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 ______________.
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?
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,
Well, those oxidation numbers seem about right to me, so I'll give you a "humerus" applause for that! As for the oxidizing and reducing agents, think of it this way: the oxidizing agent is the one that gets reduced (because it gains electrons) and the reducing agent is the one that gets oxidized (because it loses electrons). So, based on that, you might want to do a little "OCl-" dance, because it's the reducing agent, and throw some water balloons at H2O because it's the oxidizing agent in this reaction! Keep up the "pun-tastic" work!
Yes, your answers are correct.
1.) O in H2O2: According to Rule 8, the oxidation state of oxygen in compounds is always -2. Since there are two oxygen atoms in H2O2, the total oxidation state contributed by oxygen is -4. The overall charge of H2O2 is 0, so the sum of the oxidation states of all atoms must be 0. Therefore, the oxidation state of oxygen in H2O2 is -1.
2.) Cl in OCl-: According to Rule 8, the oxidation state of oxygen in compounds is always -2. The overall charge of OCl- is -1, so the sum of the oxidation states of all atoms must be -1. Therefore, the oxidation state of chlorine in OCl- is +1.
3.) The oxidizing agent for the redox reaction is H2O(l). H2O2 is being reduced to H2O(l), meaning it is gaining electrons. Therefore, H2O(l) is the oxidizing agent.
4.) The reducing agent is OCl-. OCl- is being oxidized to Cl-, meaning it is losing electrons. Therefore, OCl- is the reducing agent.
Yes, your answers are correct.
To assign oxidation numbers, we follow the given rules.
1. For O in H2O2, we use Rule 7 (hydrogen in compounds is assigned an oxidation state of +1) and Rule 8 (oxygen in compounds is assigned an oxidation state of -2). Since there are two hydrogen atoms, the oxidation state of oxygen can be calculated as follows:
2(+1) + oxidation state of oxygen = 0
2 + oxidation state of oxygen = 0
Oxidation state of oxygen = -2
Therefore, the oxidation state of O in H2O2 is -2.
2. For Cl in OCl-, we use Rule 8 (oxygen in compounds is assigned an oxidation state of -2). Since there is one oxygen atom, the oxidation state of chlorine can be calculated as follows:
oxidation state of chlorine + 1(-2) = -1
oxidation state of chlorine - 2 = -1
oxidation state of chlorine = +1
Therefore, the oxidation state of Cl in OCl- is +1.
For the identification of the oxidizing and reducing agents, we look at the changes in oxidation numbers.
3. The oxidizing agent is the species that gets reduced. In this case, OCl- gains electrons and its oxidation state decreases from +1 to -1. Therefore, the oxidizing agent is OCl-.
4. The reducing agent is the species that gets oxidized. In this case, H2O2 loses electrons and its oxidation state increases from -2 to 0. Therefore, the reducing agent is H2O2.