Calculate the mol of MnO4– added based on the concentration and volume of

KMnO4(aq).Use the mol ratio if the balanced equation to determine the mol of H2O2 that reactswith the MnO4

–. Use the equation m = n × M to determine the mass of H2O2. Usethe mass of H2O2 solution to express the concentration of the H2O2 as mass percent.

Act on Your StrategyThe balanced equation for this reaction is2MnO4

– + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2Omol MnO4

– = C × V = 0.02045 mol/L × 0.03895 L = 0.0007965 mol0.0007965 mol MnO4

– × = 0.001991 mol H2O2

mass H2O2 = n × M = 0.001991 mol × 34.02 g/’mol = 0.06774 g

Check Your SolutionThe mass percent is close to the expected value of 6% and the answer has the correctunite and number of significant figures.

30. ProblemA forensic chemist wants to determine the level of alcohol in a sample of blood plas-ma. The chemist titrates the plasma with a solution of potassium dichromate. Thebalanced equation is:16H+ + 2Cr2O7

2–

+ C2H5OH → 4Cr3+ 2CO2 + 11H2OIf 32.35 mL of 0.05023 mol/L Cr2OI7

2–is required to titrate 27.00 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 2Cr2O7

2–solution and the mass of

the plasma sample.

Plan Your StrategyUse the formula n = C × V to find the mol of 2Cr2O7

2–and use the mol ratio from

the balanced equation to determine the mol of C2H5OH that react. Use the formulam = n × M to calculate the mass of C2H5OH. Use this value and the mass of theplasma sample to calculate the mass percent of C2H5OH.

Act on Your Strategyn = C × V = 0.05023 mol/L0.03235 L = 0.001625 mol

0.001625 mol Cr2OI72– × = 0.0008125 mol C2H5OH

1 mol C H OH

2 mol 2Cr O

2 5

2 72−

mass percentmass H O

mass solution0.06774 g1.284 g

2 2= × = × =100 100 5 276% % . %

5 mol H O

2 mol MnO2 2

4


mass of C2H5OH = 0.0008125 mol 46.08 g/mol = 0.03744 g

Check Your SolutionA small mass percent would be expected this answer seems reasonable. The units andnumber of significant figures are correct.

31. ProblemAn analyst titrates an acidified solution containing 0.153 g of purified sodiumoxalate, Na2C2O4, with potassium permanganate solution, KMnO4(aq). The purpleendpoint is reached when the chemist has added 41.45 mL of potassium perman-ganate solution. What is the molar concentration of the potassium permanganatesolution? The balanced equation is 2MnO4

– + 5C2O42– + 16H+ → 2Mn2+ + 10CO2 + 8H2O

What is Required?You must find the molar concentration of the titrating solution, potassium perman-ganate.

What is Given?You know the mass of sodium oxalate and can find its molar mass using the periodictable. You also know the volume of potassium permanganate.

Plan Your StrategyFind the molar mass of Na2C2O4 and determine the mol of using . Use themol ratio in the balanced equation to find the mol of KMnO4 that has reacted.Calculate the concentration of the KMnO4 using the formula .

Act on Your StrategyM = 134.0 g/mol

mol Na2C2O4 is the same as mol C2O42–

and mol KMnO4 is the same as mol MnO4

Check Your SolutionThe answer has the correct units and number of significant figures and seems to bereasonable.

32. Problem25.00 mL of a solution containing iron(II) ions was titrated with a 0.02043 mol/Lpotassium dichromate solution. The endpoint was reached when 35.55 mL of potas-sium dichromate solution had been added. What was the molar concentration ofiron(II) ions in the original, acidic solution? The unbalanced equation is:Cr2O7

2– + Fe2+ → Cr3+ + Fe3+

What is Required?You must find the molar concentration of the Fe2+ ions in the original solution.

What is Given?You know the volume and concentration of the potassium dichromate solution andthe volume of Fe2+ ions.

CnV

= = =0.000456 mol0.04145 L

mol/L0 0110.

