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Chem 12 Practice Worksheet - Answer Key
Key page 1
Redox #1 (KEY)
1. Explain the meaning of each of the following terms:
a) oxidation a half-reaction that involves the loss of electron(s)
b) reduction a half-reaction that involves the gain of electron(s)
c) reducing agent a species that causes another to be reduced; it itself is oxidized
d) oxidizing agent a species that causes another to be oxidized; it itself is reduced
2. Identify each of the following as examples of oxidation of reduction:
a) Li(s) � Li+(aq) + e
− = oxidation
b) S(s) + 2e− � S
2−(aq) = reduction
c) Cr3+
(aq) + e− � Cr
2+(aq) = reduction
d) Sn2+
(aq) � Sn4+
(aq) + 2e− = oxidation
3. Given the following redox reaction: Fe(s) + 2 Ag+(aq) � Fe
2+(aq) + 2 Ag(s)
a) Which species is the reducing agent? Fe
b) Which species is the oxidizing agent? Ag+
c) Which species is the being oxidized? Fe
d) Which species is the being reduced? Ag+
4. Given the following redox reaction: 3 Zn2+
(aq) + 2 Al(s) � 3 Zn(s) + 2 Al3+
(aq)
a) Identify, and write the half-
reaction for oxidation and the half-
reaction for reduction.
oxidation = Al(s) � Al3+
(aq) + 3e−
reduction = Zn2+
(aq) + 2e− � Zn(s)
b) Name the reducing agent and the
oxidizing agent for the reaction
RA = Al OA = Zn2+
Chem 12
Key page 2
5. Identify the material being reduced, the material being oxidized, the reducing agent, and
the oxidizing agent in each of the following redox reactions. Try to write the half-cell
reactions.
a) 2 Na(s) + ½ O2(g) � Na2O(s)
reduction = ½ O2(g) + 2e− � O
2− O2 = reduced; OA
oxidation = Na(s) � Na+(aq) + e
− Na = oxidized; RA
b) Fe(s) + Cu2+
(aq) � Fe2+
(aq) + Cu(s)
reduction = Cu2+
(aq) + 2e− � Cu(s) Cu
2+ = reduced; OA
oxidation = Fe(s) � Fe2+
(aq) + 2e− Fe = oxidized; RA
c) Sn4+
(aq) + Fe2+
(aq) � Sn2+
(aq) + Fe3+
(aq)
reduction = Sn4+
(aq) + 2e− � Sn
2+(aq) Sn
4+ = reduced; OA
oxidation = Fe2+
(aq) � Fe3+
(aq) + e− Fe
2+ = oxidized; RA
d) 2 Mg(s) + O2(g) � 2 MgO(s)
reduction = O2 + 4e− � 2 O
2− O2 = reduced; OA
oxidation = Mg(s) � Mg2+
+ 2e− Mg = oxidized; RA
e) Ca(s) + Cl2(g) � CaCl2(s)
reduction = Cl2(g) + 2e− � 2 Cl
− Cl2 = reduced; OA
oxidation = Ca(s) � Ca2+
(aq) + 2e− Ca = oxidized; RA
f) 2 H2(g) + O2(g) � 2 H2O(l)
reduction = O2(g) + 4e− � 2 O
2− O2 = reduced; OA
oxidation = H2(g) � 2 H +
(aq) + 2e− H2 = oxidized; RA
6. When copper wire is placed in a solution of silver nitrate, what colour does the solution turn
after some time? Without looking at your notes, write the balanced half-reactions for the
oxidation and reduction processes that are occurring.
The solution turns blue, because Cu2+
is being formed.
Reduction = Ag+(aq) � Ag(s) + e
−
Oxidation = Cu(s) � Cu2+
(aq) + 2e−
Chem 12
Key page 3
Redox #2 (KEY)
1. Determine the oxidation number of all atoms in each of the following compounds and
polyatomic ions.
a) SO2 S = +4
O = –2
h) MnO2− Mn = +3
O = –2
o) N2O3 N = +3
O = –2
b) CaF2 Ca = +2
F = –1
i) KHSO3 K = +1
H = +1
S = +4
O = –2
p) NO+ N = +3
O = –2
c) PO33−
P = +3
O = –2
j) S2O72−
S = +6
O = –2
q) NH3 N = –3
H = +1
d) PO43−
P = +5
O = –2
k) NiSO4 Ni = +2
S = +6
O = –2
r) N2O N = +1
O = –2
e) H2O2 H = +1
O = –1
**peroxide
l) SO3 S = +6
O = –2
s) H2CrO4 H = +1
Cr = +6
O = –2
f) NaSiO4 Na = +1
Si = +7
O = –2
m) S2O7 S = +7
O = –2
t) P2O5 P = +5
O = –2
g) KMnO4 K = +1
Mn = +7
O = –2
n) H2S2O3 H = +1
S = +2
O = –2
u) O3 O = zero
2. Identify the species oxidized and reduced, as well as the oxidizing agent and the reducing
agent for the following reactions:
