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Balancing Redox Reactions and Identifying Oxidizing and Reducing Agents

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Instructor: Amy Meyers

Amy holds a Master of Science. She has taught science at the high school and college levels.

A redox reaction is a chemical reaction that involves a transfer of electrons and changes in oxidation number. Learn about redox reactions, and identify oxidizing and reducing agents, and explore the process to write and balance a redox reaction equation. Updated: 08/23/2022

Identifying Oxidizing and Reducing Agents

A redox reaction occurs when an electron is transferred from one substance to another.
Oxydation Reduction Ion Image

An oxidation reaction is a reaction that takes an electron from one substance. A reduction reaction is a reaction that gives an electron to a substance. A redox reaction is a reaction in which one substance gives up an electron and another substance takes that electron. I know it is strange that 'oxidation' means 'giving up' and 'reduction' means 'taking in,' but remember that an electron is negatively charged, so everything is kind of backwards. To give up an electron means to become more positively charged. To take in an electron means becoming more negatively charged.

An oxidizing agent is the substance that causes the oxidation in another substance. Common oxidizing agents include oxygen, hydrogen peroxide and halogens. A reducing agent is a substance that causes another substance to reduce. So to identify an oxidizing agent, simply look at the oxidation number of an atom before and after the reaction. If the oxidation number is greater in the product, then it lost electrons and the substance was oxidized. If the oxidation number is less, then it gained electrons and was reduced. The substance that is reduced in a reaction is the oxidizing agent because it gains electrons. The substance that is oxidized in a reaction is the reducing agent because it lost electrons.

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  • 0:05 Identifying Oxidizing…
  • 1:47 Balancing a Redox Reaction
  • 2:52 Examples
  • 5:39 Strength of the Agents
  • 6:05 Lesson Summary
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Name Oxidation Number Change in Electrons
Oxidation increases electrons are lost
Reduction decreases electrons are gained
Oxidizing agent decreases electrons are gained
Reducing agent increases electrons are lost

Balancing a Redox Reaction

Balancing an oxidation-reduction reaction can be a bit tricky. You can use the steps you used previously to balance other equations to start, but then you have to take into account the key ions and oxidation numbers. Here's a list of steps to help you balance a redox equation.

  1. Identify the reactants and the products.
    1. Write the unbalanced equation in ionic form and exclude any spectator ions.
    2. Give each atom an oxidation number and identify each atom that changes its oxidation number. Ignore any atom that doesn't change.
  2. Write and balance the half-reactions.
    1. Separate the equation into its half-reactions.
    2. Balance the atoms other than hydrogen and oxygen.
    3. Balance the oxygen atoms by adding water molecules.
    4. Balance the hydrogen atoms by adding a hydronium ion for each hydrogen atom and adding an equal number of water molecules to the other side of the equation.
    5. Balance the charge by adding electrons if needed.
  3. Make the electrons equal and combine the half-reactions.


Balance the equation of hydrogen sulfide in air reacting to form sulfur dioxide and water.

H2 S + O2 = SO2 + H2 O

1. Identify the reactants and the products and assign oxidation numbers.


2. Write and balance the half-reactions.

Note that sulfur is changing from a -2 oxidation number to a +4 oxidation number, so it had to lose six electrons. -2 + (-4) = -6.

Oxygen is changing from a zero oxidation number to a -2, gaining two electrons. 0 + (-2) = -2. Since free oxygen atoms travel in pairs, the actual number of electrons gained is -4.


3. The two half-reactions need to be balanced so that the electrons lost by sulfur are the same number gained by oxygen.


So from these half-reactions, you know that a 2 has to go in front of the H2 S and a 3 has to go in front of the O2 on the reaction side. On the product side, the six oxygen atoms needed for balance are divided between the SO2 and the H2 O.

The final equation is:

2H2 S + 3O2 = 2SO2 + 2H2 O

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