0.00114 mol C O mol MnO

mol C O mol MnO2 4

4

2 4

4× =−−2

50 000456

2

_

.

n = =0.153 g134.0 g/mol

mol Na C O2 2 40 00114.

CnV

=

nmM

=

mass percent of C H OHmass C H OH

mass of plasma0.03744 g27.00 g2 5

2 5= × = × =100 100 0 1387% % . %


Plan Your StrategyBalance the equation in acidic solution. Calculate the mol of potassium dichromateusing the formula n = C × V. Use the mole ratio in the balanced equation to find themol of Fe2+. Find the molar concentration of Fe2+ the formula .

Act on Your StrategyThe balanced equation is Cr2O7

2– + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2Omol Cr2O4

2–= 0.02043 mol/L × 0.03555 L = 0.000726 mol Cr2O7

2–

Check Your SolutionThe answer seems reasonable and has the correct units and number of significant fig-ures.


33. ProblemUse the oxidation number method to balance the following equation for the combustion of carbon disulfide.CS2 + O2 → CO2 + SO2

SolutionStep 1 The unbalanced equation is given.

CS2 + O2 → CO2 + SO2

Step 2 Assign oxidation numbers to all the elements.CS2 + O2 → CO2 + SO2

+4 −2 0 +4 −2 +4 −2

Step 3 Sulfur undergoes an increase in oxidation number.Oxygen undergoes a decrease in oxidation number.

Step 4 The oxidation number of sulfur increases from −2 to +4, an increase of 6.The oxidation number of oxygen decreases from 0 to −2, a decrease of 2.

Step 5 A 3:1 ratio of oxygen atoms to sulfur atoms ensures that the total increase inoxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 Use the 3:1 ratio to balance the numbers of atoms of oxygen and sulfur.Make sure that there are three oxygen atoms for every sulfur atom. Theunbalanced equation includes 2 sulfur atoms on the left side, so 6 oxygenatoms are needed on the left side.CS2 + 3O2 → CO2 + SO2

Step 7 Balance the sulfur atoms by inspection.CS2 + 3O2 → CO2 + 2SO2 (balanced)

34. ProblemUse the oxidation number method to balance the following equations.(a) B2O3 + Mg → MgO + Mg3B2 (b) H2S + H2O2 → S8 + H2O

Solution(a) Step 1 The unbalanced equation is given.

B2O3 + Mg → MgO + Mg3B2

Step 2 Assign oxidation numbers to all the elements.

0.000726 mol Cr O mol Fe

mol Cr O mol Fe

0.00436 mol Fe0.02500 L

mol/L

2 7

2+

2 72+

2

2

26

10 00436

0 1743

−

−× =

= = =

+.

.CnV

CnV

=


B2O3 + Mg → MgO + Mg3B2+3 −2 0 +2 −2 +2 −3

Step 3 Magnesium undergoes an increase in oxidation number.Boron undergoes a decrease in oxidation number.

Step 4 The oxidation number of magnesium increases from 0 to +2, an increase of 2.The oxidation number of boron decreases from +3 to −3, a decrease of 6.

Step 5 A 6:2 ratio of magnesium atoms to boron atoms ensures that the totalincrease in oxidation numbers and the total decrease in oxidation num-bers are equal. Therefore, the smallest whole-number ratio of magnesiumatoms to boron atoms is 3:1.

Step 6 Use the 3:1 ratio to balance the numbers of atoms of magnesium andboron in the reactants. Make sure that there are three magnesium atomsfor every boron atom. The unbalanced equation includes 2 boron atomson the left side, so 6 magnesium atoms are needed on the left side.B2O3 + 6Mg → MgO + Mg3B2

Step 7 Because magnesium is found in one reactant and both products, balancethe oxygen atoms by inspection.B2O3 + 6Mg → 3MgO + Mg3B2 (balanced)

(b) Step 1 The unbalanced equation is given.H2S + H2O2 → S8 + H2O

Step 2 Assign oxidation numbers to all the elements.H2S + H2O2 → S8 + H2O

+1 −2 +1 −1 0 +1 −2

Step 3 Sulfur undergoes an increase in oxidation number.Oxygen undergoes a decrease in oxidation number.

