Chapter 6: Oxidation-Reduction Reactions Chemistry: The Molecular Nature of Matter, 6E...

Chapter 6: Oxidation-Reduction

Reactions

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

2

Oxidation-Reduction Reactions Electron transfer reactions

Electrons transferred from one substance to another

Originally only combustion of fuels or reactions of metal with oxygen

Important class of chemical reactions that occur in all areas of chemistry & biology

Also called redox reactions


3

Oxidation–Reduction ReactionsInvolves 2 processes:

Oxidation = Loss of Electrons (LEO)Na Na+ + e Oxidation Half-Reaction

Reduction = Gain of electrons (GER)Cl2 + 2e 2Cl Reduction Half-Reaction

Net reaction: 2Na + Cl2 2Na+ + 2Cl

Oxidation & reduction always occur together

Can't have one without the other


Oxidation Reduction ReactionOxidizing Agent Substance that accepts e's

Accepts e's from another substance Substance that is reduced Cl2 + 2e 2Cl–

Reducing Agent Substance that donates e's

Releases e's to another substance Substance that is oxidized Na Na+ + e–

4


Redox Reactions Very common

Batteries—car, flashlight, cell phone, computer

Metabolism of food Combustion

Chlorine Bleach Dilute NaOCl solution Cleans through redox

reaction Oxidizing agent Destroys stains by oxidizing them

5


Redox ReactionsEx. Fireworks displays

Net: 2Mg + O2 2MgO

Oxidation:

Mg Mg2+ + 2e Loses electrons = Oxidized Reducing agent

Reduction:

O2 + 4e 2O2 Gains electrons = Reduced Oxidizing agent

6


Your Turn!Which species functions as the oxidizing agent in the following oxidation-reduction reaction?

Zn(s) + Pt2+(aq) Pt(s) + Zn2+(aq)

A. Pt(s)

B. Zn2+(aq)

C. Pt2+(aq)

D. Zn(s)

E. None of these, as this is not a redox reaction.

7


8

Guidelines For Redox Reactions

Oxidation & reduction always occur simultaneously

Total number of electrons lost by one substance = total number of electrons gained by second substance

For a redox reaction to occur, something must accept electrons that are lost by another substance


Oxidation Numbers Bookkeeping Method Way to keep track of electrons

Not all redox reactions contain O2 & give ions

Covalent molecules & ions often involvedEx. CH4, SO2, MnO4

–, etc.

Defined by set of rules How to divide up shared electrons in

compounds with covalent bonds Change in oxidation number of element

during reaction indicates redox reaction has occurred

9


Hierarchy of Rules for Assigning Oxidation Numbers

1. Oxidation numbers must add up to charge on molecule, formula unit or ion.

2. Atoms of free elements have oxidation numbers of zero.

3. Metals in Groups 1A, 2A, and Al have +1, +2, and +3 oxidation numbers, respectively.

4. H & F in compounds have +1 & –1 oxidation numbers, respectively.

5. Oxygen has –2 oxidation number.

6. Group 7A elements have –1 oxidation number.

10


Hierarchy of Rules for Assigning Oxidation Numbers



9. When there is a conflict between 2 of these rules or ambiguity in assigning an oxidation number, apply rule with lower oxidation number & ignore conflicting rule.

Oxidation State Used interchangeably with oxidation number Indicates charge on monatomic ions Iron (III) means +3 oxidation state of Fe or Fe3+

11


Ex. Assigning Oxidation Number1. Li2O

Li (2 atoms) × (+1) = +2 (Rule 3)O (1 atom) × (–2) = –2 (Rule 5) sum = 0 (Rule 1)+2 –2 = 0 so the charges are balanced to zero

2. CO2

C (1 atom) × (x) = xO (2 atoms) × (–2) = –4 (Rule 5) sum = 0 (Rule 1)x 4 = 0 or x = +4

C is in +4 oxidation state 12


13

Learning CheckAssign oxidation numbers to all atoms: Ex. ClO4

O (4 atoms) × (–2) = –8Cl (1 atom) × (–1) = –1(molecular ion) sum ≠ –1 (violates Rule 1)

Rule 5 for O comes before Rule 6 for halogens

O (4 atoms) × (–2) = –8Cl (1 atom) × (x) = x sum = –1 (Rule 1)–8 + x = –1 or x = 8 –1So x = +7; Cl is oxidation state +7


Learning CheckAssign Oxidation States To All Atoms: MgCr2O7

Mg =+2; O = –2; and Cr = x (unknown)+2 + 2x + {7 × (–2)} = 02x – 12 = 0 x = +3Cr is oxidation # of +3

KMnO4

K =+1; O = – 2; so Mn = x+1 + x + {4 × (–2)} = 0x – 7 = 0 x = +7Mn is oxidation # of +7

14


Your Turn!What is the oxidation number of each atom in H3PO4?

