Download Electrochemistry Worksheet With Answers

Assignment #6
Intro to REDOX Reactions
1. State whether the following changes are oxidation or reduction.
Reactions
Classification
oxidation or reduction
a)
b)
c)
d)
e)
f)
g)
h)
Fe2+
2O22N3X5Sn4+
Cu
Y1Br2








Fe3+
O2
N2
X7Sn2+
Cu2+
Y32Br-
Oxidation
Oxidation
Oxidation
Reduction
Reduction
Oxidation
Reduction
Reduction
2. State whether the following changes are oxidation or reduction and write the electrons in the
equation on the appropriate side.
Reactions
Classification
oxidation or reduction
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
Reduction
Reduction
Reduction
Oxidation
Oxidation
Oxidation
Oxidation
Reduction
Oxidation
Reduction
Reduction
Oxidation
(place your electrons on the
appropriate side)
X2 + 2e  2XMn4+ + 2e  Mn2+
P4 + 12e  4P3Z5 Z2- + 3e
8S2 S8 + 16e
+
Cu
 Cu2+ + 1e
Al
 Al3+ + 3e
3S8 + 48e  24S2Q2 Q2+ + 4e
5P4 + 60e  20P3J4+ + 6e  J26N3 3N2 + 18e
3. Write the two half-reactions for each of the following reactions and come up with the net
ionic equation.
a) Ba(s) + 2HCl(aq)  BaCl2(aq) + H2(g)
Ox ½ rxn:
Ba(s)  Ba2+(aq) + 2e
Red ½ rxn:
2H+(aq) + 2e  H2(g)
Net equation:
Ba(s) + 2H+(aq)  Ba2+(aq) + H2(g)
b) 2Al(s) + Zn3(PO4)2(aq)  2AlPO4(aq) + 3Zn(s)
Ox ½ rxn:
2(Al(s)  Al3+(aq) + 3e)
Red ½ rxn:
3(Zn2+(aq)+ 2e  Zn(s))
Net equation:
2Al(s) + 3Zn2+(aq)  2Al3+(aq) + 3Zn(s)
4. Write the two half-reactions for each of the following net ionic equations.
a) 2Al(s) + 6H+(aq)  2Al3+(aq) + 3H2(g)
Ox ½ rxn:
2(Al(s)  Al3+(aq) + 3e)
Red ½ rxn:
3(2H+(aq) + 2e  H2(g))
b) Cl2(g) + 2Br-(aq)  2Cl-(aq) + Br2(g)
Ox ½ rxn:
2Br-(aq)  Br2(g) + 2e
Red ½ rxn:
Cl2(g) + 2e  2Cl-(aq)
c) 3Zn(s) + 2Fe3+(aq)  3Zn2+(aq) + 2Fe(s)
Ox ½ rxn:
3(Zn(s)  Zn2+(aq) + 2e)
Red ½ rxn:
2(Fe3+(aq)+ 3e  Fe(s))
d) 2Mn7+(aq) + 10Cl-(aq)  2Mn2+(aq) + 5Cl2(g)
Ox ½ rxn:
5(2Cl-(aq)  Cl2(g) + 2e)
Red ½ rxn:
2(Mn7+(aq)+ 5e  Mn2+(aq))
5. For each of the reactions below indicate the following in the spaces provided:
 The substance being reduced using the letter R
 The substance being oxidized using the letter O
 The reducing agent using the letters RA
 The oxidizing agent using the letters OA
a)
Br2(aq)
+
R / OA
Sn2+(aq) 
O / RA
2Br-(aq)
______
+
Sn4+(aq)
______
b)
Al3+(aq)
R / OA
3Fe2+(aq) 
O / RA
Al(s)
+
______
3Fe3+(aq)
______
c)
2Ag(s) +
O / RA
Cu2+(aq) 
R / OA
2Ag+(aq)
______
+
d)
2Al(s) +
O / RA
6H+(aq) 
R / OA
2Al3+(aq)
______
+ 3H2(g)
______
e)
Br2(aq)
+
R / OA
2I2-(aq) 
O / RA
2Br-(aq)
______
+
I2(l)
______
f)
Pb2+(aq)
R / OA
+
2K(s)

O / RA
2K+(aq) +
______
Pb(s)
______
g)
Ni2+(aq)
R / OA
+
Ca(s)

