chemistry 30 – unit 2 electrochemical changes to accompany inquiry into chemistry powerpoint...
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Chemistry 30 – Unit 2 Electrochemical Changes
To accompany Inquiry into Chemistry
PowerPoint Presentation prepared by Robert Schultz
Section 12.1 Characterizing Oxidation Reduction
Chapter 12, Section 12.1• Oxidation: Historically, the reaction
of a substance with O2
• Today, oxidation: loss of electrons
• Reduction: Historically, the reaction of a metal ore to produce a smaller mass of pure metal
• Today, reduction: gain of electrons
Chapter 12, Section 12.1
LEO the Lion says GER
lose
electrons
oxidation gain
electrons
reduction
Memory Tools:
Oil Rig
oxidation is
losing
reduction
is gaining
Chapter 12, Section 12.1• Oxidation and reduction cannot
happen individually – if one substance loses electrons (oxidation), another substance must gain electrons (reduction)
• Oxidation/reduction reactions called “redox reactions”
• Redox Song
Chapter 12, Section 12.1• Consider the example:
Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s)This equation can be rewritten as a total
ionic equation:
Zn(s) + Cu2+(aq) + 2 NO3ˉ(aq) Zn2+(aq) + 2 NO3ˉ(aq)
+ Cu(s)spectators
and finally reduced to a net-ionic equation:Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Chapter 12, Section 12.1
• Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
electrons lostoxidatio
n
electrons gainedreductio
ncauses reduction
called “reducing agent”
causes oxidation
called “oxidizing agent”
Note: the oxidizing agent gets reduced; the reducing agent gets oxidized!!!
Chapter 12, Section 12.1• The net-ionic equation:
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
can be written as two half-reactions,
Cu2+(aq) + 2 e– Cu(s) reduction
Zn(s) Zn2+(aq) + 2 e– oxidation
Chapter 12, Section 12.1• Half-Reactions WS
• Discuss questions 3 and 4, page 437
• Spontaneity of redox reactions – Nelson Lab or Investigation 12.A, page 438
Chapter 12, Section 12.1• Example: Cu(s), Pb(s), Ag(s) and
Zn(s) are reacted with solutions of Cu(NO3)2(aq), Pb(NO3)2(aq), AgNO3(aq), and Zn(NO3)2(aq)
• The experimental results are summarized in the chart:
Cu2+(aq)
Chapter 12, Section 12.1
Cu(s) Pb(s) Ag(s) Zn(s)
spontaneous; non-spontaneous
red
ucin
g
ag
en
tsstrongest oxidizing agent
strongest reducing agent
oxidizing agents
Cu(NO3)2(aq) Pb(NO3)2(aq)
AgNO3(aq)
Zn(NO3)2(aq)
Pb2+(aq) Ag+(aq) Zn2+(aq)
Chapter 12, Section 12.1• Generalizations to this point:
• Oxidizing agents: metal ions, e.g. Ag+(aq), and non-metal elements, e.g. Cl2(g)
• Reducing agents: non-metal ions, e.g. Br –(aq), and metal elements, e.g. Fe(s)
Chapter 12, Section 12.1
“SRA”
Where are the weakest OA and RA?Examine data chart from previous slide. How can chart of half-reactions be used to predict whether or not a reaction is spontaneous?
“SOA”
• Results of experiment can be summarized in a table of redox half-reactions:
Ag+(aq) + e– Ag(s)Cu2+(aq) + 2 e– Cu(s)Pb2+(aq) + 2 e– Pb(s)Zn2+(aq) + 2 e– Zn(s)
Chapter 12, Section 12.1• Spontaneity Generalization:
OAspontaneous
RA
RA
non-spontaneousOA
Chapter 12, Section 12.1Take the chart of half-reactions with Zn, Cu, Pb, and Ag elements and ions and combine with the two data charts on the next page to produce a single chart of half-reactions formatted in the standard way
Chapter 12, Section 12.1
Chapter 12, Section 12.1Ag+(aq) + e– Ag(s)
Cu2+(aq) + 2 e– Cu(s)Pb2+(aq) + 2 e– Pb(s)Zn2+(aq) + 2 e– Zn(s)
Chart 1:
Cd2+(aq) + 2 e– Cd(s)
V2+(aq) + 2 e– V(s)
Be2+(aq) + 2 e– Be(s)
Ra2+(aq) + 2 e– Ra(s)
Chart 2:
Pb2+(aq) + 2 e–
Pb(s)
Cd2+(aq) + 2 e–
Cd(s)
Zn2+(aq) + 2 e–
Zn(s)
V2+(aq) + 2 e–
V(s)
Chapter 12, Section 12.1Ag+(aq) + e– Ag(s)Cu2+(aq) + 2 e– Cu(s)Pb2+(aq) + 2 e– Pb(s)Cd2+(aq) + 2 e– Cd(s)Zn2+(aq) + 2 e– Zn(s)
V2+(aq) + 2 e– V(s)Be2+(aq) + 2 e– Be(s)Ra2+(aq) + 2 e– Ra(s)
Chapter 12, Section 12.1Chart on page 7 of Data Booklet has been prepared in this manner
You can use the chart and the spontaneity generalization developed earlier to predict spontaneity of redox reactions
Chapter 12, Section 12.1• Examples:• #5a, 6a, and Review 6a page 440
purple box bottom of page
Chapter 12, Section 12.2• Writing balanced half-reactions in
acidic or neutral media:
• Step 1: balance elements other than O or H by inspection
• Step 2: balance O’s using H2O(l)’s
• Step 3: balance H’s using H+(aq)’s• Step 4: balance total charge using
electrons (e–)
Must be done in this order!