A. H = –1; P = +5; O = –2

B. H = 0; P = +3; O = –2

C. H = +1; P = +7; O = –2

D. H = +1; P = +1; O = –1

E. H = +1; P = +5; O = –2

15


Redefine Oxidation-Reduction in Terms of Oxidation Number

A redox reaction occurs when there is a change in oxidation number.

Oxidation Increase in oxidation number e loss

Reduction Decrease in oxidation number e gain

16


Using Oxidation Numbers to Recognize Redox Reactions

Sometimes literal electron transfer:

Cu: oxidation number decreases by 2 reduction

Zn: oxidation number increases by 2 oxidation

17

+ ++2 +20 0

increase oxidation

decrease reduction

Cu2+ Zn Zn2+ Cu


Using Oxidation Numbers to Recognize Redox Reactions

Sometimes electron transferred in "formal" sense.

O: oxidation number decreases by 2 reduction

C: oxidation number increases by 8 oxidation

18

2H2O2O2+ +-4 +40+1 -2 +1 -2

C: increase oxidation

O: decrease reduction

CH4 CO2


Ion Electron Method Way to balance redox equations Must balance both mass & charge Write skeleton equation

Only ions & molecules involved in reaction

Break into 2 half-reactions Oxidation Reduction

Balance each half-reaction separately Recombine to get balanced net ionic

equation19


Balancing Redox ReactionsSome Redox reactions are simple:Ex. 1 Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

Break into half-reactionsZn(s) Zn2+(aq) + 2e oxidation

LEOReducing agent

Cu2+(aq) + 2e Cu(s) reduction GEROxidizing agent

20


Example 1Zn(s) Zn2+(aq) + 2e oxidation

Cu2+(aq) + 2e Cu(s) reduction Each half-reaction is balanced for atoms

Same # atoms of each type on each side

Each half-reaction is balanced for charge Same sum of charges on each side

Add both equations algebraically, canceling e’s

NEVER have e's in net ionic equation

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)

21


Balancing Redox Equations in Aqueous Solutions

Many redox reactions in aqueous solution involve H2O and H+ or OH

Balancing the equation cannot be done by inspection.

Need method to balance equation correctly

Start with acidic solution then work to basic conditions

22


Redox in Aqueous SolutionEx. 2 Mix solutions of K2Cr2O7 & FeSO4

Dichromate ion, Cr2O72–, oxidizes Fe2+ to Fe3+

Cr2O72– is reduced to form Cr3+

Acidity of mixture decreases as H+ reacts with oxygen to form water

Skeletal Eqn. Cr2O72– + Fe2+ Cr3+ + Fe3+

23

Ox. # Cr = +6 Fe = +2 Cr = +3 Fe = +3


Ion-Electron Method – Acidic Solution

1. Divide equation into 2 half-reactions

2. Balance atoms other than H & O

3. Balance O by adding H2O to side that needs O

4. Balance H by adding H+ to side that needs H

5. Balance net charge by adding e–

6. Make e– gain equal e– loss; then add half-reactions

7. Cancel anything that is the same on both sides

24


Ion Electron MethodEx. 2 Balance in Acidic Solution

Cr2O72– + Fe2+ Cr3+ + Fe3+

1. Break into half-reactionsCr2O7

2 Cr3+

Fe2+ Fe3+


Cr2O72 2Cr3+

Put in 2 coefficient to balance Cr

Fe2+ Fe3+

Fe already balanced25


Ex. 2 Ion-Electron Method in Acid3. Balance O by adding H2O to the side

that needs O.

Cr2O72 2Cr3+

Right side has 7 O atoms Left side has none Add 7 H2O to left side

Fe2+ Fe3+

No O to balance

26

+ 7 H2O


Ex. 2 Ion-Electron Method in Acid4. Balance H by adding H+ to side that

needs H

Cr2O72 2Cr3+ + 7H2O

Left side has 14 H atoms Right side has none Add 14 H+ to right side

Fe2+ Fe3+

No H to balance

27

14H+ +


Ex. 2 Ion-Electron Method in Acid5. Balance net charge by adding electrons.

