1.8: Balancing Redox Reactions

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In any reaction, it is useful to quantify things. How much product will there be? How much of each reactant do we need to add? What ratios do we need? There are an awful lot of reactions for which this process is straightforward, but sometimes it can be tricky. Redox reactions are sometimes on the tricky side (although certainly not always). For that reason, it's good to have a reliable method for balancing redox reactions: determining the ratios of reactants needed to give the products in the proper amounts.

Suppose, for example, we have a reaction in which silver oxide (Ag₂O) reacts with manganese ion (Mn³⁺) to produce manganese dioxide and silver. That's:

Mn³⁺ + Ag₂O --> MnO₂ + Ag

What would the balanced reaction look like?

The first thing to do is make sure you are working with one half-reaction at a time. So that's:

Mn³⁺ --> MnO₂

and

Ag₂O --> Ag

We start out by balancing the atoms involved, one at a time. First we look at the atoms other than hydrogen or oxygen, and we just balance those by adding the right coefficient.

Mn³⁺ --> MnO₂

and

Ag₂O --> 2 Ag

Second, we balance the oxygen atoms by adding water to one side or the other.

Mn³⁺ + 2 H₂O --> MnO₂

and

Ag₂O --> 2 Ag + H₂O

Third, we balance any hydrogens by adding protons.

Mn³⁺ + 2 H₂O --> 4H⁺ + MnO₂

and

Ag₂O + 2 H⁺ --> 2 Ag + H₂O

Fourth, we balance the charge by adding electrons.

Mn³⁺ + 2 H₂O --> 4H⁺ + MnO₂ + e^-

and

Ag₂O + 2 H⁺ + 2 e^- --> 2 Ag + H₂O

Fifth, we multiply so that the number of electrons is the same in both reactions.

2 x (Mn³⁺ + 2 H₂O --> 4H⁺ + MnO₂ + e^- )

or

2 Mn³⁺ + 4 H₂O --> 8 H⁺ + 2 MnO₂ + 2 e^-

and

Ag₂O + 2 H⁺ + 2 e^- --> 2 Ag + H₂O

Sixth, we simply add these two reactions together. The reaction arrow functions like an equals sign. The left side adds to the left side, and the right side adds to the right.

2 Mn³⁺ + 4 H₂O + Ag₂O + 2 H⁺ + 2 e^- --> 8 H⁺ + 2 MnO₂ + 2 e^- + 2 Ag + H₂O

At that point, gratifyingly, the equation simplifies. Notice that we have added the same number of electrons to each side; they cancel out. That's perfect, because it means we have supplied just the right number of electrons from one half reaction to satisfy the other half reaction.

2 Mn³⁺ + 4 H₂O + Ag₂O + 2 H⁺ --> 8 H⁺ + 2 MnO₂ + 2 Ag + H₂O

Also if we subtract one water from each side, things get slightly simpler.

2 Mn³⁺ + 3 H₂O + Ag₂O + 2 H⁺ --> 8 H⁺ + 2 MnO₂ + 2 Ag

Subtracting two protons from each side makes it simpler still.

2 Mn³⁺ + 3 H₂O + Ag₂O --> 6 H⁺ + 2 MnO₂ + 2 Ag

This method works for any redox reaction, no matter how complicated.

Problem RO8.1.

Balance the following reactions.

1. Cu + MoO₂ --> Cu₂O + Mo

2. NH₂OH + Ag₂O --> N₂ + Ag

3. Fe₃O₄ + CO --> Fe + CO₂

4. I₂ + MnO₄^- --> IO₃^- + MnO₂

5. H₃Mo₇O₂₄ + S₂O₃^2- --> Mo + SO₃^2-

In the event that the reaction is described as occuring under basic conditions, we can simply "neutralize" our protons at the end, by adding hydroxide to both sides.

2 Mn³⁺ + 3 H₂O + Ag₂O + 6 ^-OH --> 6 H⁺ + 6 ^-OH + 2 MnO₂ + 2 Ag

Which of course means

2 Mn³⁺ + 3 H₂O + Ag₂O + 6 ^-OH --> 6 H₂O + 2 MnO₂ + 2 Ag

Simplifying to

2 Mn³⁺ + Ag₂O + 6 ^-OH --> 3 H₂O + 2 MnO₂ + 2 Ag

Problem RO8.2.

Balance the following reactions under basic conditions.

1. Fe(OH)₂ + N₂H₄ --> Fe₂O₃ + NH₄⁺

2. MnO₄^- + V³⁺ --> HMnO₄^- + VO₂⁺