Chapter 12, Section 12.2• Examples:• Zn2+(aq) Zn(s)
• Cl–(aq) Cl2(g)
Step 4
+ 2 e–
2
Step 1
+ 2 e–
Step 4
neutral media
Chapter 12, Section 12.2• Example: acidic medium
• Cr2O72-(aq) Cr3+(aq)2
Step 1
+ 7 H2O(l)
Step 2
+ 14 H+
(aq)
Step 3
+ 6 e–
Step 4
You’ll find this half-reaction on page 7 of your data booklet
Chapter 12, Section 12.2• You are not required to write half-
reactions in basic media, though you should be able to use given half-reactions if they are in basic media
Chapter 12, Section 12.2• Example: Practice Problem 1a,
page 448• Both of the required half-reactions
are in your data booklet• Half-reactions, those you write
yourself, and those from your data chart can be used to balance redox reaction equations
MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4
H2O(l)
(Ag(s) Ag+(aq) + e–)MnO4
–(aq) + 8 H+(aq) + 5 Ag(s) Mn2+(aq) + 4 H2O(l) + 5 Ag+(aq)
OA RA
5 x
Chapter 12, Section 12.2• Practice Problem 1b, page 448• Neither of these half-reactions are
in your data booklet you must write each yourself
Hg(l) + Cl–(aq) HgCl42–(s)
NO3–(aq) NO2(g)
4 + 2 e–+ H2O(l)
+ 2 H+(aq) + e–
2 x ( )Hg(l) + 4 Cl–(aq) + 2 NO3
–(aq) + 4 H+(aq) HgCl42–(s) + 2 NO2(g) + 2 H2O(l)
Chapter 12, Section 12.2• Do Practice Problems 1c, 1d page
448AsH3(s) + 4 H2O(l) H3AsO4(aq) + 8 H+(aq) + 8 e– 4 x (Zn2+(aq) + 2 e– Zn(s))
AsH3(s) + 4 H2O(l) + 4 Zn2+(aq) 4 Zn(s) + H3AsO4(aq) + 8 H+(aq)
1c
1d
I2(s) + 6 H2O(l) 2 IO3–(aq) + 12 H+(aq)
+ 10 e–5 x (OCl–(aq) + H2O(l) + 2 e– Cl–(aq) + 2 OH−
(aq))I2(s) + 11 H2O(l) + 5 OCl–(aq) 2 IO3
–(aq) + 12 H+(aq) + 5 Cl–(aq) + 10 OH−(aq)
Data Booklet (basic)
Chapter 12, Section 12.2• Half-reactions WS
• Do question 14 a,b,c (acidic conditions) on page 475 (yes I do mean page 475)
• Time to work
Chapter 12, Section 12.2• 14 a• (both half-reactions are in the Data
Booklet)
MnO2(s) + 4 H+(aq) + 2 e– Mn2+(aq) + 2 H2O(l) 2 Cl–(aq) Cl2(g) + 2
e–
MnO2(s) + 4 H+(aq) + 2 Cl–(aq) Mn2+(aq) + 2 H2O(l) + Cl2(g)
Chapter 12, Section 12.2• 14 b
2 x (NO(g) + 3 H+(aq) + 3 e– NH2OH(aq)) 3 x (Sn(s) Sn2+(aq) + 2
e–)2 NO(g) + 6 H+(aq) + 3 Sn(s) 3 Sn2+(aq) + 2 NH2OH(aq)
Chapter 12, Section 12.2• 14 c
3 x (Cd2+(aq) + 2 e– Cd(s))2 x (V2+(aq) + 3 H2O(l) VO3
–(aq) + 6 H+(aq) + 3 e–) 3 Cd2+(aq) + 2 V2+(aq) + 6 H2O(l) 3 Cd(s) + 2 VO3
–(aq) + 12 H+(aq)
Chapter 12, Section 12.2• Read Reducing Iron (to make
steel), pages 452-4
Chapter 12, Section 12.3• Assigning oxidation numbers is a
method of electron bookkeeping – they make it possible to see electron gain and loss without writing half-reactions
• Oxidation Number Rules:1. Pure elements have an oxidation
number of 0. e.g. in Cl2 each Cl atom has an oxidation number of 0; in Fe each Fe atom has an oxidation number of 0.