14H+ + Cr2O72 2Cr3+ + 7H2O

6 electrons must be added to reactant side

Fe2+ Fe3+

1 electron must be added to product side Now both half-reactions balanced for mass

& charge

28

6e +

+ e

Net Charge = 2(+3)+7(0) = 6

Net Charge = 14(+1) (–2) = 12


Ex. 2 Ion-Electron Method in Acid6. Make e– gain equal e– loss; then add

half-reactions 6e + 14H+ + Cr2O7

2– 2Cr3+ + 7H2O

Fe2+ Fe3+ + e

7. Cancel anything that's the same on both sides

6[ ]

29

6e + 6Fe2+ + 14H+ + Cr2O7

2 6Fe3+ + 2Cr3+

+ 7H2O + 6e

6Fe2+ + 14H+ + Cr2O7

2

6Fe3+ + 2Cr3+

+ 7H2O


Ion-Electron in Basic Solution The simplest way to balance an

equation in basic solution

Use steps 1-7 above, then

8. Add the same number of OH– to both sides of the equation as there are H+.

9. Combine H+ & OH– to form H2O

10. Cancel any H2O that you can from both sides

30


Ex.2 Ion-Electron Method in BaseReturning to our example of Cr2O7

2 & Fe2+

8. Add to both sides of equation the same number of OH– as there are H+.

9. Combine H+ and OH– to form H2O.

10. Cancel any H2O that you can

31

6Fe2++ 14H+

+ Cr2O72

6Fe3+ + 2Cr3+ + 7H2O+ 14 OH– + 14 OH–

6Fe2+ + 14H2O+ Cr2O7

2 6Fe3+ + 2Cr3+

+ 7H2O + 14OH

7

6Fe2+ + 7H2O+ Cr2O7

2 6Fe3+ + 2Cr3+

+ 14OH


Your Turn!Which of the following is a correctly balanced reduction half-reaction?

A. Fe3+ + e– Fe°

B. 2Fe + 6HNO3 2Fe(NO3)3 + 3H2

C. Mn2+ + 4H2O MnO4– + 8H+ + 5e–

D. 2O2– O2 + 4e–

E. Mg2+ + 2e– Mg°

32


Ex. 3 Ion-Electron MethodBalance the following equation in basic solution:MnO4

– + HSO3– Mn2+ + SO4

2

1. Break it into half-reactions

MnO4– Mn2+

HSO3– SO4

2–


MnO4 Mn2+

Balanced for Mn

HSO3 SO4

2

Balanced for S

33


Ex. 3 Ion-Electron Method3. Add H2O to balance O

MnO4 Mn2+

HSO3 SO4

2

4. Add H+ to balance H

MnO4 Mn2+ + 4H2O

H2O + HSO3 SO4

2

34

+ 4H2O

H2O +

8H+ + + 3H+


Ex. 3 Ion-Electron Method5. Balance net charge by adding e–.

8H+ + MnO4 Mn2+ + 4H2O

8(+1) + (–1) = +7 +2 + 0 = +2

Add 5 e– to reactant side

H2O + HSO3 SO4

2 + 3H+

0 + (–1) = –1 –2 + 3(+1) = +1

Add 2 e– to product side

5e– +

+ 2 e–

35


Ex. 3 Ion-Electron Method6. Make e– gain equal e– loss

5e– + 8H+ + MnO4 Mn2+ +

4H2O

H2O + HSO3 SO4

2 + 3H+ + 2e–

Must multiply Mn half-reaction by 2 Must multiply S half-reaction by 5 Now have 10 e– on each side

36

2[ ]

5[ ]


10e– + 16H+ + 2MnO4

+ 5H2O + 5HSO3

2Mn2+ + 8H2O +

5SO42 + 15H+ + 10e

Ex. 3 Ion-Electron Method6. Then add the two half-reactions10e– + 16H+ + 2MnO4

2Mn2+ + 8H2O

5H2O + 5HSO3 5SO4

2 + 15H+ + 10e–

7. Cancel anything that is the same on both sides.

Balanced in acid.37

31

H+ + 2MnO4

+ 5HSO3

2Mn2+ + 3H2O + 5SO4

2


Ex.3 Ion-Electron Method in Base8. Add same number of OH– to both sides

of equation as there are H+

9. Combine H+ and OH– to form H2O

10. Cancel any H2O that you can

2MnO4 + 5HSO3

2Mn2+ + 2H2O + OH + 5SO4

2

+ OH–

38

H+ + 2MnO4

+ 5HSO3

2Mn2+ + 3H2O + 5SO4

2

+ OH–

H2O + 2MnO4

+ 5HSO3

2Mn2+ + 3H2O + 5SO4

2 + OH

2


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Balance each equation in Acid & Base using the Ion Electron Method.

