The document provides a step-by-step guide for balancing a redox reaction involving ClO2 and H2O2. It explains how to set up the half-reactions, equalize the number of electrons, and combine the equations to achieve balance. It identifies ClO2 as the oxidizing agent and H2O2 as the reducing agent.

1. You need to put your electrons in the first one: ClO2 + e- => ClO2- In the second
one, you have done nothing with the hydrogen. H2O2 => O2 + 2H+ + 2e- Now you must
multiply the first one by 2 to get the same number of electrons on both sides: 2ClO2 + 2e- =>
2ClO2- Now add the equations and eliminate the electrons: H2O2 + 2ClO2 => 2ClO2- + O2 +
2H+ This is now balanced. The charges are the same on both sides and the number of atoms of
each element is the same on both sides. The oxidizing agent is the one which gains electrons and
is itself reduced. So, yes, it is the ClO2. The reducing agent is the H2O2. This is itself oxidized
because it loses electrons to form H+.
Solution
You need to put your electrons in the first one: ClO2 + e- => ClO2- In the second
one, you have done nothing with the hydrogen. H2O2 => O2 + 2H+ + 2e- Now you must
multiply the first one by 2 to get the same number of electrons on both sides: 2ClO2 + 2e- =>
2ClO2- Now add the equations and eliminate the electrons: H2O2 + 2ClO2 => 2ClO2- + O2 +
2H+ This is now balanced. The charges are the same on both sides and the number of atoms of
each element is the same on both sides. The oxidizing agent is the one which gains electrons and
is itself reduced. So, yes, it is the ClO2. The reducing agent is the H2O2. This is itself oxidized
because it loses electrons to form H+.