Chapter 12, Section 12.32. The oxidation number of hydrogen
in compounds is +1. This is true in molecular compounds and acids. Exception: in metal hydrides, e.g. LiH, H is -1
3. The oxidation number of oxygen in compounds is -2. Exceptions: peroxides, e.g. H2O2, where oxygen is -1 and the compound OF2, where oxygen is +2
Chapter 12, Section 12.34. In monatomic ions, e.g. Cl– and
Fe3+, the oxidation number equals the ion charge
5. Oxidation numbers of atoms in compounds add to zero. Oxidation numbers of atoms in polyatomic ions add to the ion charge
Note oxidation numbers can be fractionsDo Oxidation State WS
Chapter 12, Section 12.3• Uses for oxidation numbers:1. Determining whether a substance is
oxidized or reduced – e.g. in the reaction:
Fe changes from 0 to +2; it is oxidizedMn in MnO4
–(aq) changes from +7 to +2 in Mn2+; it is reduced
an increase in oxidation number is oxidation; a decrease in oxidation number is reduction
5 Fe(s) + 2 MnO4–(aq) + 16 H+(aq) 5 Fe2+(aq) + 2 Mn2+(aq) + 8
H2O(l)
0 +7 -2 +1 +2 +2 +1-2
(oxidation numbers in red)
Chapter 12, Section 12.32. Determining the OA and RA in a
redox reaction. The OA is reduced (has a decrease in oxidation number), the RA is oxidized (has an increase in oxidation number)
3. Determining whether or not a reaction is redox. If it is, one* substance will increase in oxidation number (oxidation) and one* will decrease (reduction). If no change in oxidation numbers, the reaction is not redox.
Chapter 12, Section 12.3• Do Practice Problem 11, page 461
H2O2(aq) + 2 Fe(OH)2(s) 2 Fe(OH)3(s)
+1 -1 +2 -2 +1 +3 -2+1Redox
PCl3(l) + 3 H2O(l) H3PO3(aq) + 3 HCl(aq)
+3-1 +1 -2 +1 +3-2 +1-1 Not Redox
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)
-3 +1 0 +4 -2 +1 -2 Redo
x
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)
+4-2 +1 -2 +1+5-2 +2-2Redox
Chapter 12, Section 12.34. Determining the number of
electrons transferred in a redox reaction or half-reaction. e.g.
Note that Mn changes from +7 to +2, a change of 5 electrons. The number of electrons transferred can be determined without a half-reaction!
MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4
H2O(l)
+7 -2 +1 +2 +1 -2
Chapter 12, Section 12.3• Another example:
• Note that Cr changes from +6 to +3, a change of 3, yet the half-reaction has 6 e–. Why???