MnO4– + C2O4

2– MnO2 + CO32–

Acid: 2MnO4– + 3C2O4

2– + 2H2O 2MnO2 + 4H+ + 6CO32–

Base: 2MnO4– + 3C2O4

2– + 4OH– 2MnO2 + 2H2O + 6CO32–

ClO– + VO3– ClO3

– + V(OH)3

Acid: ClO– + 2H2O + 2VO3– + 2H+ ClO3

–+ 2V(OH)3

Base: ClO– + 4H2O + 2VO3– ClO3

–+ 2V(OH)3 + 2OH–

Your Turn!


Acids as Oxidizing Agents Metals often react with acid

Form metal ions & Molecular hydrogen gas

Molecular Equation Zn(s) + 2HCl(aq) H2(g) +

ZnCl2(aq)

Net Ionic Equation Zn(s) + 2H+(aq) H2(g) + Zn2+(aq)

M oxidized H+ reduced H+ oxidizing reagent Zn reducing reagent

40


Oxidation of Metals by Acids Ease of oxidation process depends on

metal Metals that react with HCl or H2SO4

Easily oxidized by H+

More active than hydrogen (H2)

Ex. Mg, Zn, alkali metals

Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)

2Na(s) + 2H+(aq) 2Na+(aq) + H2(g)

Metals that don’t react with HCl or H2SO4

Not oxidized by H+

Less active than H2

Ex. Cu, Pt 41


Anion Determines Oxidizing Power Acids are divided into 2 classes:

1.Nonoxidizing Acids Anion is weaker oxidizing agent than H3O+

Only redox reaction is 2H+ + 2 e– H2 or

2H3O+ + 2 e– H2 + 2H2O

HCl(aq), HBr(aq), HI(aq)

H3PO4(aq)

Cold, dilute H2SO4(aq)

Most organic acids (e.g., HC2H3O2)42


2. Oxidizing Acids Anion is stronger oxidizing agent than H3O+

Used to react metals that are less active than H2

No H2 gas formed

HNO3(aq)

Concentrated Dilute Very dilute, with strong reducing agent

H2SO4(aq)

Hot, conc’d, with strong reducing agent Hot, concentrated

43


Nitrate Ion as Oxidizing AgentA. Concentrated HNO3

NO3– more powerful oxidizing agent than H+

NO2 is product Partial reduction of N (+5 to +4) NO3

–(aq) + 2H+(aq) + e– NO2(g) + H2O

Ex.

44

oxidation

reduction

Oxidizing agent

Reducing agent

Cu(s) + 2NO3–(aq) + 4H+(aq) Cu2+(aq) + 2NO2(g) +

2H2O

0 +5 +2 +4


Nitrate Ion as Oxidizing AgentB. Dilute HNO3

NO3– is more powerful oxidizing agent than

H+

NO is product Partial reduction of N (+5 to +2) NO3

–(aq) + 4H+(aq) + 3e– NO(g) + 2H2O

Used to react metals that are less active than H2

Ex. Reaction of copper with dilute nitric acid3Cu(s) + 8HNO3(dil, aq) 3Cu(NO3)2(aq) + 2NO(g) +

4H2O 45


Reactions of Sulfuric Acid A. Hot, Concentrated H2SO4

Becomes potent oxidizer SO2 is product

Partial reduction of S (+6 to +4) SO4

2– + 4H+ + 2e– SO2(g) + 2H2O

Ex. Cu + 2H2SO4(hot, conc.) CuSO4 + SO2 + 2H2O

B. Hot, conc’d, with strong reducing agent H2S is product

Complete reduction of S (+6 to –2) SO4

2– + 10H+ + 8e– H2S(g) + 4H2O

Ex. 4Zn + 5H2SO4(hot, conc.) 4ZnSO4 + H2S + 4H2O

46


Your Turn!Which of the following statements about oxidizing acids is false?