O / RA
Ni(s) +
______
Ca2+(aq)
______
+
Cu(s)
______
Assignment #7
Writing Complex REDOX Reactions
1. Write out the half-reaction for the oxidation of manganese(IV) oxide to form the
permanganate ion in an acidic solution.
MnO2(s) + 2 H2O(l) → MnO4– (aq) + 4 H+(aq) + 3e–
2.
Write out the half-reaction for the oxidation of nitrite ions to nitrate ions under basic
conditions.
NO2–(aq) + 2 OH–(aq) → NO3– (aq) + H2O(l) + 2e–
3. Write out the half-reaction for the reduction of the IO3–(aq) ion to iodine in an acidic solution.
2 IO3–(aq) + 12 H+(aq) + 10e–→ I2 (s) + 6 H2O(l)
4. Write out the half-reaction for the oxidation of nitrogen monoxide to form the nitrate ion in
an acidic solution.
NO(g) + 2 H2O(l) → NO3– (aq) + 4 H+(aq) + 3e–
5.
Write out the half-reaction for the reduction of the nitrite ion to nitrogen monoxide gas
under basic conditions.
NO2– (aq) + H2O(l) + 1e–→ NO(g) + 2 OH–(aq)
6. Write out the half-reaction of hydrogen peroxide to water in an acidic solution.
H2O2(aq) + 2 H+(aq) + 2e–→ 2 H2O(l)
7. Write out the half-reaction of lead(IV) oxide being reduced to lead(II) ions in an acidic
solution.
PbO2(s) + 4 H+(aq) + 2e–→ Pb2+(aq) + 2 H2O(l)
8. Write out the half-reaction for the reduction of the VO2+(aq) ion to the VO2+(aq) ion in an acidic
solution.
VO2+(aq) + 2 H+(aq) + 1e–→ VO2+ (aq) + H2O(l)
9.
Write out the half-reaction for the reduction of the permanganate ion to manganese(IV)
oxide in a basic solution.
MnO4– (aq) + 2 H2O(l) + 3e– → MnO2(s) + 4 OH–(aq)
10. Write out the half-reaction for the reduction of the IO4–(aq) ion to the IO3–(aq) ion in an acidic
solution.
IO4–(aq) + 2 H+(aq) + 2e–→ IO3– (aq) + H2O(l)
11. Write out the half-reaction of oxalate ion (C2O42–(aq)) to carbon monoxide gas under basic
conditions.
C2O42–(aq) + 2 H2O(l) + 2e– → 2 CO(g) + 4 OH–(aq)
12. Write out a balanced REDOX reaction for the reaction between permanganate and iron(II)
ions in an acidic solution.
MnO4–(aq) + Fe2+(aq) → Fe3+(aq) + Mn2+(aq)
Red ½ rxn: MnO4–(aq) + 8 H+(aq) + 5e– → Mn2+ (aq) + 4 H2O(l)
Oxid ½ rxn: (Fe2+(aq) → Fe3+(aq) + 1e–) x5
Net ionic:
MnO4–(aq) + 8 H+(aq) + 5 Fe2+(aq) → Mn2+ (aq) + 5 Fe3+(aq) + 4 H2O(l)
13. Cyanide ions are oxidized by permanganate ions in a basic solution.
CN–(aq) + MnO4–(aq) → CNO–(aq) + MnO2(s)
Red ½ rxn: (MnO4–(aq) + 2 H2O(l) + 3e– → MnO2(s) + 4 OH–(aq))x2
Oxid ½ rxn: (CN–(aq) + 2 OH–(aq) → CNO–(aq) + H2O(l) + 2e–) x3
Net ionic:
2 MnO4–(aq) + H2O(l) + 3 CN–(aq) → 2 MnO2(s) + 3 CNO–(aq) + 2 OH–(aq)
14. Silver is oxidized by permanganate ions in an acidic solution.
Ag(s) + MnO4–(aq) → Ag+(aq) + Mn2+(aq)
Red ½ rxn: MnO4–(aq) + 8 H+(aq) + 5e– → Mn2+(aq) + 4 H2O(l)
Oxid ½ rxn: (Ag(s) → Ag+(aq) + 1e–) x5
Net ionic:
MnO4–(aq) + 5 Ag(s) + 8 H+(aq) → 5 Ag+ (aq) + Mn2+(aq) + 4 H2O(l)
15. Balance the following reaction under acidic conditions.
Hg(l) + NO3–(aq) + Cl–(aq) → HgCl4–(aq) + NO2(g)
Red ½ rxn: (NO3–(aq) + 2 H+(aq) + 1e– → NO2(g) + H2O(l))x3
Oxid ½ rxn: Hg(l) + 4 Cl–(aq) → HgCl4–(aq) + 3e–
Net ionic:
Hg(l) + 4 Cl–(aq) + 3 NO3–(aq) + 6 H+(aq) → HgCl4–(aq) + 3 NO2(g) + 3 H2O(l)
16. Balance the REDOX reaction between dichromate ion and iodide ion to form chromium(III)
ion and solid iodine, which occurs in acidic solutions.
I–(aq) + Cr2O72–(aq) → Cr3+(aq) + I2(s)
Red ½ rxn: Cr2O72–(aq) + 14 H+(aq) + 6e– → 2 Cr3+(s) + 7 H2O(l)
Oxid ½ rxn: (2 I–(aq) → I2(s) + 2e–) x3
Net ionic:
Cr2O72–(aq) + 14 H+(aq) + 6 I–(aq) → 3 I2(s) + 2 Cr3+(aq) + 7 H2O(l)
17. Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an
indicator in REDOX titrations. It reacts in basic solution with oxalate ion to form carbonate
ion and solid manganese dioxide. Write a balanced equation that represents this process.
C2O42–(aq) + MnO4–(aq) → CO32–(aq) + MnO2(s)
Red ½ rxn: (MnO4–(aq) + 2 H2O(l) + 3e– → MnO2(s) + 4 OH–(aq))x2
Oxid ½ rxn: (C2O42–(aq) + 4 OH–(aq) → 2 CO32–(aq) + 2 H2O(l) + 2e–)x3
Net ionic:
2 MnO4–(aq) + 4 OH–(aq) + 3 C2O42–(aq) → 2 MnO2(s) + 6 CO32–(aq) + 2 H2O(l)
18. Balance the following reaction under basic conditions.
BrO3-(aq) + I-(aq) → I2(l) + Br-(aq)
Red ½ rxn: BrO3–(aq) + 3 H2O(l) + 6e– → Br–(aq) + 6 OH–(aq)
Oxid ½ rxn: (2 I–(aq) → I2(s) + 2e–) x3
Net ionic:
19.
BrO3-(aq) + 6 I-(aq) + 3 H2O(l)  3 I2(l) + Br-(aq) + 6 OH-(aq)
Copper(II) oxide reacts with ammonia to produce copper solid and
nitrogen gas. Write out a balanced REDOX reaction under acidic conditions.
CuO(s) + NH3(aq) → Cu(s) + N2(g)
Red ½ rxn: (CuO(s) + 2 H+(aq) + 2e– → Cu(aq) + H2O(l))x3
Oxid ½ rxn: 2 NH3(aq) → N2(s) + 6 H+(aq) + 6e–
Net ionic:
20.
3 CuO(s) + 2 NH3(aq)  3 Cu(s) + N2(g) + 3 H2O(l)
Write out the balanced REDOX reaction for the reaction of perchlorate
ion with chromium metal to produce chromite ion and chlorate ion in basic conditions.
ClO4-(aq) + Cr(s) → CrO2-(aq) + ClO3-(aq)
Red ½ rxn: (ClO4-(aq) + H2O(l) + 2e– → ClO3-(aq) + 2 OH–(aq))x3
Oxid ½ rxn: (Cr(s) + 4 OH–(aq) → CrO2-(aq) + 2 H2O(l) + 3e–)x2
Net ionic:
3 ClO4-(aq) + 2 Cr(s) + 2 OH-(aq)  2 CrO2-(aq) + 3 ClO3-(aq) + H2O(l)
Assignment #8
Building REDOX Tables
1. Create a table of REDOX from the following data collected from an experiment.
Be2+(aq) Cd2+(aq) Ra2+(aq) V2+(aq)
―
√
―
√
―
―
―
―
√
√
―
√
―
√
―
―
Be(s)
Cd(s)
Ra(s)
V(s)
Cd2+(aq) + 2e → Cd(s)
V2+(aq) + 2e → V(s)
Be2+(aq) + 2e → Be(s)
Ra2+(aq) + 2e → Ra(s)
2. Create a table of REDOX from the following data collected from an experiment.
Zn2+(aq)
V2+(aq)
Cd2+(aq)
Pb2+(aq)
Zn(s)
―
―
√
√
V(s)
√
―
√
√
Cd(s)
―
―
―
√
Pb(s)
―
―
―
―
Pb2+(aq) + 2e → Pb(s)
Cd2+(aq) + 2e → Cd(s)
Zn2+(aq) + 2e → Zn(s)
V2+(aq) + 2e → V(s)
3. Combine your answers from question 1 and 2 along with activity 13.0 to create a larger
REDOX table.
Ag+(aq) + 1e → Ag(s)
Cu2+(aq) + 2e → Cu(s)
Pb2+(aq) + 2e → Pb(s)
Cd2+(aq) + 2e → Cd(s)
Zn2+(aq) + 2e → Zn(s)
V2+(aq) + 2e → V(s)
Be2+(aq) + 2e → Be(s)
Ra2+(aq) + 2e → Ra(s)
4. Using the evidence from the following REDOX reactions, set up a table of relative strengths
of oxidizing and reducing agents.
Co(s) + Pd2+(aq) → Co2+(aq) + Pd(s)
Pd(s) + Pt2+(aq) → Pd2+(aq) + Pt(s)
Mg(s) + Co2+(aq) → Mg2+(aq) + Co(s)
Pt2+(aq) + 2e → Pt(s)
Pd2+(aq) + 2e → Pd(s)
Co2+(aq) + 2e → Co(s)
Mg2+(aq) + 2e → Mg(s)
5. The following equations are interpretations of the evidence from the reactions of four metals
with various solutions. Make a table of REDOX half-reactions.
Cd(s) + 2 H+(aq) → Cd2+(aq) + H2(g)
Hg(l) + 2 H+(aq) → no evidence
Be(s) + Cd2+(aq) → Be2+(aq) + Cd(s)
Be(s) + Ca2+(aq) → no evidence
Hg2+(aq) + 2e → Hg(l)
2 H+(aq) + 2e → H2(g)
Cd2+(aq) + 2e → Cd(s)
Be2+(aq) + 2e → Be(s)
Ca2+(aq) + 2e → Ca(s)
6. In an experiment four metals were placed into test tubes containing various ion solutions.
Their resulting behavior is communicated by the equations below. Create a REDOX halfreaction table.
2 Ce(s) + 3 Ni2+(aq) → 2 Ce3+(aq) + 3 Ni(s)
Pt(s) + 2 H+(aq) → no evidence
3 Sr(s) + 2 Ce3+(aq) → 3 Sr2+(aq) + 2 Ce(s)
Ni(s) + 2 H+(aq) → Ni2+(aq) + H2(g)
Pt2+(aq) + 2e → Pt(s)
2 H+(aq) + 2e → H2(g)
Ni2+(aq) + 2e → Ni(s)
Ce3+(aq) + 3e → Ce(s)
Sr2+(aq) + 2e → Sr(s)
7. An analytical chemist reacts an unknown metal X with a copper(II) sulfate solution, plating
out copper metal. Metal X does not react with a zinc nitrate solution. What is the REDOX
table for these reactions?
Cu2+(aq) + 2e → Cu(s)
X2+(aq) + 2e → X(s)
Zn2+(aq) + 2e → Zn(s)
8. What groups of metals are eliminated as a possible identity of metal X?
Metal X can not be group 1, 2, or 3
Assignment #9
Predicting REDOX Reactions
For the following assignment you are to write the net ionic equations for the chemical combinations as well you are
to state whether or not the reactions are spontaneous or non-spontaneous.
1. Iron is used in an environment containing aqueous magnesium chloride. An example of this
is when one is working under ocean waters.
Entities
Mg2+
ClFe SRA
H2O SOA
