Best Answer:
reduction half eqaution

Cl2O7 -----> ClO2^-

balance the Cl

Cl2O7 ----> 2ClO2^-

balance the O by adding H2O to the side which has less O

Cl2O7 ---> 2ClO2^- + 3H2O

balance the H by adding H+ to the other side

Cl2O7 + 6H+ ---> 2ClO2^- + 3H2O

Now, this has used acid to this point. convet to basic by adding the same number of OH- as you have H+ to each side of the arrow

Cl2O7 + 6H+ + 6OH- ---> 2ClO2^- + 3H2O + 6OH-

where there is H+ and OH- on the same side you will get H2O

Cl2O7 + 6H2O ---> 2ClO2^- + 3H2O + 6OH-

cancel out H2O that occurs on both sides

Cl2O7 + 3H2O ---> 2ClO2^- + 6OH-

Now, we need to work out how many electrons are transferred. To do this add up the total charge on both sides of the arrow. Then add as many electrons as needed to baalnce the charge to the side that has the most positive charge

LHS = 2 x ClO2^- + 6 x OH- = 8-

RHS = no charge

add 8 e to the LHS

Cl2O7 + 3H2O + 8e ------------> 2ClO2^- + 6OH

this is the balanced oxidation half equation

Reduction half equation, follow the same method

H2O2 -----> O2

H2O2 ----> O2 + 2H+

H2O2 + 2OH- ---> O2 + 2H+ + 2OH-

H2O2 + 2OH- ---> O2 + 2H2O

H2O2 + 2OH- ---> O2 + 2H2O + 2e

balanced reduction half equation

Now, add the two together. But first, the oxidation half equation transfers 8 electrons, wheras the reduction half equation transfers only 2. So you need 4 x the reduction half equation to balance the oxidation half equation

4 x (H2O2 + 2OH- ---> O2 + 2H2O + 2e)

gives

4H2O2 + 8OH- ----> 4O2 + 8H2O + 8e

Now

Cl2O7 + 3H2O + 8e ------------> 2ClO2^- + 6OH-

4H2O2 + 8OH- -----> 4O2 + 8H2O + 8e

--------------------------------------...

Cl2O7 + 3H2O + 4H2O2 + 8OH- + 8e ---> 2ClO2^- + 6OH- + 4O2 + 8H2O + 8e

cancel out the electons and anything else that occurs on both sides.

Cl2O7 + 4H2O2 + 2OH- ---> 2ClO2^- + 4O2 + 5H2O

balanced equation

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