A. H2SO4 can behave as either an oxidizing or nonoxidizing acid, depending on the solution conditions.

B. Oxidizing acids can oxidize metals that are less active than hydrogen.

C. The anions of oxidizing acids are reduced in their reactions with metals.

D. Most strong acids are oxidizing acids.

E. Oxidizing acids are acids whose anions are stronger oxidizing agents than H+.

47


Redox Reactions of Metals Acids reacting with metal

Special case of more general phenomena

Single Replacement Reaction Reaction where one element replaces

another A + BC → AC + B

1.Metal A can replace metal B If A is more active metal, or

2.Nonmetal A can replace nonmetal C If A is more active than C

48


Single Replacement Reaction Left = Zn(s) + CuSO4(aq)

Center = Cu2+(aq) reduced to Cu(s); Zn(s) oxidized to Zn2+(aq)

Right = Cu(s) plated out on Zn bar

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

49


Single Replacement Reaction

Zn2+ ions take place of Cu2+ ions in solution

Cu atoms take place of Zn atoms in solid Cu2+ oxidizes Zn° to Zn2+

Zn° reduces Cu2+ to Cu°

More active Zn° replaces less active Cu2+

Zn° is easier to oxidize!

50


Activity Series of Metals Cu less active, can't replace Zn2+

Can't reduce Zn2+

Cu(s) + Zn2+(aq) No reaction

General phenomenon Element that is more easily oxidized will displace

one that is less easily oxidized from its compounds

Activity Series (Table 6.3) Metals at bottom more easily oxidized (more

active) than those at top

This means that given element will be displaced from its compounds by any metal below it in table

51


How Activity Series Generated

2H+(aq) + Sr(s) Sr2+(aq) + H2(g)

H+ oxidizes Sro to Sr2+

Sro reduces H+ to H2

More active Sro replaces less active H+

Sro is easier to oxidize!

H2 (g) + Sr2+(aq) NO REACTION!

Why? H2 less active, can't replace Sr2+

Can't reduce Sr2+ 52


53

Learning Check: Metal Activity

Mg > Zn > H > Cu

Using the following observations, rank these metals from most reactive to least reactive:

Cu(s) + HCl(aq) → no reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

Mg(s) + ZnCl2(aq) → MgCl2(aq) + Zn(s)


Table 6.3 Activity Series of Some Metals

54


Reactivity Varies by Metal M at very bottom of Table

Very strong reducing agents Very easily oxidized Na down to Cs

Alkali & alkaline earth metals

React with H2O as well as H+

55

2Na(s) + 2H2O H2(g) + 2NaOH(aq)


Reactivity Varies by Metal Ag = no reaction (top of activity series)

2HCl(aq) + Ag(s) 2AgCl(aq) + H2(g)

Zn= somewhat reactive (middle of activity series) 2HCl(aq) + Zn(s) ZnCl2(aq) + H2(g)

Mg = very reactive (bottom of activity series) 2HCl(aq) + Mg(s) MgCl2(aq) + H2(g)

56


Using Activity Series to Predict Reactions

If M is below H Can displace H from solutions

containing H+

2H+ H2(g) If M is above H Doesn't react with Nonoxidizing acids

HCl, H3PO4, etc.In general Metal below replaces ion above

57


Uses of Activity Series

Predictive tool for determining outcome of single replacement reactions

Given M & M'n+ Look at chart & draw arrow from M to M'n+ Arrow that points up from bottom left to

top right will occur Arrow that points down from top left to

bottom right will NOT occur

58


Learning Check

2Au3+(aq) + 3Ca(s)

Au(s) + Ca2+(aq)

Sn(s) + Na+(aq)

Mn(s) + Co2+(aq)

Cu(s) + H+(aq)

2Au(s) + 3Ca2+(aq)rxn occurs

NO reaction

NO reaction

Co(s) + Mn2+(aq)rxn occurs

NO reaction59


Your Turn!The activity series of metals is

Au < Ag < Cu < Sn < Cd < Zn < Al < Mg < Na < Cs

(least active) (most active)

Based on this list, which element would undergo reduction most readily?