Step 4 The oxidation number of sulfur increases from −2 to 0, an increase of 2.The oxidation number of oxygen decreases from −1 to −2, a decrease of 1.

Step 5 A 2:1 ratio of oxygen atoms to sulfur atoms ensures that the total increase in oxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 Use the 2:1 ratio to balance the numbers of atoms of oxygen and sulfur. Make sure that there are two oxygen atoms for every sulfur atom.The unbalanced equation includes 8 sulfur atoms on the right side, so 16 oxygen atoms are needed on the right side.H2S + H2O2 → S8 + 16H2O

Step 7 Because hydrogen is found in both reactants and one product, balancethe sulfur atoms and the oxygen atoms by inspection.8H2S + H2O2 → S8 + 16H2O8H2S + 8H2O2 → S8 + 16H2O (balanced)

35. ProblemUse the oxidation number method to balance each ionic equation in acidic solution.(a) Cr2O7

2− + Fe2+ → Cr3+ + Fe3+

(b) I2 + NO3− → IO3

− + NO2

(c) PbSO4 → Pb + PbO2 + SO42−


Cr2O72− + Fe2+ → Cr3+ + Fe3+

Step 2 Assign oxidation numbers to all the elements.Cr2O7

2− + Fe2+ → Cr3+ + Fe3+

+6 −2 +2 +3 +3


Step 3 Iron undergoes an increase in oxidation number.Chromium undergoes a decrease in oxidation number.

Step 4 The oxidation number of iron increases from +2 to +3, an increase of 1.The oxidation number of chromium decreases from +6 to +3, a decreaseof 3.

Step 5 A 3:1 ratio of iron atoms to chromium atoms ensures that the totalincrease in oxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 Use the 3:1 ratio to balance the numbers of atoms of iron and chromium.Make sure that there are three iron atoms for every chromium atom. Theunbalanced equation includes 2 chromium atoms on the left side, so 6iron atoms are needed on the left side.Cr2O7

2− + 6Fe2+ → Cr3+ + Fe3+

Step 7 Balance the chromium atoms by inspection.Cr2O7

2− + 6Fe2+ → 2Cr3+ + Fe3+

Balance the iron atoms by inspection.Cr2O7

2− + 6Fe2+ → 2Cr3+ + 6Fe3+

Step 8 Balance for acidic conditions.Add water molecules to balance the oxygen atoms.Cr2O7

2− + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2OAdd hydrogen ions to balance the hydrogen atoms.Cr2O7

2− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O (balanced)

(b) Step 1 The unbalanced equation is given.I2 + NO3

− → IO3− + NO2

Step 2 Assign oxidation numbers to all the elements.I2 + NO3

− → IO3− + NO2

0 +5 −2 +5 −2 +4 −2

Step 3 Iodine undergoes an increase in oxidation number.Nitrogen undergoes a decrease in oxidation number.

Step 4 The oxidation number of iodine increases from 0 to +5, an increase of 5.The oxidation number of nitrogen decreases from +5 to +4, a decrease of 1.

Step 5 A 5:1 ratio of nitrogen atoms to iodine atoms ensures that the totalincrease in oxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 Use the 5:1 ratio to balance the numbers of atoms of nitrogen and iodine. Make sure that there are five nitrogen atoms for every iodineatom. The unbalanced equation includes 2 iodine atoms on the left side,so 10 nitrogen atoms are needed on the left side.I2 + 10NO3

− → IO3− + NO2

Step 7 Balance the iodine atoms by inspection.I2 + 10NO3

− → 2IO3− + NO2

Balance the nitrogen atoms by inspection.I2 + 10NO3

− → 2IO3− + 10NO2

Step 8 Balance for acidic conditions.Add water molecules to balance the oxygen atoms.I2 + 10NO3

