Achieve wk 7/8 #5

[Janelle2E]

In the reaction Cl2O7 (g) + H2O2 (aq) --> ClO2- (aq) + O2 (g), why is Cl2O7 the oxidizing agent and H2O2 the reducing agent? I am also having trouble balancing the equation. Could someone explain the steps to balance it pls? Thanks!

[duketh10]

In the reaction Cl2O7 (g) + H2O2 (aq) --> ClO2- (aq) + O2 (g), why is Cl2O7 the oxidizing agent and H2O2 the reducing agent? I am also having trouble balancing the equation. Could someone explain the steps to balance it pls? Thanks!
Hi! First we need to identify the oxidation and reduction reactions.
Oxidation: H202 -> O2 (this is oxidation since H2O2 is losing electrons to become O2.)
Reduction: Cl2O7 -> ClS2- (this is reduction since ClO7 is gaining electrons and we can see this as the charge of ClS2 is negative.)
When it comes to balancing think of it as balancing regular chemical reactions.
Oxidation:
H2O2 -> O2
H2O2 -> O2 + 2H⁺(to have 2 hydrogens on both sides)
H2O2 -> O2 + 2H⁺+2e^- (you add electrons to balance the charges and make the reaction neutral)
Reduction:
Cl2O7 -> ClO2^-
Cl2O7 -> 2ClO2^- +3H2O (2 in front of ClO2 to balance the Cl, and add 3H2O's to balance the amount of Oxygens on both sides)
Cl2O7 +6H⁺-> 2ClO2^- +3H2O (add 6 hydrogens to balance H on both sides)
Cl2O7 +6H⁺+8e--> 2ClO2^- +3H2O (add 8 electrons to balance the charge on the left side and ensure neutrality.)
Then we compare electrons and we see that there is more in the red rxn, so we multiply the ox rxn to balance which is 4.
After this, we can cancel what is present on both sides, and add like terms to get our overall reaction. :)