A. Ag

B. Al

C. Cu

D. Cd

E. Zn

60


Oxygen as an Oxidizing Agent Oxygen Reacts With Many Substances

Combustion Rapid reaction of substance with oxygen that

gives off both heat and light Hydrocarbons are important fuels Products depend on how much O2 is available

1. Complete Combustion O2 plentiful

CO2 & H2O products

Ex. CH4(g) + 2 O2(g) CO2(g) + 2 H2O

2 C8H18(g) + 25 O2(g) 16 CO2(g) + 18 H2O61


Oxidation of Organic Compounds2. Incomplete Combustion

Not enough O2

a. Limited O2 supply CO is carbon product

2CH4(g) + 3O2(g) 2CO(g) + 4H2O

b. Very limited O2

C(s) is carbon product

CH4(g) + O2(g) C(s) + 2H2O Gives tiny black particles Soot—lamp black Component of air pollution

62


Oxidation of Organic Compounds3. Combustion of Organics containing O

Still produce CO2 & H2O

Need less added O2

C12H22O11(s) + 12 O2(g) 12 CO2(g) + 11 H2O

4. Combustion of Organics containing S Produce SO2 as product

2C4H9SH + 15O2(g) 8CO2(g) + 10H2O + 2SO2(g)

SO2 turns into acid rain when mixed with water

SO2 oxidized to SO3

SO3 reacts with H2O to form H2SO4

63


B. Reaction of Metals with O2 Corrosion

Direct reaction of metals with O2

Many metals corrode or tarnish when exposed to O2

Ex.2Mg(s) + O2(g) 2MgO(s)

4Al(s) + 3O2(g) 2Al2O3(s)

4Fe(s) + 3O2(g) 2Fe2O3(s)

4Ag(s) + O2(g) 2Ag2O(s)

64


C. Reaction of Nonmetals with O2

Many nonmetals react directly with O2 to form nonmetal oxides

Sulfur reacts with O2

Forms SO2

S(s) + O2(g) 2SO2(g)

Nitrogen reacts with O2 Forms various oxides NO, NO2, N2O, N2O3, N2O4, and N2O5

Dinitrogen oxide, N2O

Laughing gas used by dentists Propellant in canned whipped cream 65


66

Learning Check: Complete Following Reactions

Aluminum metal and oxygen gas forms aluminum oxide solid

Solid sulfur (S8) burns in oxygen gas to make gaseous sulfur trioxide

Copper metal is heated in oxygen to form black copper(II) oxide solid

S8(s) + 12O2(g) → 8SO3(g)

2Cu(s) + O2(g) → 2CuO(s)

4Al(s) + 3O2(g) → 2Al2O3(s)


Your Turn! Which of the following reactions is not a redox reaction?

A. Na2S(aq) + MnCl2(aq) 2NaCl(aq) + MnS(s)

B. CH4(g) + O2(g) C(s) + 2H2O

C. 2Zn(s) + O2(g) 2ZnO(s)

D. Cu(s) + 4H+(aq) + 2NO3–(aq) Cu2+(aq) +

2NO2(g) + 2H2O

E. Sr(s) + 2H+(aq) Sr2+(aq) + H2(g)

67


Stoichiometry in Redox Reactions Like any other stoichiometry problem

Balance redox reaction Use stoichiometric coefficients to relate

mole of 1 substance to moles of another

Types of problems Start with mass or volume of one reactant

& find mass or volume of product Perform titrations Have limiting reactant calculations Calculate % yields

68


Stoichiometry in Redox Reactions Ex. How many grams of Na2SO3 (126.1

g/mol) are needed to completely react with 12.4 g of K2Cr2O7 (294.2 g/mol)?

1st need balanced redox equation8H+(aq) + Cr2O7

2(aq) + 3SO32(aq) 3SO4

2(aq)

+ 2Cr3+(aq) + 4H2O

Then do calculations1. g K2Cr2O7 moles K2Cr2O7 moles Cr2O7

2(aq)

2. moles Cr2O72(aq) moles 3SO4

2(aq)

3. moles SO32(aq) moles Na2SO3 g Na2SO3

69


Stoichiometry Example (cont)grams K2Cr2O7 moles K2Cr2O7 moles Cr2O7

2

(aq)

moles Cr2O72 (aq) moles 3SO3

2 (aq)

moles SO32 (aq) moles Na2SO3 g Na2SO3

722

272

722

722722 OCrK mol 1

OCr 1molOCrK g 294.2

OCrK mol 1OCrK g 12.4

272OCr mol 0.0421

272

232

72OCr mol 1

SO mol 3OCr mol 0.0421 2

3SO mol 0.126

70

32

3223

3223 SONa mol 1

SONa g 126.1

SO mol 1

SONa mol 1SO mol 0.126

32SONa g 9.15


Redox Titrations Equivalence point reached when # of moles of

oxidizing & reducing agents have been mixed in the correct stoichiometric ratio

No simple indicators to detect endpoints 3 very useful oxidizing agents that change

color 1. KMnO4: Deep purple of MnO4

fades to almost colorless Mn2+ (very pale pink)