a) CH4 + 2 O2 � CO2 + 2 H2O
oxidized/RA = CH4 (Ox # of C goes from –4 to +4)
reduced/OA = O2 (Ox # of O goes from 0 to –2)
b) Cr2O72−
+ 2 OH− � 2 CrO4
2− + H2O
not a redox (oxidation number of Cr stays at +6)
c) O3 + NO � O2 + NO2
oxidized/RA = NO (Ox # of N goes from +2 to +4)
reduced/OA = O3 (Ox # of O goes from 0 to –2)
Chem 12
Key page 4
d) 2 H2O2 � 2 H2O + O2
oxidized/RA = O in H2O2 (Ox # of O goes from –1 to 0 in O2)
reduced/OA = O in H2O2 (Ox # of O goes from –1 to –2 in H2O)
e) 2 CuCl � CuCl2 + Cu
oxidized/RA = Cu in CuCl (Ox # of Cu goes from +1 to +2)
reduced/OA = Cu in CuCl (Ox # of Cu goes from +1 to 0)
f) HCl + NH3 � NH4Cl
not a redox (oxidation numbers did not change for all atoms)
g) SiCl4 + 2 H2O � 4 HCl + SiO2
not a redox (oxidation number of Si stays at +4)
h) I− + ClO
− � I3
− + Cl
−
oxidized/RA = I− (Ox # of I goes from –1 to −
�
� )
reduced/OA = ClO− (Ox # of Cl goes from +1 to –1)
i) As2O3 + MnO3− � H3AsO4 + Mn
2+
oxidized/RA = As2O3 (Ox # of As goes from +3 to +5)
reduced/OA = MnO3− (Ox # of Mn goes from +5 to +2)
j) Mn2+
+ NaBiO3 � Bi3+
+ MnO4−
oxidized/RA = Mn2+
(Ox # goes from +2 to +7)
reduced/OA = NaBiO3 (Ox # of Bi goes from +5 to +3)
Chem 12
Key page 5
Redox #3 (KEY)
1. Balance the following half-reactions
ACIDIC
a) Ce4+
↔ Ce2+
Ce4+
+ 2e– ↔ Ce
2+
b) I2 ↔ I– I2 + 2e
– ↔ 2 I
–
c) I2 ↔ I3– 3 I2 + 2e
– ↔ 2 I3
–
d) Mn2+
↔ MnO2 Mn2+
+ 2 H2O ↔ MnO2 + 4 H+ + 2e
–
e) O2 ↔ H2O2 O2 + 2 H+ + 2e
– ↔ H2O2
f) S2O82–
↔ HSO4– S2O8
2– + 2 H
+ + 2e
– ↔ 2 HSO4
–
g) H3AsO4 ↔ HAsO2 H3AsO4 + 2 H+ + 2e
– ↔ HAsO2 + 2 H2O
h) H2SeO3 ↔ Se H2SeO3 + 4 H+ + 4e
– ↔ Se + 3 H2O
i) CH3COOH ↔ C2H5OH CH3COOH + 4 H+
+ 4e– ↔ C2H5OH + H2O
j) I– + ClO
– ↔ I3
– + Cl
– 3 I
– + ClO
– + 2 H
+ ↔ I3
– + Cl
– + H2O
k) Br– + MnO4
– ↔ Br2 + Mn
2+ 2 Br
– + MnO4
– + 8 H
+ + 3e
– ↔ Br2 + Mn
2+ + 4 H2O
*** reaction (j) is a full redox.
BASIC
l) N2H4 ↔ N2 N2H4 + 4 OH– ↔ N2 + 4 H2O + 4e
–
m) HO2– ↔ O2 HO2
– + OH
– ↔ O2 + H2O + 2e
–
n) HXeO4– ↔ HXeO6
3– HXeO4
– + 4 OH
– ↔ HXeO6
3– + 2 H2O + 2e
–
o) Cr(OH)3 ↔ CrO42–
Cr(OH)3 + 5 OH– ↔ CrO4
2– + 4 H2O + 3e
–
p) CH3CHO ↔ CH2CH2 CH3CHO + H2O + 2e– ↔ CH2CH2 + 2 OH
–
q) Al + MnO4– ↔ MnO2 + Al(OH)4
– Al + MnO4
– + 2 H2O ↔ MnO2 + Al(OH)4
–
r) Cl2 ↔ Cl– + OCl
– 2 OH
– + Cl2 ↔ Cl
– + OCl
– + H2O
s) NO2– + Al ↔ NH3 + AlO2
– NO2
– + Al + 3 H2O + 3e
– ↔ NH3 + AlO2
– + 3 OH
–
*** reactions (q) & (r) are full redox.
2. Balance the following and calculate the change in oxidation number for the species oxidized
or reduced. Compare this to the number of electrons in the balanced reaction.
a) NO2– � NO3
– balanced: NO2
– + H2O � NO3
– + 2 H
+ + 2e
–
+3 +5
loss of 2e–
b) MnO4– � MnO2 balanced: MnO4
– + 4 H
+ + 3e
– � MnO2 + 2 H2O
+7 +4
gain of 3e–
Chem 12
Key page 6
Redox #4 (KEY)
1. Predict whether the following chemicals with react spontaneously.
a) calcium metal mixed with potassium ions non-spontaneous
b) lithium metal mixed with tin ions spontaneous
c) iron metal mixed oxygen gas spontaneous
d) aluminum ions with zinc metal non-spontaneous
2. Data for question 2:
Element C is the best reducing agent forming C+.
D2+
will oxidize element A forming A+ and D.
B2+
+ 2 A ↔ B + 2 A+ (not spontaneous)
Using the data given above, write reduction half-reactions in order of DECREASING strength
of oxidizing agent.
OA RA
D2+
↔ D
A+ ↔ A
B2+
↔ B
C+ ↔ C
3. Data for question 3:
Element B will oxidize element C.
Element A is the best oxidizing agent.
B– will reduce D
3+.
Using the data given above, write reduction half-reactions in order of DECREASING strength
of oxidizing agent.
