Humboldt State University ® Department of Chemistry

Richard A. Paselk

	Introductory Chemistry	Spring 2007
Exercise: Redox Balancing	© R. Paselk 2005
	Worked Examples
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Redox Equation Balancing Practice Problems

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).

1. Reactions balanced in aqueous acidic solutions:

a) MnO₄^- + Fe²⁺ Fe³⁺ + Mn²⁺

First break equation into two half reactions, keeping elements other than O and H together, e.g. Mn vs. Fe and balance them for elements other than H and O:

Now balance the two half reactions by:

1. Balance the oxygens (O) for each half-reaction with waters added to the opposite side:

2. Balance the hydrogens by adding protons (H⁺) to the deficient sides:

3. Balance the charge by adding electrons ( e^-) to the more positive sides, e.g. (8+) + (-) + (5e^-) = (2+):

Now multiply the appropriate half-reactions by appropriate factors to give the same number of electrons in each:

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

5 e^- + 8 H⁺ + MnO₄^- + 5 Fe²⁺ 5 Fe³⁺ + Mn²⁺ + 4 H₂O + 5 e^-

8 H⁺ + MnO₄^- + 5 Fe²⁺ 5 Fe³⁺ + Mn²⁺ + 4 H₂O

Which may also be expressed as: 1,5,5,1          Left side: 8 H⁺            Right side: 4 H₂O

b) H₂O₂+ MnO₄^- Mn²⁺ + O_2(g)

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

Next balance each of the half reactions for as above:

Now multiply the appropriate half-reactions by appropriate factors to give the same number of electrons in each:

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

10 e^- + 16 H⁺ + 5 H₂O₂+ 2 MnO₄^- 2 Mn²⁺ + 5 O_2(g) + 8 H₂O + 10 H⁺ + 10 e^-

6 H⁺ + 5 H₂O₂+ 2 MnO₄^- 2 Mn²⁺ + 5 O_2(g) + 8 H₂O

Which may also be expressed as: 5,2,2,5         Left side: 6 H⁺            Right side: 8 H₂O

c) 2,1,1            Left side: 2 H⁺            Right side: 2 H₂O

d) 5,3,10,6       Left side: 34 H+          Right side: 17 H₂O

e) 2,1,1,2

f) 3,2,3,3,2       Left side: 8 H⁺           Right side: 4 H₂O

g) 3,1,2            Left side: 4 H⁺            Right side: 2 H₂O

h) 1,1,2,2,        Left side: 4 H⁺            Right side: 2 H₂O

2) Reactions balanced in aqueous basic solutions:

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

Balance metals:

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

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!):

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

6 e^- + 4 H₂O + 6 OH^- + 2 MnO₄^- + 6 Fe(OH)_{2 (s)} 3 Fe₂O_{3 (s)} + 2 MnO_{2 (s)} + 9 H₂O + 8 OH^- + 6 e^-

2 MnO₄^- + 6 Fe(OH)_{2 (s)} 3 Fe₂O_{3 (s)} + 2 MnO_{2 (s)} + 5 H₂O + 2 OH^-

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

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

c) 2,1,1            Left side: 2 OH^-        Right side: 1 H₂O

d) 4,3,4,3         Right side: 4 H₂O, 4 OH^-

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

f) 1,1,1,1         Left side: H₂O