2. K2Cr2O7: Bright yellow orange of Cr2O72 changes

to pale blue green of Cr3+

3. IO3 : When reduced to I2(s) in presence of I,

forms I3 which forms dark blue complex with starch

71


Redox Titration ExampleI reacts with IO3

in acidic solution to form I2(s). If 12.34 mL of 0.5678M I is needed to titrate 25.00 mL of a solution containing IO3

, what is the M of the solution?

1. Write Unbalanced Equation

1 +5 0

I(aq) + IO3(aq) I2(s)

I(aq) is oxidized to I2 IO3

(aq) is reduced to I272


Redox Titration Example (cont)2. Balance Equation

Note: we are in acidic solution

2I(aq) I2(s) + 2e

Not done as not lowest whole number coefficients

5I(aq) + IO3(aq) + 6H+(aq) 3I2(s) + 3H2O

]5 [2IO3

(aq) + 12H+(aq) + 10e I2(s) + 6H2O

10I(aq) + 2IO3(aq) + 12H+(aq) 6I2(s) + 6H2O

2 2 2 2 2

73


3. Now for the Calculations Calculate mmol of I– titrated

Convert to mmol of IO3– present

Convert to M of IO3– solution

ImL 1

I mmol 0.5678ImL 2.341

I mmol 5

IO mmol 1I mmol 7.007 3

3

3

IOmL 25.00

IO mmol 1.401

74

= 0.0561 M IO3–

I mmol 7.007

3IO mmol 1.401


75

A 0.3000 g sample of tin ore was dissolved in acid solution converting all the tin to tin(II). In a titration, 8.08 mL of 0.0500 M KMnO4 was required to oxidize the tin(II) to tin(IV). What was the percentage tin in the original sample?

M of KMnO4 V = mol KMnO4

mol KMnO4 mol Sn/mol KMnO4 = mol Sn2+

mol Sn2+ MM = g Sn2+ in sample

%Sn = g Sn/g sample 100 %

Ore Analysis

3Sn2+(aq) + 2MnO4(aq) + 8H+(aq)

3Sn4+(aq) + 2MnO2(s) + 4H2O


Tin Ore Analysis ContinuedM of KMnO4 V = mmol KMnO4

0.0500 M KMnO4 8.08 mL = 0.404 mmol KMnO4

mmol KMnO4 mmol MnO4 mmol Sn2+

Mol Sn2+ g/mol = g Sn in original sample

%Sn = g Sn/ g sample 100 %

4

2

4

44

MnO mmol 2

Sn mmol 3×

KMnO mmol 1MnO mmol 1

× KMnO mmol 0.404

mg 1000g 1

Sn mmol 1Sn mg 118.7

Sn mmol 1

Sn mmol 1Sn mmol 0.606

22

100ore g 0.3000Sn g 0.07194

76

= 0.606 mmol Sn2+

= 23.97% Sn

= 0.07194 g Sn


Your Turn!The amount of hydrogen peroxide (H2O2, MM = 34.01 g/mol) in hair bleach was determined by titration with a standard KMnO4 (MM = 158.0 g/mol) solution:2MnO4

–(aq) + 5H2O2(aq) + 6H+(aq) 5O2(g) + 2Mn2+(aq) + 8H2O

If 43.2 mL of 0.105 M MnO4– was needed to reach the

endpoint, how many grams of H2O2 are in the sample of hair bleach?

A. 0.771 gB. 0.386 gC. 0.0771 gD. 386 gE. 154 g

77

OH mol 1OH g 10.34

MnO mol 2

OH mol 5L mL/1 1000

mL 43.2MnO 105.0

22

22

4

224

M

= 0.386 g H2O2

Chapter 6: Oxidation-Reduction Reactions Chemistry: The Molecular Nature of Matter, 6E...

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Transcript of Chapter 6: Oxidation-Reduction Reactions Chemistry: The Molecular Nature of Matter, 6E...