OA RA
A ↔ A–
D3+
↔ D
B ↔ B–
C+ ↔ C
Chem 12
Key page 7
4. Predict whether the following mixtures will react spontaneously:
a) Zn and H2 non-spontaneous (both RA’s)
b) Sn and Sn4+
spontaneous
c) Cu+ spontaneous (disproportionation!)
d) H+ and Mn spontaneous
e) Fe2+
and Cr2O72−
non-spontaneous (No H+, see (f))
f) Fe2+
and Cr2O72−
and H+ spontaneous
g) Cu and H+ non-spontaneous (copper only dissolves in HNO3)
h) MnO2 and H+ and I
− spontaneous
i) SO42−
and Sn non-spontaneous (both RA’s)
5. What substances are in the following categories:
a) can be oxidized by I2 but not acidic SO42+
Cu(s), S(s)
b) can be reduced by I− but not by Fe
2+
O2(g) + 2H+, MnO4
−(aq)
c) can be an oxidizing agent for Pb but not Sn2+
S(s) + 2H+, H
+
d) can oxidize Co and reduce H+
none
6. Why is gold used in so many high end electrical applications?
First off, gold is an excellent electrical conductor, behind copper, aluminum and iron. But
in terms of reducing strength, gold is a very weak reducing agent compared to these
metals. Gold cannot be oxidized by oxygen gas (i.e. gold does NOT “rust” in air), and it is
also resistant towards most acids. Very few substances can actually react with gold,
which makes it a “noble” metal.
Chem 12
Key page 8
Redox #5 (KEY)
1. Balance each of the following redox reactions using half-reactions.
a) U4+
+ MnO4–
� Mn2+
+ UO22+
5U4+
+ 2MnO4–
+ 2H2O � 2Mn2+
+ 5UO22+
+ 4H+
b) Zn + As2O3 � AsH3 + Zn2+
6Zn + As2O3 + 12H+ � 2AsH3 + 6Zn
2+ + 3H2O
c) Fe2+
+ Cr2O72–
� Cr3+
+ Fe3+
6Fe2+
+ Cr2O72–
+ 14H+ � 2Cr
3+ + 6Fe
3+ + 7H2O
d) Cl2 + SO2 � Cl– + SO4
2–
Cl2 + SO2 + 2H2O � 2Cl– + SO4
2– + 4H
+
e) Cu + NO3– � Cu
2+ + NO
3Cu + 2NO3– + 8H
+ � 3Cu
2+ + 2NO + 4H2O
f) Br2 � Br– + BrO3
– (basic)
3Br2 + 6 OH– � 5Br
– + BrO3
– + 3H2O
g) CN–
+ IO3–
� I–
+ CNO–
3CN–
+ IO3–
� I–
+ 3CNO–
h) Sn2+
+ H2O2 � Sn4+
(basic)
Sn2+
+ H2O2 � Sn4+
+ 2 OH–
i) Mn2+
+ HBiO3 � Bi3+
+ MnO4–
2Mn2+
+ 5HBiO3 + 9H+ � 5Bi
3+ + 2MnO4
– + 7H2O
j) Cr2O72–
+ I– � Cr
3+ + I2
Cr2O72–
+ 6 I– + 14H
+ � 2Cr
3+ + 3 I2 + 7H2O
Chem 12
Key page 9
k) Cr3+
+ ClO3– � ClO2 + Cr2O7
2–
2Cr3+
+ 6ClO3– + H2O � 6ClO2 + Cr2O7
2– + 2H
+
l) CHCl3 + MnO4– � Cl2 + CO2 + Mn
2+
2CHCl3 + 2MnO4– + 6H
+ � 3Cl2 + 2CO2 + 2Mn
2+ + 4H2O
m) MnO4– + NO2
– � MnO2 + NO3
– (basic)
2MnO4– + 3NO2
– + H2O � 2MnO2 + 3NO3
– + 2 OH
–
n) SnS2O3 + MnO4– � Mn
2+ + SO4
2– + Sn
4+
SnS2O3 + 2MnO4– + 6H
+ � 2Mn
2+ + 2SO4
2– + Sn
4+ + 3H2O
o) O2 + H2O + I– � I2 + OH
– (basic)
O2 + 2 H2O + 4 I– � 2 I2 + 4 OH
–
p) S2–
+ ClO3– � Cl
– + S (basic)
3S2–
+ ClO3– + 3H2O � Cl
– + 3S + 6 OH
–
q) HSO3–
+ IO3– � I2 + SO4
2–
5HSO3–
+ 2IO3– � I2 + 5SO4
2– + 3H
+ + H2O
Chem 12
Key page 10
Redox #6 (KEY)
1. Balance each of the following redox reactions using the oxidation number method.
a) S2–
+ NO3–
� S + NO (acidic)
3S2–
+ 2NO3–
+ 8H+ � 3S + 2NO + 4H2O
b) Pb + NO3– � Pb
2+ + NO2 (basic)
Pb + 2NO3– + 2H2O � Pb
2+ + 2NO2 + 4 OH
–
c) Cd + NO3– � Cd
2+ + NO (acidic)
3Cd + 2NO3– + 8H
+ � 3Cd
2+ + 2NO + 4H2O
d) MnO4– + Cl
– � Mn
2+ + Cl2 (basic)
2MnO4– + 10Cl
– + 8H2O � 2Mn
2+ + 5Cl2 + 16 OH
–
e) Cr2O72–
+ S2–
� S + Cr3+
(acidic)
Cr2O72–
+ 3S2–
+ 14H+ � 3S + 2Cr
3+ + 7H2O
f) I– + SO4
2– � I2 + S
2– (basic)
8 I– + SO4
2– + 4H2O � 4 I2 + S
2– + 8 OH
–
g) Br– + SO4
2– � Br2 + SO2
(basic)
2Br– + SO4
2– + 2H2O � Br2 + SO2
+ 4 OH
–
h) I2 + ClO3– � IO3
– + Cl
– (acidic)
3 I2 + 5ClO3– + 3H2O � 6 IO3
– + 5Cl
– + 6H
+
i) BrO3– + I
– � Br
– + I2 (basic)
BrO3– + 6 I
– + 3H2O � Br
– + 3 I2 + 6 OH
–
j) MnO4– + S
2– � Mn
2+ + S (acidic)
2MnO4– + 5 S
2– + 16H
+ � 2Mn
2+ + 5 S + 8H2O
Chem 12
Key page 11
k) (see part (d))
l) (see part (c))
m) (see part (b))
n) (see part (a))
o) MnO4– + NH3 � MnO2 + NO3
– (basic)
8MnO4– + 3NH3 � 8MnO2 + 3NO3
– + 5 OH
– + 2H2O
p) Cl2 � Cl– + ClO3
– (basic)
3Cl2 + 6 OH– � 5Cl
– + ClO3
– + 3H2O
q) Cr2O72–
+ SO2 � Cr3+
+ HSO4– (acidic)
Cr2O72–
+ 3SO2 + 5H+ � 2Cr
3+ + 3HSO4
– + H2O
r) NH3 + O2 � NO2 + H2O (basic)
4NH3 + 7 O2 � 4NO2 + 6H2O
s) NF3 + H2O � HF + NO + NO2 (acidic)
2NF3 + 3H2O � 6HF + NO + NO2
t) BrO3– + Br
– � Br2 (basic)
5Br– + BrO3
– + 3H2O � 3Br2 + 6 OH
–
u) I2 + Cl2 + H2O � HIO3 + HCl
I2 + 5Cl2 + 6H2O � 2HIO3 + 10HCl
v) (see part (e))
2. When a solution containing iodide ion is mixed with manganese dioxide (MnO2) and acid, a
cloud of purple I2(g) is given off.
