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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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Hierarchy of Rules for Assigning Oxidation Numbers
7. Group 6A elements have –2 oxidation number.
8. Group 5A elements have –3 oxidation number.
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ion Electron MethodEx. 2 Balance in Acidic Solution
Cr2O72– + Fe2+ Cr3+ + Fe3+
1. Break into half-reactionsCr2O7
2 Cr3+
Fe2+ Fe3+
2. Balance atoms other than H & O
Cr2O72 2Cr3+
Put in 2 coefficient to balance Cr
Fe2+ Fe3+
Fe already balanced25
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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+ +
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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–
2. Balance atoms other than H & O
MnO4 Mn2+
Balanced for Mn
HSO3 SO4
2
Balanced for S
33
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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+
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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[ ]
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
39
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!
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Table 6.3 Activity Series of Some Metals
54
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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