− → 2IO3− + 10NO2 + 4H2O

Add hydrogen ions to balance the hydrogen atoms.I2 + 10NO3

− + 8H+ → 2IO3− + 10NO2 + 4H2O (balanced)

(c) Step 1 The unbalanced equation is given.PbSO4 → Pb + PbO2 + SO4

2−


Step 2 Assign oxidation numbers to all the elements.PbSO4 → Pb + PbO2 + SO4

2−

+2 +6 −2 0 +4 −2 +6 −2

Step 3 Lead undergoes both an increase and a decrease in oxidation number.Step 4 The oxidation number of some lead atoms increases from +2 to +4,

an increase of 2. The oxidation number of other lead atoms decreases from +2 to 0, adecrease of 2.

Step 5 A 1:1 ratio of lead atoms that undergo an increase in oxidation numberto lead atoms that undergo a decrease in oxidation number ensures that the total increase in oxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 From the 1:1 ratio, the number of lead atoms in the products elementallead and lead(IV) oxide must be equal.PbSO4 → Pb + PbO2 + SO4

2−

Step 7 Keeping the 1:1 ratio of lead atoms on the right side, balance the leadatoms by inspection.2PbSO4 → Pb + PbO2 + SO4

2−

Balance the sulfur atoms by inspection.2PbSO4 → Pb + PbO2 + 2SO4

2−

Step 8 Balance for acidic conditions.Add water molecules to balance the oxygen atoms.2PbSO4 + 2H2O → Pb + PbO2 + 2SO4

2−

Add hydrogen ions to balance the hydrogen atoms.2PbSO4 + 2H2O → Pb + PbO2 + 2SO4

2− + 4H+ (balanced)

36. ProblemUse the oxidation number method to balance each ionic equation in basic solution.(a) Cl− + CrO4

2− → ClO− + CrO2−

(b) Ni + MnO4− → NiO + MnO2

(c) I− + Ce4+ → IO3− + Ce3+


Cl− + CrO42− → ClO− + CrO2

−

Step 2 Assign oxidation numbers to all the elements.Cl− + CrO4

2− → ClO− + CrO2−

−1 +6 −2 +1 −2 +3 −2

Step 3 Chlorine undergoes an increase in oxidation number.Chromium undergoes a decrease in oxidation number.

Step 4 The oxidation number of chlorine increases from −1 to +1, an increase of 2.The oxidation number of chromium decreases from +6 to +3, a decrease of 3.

Step 5 A 3:2 ratio of chlorine atoms to chromium atoms ensures that the total increase in oxidation numbers and the total decrease in oxidationnumbers are equal.

Step 6 Use the 3:2 ratio to balance the numbers of atoms of chlorine andchromium. Make sure that there are three chlorine atoms for every two chromium atoms.3Cl− + 2CrO4

2− → ClO− + CrO2−


Step 7 Balance the chlorine atoms by inspection.3Cl− + 2CrO4

2− → 3ClO− + CrO2−

Balance the chromium atoms by inspection.3Cl− + 2CrO4

2− → 3ClO− + 2CrO2−

Step 8 Balance for basic conditions.Add water molecules to balance the oxygen atoms.3Cl− + 2CrO4

2− → 3ClO− + 2CrO2− + H2O

Add hydrogen ions to balance the hydrogen atoms.3Cl− + 2CrO4

2− + 2H+ → 3ClO− + 2CrO2− + H2O

Adjust for basic conditions by adding two hydroxide ions to each side.3Cl− + 2CrO4

2− + 2H+ + 2OH− → 3ClO− + 2CrO2− + H2O + 2OH−

Combine the hydrogen ions and hydroxide ions on the left side intowater molecules.3Cl− + 2CrO4

2− + 2H2O → 3ClO− + 2CrO2− + H2O + 2OH−

Remove one water molecule from each side.3Cl− + 2CrO4

2− + H2O → 3ClO− + 2CrO2− + 2OH− (balanced)

(b) Step 1 The unbalanced equation is given.Ni + MnO4

− → NiO + MnO2

Step 2 Assign oxidation numbers to all the elements.Ni + MnO4

− → NiO + MnO20 +7 −2 +2 −2 +4 −2

Step 3 Nickel undergoes an increase in oxidation number.Manganese undergoes a decrease in oxidation number.

