18.4: Balancing Redox Reactions in Solution

Many redox reactions occur in aqueous solution—in water. Because of this, in many cases H₂O or a fragment of an H₂O molecule (H⁺ or OH⁻, in particular) can participate in the redox reaction. As such, we need to learn how to incorporate the solvent into a balanced redox equation.

Consider the following oxidation half reaction in aqueous solution, which has one Cr atom on each side:

Cr³⁺ → CrO₄⁻

Here, the Cr atom is going from the +3 to the +7 oxidation state. To do this, the Cr atom must lose four electrons. Let us start by listing the four electrons as products:

Cr³⁺ → CrO₄⁻ + 4e⁻

But where do the O atoms come from? They come from water molecules or a common fragment of a water molecule that contains an O atom: the OH⁻ ion. When we balance this half reaction, we should feel free to include either of these species in the reaction to balance the elements. Let us use H₂O to balance the O atoms; we need to include four water molecules to balance the four O atoms in the products:

4H₂O + Cr³⁺ → CrO₄⁻ + 4e⁻

This balances the O atoms, but now introduces hydrogen to the reaction. We can balance the H atoms by adding an H⁺ ion, which is another fragment of the water molecule. We need to add eight H⁺ ions to the product side:

4H₂O + Cr³⁺ → CrO₄⁻ + 4e⁻ + 8H⁺

The Cr atoms are balanced, the O atoms are balanced, and the H atoms are balanced; if we check the total charge on both sides of the chemical equation, they are the same (3+, in this case). This half reaction is now balanced, using water molecules and parts of water molecules as reactants and products.

Reduction reactions can be balanced in a similar fashion. When oxidation and reduction half reactions are individually balanced, they can be combined in the same fashion as before: by taking multiples of each half reaction as necessary to cancel all electrons. Other species, such as H⁺, OH⁻, and H₂O, may also have to be canceled in the final balanced reaction.

Unless otherwise noted, it does not matter if you add H₂O or OH⁻ as a source of O atoms, although a reaction may specify acidic solution or basic solution as a hint of what species to use or what species to avoid. OH⁻ ions are not very common in acidic solutions, so they should be avoided in those circumstances.

Example \(\PageIndex{1}\):

Balance this redox reaction. Assume a basic solution.

MnO₂ + CrO₃⁻ → Mn + CrO₄⁻

Solution

We start by separating the oxidation and reduction processes so we can balance each half reaction separately. The oxidation reaction is as follows:

CrO₃⁻ → CrO₄⁻

The Cr atom is going from a +5 to a +7 oxidation state and loses two electrons in the process. We add those two electrons to the product side:

CrO₃⁻ → CrO₄⁻ + 2e⁻

Now we must balance the O atoms. Because the solution is basic, we should use OH⁻ rather than H₂O:

OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻

We have introduced H atoms as part of the reactants; we can balance them by adding H⁺ as products:

OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻ + H⁺

If we check the atoms and the overall charge on both sides, we see that this reaction is balanced. However, if the reaction is occurring in a basic solution, it is unlikely that H⁺ ions will be present in quantity. The way to address this is to add an additional OH⁻ ion to each side of the equation:

OH⁻ + CrO₃⁻ + OH⁻ → CrO₄⁻ + 2e⁻ + H⁺ + OH⁻

The two OH⁻ ions on the left side can be grouped together as 2OH⁻. On the right side, the H⁺ and OH⁻ ions can be grouped into an H₂O molecule:

2OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻ + H₂O

This is a more appropriate form for a basic solution.

Now we balance the reduction reaction:

MnO₂ → Mn

The Mn atom is going from +4 to 0 in oxidation number, which requires a gain of four electrons:

4e⁻ + MnO₂ → Mn

Then we balance the O atoms and then the H atoms:

4e⁻ + MnO₂ → Mn + 2OH⁻2H⁺ + 4e⁻ + MnO₂ → Mn + 2OH⁻

We add two OH⁻ ions to each side to eliminate the H⁺ ion in the reactants; the reactant species combine to make two water molecules, and the number of OH⁻ ions in the product increases to four:

2H₂O + 4e⁻ + MnO₂ → Mn + 4OH⁻

This reaction is balanced for a basic solution.

Now we combine the two balanced half reactions. The oxidation reaction has two electrons, while the reduction reaction has four. The least common multiple of these two numbers is four, so we multiply the oxidation reaction by 2 so that the electrons are balanced:

2 × [2OH⁻ + CrO₃⁻ → CrO₄⁻ + 2e⁻ + H₂O]2H₂O + 4e⁻ + MnO₂ → Mn + 4OH⁻

Combining these two equations results in the following equation:

4OH⁻ + 2CrO₃⁻ + 2H₂O + 4e⁻ + MnO₂ → 2CrO₄⁻ + 4e⁻ + 2H₂O + Mn + 4OH⁻

The four electrons cancel. So do the two H₂O molecules and the four OH⁻ ions. What remains is

2CrO₃⁻ + MnO₂ → 2CrO₄⁻ + Mn

which is our final balanced redox reaction.

Exercise \(\PageIndex{1}\)

Balance this redox reaction. Assume a basic solution.

Cl⁻ + MnO₄⁻ → MnO₂ + ClO₃⁻

Answer

H₂O + Cl⁻ + 2MnO₄⁻ → 2MnO₂ + ClO₃⁻ + 2OH⁻

Key Takeaways

• Redox reactions can be balanced by inspection or by the half reaction method.
• A solvent may participate in redox reactions; in aqueous solutions, H₂O, H⁺, and OH⁻ may be reactants or products.