# H2O2+ClO2–>ClO2+O2 (basic solution)?

I can’t figure out how to balance the half reactions because the

ClO2 –> ClO2 is already balanced. Anyone got any idea how to work this? I’m sure that it’s not ClO2 –> ClO2- because I tried it like that and got it wrong.

Thanks!

• 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 oxidising 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 oxidised because it loses electrons to form H+.

• Balancing Redox Reactions Calculator

• Separate right into a million/2 reactions: ClO2 + e- —> ClO2- (extra e- to stability the charge) H2O2- —> O2 a million. stability H2O with H+ ions 2. upload OH- to the edge of the rxn with extra H+ (in case you upload OH- to a minimum of one edge of the rxn, upload it to the different component to boot) 3. stability out expenses with e- 2OH- + H2O2 —> O2 + 2H+ + 2e- + 2OH- ClO2 + e- —> ClO2- Multiply the 2d rxn (ClO2 + e- —> ClO2-) by making use of two so as which you will cancel out the electrons 2OH- + H2O2 —> O2 + 2H+ + 2e- + 2OH- 2 ClO2 + 2e- —> 2ClO2- upload the two rxn jointly and cancel out basic products BTW 2 adjoining H+ and OH- = H20 answer: H2o2 + 2ClO2 + 2OH- —> O2 + 2ClO2- + 2H2O wish this facilitates and sturdy success 