SRA: Fe(s)  Fe2+(aq) + 2e
SOA: 2H2O(l) + 2e  H2(g) + 2OH-(aq)
net: Fe(s) + 2H2O(l)  Fe2+(aq) + H2(g) + 2OH-(aq)
RA above OA on the list  non-spontaneous
2. A lab technician stores an aqueous solution of iron(III) chloride in a nickel plated container.
Entities
Fe3+ SOA
ClNi SRA
H2O
SRA: Ni(s)  Ni2+(aq) + 2e
SOA: 2(Fe3+(aq) + 1e  Fe2+(aq))
net: 2Fe3+(aq) + Ni(s)  2Fe2+(aq) + Ni2+(aq)
OA above RA on the list  spontaneous
3. A student uses hydrobromic acid to acidify a solution of potassium dichromate for later use
in a REDOX titration experiment.
Entities
H+
Cr2O72- SOA
Br- SRA
K+
SRA: 3(2Br-(aq)  Br2(l) + 2e)
SOA: Cr2O72-(aq) + 14H+(aq) + 6e  2Cr3+(aq)+ 7H2O(l)
net: Cr2O72-(aq) + 14H+(aq) + 6Br-(aq)  2Cr3+(aq) + 3Br2(l) + 7H2O(l)
OA above RA on the list  spontaneous
4. A solution of tin(II) chloride is added to a solution of lead(IV) chloride.
Entities
Sn2+ SOA
ClPb4+
H2O SRA
SRA: 2H2O(l)  O2(g) + 4H+(aq) + 4e
SOA: 2(Sn2+(aq) + 2e  Sn(s))
net: 2Sn2+(aq) + 2H2O(l)  2Sn(s) + O2(g) + 4H+(aq)
RA above OA on the list  non-spontaneous
5. A student uses copper electrodes to test the conductivity of a nitric acid solution.
Entities
Cu SRA
H+
NO3- SOA
H2O
SRA: Cu(s)  Cu2+(aq) + 2e
SOA: 2NO3-(aq) + 4H+(aq) + 2e  N2O4 (g) + 2H2O(l)
net: 2NO3-(aq) + 4H+(aq) + Cu(s)  Cu2+(aq) + N2O4 (g) + 2H2O(l)
OA above RA on the list  spontaneous
6. An iron bolt is exposed to air and water. (assume the air to be just oxygen)
Entities
Fe SRA
O2 SOA
H2O
SRA: 2(Fe(s)  Fe2+(aq) + 2e)
SOA: O2(g) + 2H2O(l) + 4e  4OH-(aq)
net: 2Fe(s) + O2(g) + 2H2O(l)  2Fe2+(aq) + 4OH-(aq)
OA above RA on the list  spontaneous
7. A solution of acidic potassium dichromate is mixed with an aqueous solution of hydrogen
peroxide.
Entities
H+
Cr2O72- SOA
K+
H2O2 SRA
SRA: 3(H2O2(l)  O2(g) + 2H+(aq) + 2e)
SOA: Cr2O72-(aq) + 14H+(aq) + 6e  2Cr3+(aq)+ 7H2O(l)
net: Cr2O72-(aq) + 8H+(aq) + 3H2O2(l)  2Cr3+(aq) + 3O2(g) + 7H2O(l)
OA above RA on the list  spontaneous
8. A chemistry teacher demonstrates the test for bromide ions by bubbling some chlorine gas
through a sodium bromide solution.
Entities
Na+
Br- SRA
Cl2 SOA
H2O
SRA: 2Br- (aq)  Br2(l) + 2e
SOA: Cl2(g) + 2e  2Cl-(aq)
net: 2Br-(aq) + Cl2(g)  2Cl-(aq) + Br2(l)
OA above RA on the list  spontaneous
9. Aqueous solutions of tin(II) bromide and iron(III) nitrate are mixed.
Entities
Sn2+ SRA
BrFe3+ SOA
NO3H2O
SRA: Sn2+(aq)  Sn4+(aq) + 2e
SOA: 2(Fe3+(aq) + 1e  Fe2+(aq))
net: 2Fe3+(aq) + Sn2+(aq)  2Fe2+(aq) + Sn4+(aq)
OA above RA on the list  spontaneous
10. Two students try etching their initials on a copper plate using hydrochloric acid.
Entities
H+ SOA
ClCu SRA
H2O
SRA: Cu(s)  Cu2+(aq) + 2e
SOA: 2H+(aq) + 2e  H2(g)
net: Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)
RA above OA on the list  non-spontaneous
11. Sodium metal is added to some water in a typical demonstration of the reactivity of alkali
metals.
Entities
Na SRA
H2O SOA
SRA: 2(Na(s)  Na+(aq) + 1e)
SOA: 2H2O(l) + 2e  H2(g) + 2 OH-(aq)
net: 2Na(s) + 2H2O(l)  2Na+(aq) + H2(g) + 2OH-(aq)
OA above RA on the list  Spontaneous
Assignment #10
Assigning Oxidation Numbers
1. Find the oxidation number for the sulfur atoms in the following compounds.
0
+4
+2
+6
+6
+6
+4
S8
SO2
S2O32-
H2SO4
MgSO4
SO42-
H2SO3
2. Find the oxidation number for all the phosphorous atoms in the following compounds.
+5
+3
P2O5
0
PCl3
+3
P4
+5
H3PO3
Na3PO4
+5
+5
PO43-
H3PO4
3. Find the oxidation number for all the carbon atoms in the following compounds.
+4
+3
+4
+4
+4
+4
-2
CO2
C2O42-
CO32-
NaHCO3
H2CO3
CaCO3
C2H5OH
+2
+3
0
MnO
ClNO
C6H12O6
-1
+5
+1
+4
NaH
S2O62-
HXeO
SnCl4
+6
+5
4. Find the oxidation number of each of the underlined species.
+3
-4
+1
H3BO3
CH4
OBr3+
+3
+2
+5
VO2
+
HNO2
XeO4
+4
6-
HCO3-
+3
+6
-1
-1
-1
NO+
Cr2O72-
CaH2
NH3OH+
Na2O2
MnO42-
Na2HPO4
Assignment #11
Using Oxidation Numbers
1.
State whether the following changes are Oxidation (OX) or Reduction (RED) in nature.
a) AsO3- 
c) H2O2

