Question

balance each of the following oxidation/reduction reactions utilizing the half reaction method,

H2O2 (aq) + Cl2O7 (aq) → ClO2- (aq) + O2 (g) in basic solution

Answer 1

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