Humboldt State University ® Department of Chemistry

Richard A. Paselk

 

General Chemistry

Spring 2011

Exercise: Redox Balancing

© R. Paselk 2008
 
 

Discussion Modules

 

 

Basic Redox Equation Balancing

Answers

The first couple of answers are worked in detail as examples. All answers list the coefficients for the species other than protons, water, and/or hydroxide in the order they appear in the completed redox equation. The protons, water, and/or hydroxide are then listed along with the side of the equation where they appear (e.g.: 2,1,1            Left side: 2 H+            Right side: 2 H2O).

Reactions balanced in aqueous basic solutions:

(Basic reactions may be balanced a variety of ways, including by balancing as if in acid first, then canceling with base as is done in Zumdahl. My personal favorite jumps right in with hydroxide ion as below in example a.)

a) MnO4- + Fe(OH)2 (s) Fe2O3 (s) + MnO2 (s)

First break equation into two half reactions, keeping elements together, e.g. Mn vs. Fe:

MnO4- MnO2 (s)

Fe(OH)2 (s) Fe2O3 (s)

Balance metals:

MnO4- MnO2 (s)

2 Fe(OH)2 (s) Fe2O3 (s)

Now balance the two half reactions by:

1. Balance the oxygens (O) for each half-reaction with hydroxide ion (OH-) added to the side deficient in oxygen:

MnO4- MnO2 (s) + 2 OH-
2 Fe(OH)2 (s) Fe2O3 (s) + OH-

2. Balance the hydrogens by adding water (H2O) to the deficient sides and hydroxide (OH-) to the opposite side to maintain oxygen balance (Adding a water to one side and a hydroxide to the other has the net effect of adding one proton to the water side!):

2 H2O + MnO4- MnO2 (s) + 2 OH- + 2 OH-
3OH- + 2 Fe(OH)2 (s) Fe2O3 (s) + OH- + 3 H2O

3. Balance the charge by adding electrons (e-) to the more positive sides, e.g. (3e-) + (-) = (4-):

3 e- + 2 H2O + MnO4- MnO2 (s) + 2 OH- + 2 OH-
3OH- + 2 Fe(OH)2 (s) Fe2O3 (s) + OH- + 3 H2O + 2 e-

Now combine and cancel terms in each half-reaction, then multiply the appropriate half-reactions by appropriate factors to give the same number of electrons in each:

(3 e- + 2 H2O + MnO4- MnO2 (s) + 4 OH- ) x 2 =
6 e- + 4 H2O + 2 MnO4- 2 MnO2 (s) + 8 OH-
(2 OH- + 2 Fe(OH)2 (s) Fe2O3 (s) + 3 H2O + 2 e- ) x 3 =
6 OH- + 6 Fe(OH)2 (s) 3 Fe2O3 (s) + 9 H2O + 6 e-

Finally, add the two half reactions and cancel species appearing on both sides:

6 e- + 4 H2O + 6 OH- + 2 MnO4- + 6 Fe(OH)2 (s) 3 Fe2O3 (s) + 2 MnO2 (s) + 9 H2O + 8 OH- + 6 e-

2 MnO4- + 6 Fe(OH)2 (s) 3 Fe2O3 (s) + 2 MnO2 (s) + 5 H2O + 2 OH-

Which may also be expressed as: 2,6,3,2         Right side: 5 H2O, 2 OH-

 

b) MnO4- + HO2- MnO2(s) + O2(g)

2,3,2,3         Left side: 1 H2O         Right side: 5 OH-

c) ClO2 ClO2- + ClO3-

2,1,1            Left side: 2 OH-        Right side: 1 H2O

d) CrO4- + N2H4 CrO2- + N2(g)

4,3,4,3         Right side: 4 H2O, 4 OH-

e) Ag(s) + CN- + O2(g) Ag(CN)2-

4,8,1,4         Left side: 2 H2O         Right side: 4 OH-

f) Co(s) + ClO- Co(OH)2(s) + Cl-

1,1,1,1         Left side: H2O


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© R A Paselk

Last modified 11 February 2009