e) NH4+

3+
g) Cr

i) Au

+
k) NH3OH 
2.
AsO2
O2
N2O5
Cr2O72Au(CN)4NO2
Ox -1 ot 0
Ox -3 to +5
Ox +3 to +6
Ox 0 to +3
Ox -1 to +4






MnO2
CO2
CH4
I2
H2S
P2O5
Red +7 to +4
Ox -2 to +4
Red +3 to –4
Red +7 to 0
Red +4 to –2
Neither
State whether the following changes are Oxidation (OX) or Reduction (RED) in nature and
indicate how many electrons must be lost or gained in the process.
Ox or
Red
a) C2O42- 
c) ClO3
e) Na2HPO4 
g) XO54
i) 2P2O5 
k) C12H22O11
3.
b) MnO4d) C2H3OH
f) C2O42h) H4IO6j) H2SO3
l) Na2HPO4
Red +5 to +4
2CO2
ClPO2
H3 X
P4
12C
Ox
Red
Red
Red
Red
none
e- lost or
gianed
Lost 2
Ox or
Red
b) NiO2
 Ni(OH)2
d) C2H5OH  2CO2
f) N2O3
 2NH3
h) KMnO4  MnO2
j) S8
 8SO42-
Gained 6
Gained 1
Gained 9
Gained 20
Red
Ox
Red
Red
Ox
e- lost or
gianed
Gained 2
Lost 12
Gained 12
Gained 3
Lost 48
No change
For each of the following reactions identify the oxidizing agent and the reducing agent by
circling each and labeling it OA or RA respectively.
a) Sn
+
NO3- 
SnO2 +
NO
RA
OA
O2RA

b) PbCl4
OA
+
PbCl2 +
O2
c) Fe3+
OA
+
NH3OH+ 
RA
Fe2+
+
N2O
d) H3PO2
RA
+
Cr2O72- 
OA
Cr3+
+
H3PO4
e) VO2+
OA
+
Sn2+
RA

VO2+ +
Sn4+
f) N2O4
OA
+
H2
RA

HOH +
N2
g) CrO2RA
+
S2O8
OA

CrO42- +
SO42-
h) XeO3
OA
+
IRA

Xe
I2
+
4.
In the balanced equations below, establish the oxidation numbers for all atoms involved, and
use them to identify the substance being oxidized (RA) and the substance being reduced
(OA) by circling it and labeling it.
+1 +5 –2
-2 +1 -2 +1
OA
RA
+4 –2
+1 -1
+1 -2
a) 2KClO3(s) + C2H5OH(l)  2CO2(g) + 2KCl(s) + 3H2O(l)
+1–2+1
-1
-
0
0
-2+1
b) 2HOH(l) + 2I (aq)  I2(s) + H2(g) + 2OH-(aq)
OA
RA
+4 –2
0
-3 +1
+1-2+1
0
+7 –2
c) 2NO2(g) + 7H2(g)  2NH3(g) + 4HOH(g)
OA
RA
+5 –2
+1
-
d) 7ClO3 (aq) +
2H+(aq)
OA/RA
Example of disproportionation
+1 -2
 Cl2(g) + 5ClO4 (aq) + H2O(l)
-
Assignment #12
Balancing Equations Using Oxidation Numbers
For the following reaction equations, you are to use oxidation numbers to identify the OA, RA,
and to balance the equation.
Remember to show all your work.
+3 -2
+5-2
+5 –2
-1
+5
-1
1. AsO33–(aq) + IO3–(aq)  AsO43–(aq) + I–(aq)
+3
lost 2e
tot lost 2e
x3
+5
gained 6e
tot gained 6e
x1
3 AsO33–(aq) + IO3–(aq)  3 AsO43–(aq) + I–(aq)
+7 -2
2.
+2
MnO4-(aq)
+7
gained 5e
tot gained 5e
+1
2+
+ Fe
(aq)
+ H
+2
+
(aq)
 Mn
2+
+2
lost 1e
tot lost 1e
x5
+3
(aq)
+1 -2
3+
+ Fe
+2
(aq)
+ H2O(l)
+3
MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq)  Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
+3 -2
+5 -2
+5 -2
-1
+5
-1
3. AsO33-(aq) + BrO3-(aq)  AsO43-(aq) + Br-(aq)
+3
lost 2e
tot lost 2e
x3
+5
gained 6e
tot gained 6e
3 AsO33-(aq) + BrO3-(aq)  3 AsO43-(aq) + Br-(aq)
+1+4-2
4.
HPO3-(aq)
+4
lost 1e
tot lost 1e
x2
0
+1 –2
+ I2(l) + H2O(l) 
0
gained 1e
tot gained 2e
+1 +5 –2
-1
H2PO4-(aq)
I-(aq)
+5
+
+1
+ H+(aq)
-1
2 HPO3-(aq) + I2(l) + 2 H2O(l)  2 H2PO4-(aq) + 2 I-(aq) + 2 H+(aq)
+2 -2
-3+1
0
0
-3
lost 3e
tot lost 3e
x2
0
0
+1 -2
5. CuO(s) + NH3(aq)  Cu(s) + N2(g) + H2O(l)
+2
gained 2e
tot gained 2e
x3
3 CuO(s) + 2 NH3(aq)  3 Cu(s) + N2(g) + 3 H2O(l)
+4 –2
6.
SeO32-(aq)
+4
lost 2e
tot lost 2e
0
+ Cl2(g) +
-2+1
OH-(aq)
0
gained 1e
tot gained 2e