a) Write the net ionic
equation for the reaction.
2 I−(aq) + MnO2(s) + 4H
+(aq) � I2(g) + Mn
2+(aq)
+ 2 H2O(l)
b) Write the half-reactions
that occur.
2 I−(aq) � I2(g) + 2e
−
MnO2(s) + 4H+(aq) + 2e
− � Mn
2+(aq) + 2 H2O(l)
c) Name the reducing agent
and the oxidizing agent
for this reaction.
OA = MnO2(s)
RA = I−(aq)
Chem 12
Key page 12
3. When hot cesium metal is exposed to chlorine gas, a bright flash occurs as the elements
react. The product is a white solid composed of cesium ions and chloride ions.
a) Write the net ionic equation for the
reaction.
2 Cs(s) + Cl2(g) � 2 CsCl(s)
b) Write the half-reactions that occur. Cl2(g) + 2e− � 2 Cl
−
Cs(s) � Cs+ + e
−
c) Name the reducing agent and the
oxidizing agent for this reaction.
OA = Cl2(g)
RA = Cs(s)
4. Balance each of the following redox reactions using half-reactions or oxidation numbers.
a) HNO3 + H2S � H2O + NO + S
2 HNO3 + 3 H2S � 4 H2O + 2 NO + 3 S
b) K2Cr2O7 + HCl � KCl + CrCl3 + H2O + Cl2
K2Cr2O7 + 14 HCl � 2 KCl + 2 CrCl3 + 7 H2O + 3 Cl2
c) Na2TeO3 + NaI + HCl � NaCl + H2O + Te + I2
Na2TeO3 + 4 NaI + 6 HCl � 6 NaCl + 3 H2O + Te + 2 I2
d) PbO2 + HCl � PbCl2 + Cl2 + H2O
PbO2 + 4 HCl � PbCl2 + Cl2 + 2H2O
e) K2Cr2O7 + NO2 + HNO3 � KNO3 + Cr(NO3)3 + H2O
K2Cr2O7 + 6 NO2 + 2 HNO3 � 2 KNO3 + 2 Cr(NO3)3 + H2O
f) Cu + HNO3 � Cu(NO3)2 + H2O + NO
3 Cu + 8 HNO3 � 3 Cu(NO3)2 + 4 H2O + 2 NO
g) KMnO4 + H2SO4 + H2S � H2O + MnSO4 + S + K2SO4
2 KMnO4 + 3 H2SO4 + 5 H2S � 8 H2O + 2 MnSO4 + 5 S + K2SO4
h) CdS + I2 + HCl � CdCl2 + HI + S
CdS + I2 + 2 HCl � CdCl2 + 2 HI + S
i) HNO3 + I2 � HIO3 + NO2 + H2O
10 HNO3 + I2 � 2 HIO3 + 10 NO2 + 4 H2O
j) H2S + NO3− � NO + S (acidic)
3 H2S + 2 NO3− + 2 H
+ � 2 NO + 3 S + 4 H2O
k) CuS + NO3− � Cu(NO3)2 + S + NO (acidic)
3 CuS + 8 NO3− + 8 H
+ � 3 Cu(NO3)2 + 3 S + 2 NO + 4 H2O
l) CuS + NO3− � Cu(NO3)2 + S + NO (basic)
3 CuS + 8 NO3− + 4 H2O � 3 Cu(NO3)2 + 3 S + 2 NO + 8 OH
−
m) MnO4− + SO4
2− + H2S � MnSO4 + S (acidic) see part (g)
2 MnO4− + 2 SO4
2− + 5 H2S + 6 H
+ � 2 MnSO4 + 5 S + 8 H2O
n) S2−
+ I2 + H+ � HI + S
S2−
+ I2 + 2 H+ � 2 HI + S
o) NO3− + I2 � IO3
− + NO2 (acidic)
10 NO3− + I2 + 8 H
+ � 2 IO3
− + 10 NO2 + 4 H2O
p) Cl2 � Cl− + ClO
− (basic)
Cl2 + 2 OH− � Cl
− + ClO
− + H2O
Chem 12
Key page 13
5. Write balanced chemical equations for each of the following reactions. If no reaction
occurs, write “No Reaction”.
