Chem 1612009 Homework 5
th
Week
Chapter 4: Reading: 4.34.6.
Homework: 65,69,75,79,83,91,104
Chapter 5: Reading: 5.15.6.
Homework: 5,21,23,25,31,37,41,47,95
65.
Indicate whether the firstnamed substance in each change undergoes an oxidation, a reduction, or
neither.
Explain your reasoning.
(a) Blue CrCl
2
(aq) changes to green CrCl
3
(aq) when exposed to air.
(b) Yellow K
2
CrO
4
(aq) changes to orange K
2
Cr
2
O
7
(aq) when acidified.
(c) Dinitrogen pentoxide produces nitric acid when it reacts with water.
(a) Cr
2+
(O.N. +2) → Cr
3+
(O.N. +3)= oxidation, because the oxidation number increases.
(b) CrO
4
2
(Cr O.N. +6) → Cr
2
O
7
2
(Cr O.N. +6) = Neither because the oxidation number doesn’t change.
(c) N
2
O
5
(N O.N. +5) → HNO
3
(N O.N. +5) = Neither because the oxidation number doesn’t change.
69.
Identify the oxidizing and reducing agents in Problem 67.
Problem 67:
(a) HCl + O
2
→ Cl
2
+ H
2
O
(b) NO + H
2
→ NH
3
+ H
2
O
(c) CH
4
+ NO → N
2
+ CO
2
+ H
2
O
(d) Ag + H
+
+ NO
3

→ Ag
+
+ H
2
O + NO
(e) IO
4

+ I

+ H
+
→ I
2
+ H
2
O
+1
1
0
0
+1
2
(a)
H
Cl
+
O
2
→
Cl
2
+
H
2
O
Cl

oxidation state increased from 1 to 0; therefore oxidized; therefore, HCl is the reducing agent.
O oxidation state reduced from 0 to 2; therefore reduced; therefore, O
2
is the oxidizing agent.
+2
2
0
3
+1
+1
2
(b)
N
O
+
H
2
→
N
H
3
+
H
2
O
N oxidation state reduced from +2 to 3; therefore, reduced; therefore, NO is the oxidizing agent.
H oxidation state increased from 0 to +1; therefore, oxidized; therefore, H
2
is the reducing agent.
4
+1
+2
2
0
+4
2
+1
2
(c)
C
H
4
+
N
O
→
N
2
+
C
O
2
+
H
2
O
C oxidation state increased from 4 to +4; therefore, oxidized; therefore, CH
4
is the reducing agent.
N oxidation state decreased from +2 to 0; therefore reduced; therefore, NO is the oxidizing agent.
0
+1
+5
2
+1
+1
2
+2
2
(d)
Ag
+
H
+
+
N
O
3

→
Ag
+
+
H
2
O
+
N
O
Ag oxidation state increased from 0 to +1; therefore oxidized; therefore Ag is the reducing agent.
N oxidation state reduced from +5 to +2; therefore, reduced; therefore NO
3

is the oxidizing agent.
+7
2
1
+1
0
+1
2
(e)
I
O
4

+
I

+
H
+
→
I
2
+
H
2
O
I oxidation state decreased from +7 to 0; therefore reduced; therefore IO
4

is the oxidizing agent.
I oxidation state increased from 1 to 0; therefore, oxidized; therefore, I

is the reducing agent.
Chem 1612009 homework 5
th
week
1