Welcome to balancing redox reactions. Today we will learn how to balance the decomposition of hydrogen peroxide into water and oxygen gas. This is a special type of redox reaction called a disproportionation reaction, where the same element is both oxidized and reduced.
The first step in balancing this redox reaction is to identify the oxidation states. In hydrogen peroxide, oxygen has an oxidation state of negative one. In water, oxygen has an oxidation state of negative two. In oxygen gas, oxygen has an oxidation state of zero. This means oxygen is both reduced from negative one to negative two, and oxidized from negative one to zero.
Now let's balance the equation by counting atoms. The unbalanced equation has 2 oxygen atoms on the left and 3 on the right. To balance this, we need 2 molecules of hydrogen peroxide, giving us 4 oxygen atoms on the left. On the right, we have 2 water molecules plus 1 oxygen molecule, also totaling 4 oxygen atoms. Let's check hydrogen: 4 hydrogen atoms on each side. The equation is now balanced.
Let's verify our balanced equation using electron transfer. In the reduction half-reaction, hydrogen peroxide gains 2 electrons to form water. In the oxidation half-reaction, hydrogen peroxide loses 2 electrons to form oxygen gas. The electrons gained equal the electrons lost, confirming our balanced equation is correct. This demonstrates the disproportionation nature of the reaction.
To summarize what we've learned: Hydrogen peroxide undergoes a disproportionation reaction where oxygen changes oxidation states. We balanced the equation by counting atoms and verified it through electron transfer. The final balanced equation is 2 H₂O₂ equals 2 H₂O plus O₂. This systematic approach can be applied to balance any redox reaction.