+6 –2
-1
SeO42-(aq)
Cl-(aq)
+
+6
+1 -2
+ H2O(l)
-1
SeO32-(aq) + Cl2(g) + 2 OH-(aq)  SeO42-(aq) + 2 Cl-(aq) + H2O(l)
+4 –2
-1
+1 –2
+3 –2
+5 –2
+3
+5
+1
7. N2O4(aq) + Br-(aq) + H2O(l)  NO2-(aq) + BrO3-(aq) + H+(aq)
+4
gained 1e
tot gained 2e
x3
-1
lost 6e
tot lost 6e
3 N2O4(aq) + Br-(aq) + 3 H2O(l)  6 NO2-(aq) + BrO3-(aq) + 6 H+(aq)
+1 –1
+3 –2
-1
gained 1e
tot gained 2e
x3
+3
lost 3e
tot lost 3e
x2
+6 –2
+1 –2
+1
8. H2O2(aq) + CrO2-(aq)  CrO42-(aq) + H2O(l) + H+(aq)
+6
-2
3 H2O2(aq) + 2 CrO2-(aq)  2 CrO42-(aq) + 2 H2O(l) + 2 H+(aq)
+6 –2
+4-2
+6
gained 3e
tot gained 3e
x2
+4
lost 2e
tot lost 2e
x3
+1 –2
+3 –2
+6-2
+3
+6
-2+1
9. CrO42-(aq) + SO32-(aq) + H2O(l)  CrO2-(aq) + SO42-(aq) + OH-(aq)
2 CrO42-(aq) + 3 SO32-(aq) + H2O(l)  2 CrO2-(aq) + 3 SO42-(aq) + 2 OH-(aq)
+5 –2
+1 –2
-1
0
-1
0
-1
-2+1
10. BrO3-(aq) + I-(aq) + H2O(l)  I2(l) + Br-(aq) + OH-(aq)
+5
gained 6e
tot gained 6e
-1
lost 1e
tot lost 1e
x6
BrO3-(aq) + 6 I-(aq) + 3 H2O(l)  3 I2(l) + Br-(aq) + 6 OH-(aq)
+5 –2
0
+1 –2
NO3-(aq)
11. Zn(s) +
-3+1
+ H2O(l)  NH4
0
+5
lost 2e gained 8e
tot lost 2e tot gained 8e
x4
+2
+
(aq)
+ Zn
-3
-2+1
2+
+ OH-(aq)
(aq)
+2
4 Zn(s) + NO3-(aq) + 7 H2O(l)  NH4+(aq) + 4 Zn2+(aq) + 10 OH-(aq)
+7 -2
12.
-3+1
MnO4-(aq)
+7
gained 2e
tot gained 2e
x8
+1
+ NH3(aq) + H
+
+4 -2
(aq)
 MnO2(aq) +
-3
lost 8e
tot lost 8e
x3
+4
+5-2
+1 -2
NO3-(aq)
+ H2O(l)
+5
8 MnO4-(aq) + 3 NH3(aq) + 5 H+(aq)  8 MnO2(aq) + 3 NO3-(aq) + 7 H2O(l)
+5-2
13.
IO3-(aq)
+4-2
+1 –2
0
+ SO2(g) + H2O(l)  I2(l) +
+5
gained 5e
tot gained 5e
x2
+4
lost 2e
tot lost 2e
x5
0
+6 –2
+1
SO42-(aq)
+ H+(aq)
+6
2 IO3-(aq) + 5 SO2(g) + 4 H2O(l)  I2(l) + 5 SO42-(aq) + 8 H+(aq)
+7 –2
14.
ClO4-(aq)
+7
gained 2e
tot gained 2e
x3
0
-2+1
+ Cr(s) + OH (aq) 
-
0
lost 3e
tot lost 3e
x2
+3 –2
+5 –2
CrO2-(aq)
ClO3-(aq)
+3
+
+1 -2
+ H2O(l)
+5
3 ClO4-(aq) + 2 Cr(s) + 2 OH-(aq)  2 CrO2-(aq) + 3 ClO3-(aq) + H2O(l)
+6 -2
-1
+6
gained 3e
tot gained 6e
-1
lost 1e
tot lost 1e
x6
+1
+3
0
+3
0
+1 -2
15. Cr2O72-(aq) + Cl-(aq) + H+(aq)  Cr3+(aq) + Cl2(g) + H2O(l)
Cr2O72-(aq) + 6 Cl-(aq) + 14 H+(aq)  2 Cr3+(aq) + 3 Cl2(g) + 7 H2O(l)
+1 +3 -2
+7 -2
+1
+4 –2
+2
+4
+2
+1 -2
16. H2C2O4(aq) + MnO4-(aq) + H+(aq)  CO2(g) + Mn2+(aq) + H2O(l)
+3
lost 1e
tot lost 2e
x5
+7
gained 5e
tot gained 5e
x2
5 H2C2O4(aq) + 2 MnO4-(aq) + 6 H+(aq)  10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l)
+2 –2
-3 +1
0
-3
lost 3e
tot lost 3e
x2
0
+1 –2
0
17. CuO(s) + NH3(g)  N2(g) + H2O(l) + Cu(s)
+2
gained 2e
tot gained 2e
x3
0
3 CuO(s) + 2 NH3(g)  N2(g) + 3 H2O(l) + 3 Cu(s)
+7 -2
+1 -2
+1
+7
gained 5e
tot gained 5e
x2
-2
lost 2e
tot lost 2e
x5
+2
0
+2
0
+1 -2
18. MnO4-(aq) + H2Se(g) + H+(aq)  Mn2+(aq) + Se(s) + H2O(l)
2 MnO4-(aq) + 5 H2Se(g) + 6 H+(aq)  2 Mn2+(aq) + 3 Se(s) + 8 H2O(l)
+4 –2
0
+1+6-2
+2+6-2
+1 -2
19. PbO2(s) + Pb(s) + H2SO4(aq)  PbSO4(s) + H2O(l)
+4
gained 2e
tot gained 2e
0
lost 2e
tot lost 2e
+2
PbO2(s) + Pb(s) + 2 H2SO4(aq)  2 PbSO4(s) + 2 H2O(l)
20. In a reaction of acidified potassium dichromate with zinc, 3.75 g of zinc was reacted. What
was the concentration of potassium dichromate if 28.3 mL was used?
1 Cr2O72-(aq) + 3 Zn(s) + 14 H+(aq)  2 Cr3+(aq) + 3 Zn2+(aq) + 7 H2O(l)
+6
G 3e
TG 6e
0
L 2e
TL 2e
 ng = m = 0.057348218
M
+3
+2
 nr = Coef r * ng = 0.0191160728
Coef g
 C = n = 0.675 mol/L
v
21. In the following reaction below, you are to find the mass of carbon dioxide that is produced
when 36.5 mL of a 1.50 mol/L solution of potassium permanganate was used.
2 MnO4-(aq) + 5 C2O42-(aq) + 16 H+(aq)  2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)
+7
G 5e
TG 5e
+3
L 1e
TL 2e
 ng = Cv = 0.05475
+2
+4
 nr = Coef r * ng = 0.27375
Coef g
 m = nM = 12.0 g
22. What concentration of phosphorous acid was used in the reaction below if we got 6.35 g of
nitrogen monoxide? We used 3.50 L of the acid.
2 NO3-(aq) + 3 H3PO3(aq) + 2 H+(aq)  2 NO(g) + 3 H3PO4(aq) + 1 H2O(l)
+5
G 3e
TG 3e
+3
L 2e
TL 2e
 ng = m = 0.2115961346
M
+2
+5
 nr = Coef r * ng = 0.3173942019
Coef g
 C = n = 0.0907mol/L
v
23. What mass of silver was reacted if 5.24 g of nitrogen monoxide was produced?
1 NO3-(aq) + 3 Ag(s) + 4 H+(aq)  3 Ag+(aq) + 1 NO(g) + 2 H2O(l)
+5
G 3e
TG 3e
0
L 1e
TL 1e
 ng = m = 0.1746084638
M
+1
+2
 nr = Coef r * ng = 0.5238253915
Coef g
 m = nM = 56.5 g
Assignment #13
REDOX Stoichiometry
1. The silver in a 5.90 g chunk of material was removed and then reacted with and acidified
solution of potassium dichromate. It was found that 35.9 mL of a 0.200 mol/L solution of
potassium dichromate was required for the reaction. Calculate the following:
e) The number of moles of dichromate ions used in the reaction.
Answer: 0.00718 mol
+
2K2Cr2O7(aq)  2K (aq) + Cr2O7 (aq)
C = 0.200 mol/L
v = 0.0359 L
n = Cv = (0.200 mol/L)(0.0359 L)
= 0.00718 mol
1 : 1 ratio  0.00718 mol of Cr2O72-(aq)
f) The number of grams of silver that reacted.
Answer: 4.65 g
Cr2O72-(aq)
+
14H+(aq)
+ 6Ag(s)  7H2O(l) +
n = 0.00718 mol
 nr = 6 x ng
2Cr3+(aq)
+
6Ag+(aq)
m=?
M = 107.87 g/mol
 m = nrM = (0.04308 mol)(107.87 g/mol)
1
= 4.6470396 g
 4.65 g
g) The percentage of silver that is in the chunk.
Answer: 78.8 %
Reacted x 100% = (4.65 g)(100%) = 78.813559 %
Actual
5.90 g
 78.8 %
2. A 0.400 mol/L solution of tin(II) nitrate was titrated with an acidified potassium
permanganate solution. Use the following data below to calculate the molar concentration of
the potassium permanganate solution.
Volume of tin(II) nitrate used is 10.0 mL
Volumes of potassium permanganate solution required
Trial
1
2
3
Initial buret reading (mL)
8.70
17.30
26.00
Final buret reading (mL)
17.30
26.00
34.80
2MnO4-(aq)
+
16H+(aq)
v = 0.00870 L
C=?
 ng = Cv = (0.400 mol/L)(0.0100 L)
= 0.00400 mol
 nr = 2/5ng  0.00160 mol
+
5Sn2+(aq)
 8H2O(l) +
2Mn2+(aq)
Answer: 0.184 mol/L
+ 5Sn4+(aq)
C = 0.400 mol/L
v = 0.0100 L
 C = nr = 0.00160 mol = 0.1839081 mol/L  0.184 mol/L
v
0.00870 L
3. In a titration experiment all the bromide ions in an acidic solution were oxidized to bromine
by a 0.0200 mol/L potassium permanganate solution. The volume of the bromide ion solution
used was 25.0 mL. Use the data below to calculate the concentration of the bromide ions in
the solution.
Volumes of potassium permanganate solution required
Trial
1
2
3
Final buret reading (mL)
17.30 23.00 24.80
Initial buret reading (mL)
2.30 7.00 10.80
Answer: 0.0600 mol/L
+
2+
2MnO4 (aq) + 16H (aq) + 10Br (aq)  8H2O(l) + 2Mn (aq) + 5Br2(l)
v = 0.0150 L
C = 0.0200 mol/L
 ng = Cv = (0.0200 mol/L)(0.0150 L)
= 0.000300 mol
 nr = 10/2ng  0.00150 mol
C=?
v = 0.0250 L
 C = nr = 0.00150 mol  0.0600 mol/L
v
0.0250 L
4. In a REDOX titration 12.50 mL of 0.0800 mol/L potassium dichromate was used in an acidic
solution to oxidize tin(II) ions to tin(IV) ions. The volume of tin(II) ions that was oxidized
was 10.0 mL. Calculate the concentration of the tin(II) ions in the solution. Also what color
change would indicate the endpoint of this reaction?
Answer: 0.300 mol/L
Cr2O72-(aq) + 14H+(aq) + 3Sn2+(aq)  7H2O(l) + 2Cr3+(aq) + 3Sn4+(aq)
v = 0.0125 L
C = 0.0800 mol/L
 ng = Cv = (0.0800 mol/L)(0.0125 L)
= 0.00100 mol
3
 nr = /1ng  0.00300 mol
C=?
v = 0.0100 L
 C = nr = 0.00300 mol  0.300 mol/L
v
0.0100 L
5. What volume of 0.0500 mol/L KMnO4(aq) is needed to oxidize all the bromide ions in 25.0
mL of an acidic 0.200 mol/L NaBr solution? Also what color change would indicate the
endpoint of this reaction?
Answer: 0.0200 L
2MnO4-(aq) + 16H+(aq) + 10Br-(aq)  8H2O(l) + 2Mn2+(aq) + 5Br2(l)
v=?
C = 0.0500 mol/L
 ng = Cv = (0.200 mol/L)(0.0250 L)
= 0.00500 mol
2
 nr = /10ng  0.00100 mol
C = 0.200 mol/L
v = 0.0250 L
 v = nr = 0.00100 mol  0.0200 L
C
0.0500 L
6. The mass of iodine produced by the reaction of 400 mL of 0.200 mol/L magnesium iodide
solution with excess chlorine gas is?
Answer: 20.3 g
2+
Cl2(g) + 2I (aq)  I2(l) + 2Cl (aq)
MgI2  Mg + 2I
v = 0.400 L
m =?
C=0.200
CI = 2C =0.400
C = 0.400 mol/L
 ng = Cv = (0.400 mol/L)(0.400 L)
 m = nrM = (0.0800 mol)(253.80 g/mol)
= 0.160 mol
= 20.304 g  20.3 g
 nr = 1/2ng  0.0800 mol
7. 200 mL of iron(III) sulfate solution is required to react with 50.0 g of zinc. What is the
concentration of the iron(III) ions in the solution?
Answer: 7.65 mol/L
3+
2+
2+
Zn(s) + 2Fe (aq)  2Fe (aq) + Zn (aq)
m = 50.0 g v = 0.200 L
C=?
 ng = m = 50.0 g = 0.7647599 mol
M 65.38 g/ mol
 nr = 2/1ng  1.5295197 mol
 C = nr = 1.5295197 mol = 7.6475987 mol/L  7.65 mol/L
v
0.200 L
8. If 20.0 mL of a calcium iodide solution is required to reduce 100 g of bromine, calculate the
concentration of the calcium iodide solution.
Answer: 31.3 mol/L
Br2(l) + 2I (aq)  2Br (aq) + I2(l)
m = 100 g v = 0.0200 L
C=?
 ng = m = 100 g = 0.6257822 mol
M 159.80 g/ mol
 nr = 2/1ng  1.2515645 mol
 C = nr = 1.2515645 mol = 62.578223 mol/L  62.6 mol/L
v
0.0200 L
CaI2  Ca2+ + 2IC = ½CI  31.3 mol/L
CI = 62.6 mol/L
9. If 100 mL of a 0.500 mol/L solution of magnesium bromide is used to reduce
50.0 mL of
a solution of acidic potassium permanganate. Calculate the concentration of the
permanganate ion in the solution.
Answer: 0.400 mol/L
+
2+
2MnO4 (aq) + 16H (aq) + 10Br (aq)  8H2O(l) + 2Mn (aq) + 5Br2(l)
v = 0.0500 L
C=1.00 mol/L
MgBr2  Mg2+ + 2BrC=?
 ng = Cv = (1.00 mol/L)(0.100 L)

