Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Fall 2008
Exercise: Redox Balancing	© R. Paselk 2008
	Discussion Modules

Acid 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 and water in the order they appear in the completed redox equation. The protons and water 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 acidic solutions:

1) 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:

MnO4- Mn2+ + 4 H2O

...

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

8 H+ + MnO4- Mn2+ + 4 H2O

...

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

5 e- + 8 H+ + MnO4- Mn2+ + 4 H2O

Fe2+ Fe3+ + e-

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

(5 e- + 8 H+ + MnO4- Mn2+ + 4 H2O) x 1 = 5 e- + 8 H+ + MnO4- Mn2+ + 4 H2O

(Fe2+ Fe3+ + e- ) x 5 = 5 Fe2+ 5 Fe3+ + 5 e-

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

2) 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

3) I ^- + H₂O₂ I₂

2 H⁺ + 2 I ^- + H₂O₂ I₂ + 2 H₂O

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

4) Cr₂O_7^2- + I ^- Cr³⁺ + IO_3^-

8 H⁺ + Cr₂O_7^2- + I ^- 2 Cr³⁺ + IO_3^- + 4 H₂O

1,1,2,1       Left side: 8 H+          Right side: 4 H₂O

5) S₂O_3^2- + I₂ S₄O_6^2- + I ^_-

2 S₂O_3^2- + I₂ S₄O_6^2- + 2 I ^_-

2,1,1,2

6) CuS_(s) + NO₃^- Cu^₂₊ + S_(s) + NO_(g)

8 H⁺ + 3 CuS_(s) + 2 NO₃^- 3 Cu_²⁺ + 3 S_(s) + 2 NO_(g) + 4 H₂O

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

7) MnO₄^2- MnO_2(s) + MnO₄^-

4 H⁺ + 3 MnO₄^2- MnO_2(s) + 2 MnO₄^- + 2 H₂O

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

8) Pb_(s) + PbO_2(s) + SO₄^2- PbSO_4(s)

4 H⁺ + Pb_(s) + PbO_2(s) + 2 SO₄^2- 2 PbSO_4(s) + 2 H₂O

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

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© R A Paselk

Last modified 6 November 2008