Question

Balance each of the following redox reactions occurring in basic solution.
H2O2+ClO2-->ClO2- +O2


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broke it down to
H2O2(aq) --->O2(g)
ClO4- (aq)_____>ClO2-(aq)

4e-+4H+ClO4-(aq)____>ClO2- +2H2O
H2O2(aq)---->O2+2H


am i right so far? Do I multiply H2O2(aq)---->O2+2H by 2. that seems to be balanced don't see where to add e-.

Answer 1

broke it down to
H2O2(aq) --->O2(g)
ClO4- (aq)_____>ClO2-(aq)

4e-+4H+ClO4-(aq)____>ClO2- +2H2O
H2O2(aq)---->O2+2H

Answer 2

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+.