a) Chlorine gas reacting with aqueous tin(II) nitrate
2 Cl2(g) + 2 Sn(NO3)2(aq) � Sn(NO3)4(aq) + SnCl4(aq)
b) Aqueous iron(II) nitrate reacting with aqueous potassium iodide
Fe(NO3)2(aq) + KI(aq) � No Reaction
c) Silver metal reacting with hydrochloric acid
Ag(s) + 2 HCl(aq) � No Reaction
d) Cadmium metal reacting with sulfuric acid (check online)
Cd(s) + H2SO4(aq) � CdSO4(aq) + H2(g)
e) Zinc metal reacting with aqueous nickel(II) nitrate
Zn(s) + Ni(NO3)2(aq) � Ni(s) + Zn(NO3)2(aq)
f) Aqueous iron(II) nitrate reacting with aqueous aluminum nitrate
Fe(NO3)2(aq) + Al(NO3)3(aq) � No Reaction
g) Liquid bromine reacting with magnesium metal
Br2(l) + Mg(s) � MgBr2(s)
Chem 12
Key page 14
Redox #7 (KEY)
1. A 3.00 g sample of pure iron is dissolved in HCl and treated to produce Fe2+
. The solution is
diluted to 500.0 mL. A 25.0 mL sample of this solution is titrated with 0.0500 M KMnO4.
What volume of KMnO4 is needed in this titration?
Rxn: 5 Fe2+
+ MnO4− + 8 H
+ � 5 Fe
3+ + Mn
2+ + 4 H2O
Moles of iron per 25.0 mL = 5.376×10−2
mol Fe × � ��.������.��� = 2.688×10−3
mol
Moles of MnO4− = 2.688×10
−3 mol Fe × ���� ���
�
�������� = 5.376×10−4
mol MnO4−
Volume of KMnO4 = �� = ��.����
������.���� = 1.08×10
−2 L or 10.8 mL
2. A 25.0 mL solution of HNO3 is reacted with excess KI to produce I2 and NO. The I2 produced
is titrated with Na2S2O3 solution with a starch indicator. At the end, 12.36 mL of 0.150 M of
Na2S2O3 solution is used. What is the [HNO3] in the original solution?
Rxn 1 : 2 HNO3 + 6 I− + 6 H
+ � 3 I2 + 2 NO + 4 H2O
Rxn 2 : I2 + 2 Na2S2O3 � 2 I− + Na2S4O6
Start with reaction 2:
Moles of thiosulfate = 0.150 M x 0.01236 L = 1.854×10−3
mol Na2S2O3
Moles of HNO3 = 1.854×10−3
mol Na2S2O3 × � ����������������
× ����� ���������
= 6.18×10−4
mol HNO3
[HNO3] = ��.!������
�.����� = 2.47×10−2
M
3. A sample of tin is oxidized to Sn2+
and is titrated with permanganate. The titration used
120.8 mL of 0.50 M MnO4− solution. How many grams of tin were in the original sample?
Rxn : 5 Sn(s) + 2 MnO4− + 16 H
+ � 5 Sn
2+ + 2 Mn
2+ + 8 H2O
Moles of MnO4− = 0.50 M × 0.1208 L = 0.0604 mol MnO4
−
Moles of Sn = 0.0604 mol MnO4− × � ������
���� ���� = 0.151 mol Sn
Mass of Sn = 0.151 mol Sn × �!.�"������� = 17.9 g Sn
Chem 12
Key page 15
4. A 25.00 mL sample of acidified household bleach (NaOCl) is reacted with excess NaI. The
half-reaction for NaOCl is: 2e− + 2 H
+ + OCl
− � Cl
− + H2O . The I2 produced is titrated
with 22.64 mL of 0.100 M Na2S2O3. What is the concentration of the original bleach
solution?
Rxn 1 : OCl− + 2 I
− + 2 H
+ � I2 + Cl
− + H2O
Rxn 2 : I2 + 2 Na2S2O3 � 2 I− + Na2S4O6
Start with reaction 2:
Moles of thiosulfate = 0.100 M x 0.02264 L = 2.264×10−3
mol Na2S2O3
Moles of OCl− = 2.264×10
−3 mol Na2S2O3 × � �����
����������� × �������
�
�����
= 1.132×10−3
mol OCl−
[OCl−] = �.��×�
������.����� = 4.53×10
−2 M
Chem 12
Key page 16
Redox #8 (KEY)
1. Calculate the cell voltage for each of the following reactions and indicate whether or not
the reaction is expected to be spontaneous.
Ɛ°cell (V) spontaneity
a) Cr + 3 Ag+ � Cr
3+ + 3 Ag +1.54 spontaneous
b) Cu+ + Fe
3+ � Cu
2+ + Fe
2+ +0.62 spontaneous
c) Mn2+
+ 2 H2O + I2 � MnO2 + 4 H+ + 2 I
− −0.62 non-spontaneous
d) 3 Cu + 2 NO3− + 8 H
+ � 3 Cu
2+ + 4 H2O + 2 NO +0.62 spontaneous
e) 2 Cr3+
+ 7 H2O + 3 Pb2+
� Cr2O72−
+ 14 H+ + 3 Pb −1.36 non-spontaneous
f) Co + Pb2+
� Co2+
+ Pb +0.15 spontaneous
g) Co + Cu2+
� Co2+
+ Cu +0.62 spontaneous
h) Cu + Co2+
� Cu2+
+ Co −0.62 non-spontaneous
i) Cu + 2 Ag+ � Cu
2+ + 2 Ag +0.46 spontaneous
j) 5 Fe2+
+ MnO4− + 8 H
+ � 5 Fe
3+ + Mn
2+ + 4 H2O +0.74 spontaneous
k) Mg + Cl2 � MgCl2 +3.73 spontaneous
l) 2 Na + Cl2 � 2 NaCl +4.07 spontaneous
m) Cu + 2 H+ � Cu
2+ + H2 −0.34 non-spontaneous
n) Cu2+
+ 2 Cl− � Cu + Cl2 −1.02 non-spontaneous
o) 3 PbCl2 + 2 Al � 2 AlCl3 + 3 Pb +1.53 spontaneous
2. Explain the meaning of each of the following terms:
a) electrochemical cell
A system with many connected components that is capable
of producing electricity.
b) electrode
A solid substance (metallic or nonmetallic) that facilitates
the oxidation or reduction of a species.
c) cathode
An electrode where reduction occurs.
d) anode
An electrode where oxidation occurs.
e) salt bridge
A tube containing a soluble salt that allows passages of
ions to both cathode and anode, in order to achieve charge
balance between anode and cathode. It completes the
circuit.
f) anion
a negatively charged ion.
g) cation
a positively charged ion.