• There are 2 Cr’s, each changing by 3
2 x 3 = 6
Cr2O72-(aq) + 14 H+(aq) + 6 e– 2 Cr3+(aq) + 7
H2O(l)
+6 -2 +1 +3 +1-2
Chapter 12, Section 12.35. Determining which substance in a
half-reaction is really functioning as the OA (or RA) e.g.:
Note that the Mn in the MnO4−(aq) is
reducedMnO4
−(aq) is the OA H+(aq) is a very necessary spectator
6. Balancing equations by the oxidation number method
MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4
H2O(l)
+7 -2 +1 +2 +1 -2
Chapter 12, Section 12.2• Disproportionation Reaction – a
redox reaction where the same element, is both oxidized and reduced (acts as both OA and RA)
• Example:
• Note that Cl in Cl2 goes from 0 to -1, and from 0 to +1 in NaClO(aq)
2 NaOH(aq) + Cl2(g) NaCl(aq) + NaClO(aq) + H2O(l)
+1-2+1 0 +1 -1 +1 +1-2 +1 -2
Chapter 12, Section 12.3• Another example:
• N is oxidized from +4 to +5 and reduced from +4 to +2
3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g)
+4-2 +1 -2 +1+5-2 +2-2
Chapter 12, Section 12.3• Redox reactions are balanced by
making electrons gained and lost equal
• This can be done using half-reactions or by observing change in oxidation number
• Try balancing the following equation by inspection: S8(s) + KMnO4(aq) + HCl(aq) SO2(g) + MnCl2(s) +
KCl(aq) + H2O(l)
Chapter 12, Section 12.3• Try it using change in oxidation
number:
S8(s) + KMnO4(aq) + HCl(aq) SO2(g) + MnCl2(s) + KCl(aq) + H2O(l)
0 +1+7 -2 +1 -1 +4 -2 +2 -1 +1 -1 +1 -2
4e–/S
32e– /S8
5e–/Mn
5e– /KMnO4
Insert coefficients in equation to make these numbers equal
325
Usually the other coefficients will be easily found
32 32
40
96
48
RA OA
Chapter 12, Section 12.3
H2S(g) + H2O2(aq) S8(s) + H2O(l)
+1 -2 +1 -1 0 +1 -2
2e−/S
2e−
/H2S
1e−/O
2e−
/H2O2
Because of the 8 in S8, it is necessary to put 8 in front of H2S and because of 2e−/2e− ratio, 8 must also go in front of the H2O2
88 16
Do WS 56 B,C
RA OA
Chapter 12, Section 12.4• Redox Titrations
Buret containing solution
Erlenmeyer flask containing solution
titrant
sample
Statement: (applies to all titrations: “titration ofwith ”
sampletitrant
Know this!
Chapter 12, Section 12.4• Oxidizing and reducing agents can
be titrated to find concentrations (just like acids and bases)
• KMnO4(aq) is used as the titrant in many redox titrations – acidified MnO4
–(aq) is a very strong OA in the half-reaction:
MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4
H2O(l)
dark purple
colourless
Chapter 12, Section 12.4• As long as some RA is left in the
sample solution, the purple colour of the MnO4
–(aq) disappears. Once the RA is used up the purple colour remains. Ideally the endpoint is a very light purple
Chapter 12, Section 12.4• Example Practice Problem 19, page
469
-41
-4 32
3
0.02045 0.03895 7.965 10
57.965 10 1.991 10
2
1.991 10 34.02 0.06774
0.06774% 100% 5.276%
1.284mm
moln L molL
n mol mol
gm mol gmolg
g
2 x (MnO4–(aq) + 8 H+(aq) + 5 e– Mn2+(aq) + 4
H2O(l)) 5 x (H2O2(l) O2(g) + 2 H+(aq) +2 e–)
6
n2
m=?
2 MnO4–(aq) + 16 H+(aq) + 5 H2O2(l) 5 O2(g) + 10 H+(aq) + 2 Mn2+
(aq) + 8 H2O(l)2 MnO4
–(aq) + 6 H+(aq) + 5 H2O2(l) 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)n1
0.02045 mol/L
38.95 mL
Chapter 12, Section 12.4• Example Practice Problem 22, page 469The given equation:Cr2O7
2-(aq) + Fe2+(aq) Cr3+(aq) + Fe3+(aq)
is not complete because of the O’sBoth half-reactions are in Data Booklet:
Cr2O72-(aq) + 14 H+(aq) + 6 e– 2 Cr3+(aq) + 7
H2O(l)
6 x (Fe2+(aq) Fe3+(aq) + e–Cr2O72-(aq) + 14 H+(aq) + 6 Fe2+(aq) 6 Fe3+(aq) + 2 Cr3+(aq) +
7 H2O(l)n1
0.02043 mol/L35.55 mL
n2
c=?25.00 mL
41 0.02043 0.03555 7.263 10moln L molL
Chapter 12, Section 12.4
4 32
67.263 10 4.358 10
1n mol mol
34.358 100.1743
0.02500mol molc LL
Chapter 12, Section 12.4• Practice Problems 20 and 21, page
469
• Answers:20. 0.1387% *
21. 0.0110 mol/L
Redox titration calculation worksheet
* Detailed solution at end
checked
mm
Chapter 12, Section 12.4• Vitamin C content of orange juice
experiment (Lab 12.C) Prelab
• Lab
Chapter 12, Section 12.4 Practice Problem 20, page 469
16 H+(aq) + 2 Cr2O72-(aq) + C2H5OH(l) 4 Cr3+(aq) + 2 CO2(g) +
11 H2O(l) n1
0.05023 mol/L32.35 mL
n2
m=?
3 42
11.625 10 8.125 10
2n mol mol
31 0.05023 0.03235 1.265 10moln L molL
48.125 10 46.08 0.03745gm mol gmol
0.03744% 100% 0.1387%
27.00m
m
gg
Chapter 12, Section 12.4