Balancing redox reactions– the basics

By Tutor GuyNo Comments


When balancing redox reactions, you must start by finding the number of electrons transferred in each half reaction. You usually can’t get the charges to balance if you don’t equate the electrons. In this post, I show you how the process works when both the oxidizing agent and the reducing agent are single elements.

Example (a very simple example): Balance the following reaction:

… Mg(s) + … Al3+(aq) → … Mg2+(aq) + … Al(s)

Solution: At first glance, it may appear that the reaction is already balanced because there is one magnesium atom and one aluminum atom on each side. But notice that the charges do not balance. Because aluminum gains three electrons in this reaction, but magnesium atoms lose only two electrons, you will need more magnesiums than aluminums. To find the right ratio, determine the least common multiple of 2 and 3, which is 6. Then multiply the magnesiums by 3 and the aluminums by 2 so that there is a total transfer of 6 electrons in this reaction:

3 Mg(s) + 2 Al3+(aq) → 3 Mg2+(aq) + 2 Al(s)

The charges balance, as required.

(Look for other tips on this website if you need help with balancing redox reactions in acidic or basic solutions.)

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