Chem 12
Key page 17
3. An electrochemical cell was made as follows: A weighed strip of Sn was immersed in a
beaker of 1 M SnSO4 and a weighed strip of Ag was immersed in a second beaker containing
1 M AgNO3. The metal strips were then connected by a wire and the beakers connected by
a salt bridge. After several hours, the mass of the Sn electrode was found to have
decreased.
a) Draw a labeled diagram of
the cell.
Ex.
b) Write the net ionic equation
for the reaction.
Sn(s) + 2 Ag+(aq) � Sn
2+(aq) + 2 Ag(s)
c) Which electrode is the
cathode?
Ag(s)
d) Towards which electrode do
the Ag+ ions migrate?
Ag(s)
e) Which way do the electrons
flow in the wire?
From Sn to Ag (anode to cathode)
f) Did the Ag electrode gain or
lose mass?
Ag electrode gained mass.
g) How many moles of electrons
flow through the salt bridge?
2 moles of electrons per mole of Sn
4. Given 1 M solutions of CuSO4 and AgNO3, a silver strip, and a copper strip, sketch an
electrochemical cell with a porous barrier and answer the following questions.
Diagram:
Silver strip
(cathode)
Copper strip
(anode)
wire
Porous barrier
1 M CuSO4 1 M AgNO3
Chem 12
Key page 18
a) In this electrochemical cell, electrons are flowing
from where to where?
copper to silver
b) Oxidation is occurring at which electrode? anode
c) Reduction is occurring at which electrode? cathode
d) Which metal strip is the anode in this reaction? copper
e) Which metal strip is the cathode in this reaction? silver
f) Which electrode will gain mass as the reaction
proceeds?
silver (cathode always gains
mass)
g) Which electrode will lose mass as the reaction
proceeds?
copper (anode always loses
mass, except when it is inert)
h) The SO42−
ions will travel towards which electrode? They stay at the anode
i) The NO32−
ions will travel towards which electrode? towards the anode across the
porous barrier
j) The Cu2+
ions will travel towards which electrode? towards the cathode, where
they will be reduced
k) The Ag+ ions will travel towards which electrode? towards the cathode across
the porous barrier
l) In which direction do the electrons flow in the
solution?
(trick question) Electrons do
NOT travel in the solution,
only in the wire.
m) The number of e- given up by the _____oxidation_____ half-reaction must equal the
number of electrons used by the ______reduction______ half-reaction.
5. An electrochemical cell was made as follows: A piece of Ni foil is immersed in a beaker of
NiCl2 solution, and a strip of Cu foil is immersed in a beaker of CuSO4 solution. The metal
electrodes are then connected by a wire and the beakers connected by a salt bridge.
a) Draw this cell. Ex.
b) Which electrode is the anode? Ni(s)
c) Towards which electrode do the SO42−
ions
migrate?
Cu(s)
d) Which way do the electrons flow in the
wire?
Anode to cathode (Ni to Cu)
CuSO4(aq) NiCl2(aq)
Chem 12
Key page 19
e) If 0.025 mol of Cu(s) is produced in the
reaction, how many moles of electrons flow
through the wire?
0.050 mol of electrons (1:2 ratio)
f) Towards which electrode do the Ni2+
ions
migrate?
Towards the cathode, through the
salt bridge. (cathode attracts cations)
6. Draw a labeled diagram showing the construction of an electrochemical cell designed to
deliver a potential of between 0.90 and 1.10 volts.
a) Justify the choice of materials by using half-reactions to calculate Ɛ° for your cell.
b) Which electrode is the cathode in your electrochemical cell? Give reasons for your
choice of material.
c) If the ionic concentration of the solution surrounding the cathode were reduced, explain
the changes that would occur with the cell voltage Ɛ°cell.
[many choices possible]
For example: Sn | Sn2+
|| Ag+ | Ag voltaic cell
Anode: Sn(s) � Sn2+
+ 2e− Ɛ° = +0.14 V
Cathode: Ag+ + e
− � Ag(s) Ɛ° = +0.80 V
Overall: Sn(s) + Ag+ � Sn
2+ + Ag(s)
Ɛ°cell = +0.94 V
If [ion] around the cathode is decreased, then the cell voltage will decrease as well,
because the ion around the cathode is a reactant of the overall redox reaction.
7. For any 5 spontaneous reactions shown in Question 1, construct electrochemical cells.
Construct 3 with salt bridges and 2 with porous barriers. Label the diagrams, and show all
relevant reactions and ion movement.
[Hint: Must pick from the choices of SPONTANEOUS reactions!]
See #3 for an example of a cell using a salt bridge.
See #4 for an example of a cell using a porous barrier.
Remember: The stronger OA will be at the cathode!
Chem 12
Key page 20
Redox #9 (KEY)
1. Describe what happens to the strength of a reducing agent and an oxidizing agent if they
are increased above 1 M or decreased below 1 M.
Since they are the reactants of the redox reaction, their strength (aka voltage) increases if
their concentration is greater than 1 M. Their strength decreases if their concentration is
less than 1 M.