= 0.100 mol
1
nr = /5ng  0.0200 mol
v= 0.100 L
C =0.500
 C = nr = 0.0200 mol  0.400 mol/L
v
0.0500 L
2CBr =1. 00
10. Say 20.0 g of calcium metal is required to reduce 250 mL of a solution of chromium(III)
sulfate. Calculate the concentration of the chromium(III) sulfate solution.
Answer: 2.00 mol/L
3+
2+
2+
Ca(s) + 2Cr (aq)  2Cr (aq) + Ca (aq)
m= 20.0 g v = 0.250 L
C= ?
 ng = m = 20.0 g = 0.499002 mol
M 40.08 g/ mol
 nr = 2/1ng  0.998004 mol
 C = nr = 0.998004 mol = 3.992016 mol/L  3.99 mol/L
v
0.250 L
Cr2(SO4)3  2Cr3+ + 3SO42C = ½CI  2.00 mol/L
CCr = 3.99 mol/L
11. 50.0 mL of a 0.600 mol/L aluminium chloride solution is required to reduce 30.0 mL of a
gold(III) nitrate solution. Calculate the concentration of the gold ions in the solution.
Answer: 1.00 mol/L
2Au3+(aq) + 6Cl-(aq)  3Cl2(g) + 2Au(s)
AlCl3  Al3+ + 3ClC=?
v = 0.0500 L
C=0.600 CCl=3C =1.80
v= 0.0300 L C = 1.80 mol/L
 ng = Cv = (1.80 mol/L)(0.0500 L)
 C = nr = 0.0300 mol  1.00 mol/L
= 0.0900 mol
v
0.0300 L
1
 nr = /3ng  0.0300 mol