H2O2→ O2 + 2H+ + 2e- OCl- + 2H+ + 2e-→ H2O+ Cl- H2O2(aq) + OCl-(aq) → H2O(l) + Cl- (aq) + O2(g)Assign oxidation numbers to the following atoms: O in H2O2 ________; Cl in OCl- __________The oxidizing agent for this RedOx rxn is_____________. The reducing agent is ______________.

Question

Rules are in order of priority (Rule 1 is more important than Rule 2...)1. Free elements have an oxidation state = 02. The sum of the oxidation states of all the atoms in a species must be equal to the net charge of the species3. The alkali metals (Li, Na, K, Rb and Cs) in compounds have anare always assigned an oxidation state of +1 4. Fluorine in its compounds is always assigned an oxidation state of -15. The alkaline earth metals (Be, Mg, Ca, Sr, Ba and Ra), Zn and Cd in compounds are always assigned an oxidation state of +26. Aluminum and gallium are always assigned an oxidation state of +3 in their compounds7. Hydrogen in compounds is assigned an oxidation state of +18. Oxygen in compounds is assigned an oxidation state of -2

1.) O in H2O2= Rule 7 & 8 and Cl in OCl- = Rule 8

2.) The oxidizing agent for the redox rxn is H2O (l). The reducing agent is OCl-.

Are those correct?

Answer 1

I don't think the problem is asking for you to answer by rule number. I think they want you to give the oxidation number.
O in H2O2 is -1 for EACH O atom.
Cl is +1 NaOCl

If those numbers agree with your rules you're oK.
For the other part of the problem , remember these definitions.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.
The reducing agent is the substance oxidized.
The oxidizing agent is the substance reduced.
You can look at your half equations and tell easily which is oxidized/reduced. For example, look at the H2O2 half equation. It is giving away electrons; therefore, H2O2 is oxidized(loss of electrons) and is the reducing agent,