2. Explain what will happen to the voltage of an electrochemical cell if the reactant
concentration is changed and if the product concentration is changed.
Using Nernst equation, � = �° −��
��ln � , the more product and the less reactant
there are, then the reaction quotient Q will be greater. But then, the Ɛ value is decreased.
Vice versa, if there were less product and more reactant at the beginning of a reaction, Q
will be smaller and the Ɛ value is increased.
3. Why do flashlights slowly get dimmer as their batteries die instead of just staying bright and
then going out?
The cell voltage decreases over time in gradual manner. It does not drop to zero
suddenly.
4. How can you tell an electrochemical cell is not at equilibrium?
At equilibrium, the cell voltage would be ZERO.
So if the voltage is not zero, then the cell is not at equilibrium.
5. How and why does temperature change the voltage of a cell?
Using Nernst Equation, the high the temperature, the lower the Ɛ value. Conversely, the
lower the temperature, the higher the Ɛ value.
Chem 12
Key page 21
Redox #10 (KEY)
1. Balance the reaction for the breathalyzer.
3 CH3CH2OH + 2 Cr2O72−
+ 16 H+ � 3 CH3COOH + 4 Cr
3+ + 11 H2O
2. Write the overall reaction that occurs in the lead-acid storage battery.
Pb(s) + 2 HSO4–
(aq) + PbO2(s) + 2H+ (aq) � 2 PbSO4(s) + 2H2O (l)
3. What happens to the amount of H2SO4 in the cell as the battery operates? H2SO4 solution is
more dense than water. How is this used to determine if a battery is functioning?
The amount of H2SO4 decreases over time. To check if the battery is functioning, one
can measure the density of the acid solution against a reference value. If it drops
below this value, the battery no longer operates at the optimum level.
4. What chemical properties should a battery have to be rechargeable?
The rechargeable battery should not have consumed all its reactants, and the reaction
must be reversible.
5. If a dry cell is left for a long period of time, it may leak. Why does this happen?
A reason for battery leaks (e.g. alkaline batteries, AA, etc..) is that as batteries
discharge — the chemistry of the battery changes and some hydrogen gas is
generated. This out-gassing process increases pressure in the battery. Eventually, the
excess pressure either ruptures the insulating seals at the end of the battery, or the
outer metal canister, or both.
6. What would cause a battery to have a short shelf life?
Temperature is too high, or the battery is self-discharging.
7. What could cause the different reaction in the alkaline cell even though the reactants are
similar?
In dry cells, hydrogen gas can be produced, whereas in alkaline cell, no gas is
produced.
8. What happens to the [OH−] as the battery operates? What does this do to the voltage of
the cell?
It stays the same. So the voltage does not change drastically.
Chem 12
Key page 22
9. What acts as a salt bridge for these cells?
A paper sheet soaked with potassium hydroxide.
10. Check the cells in your text (p.658). Why has the mercury cell been discontinued? Why
does the nickel-cadmium battery not work in all devices?
Mercury is a toxic heavy metal, damaging to life and environment. Nickel-cadmium
battery does not work in devices that only use non-rechargeable batteries.
11. What is the overall reaction for a hydrogen fuel cell? Why is it considered a pollution free
source of electricity?
O2(g) + 2 H2(g) � 2 H2O(l)
It is considered pollution-free because the product is just water.
12. The main source of hydrogen for fuel cells is “stripping” of methane and methanol. Propose
a reaction to get this and explain if the hydrogen is really pollution free.
CH4 + H2O → CO + 3 H2
CO + H2O → CO2 + H2
The production of hydrogen gas is accompanied by the formation of carbon monoxide
and dioxide. The latter can contribute to the greenhouse effect, leading to global
warming.
Chem 12
Key page 23
Redox #11 (KEY)
1. “Native copper” is a term used to describe copper found in its elemental form. Why is
native gold fairly common, but native copper very rare, and native iron even rarer? Why is
native sodium never found?
Gold is a relatively weak reducing agent, not able to combine with oxygen or other
reactive substance in the ground. Copper and iron tend to react with oxygen and
water more readily. Sodium, especially, is a much stronger reducing agent than gold,
copper, or iron. Thus sodium is always found as part of a larger compound on Earth.
2. Of the above elements, would any be more likely found in native form if our atmosphere
was primarily hydrogen gas?
Hydrogen gas acts a reducing agent, thus will not react with any metals. So all of the
above-mentioned metals will be found in native form.
3. In India there is an iron pillar buried partially in the ground for many years. It has remained
there mostly rust free for many years. Suggest several reasons why this may be so.
There may be a metal buried in the ground that is in contact with the iron pillar. This
metal is providing cathodic protection to the iron.
4. Materials made of aluminum and chromium do not react in the atmosphere. Find their
places in the reduction potential table, and suggest why they are stable in the atmosphere.
They are stable in the atmosphere because these metals readily form an oxide with
oxygen gas. But their oxide is a strong and stable compound that coats the surface of
the metals completely, thus preventing further corrosion from happening.
5. A tin coating is used to protect iron cans. Does the tin provide cathodic protection for the
iron? Explain.
No, tin coating simply prevents contact between iron and oxygen/water, preventing
rust from forming.
6. How does a piece of magnesium protect a boat’s hull in water?
It provides cathodic protection for the iron in the boat’s hull.
7. WWII battleships are protected in salt-water by an electric current. Explain how this can
work.
The current forces the electrons to travel back into the iron metal, preventing it from
rusting (i.e. oxidizing).
8. Cars from Ontario often have worse rust than cars form B.C. It is said that this is due to
Ontario Government’s policy of salting icy roads. Is this a reasonable suggestion?
Yes, salt increases rate of rusting by acting as a suitable electrolyte to transfer
electrons better to the oxygens molecules.
