Tutorials

How to Balance Redox Reactions

redox reaction example

Core Concepts

Redox reactions are incredibly interesting and important reactions in chemistry, particularly in electrochemistry! However, they are not of much use unless properly balanced. In this tutorial, you will learn how to balance redox reactions, and what it means for such a reaction to become balanced.

Topics Covered in Other Articles

Vocabulary

  • Oxidation: a type of chemical reaction where one or more electrons are lost.
  • Oxidation State/Number: a number assigned to an atom describing its degree of oxidation, meaning how many electrons it has gained or lost.
  • Reduction: a type of chemical reaction where one or more electrons are gained.

Review: What is a redox reaction?

A redox reaction is a reaction where both reduction and oxidation take place, meaning that the electrons lost during the oxidation of one species are then gained during the reduction of another species. In this manner, the electrons are exchanged between species, and there is no net charge on the equation as a whole.

What does it mean for a redox reaction to be balanced?

A redox reaction is balanced if all of the atoms in the reaction are balanced, meaning that there are the same number of X and Y atoms on the reactants’ side as there are on the products’ side. It also means that the electrons and charges are balanced: the number of electrons lost during oxidation is equal to the number of electrons gained during reduction. Balancing redox reactions allows for the net reaction equation to be conveyed in a concise manner.

How to balance redox reactions: step by step

  1. Write the reduction half reaction
  2. Write oxidation half reaction
  3. Balance the number of atoms within the half reactions
  4. Determine the number of electrons transferred in each half reaction
  5. Balance the number of electrons between half reactions, so the number lost in one is equal to the number gained in the other.
  6. Combine the two half reactions into a net equation. Be sure to always double check at the end if the atoms are balanced on both sides!

Examples of how to balance redox reactions

Example #1

Let’s go through an example, step by step, to understand this. Let’s look at a simple example, of a common electrochemical cell comprised of copper and zinc.

Step 1.

Write the reduction half-reaction. Copper gets reduced in this reaction, which makes the following half-reaction:

Cu2+ (aq) → Cu0 (s)

Step 2.

Write the oxidation half-reaction. Zinc gets oxidized in this reaction, which makes the following half-reaction:

Zn0 (s) → Zn2+ (aq)

Step 3.

Balance the atoms within the half-reactions. All balanced! There is one copper atom in both the products and reactants of the reduction half-reaction and only one zinc atom in both the products and reactants of the oxidation half-reaction.

Step 4.

Determine the number of electrons in each half-reaction. As the oxidation number in the reduction half-reaction decreases by two, this indicates that two electrons are transferred: Cu2+ (aq) + 2e → Cu0 (s).
In this same manner, as the oxidation number in the oxidation half-reaction increases by two, this indicates that two electrons are transferred: Zn0 (s) → Zn2+ (aq) + 2e

Step 5.

Balance the number of electrons. There are two electrons lost in the oxidation half-reaction and two electrons gained in the reduction half-reaction, so all balanced!

Step 6.

Combine the half-reactions. The full reaction is thus: Cu2+ (aq) + Zn0 (s) + 2e → Zn2+ (aq) + Cu0 (s) + 2e, or just Cu2+ (aq) + Zn0 (s) → Zn2+ (aq) + Cu0 (s) since the electrons cancel out.

Example #2

Let’s look at a less straightforward example, like the reaction involved in our own making copper powder from aluminum foil experiment. In this experiment, we start with aluminum foil, Al, and copper sulfate, CuSO4. The single displacement reaction results in the formation of copper, Cu, and aluminum sulfate, Al2(SO4)3

Step 0:

Identify oxidation states. This is an important step to take before even starting to balance the redox reaction, as this will help us identify which species undergo reduction, and which undergo oxidation.

Reactants: Al0, Cu2+, SO42-
Products: Al3+, Cu0, SO42-

Since the oxidation state of the SO4 ion does not change throughout the reaction and it appears as both a reactant and a product, it is a spectator ion. Hence, we’re going to leave it out until the end as it may make the half-reactions more confusing. 

Step 1.

Write the reduction half-reaction. Reduction, as we know, is the gaining of electrons, meaning a decrease in the oxidation number. This indicates that the copper gets reduced: Cu2+ → Cu0 

Step 2.

Write the oxidation half-reaction. Oxidation, as we know, is the loss of electrons, meaning an increase in the oxidation number. This indicates that the aluminum gets oxidized: Al0 → Al3+

Step 3.

Balance the atoms within the half-reactions:

Cu2+ → Cu0 stays as is since there is one mole of copper in both the reactants and the products.

However, we need to account for the fact that the product is actually Al2(SO4)3, meaning that for every single aluminum in the reactants, there are two aluminum atoms in the products. Hence, we place a coefficient of 2 on the reactant side: 2Al0 → Al3+. At this point, it may be helpful to bring the spectator ion back in, for clarification. In that case, we see that 2Al → Al2(SO4)3

Step 4.

Determine the number of electrons in each half reaction.

Cu2+ → Cu0 gains 2 electrons, so it becomes: Cu2+ + 2e → Cu0 

2Al0 → Al3+ each aluminum atom loses three electrons, multiplied by the coefficient of 2: 2Al0 → Al3+ + 6e, or 2Al → Al2(SO4)3 + 6e

Step 5.

Balance the number of electrons. The reduction half-reaction only gains 2 electrons, while the oxidation half-reaction loses 6 electrons total. Thus, we multiply the reduction half-reaction by a coefficient of 3 in order to obtain 6 electrons in both half-reactions, evening it out: 3Cu2+ + 6e → 3Cu0. Again, bringing the spectator ion back in for clarity, we see that 3CuSO4 + 6e → 3Cu. 

Step 6.

Combine the half-reactions. The full reaction thus becomes: 2Al + 3CuSO4 + 6e → Al2(SO4)3 + 3Cu + 6e, or just 2Al + 3CuSO4 → Al2(SO4)3 + 3Cu.

Balance redox reactions: acidic or basic conditions

When balancing a redox reaction under acidic or basic conditions, follow the same steps as outlined above, with one significant caveat.

In acidic conditions, the atoms will often not balance as they are, so you should add H2O to balance an excess of oxygen, and then H+ to balance an excess of hydrogen. This is because both H2O and H+ are found under acidic conditions, so they will be present in the reaction, so you can add as many as needed. To this extent, you should also be sure to keep track of your charges on the reactants and products sides, as the net charge must be conserved (this is particularly relevant when balancing the half-reactions). If adding H+ creates an excess of charge, you can offset it by adding electrons where appropriate. In the end, however, the electrons must still cancel out on both sides of the full redox equation, as described in step 5 above.

Balancing a redox reaction under basic conditions is very similar to the method described above, but with the use of OH rather than H+.

Further Reading

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