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+..

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+.