Chem 12
Key page 24
Redox #12 (KEY)
1. For each of the following solution that is electrolyzed, complete the following:
• Draw the cell.
• Indicate what substance is produced at the anode.
• Write the half-cell reaction that occurs at the anode.
• Indicate what substance is produced at the cathode.
• Write the half-cell reaction that occurs at the cathode.
• Calculate the minimum voltage required for this electrolytic cell to operate.
a) Molten copper(II) chloride using inert electrodes:
Anode: 2 Cl− � Cl2(g) + 2e
− Ɛ° = -1.36V
Chlorine gas is produced
Cathode: Cu2+
+ 2e− � Cu(s) Ɛ° = +0.34V
Copper metal is produced
Min voltage required = 1.02 V
b) An aqueous 1 M solution of NiSO4 using inert electrodes:
Anode: H2O � ½O2(g) + 2H+ + 2e
− Ɛ° = -0.82V
Oxygen gas is produced
Cathode: Ni2+
+ 2e− � Cu(s) Ɛ° = -0.26V
Nickel metal is produced
Min voltage required = 1.08 V
c) 1 M KCl using inert electrodes:
Anode: 2 Cl− � Cl2(g) + 2e
− Ɛ° = -1.36V
Chlorine gas is produced
Cathode: 2 H2O + 2e− � H2(g) + 2 OH
− Ɛ° = -0.41V
Hydrogen gas is produced
Min voltage required = 1.77 V
power
Cl− Cu
2+
anode cathode
power
H2O Ni2+
anode cathode
SO42−
power
Cl− H2O
anode cathode
K+
Chem 12
Key page 25
d) 1 M NaI using iron electrodes:
Anode: Fe(s) � Fe2+
(aq) + 2e− Ɛ° = +0.45V
Iron(II) ion is produced
Cathode: 2 H2O + 2e− � H2(g) + 2 OH
− Ɛ° = -0.41V
Hydrogen gas is produced
Min voltage required = -0.04 V
(actually more than 0 V, of course)
e) A solution containing 1.0 M CoI2 and HCl using inert electrodes:
Anode: 2 I− � I2(g) + 2e
− Ɛ° = -0.54V
Iodine is produced
Cathode: 2 H+ + 2e
− � H2(g) Ɛ° = 0.00V
Hydrogen gas is produced
Min voltage required = 0.54 V
f) Molten AgI using inert electrodes:
Anode: 2 I− � I2(g) + 2e
− Ɛ° = -0.54V
Iodine is produced
Cathode: Ag+ + e
− � Ag(s) Ɛ° = +0.80V
Silver metal is produced
Min voltage required = 0.26 V
g) 1 M PbSO4 and H2SO4 using inert electrodes:
Anode: H2O � ½O2(g) + 2H+ + 2e
− Ɛ° = -0.82V
Oxygen gas is produced
Cathode: Pb2+
+ 2e− � Pb(s) Ɛ° = -0.13V
Lead metal is produced
Min voltage required = 0.95 V
power
H2O
anode cathode
Na+
Fe
Fe
power
I− H
+
anode cathode
Co2+
I−
Cl−
power
anode cathode
Ag+ I
−
power
H2O
SO42−
anode cathode
Pb2+
Chem 12
Key page 26
h) Molten KBr using inert electrodes:
Anode: 2 Br− � Br2(g) + 2e
− Ɛ° = -1.09V
Bromine is produced
Cathode: K+
+ e− � K(l) Ɛ° = -2.93V
Hydrogen gas is produced
Min voltage required = 4.02 V
1. Why can an aqueous solution not produce sodium metal?
Because H2O is a stronger oxidizing agent than Na+ in aqueous solution. H2O would get
reduced first.
2. If bromothymol blue were placed in the Hoffman apparatus, what would you see?
Yellow (acidic), since the solution contains H2SO4.
power
anode cathode
Br− K
+
Chem 12
Key page 27
Redox #13 (KEY)
1. What is the overall reaction for the production of aluminum?
4 Al3+
+ 6 O2-
� 4 Al(l) + 3 O2(g)
or 2 Al2O3(l) � 4 Al(l) + 3 O2(g)
2. Where is aluminum produced in B.C.?
Kitimat, BC
3. The electrodes are not inert in aluminum production. What happens to the electrodes, and
what must be done to compensate for this?
The carbon anode is consumed by oxygen gas and becomes carbon dioxide. The
carbon anode needs to be replaced at regular intervals.
4. Write the reactions that happen in the electro-refining of copper, and the electroplating
with silver.
Electro-refining of coper: anode: Cu(s, impure) � Cu2+
+ 2e−
cathode: Cu2+
+ 2e− � Cu(s, pure)
Electroplating with silver: anode: Ag(s) � Ag+ + e
−
Cathode: Ag+ + e
− � Ag(s)
5. In electro-refining and electroplating, the voltage required is not zero. Suggest a reason
that it is not zero.
Even though overall cell voltage is zero, the actual voltage required must be greater
than zero because a power source is needed to push the electrons from anode to
cathode.
6. Draw a device that could electroplate a spoon with gold.
• Power supply
• Gold anode
• Spoon at cathode
• 1 M Au(NO3)3 solution
power
anode cathode
Au3+
Au
1 M Au(NO3)3
Chem 12
Key page 28
7. Why is a silver anode used in a silver-plating device?
The anode is made of silver because once the silver is oxidized to Ag+, it helps keep the
[Ag+] constant in the solution.
8. You have several metals and their nitrate salts. Construct a device that would contain an
electrochemical cell that could power an electroplating device.
For example:
anode cathode
Au3+
Au
1 M Au(NO3)3
1 M Mg(NO3)2
Mg
1 M AgNO3
Ag
KNO3
anode cathode
Electrochemical cell
Electrolytic cell
