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- Mn 2+ 1. Balance Mn

MnO4 1- Mn 2+ + 4 H2O 2. Balance O

8 H 1+ + MnO4 1- 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- 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- Mn 2+ + 4H2O) x 2

10 e1- + 16 H 1+ + 2 MnO4 1- 2 Mn 2+ + 8H2O

 

 

Half equation:

C2O4 2- 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- 2 CO2 + 2 e1-

This is the Oxidation Step

 

5. Make e1- Gain = e1- Loss:

2 Reductions = 5 Oxidations

( C2O4 2- 2 CO2 + 2 e1- ) x 5

 

5 C2O4 2- 10 CO2 + 10 e1-

 

Substitute half equations into net ion equation to balance

 

5 C2O4 2- 10 CO2 + 10 e1-

10 e1- + 16 H 1+ + 2 MnO4 1- 2 Mn 2+ + 8H2O

--------------------------------------------------------------------

5 C2O4 2-- + 2 MnO4 1- + 16 H 1+ 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 2 BiO3 1- 1. Balance Bi

 

3 OH 1- + Bi2O3 2 BiO3 1- 2. Balance O (add OH1-)

 

3. Balance H (add OH1- and H2O)

3 OH 1- + 3 OH 1- + Bi2O3 2 BiO3 1- + 3 HOH

 

6 OH 1- + Bi2O3 2 BiO3 1- + 3 HOH

 

 

4. Charge Balance:

-6 + 0 = -2 + 0

-6 = -2

-6 = -2 + -4

 

Loss 4 electrons:

6 OH 1- + Bi2O3 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 2 BiO3 1- + 3 HOH + 4 e1-) x 1

 

6 OH 1- + Bi2O3 2 BiO3 1- + 3 HOH + 4 e1-

 

 

2nd half equation:

OCl 1- Cl 1- 1. Balance Cl (Cl is balanced)

 

OCl 1- Cl 1- + 1 OH 1- 2. Balance O (add OH1-)

 

3. Balance H (add H2O and OH1-)

 

1 HOH + OCl 1- Cl 1- + 1 OH 1-+ 1 OH 1-

 

1 HOH + OCl 1- 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- 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- Cl 1- + 2 OH 1- ) 2

 

4 e1- + 2 HOH + 2 OCl 1- 2 Cl 1- + 4 OH 1-

 

 

Substitute half equations into net ion equation to balance

 

4 e1- + 2 HOH + 2 OCl 1- 2 Cl 1- + 4 OH 1-

6 OH 1- + Bi2O3 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)

 

 

 

Acid Media Homework:

 

Zn + NO3 1- + H 1+ Zn 2+ + NH4 1+ + H2O

 

 

MnO4 1- + C2O4 2- + H 1+ Mn 2+ + CO2 + H2O

 

 

Cr2O7 2- + C2H5OH + H 1+ Cr 3+ + HC2H3O 2 + H2O

 

 

SO4 2- + CH2O + H 1+ H2S + CO2 + H2O

 

 

FeS + NO3 1- + H 1+ NO + Fe 2+ + SO4 2- + H2O

 

 

Cr2O7 2- + Cl 1- + H 1+ Cr 3+ + Cl2 + H2O

 

 

H2O2 + MnO4 1- + H 1+ Mn 2+ + O2 + H2O

 

Basic Media Homework:

 

Cr 3+ + ClO3 1- + OH 1- CrO4 2- + Cl 1- + H2O

 

 

Mn 2- + Br2 + OH 1- MnO2 + Br 1- + H2O

 

Fe(OH)2 + O2 + H2O Fe 3+ + OH 1-

 

Zn + NO3 1- + OH 1- ZnO2 2- + NH3 + H2O

 

AsO2 1- + ClO 1- AsO3 1- + Cl 1- (basic solution)

 

MnO4 1- + C2O4 2- + OH 1- + H2O MnO2 + CO3 2-

 

N2H4 + O2 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)