CHM
2045C
Name:___Answers______
Module
8 Part H: Redox
Equations 10 points
Balance the
following REDOX equations written in net ionic form:
Acid
Media: 5 points
1. C2O4
2- (aq) +
MnO4 1- (aq) + H 1+ (aq)
→ Mn 2+ (aq) + CO2
(g) +
HOH (l)
Half equation: MnO4
1-
à Mn 2+ 1. Balance Mn
MnO4 1- à Mn 2+ + 4 H2O 2. Balance O
8 H 1+
+ MnO4 1- à Mn 2+ + 4 H2O 3. Balance H
4. Charge Balance:
+8 -1
= +2 + 0
+7
= +2
-5 +7
= +2
Gain
5 electrons :
5 e1- + 8 H 1+ +
MnO4 1- à Mn 2+ + 4H2O
This is the Reduction Step
5. Make e1- Gain = e1- Loss:
2 Reductions = 5
Oxidations
(5 e1- + 8 H 1+ + MnO4
1-
à Mn 2+ + 4H2O) x 2
10 e1- + 16 H 1+ + 2 MnO4 1- à 2 Mn 2+ + 8H2O
Half equation:
C2O4 2- à 2 CO2 1. Balance C
2. Balance O
Oxygens are balanced
3. Balance H
Hydrogens are balanced
4. Charge Balance:
-2 = 0
-2 = 0 -2
Loss
2 electrons :
C2O4 2- à 2 CO2
+ 2 e1-
This is the Oxidation Step
5. Make e1- Gain = e1- Loss:
2 Reductions = 5
Oxidations
( C2O4 2- à 2 CO2
+ 2 e1- ) x 5
5 C2O4 2- à 10 CO2
+ 10 e1-
Substitute
half equations into net ion equation to balance
5 C2O4
2- à 10 CO2
+ 10 e1-
10 e1- + 16 H 1+ + 2
MnO4 1- à 2 Mn 2+ + 8H2O
--------------------------------------------------------------------
5 C2O4 2-- + 2 MnO4 1-
+ 16 H 1+ à 2 Mn
2+ + 10 CO2 + 8 HOH
Basic
Media 5
points
2. Bi2O3
(s) + OH 1- (aq) + OCl 1-
(aq) → BiO3
1- (aq) + Cl 1-
(aq) + HOH
(l)
1st
half equation:
Bi2O3 à 2
BiO3 1- 1. Balance Bi
3
OH 1- + Bi2O3 à 2 BiO3
1- 2. Balance O (add
OH1-)
3. Balance H (add OH1- and H2O)
3 OH 1- + 3 OH
1- +
Bi2O3 à 2 BiO3
1- + 3 HOH
6 OH 1- +
Bi2O3 à 2 BiO3
1- + 3 HOH
4. Charge Balance:
-6 + 0
= -2 + 0
-6
= -2
-6
= -2 + -4
Loss 4 electrons:
6
OH 1-
+ Bi2O3 à 2 BiO3
1- + 3 HOH + 4 e1-
This is the Oxidation Step
5. Make e1- Gain = e1- Loss:
2 Reductions (2 e1 Gained) = 1 Oxidation (4 e1 Lost)
(6 OH 1-
+ Bi2O3 à 2 BiO3
1- + 3 HOH + 4 e1-)
x 1
6 OH 1-
+ Bi2O3 à 2 BiO3
1- + 3 HOH + 4 e1-
2nd
half equation:
OCl 1- à Cl 1- 1. Balance Cl (Cl
is balanced)
OCl
1- à Cl 1- + 1 OH 1- 2. Balance O (add OH1-)
3. Balance H (add H2O and OH1-)
1 HOH
+ OCl
1- à Cl 1- + 1 OH 1-+ 1 OH 1-
1 HOH +
OCl 1- à Cl 1- + 2 OH 1-
4. Charge Balance:
+0 -1
= -1 - 2
-1
= -3
-2
-1 = -3
Gain
2 electrons :
2 e1- + 1 HOH +
OCl 1- à Cl 1- + 2 OH 1-
This is the Reduction Step
5. Make e1- Gain = e1- Loss:
2 Reductions (2 e1 Gained) = 1 Oxidation (4 e1 Lost)
( 2 e1- + 1 HOH +
OCl 1- à Cl 1- + 2 OH 1- ) 2
4 e1-
+ 2 HOH
+
2 OCl 1- à 2 Cl 1- + 4 OH 1-
Substitute
half equations into net ion equation to balance
4 e1- + 2 HOH + 2 OCl 1- à 2 Cl 1- + 4 OH 1-
6 OH 1- + Bi2O3 à 2 BiO3
1- + 3 HOH
+ 4 e1-
--------------------------------------------------------------------------------
1 Bi2O3
(s) + 2 OH 1- (aq) + 2 OCl
1- (aq) → 2 BiO3
1- (aq) + 2 Cl
1- (aq) + 1 HOH
(l)
MnO4 1- + C2O4
2- + H 1+ à Mn 2+ + CO2 +
H2O
Cr2O7 2- + C2H5OH
+ H 1+ à Cr 3+ + HC2H3O
2 + H2O
SO4 2- +
CH2O + H
1+ à
H2S + CO2 +
H2O
FeS
+ NO3 1- +
H 1+ à
NO + Fe 2+ +
SO4 2-
+ H2O
Cr2O7 2- + Cl 1-
+ H 1+ à Cr 3+ + Cl2 +
H2O
H2O2 +
MnO4 1-
+ H 1+ à Mn 2+ + O2 + H2O
Mn 2- +
Br2 + OH 1- à MnO2 + Br
1- + H2O
Fe(OH)2 +
O2 + H2O à Fe 3+ +
OH 1-
Zn
+ NO3 1- +
OH 1- à
ZnO2 2- + NH3 +
H2O
AsO2 1- +
ClO 1- à AsO3 1- + Cl 1-
(basic solution)
MnO4 1- +
C2O4 2- +
OH 1- + H2O
à MnO2 +
CO3 2-
N2H4 +
O2 à
N2 + H2O2 (basic solution)
Text Reference Kotz 6th:
Section 5.7 (Electron Transfer & Oxidation
Numbers)
P 225
Q#37-#40
Section 20.1 (Ion Electron Method): p945-952
Example 20.2 Balancing REDOX in Acid Solutions p
948
Problem Solving Tips 20.1 p 950
Example 20.3 Balancing REDOX in Basic Solutions
p951
P990 Q#1-#6 (#3-#4 Acid) (#5-#6 Basic)