Step 4 The oxidation number of nickel increases from 0 to +2, an increase of 2.The oxidation number of manganese decreases from +7 to +4, a decrease of 3.

Step 5 A 3:2 ratio of nickel atoms to manganese atoms ensures that the totalincrease in oxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 Use the 3:2 ratio to balance the numbers of atoms of nickel and manganese. Make sure that there are three nickel atoms for every two manganese atoms.3Ni + 2MnO4

− → NiO + MnO2

Step 7 Balance the nickel atoms by inspection.3Ni + 2MnO4

− → 3NiO + MnO2

Balance the manganese atoms by inspection.3Ni + 2MnO4

− → 3NiO + 2MnO2

Step 8 Balance for basic conditions.Add water molecules to balance the oxygen atoms.3Ni + 2MnO4

− → 3NiO + 2MnO2 + H2OAdd hydrogen ions to balance the hydrogen atoms.3Ni + 2MnO4

− + 2H+ → 3NiO + 2MnO2 + H2OAdjust for basic conditions by adding two hydroxide ions to each side.3Ni + 2MnO4

− + 2H+ + 2OH− → 3NiO + 2MnO2 + H2O + 2OH−

Combine the hydrogen ions and hydroxide ions on the left side intowater molecules.3Ni + 2MnO4

− + 2H2O → 3NiO + 2MnO2 + H2O + 2OH−

Remove one water molecule from each side.3Ni + 2MnO4

− + H2O → 3NiO + 2MnO2 + 2OH− (balanced)


(c) Step 1 The unbalanced equation is given.I− + Ce4+ → IO3

− + Ce3+

Step 2 Assign oxidation numbers to all the elements.I− + Ce4+ → IO3

− + Ce3+

−1 +4 s+5−2 +3

Step 3 Iodine undergoes an increase in oxidation number.Cerium undergoes a decrease in oxidation number.

Step 4 The oxidation number of iodine increases from −1 to +5, an increase of 6.The oxidation number of cerium decreases from +4 to +3, a decrease of 1.

Step 5 A 6:1 ratio of cerium atoms to iodine atoms ensures that the totalincrease in oxidation numbers and the total decrease in oxidation numbers are equal.

Step 6 Use the 6:1 ratio to balance the numbers of atoms of cerium and iodine.Make sure that there are six cerium atoms for every iodine atom.I− + 6Ce4+ → IO3

− + Ce3+

Step 7 Balance the cerium atoms by inspection.I− + 6Ce4+ → IO3

− + 6Ce3+

Step 8 Balance for basic conditions.Add water molecules to balance the oxygen atoms.I− + 6Ce4+ + 3H2O → IO3

− + 6Ce3+

Add hydrogen ions to balance the hydrogen atoms.I− + 6Ce4+ + 3H2O → IO3

− + 6Ce3+ + 6H+

Adjust for basic conditions by adding six hydroxide ions to each side.I− + 6Ce4+ + 3H2O + 6OH− → IO3

− + 6Ce3+ + 6H+ + 6OH−

Combine the hydrogen ions and hydroxide ions on the right side intowater molecules.I− + 6Ce4+ + 3H2O + 6OH− → IO3

− + 6Ce3+ + 6H2ORemove three water molecules from each side.I− + 6Ce4+ + 6OH− → IO3

− + 6Ce3+ + 3H2O (balanced)


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Chapter 18 Oxidation-Reduction Reactions - …hrsbstaff.ednet.ns.ca/benoitn/chem12/electrochemistry/NF...Write balanced half-reactions from the net ionic equation for the reaction