Redox reactions are chemical reactions that involve the transfer of electrons between two substances. Examples of redox reactions can be found everywhere in the world and in every aspect of life. From the combustion of fires to the electrochemical exchange in your batteries, they form the basis for many important reactions that we depend on.

"Redox" is a combination of "reduction" and "oxidation". An oxidation reaction involves the loss of electrons. A reduction reaction involves a substance gaining electrons. Because of the extra emphasis on electrons, balancing a redox reaction is slightly different than balancing a standard reaction. Luckily, it still follows a fairly simple set of rules. This instructable will hopefully serve to guide and illustrate how to balance a redox reaction.

Requirements:

Writing Utensil

A redox reaction to balance

A small amount of chemistry know-how

Step 1: The Redox Reaction

In this example, we have a reaction of Methanol and Chromium Trioxide. The products of this reaction are Formaldehyde and a Chromium Ion.

Step 2: Splitting the Reaction

So the first step is the separate the reaction into two halves: the reduction and the oxidation. The methanol (CH3OH) is the one being oxidized in the reaction, while the chromium trioxide (CrO3) is the one that will be reduced. A good rule of thumb when determining whether a particular molecule is being reduced or oxidized is looking at the oxygen(s) attached to it. The chromium trioxide in this reaction is turned into Cr +4. Since it lost some oxygen, we can call it "reduced". In the methanol, a double bond was formed between the carbon and the oxygen. Since the carbon gained an additional bond to oxygen, we can say it was "oxidized". Granted this rule of thumb doesn't actually fit the strict definitions of reduction and oxidation, but it's useful when one is pressed for time.

Step 3: Balancing Pt. 1

The next step will be to balance all the elements besides oxygen and hydrogen. In this example, carbon and chromium are already equal to each other in both sides of the reaction so no balancing is necessary.

Afterwards, it is necessary balance the oxygens by adding water to the other side of the half reactions. In the methanol half reaction the oxygens are equal so no water is needed there. In the chromium trioxide reaction though there are 3 more oxygens on the right than there are on the left, so 3 waters are needed on the right to balance it out.

Step 4: Balancing Pt. 2

Next we must balance the hydrogens by adding protons (H+) to the reactions. The methanol half reaction has 2 more hydrogens on the left than the right, so we add 2 protons to the right. The chromium trioxide reaction now has 6 more hydrogens on the right than the left due to the addition of water in the last step, so we add 6 protons on the left.

Step 5: Balancing the Charges

The next step is to balance the charges by adding electrons. Remember that the protons we added in the last step are positive and therefore also affect the charges. The 2 positive protons on the right side of the methanol reaction must be balanced by adding in 2 electrons to achieve a neutral charge.

The right side of the chromium trioxide reaction is has a positive charge of 4 due to the chromium ion. The left side has a positive charge of 6 due to the 6 protons added in. Both sides must be made equal to each other by adding in 2 electrons to the right, getting a charge of +4 on both sides.

Step 6: Canceling Out and Finishing

Before the half reactions are combined again, the amount of electrons in both must be equal to one another. This can be done by multiplying the entire half reaction so that the electrons in both are equal. In this example, the amount of electrons are already equal to one another in both half reactions. Then the electrons will cancel each other out.

In the example, we have crossed out the electrons since there are 2 on both sides of the reaction. There are also protons (H+) on both sides of the reaction which will start canceling each other out. With 6 H+ on the left side and 2 H+ on the right, we end up with 4 H+ on the left side.

Afterwards, the half reactions can be combined back into one full reaction and you will have had a fully balanced redox reaction.