1 / 17

SKILLS Project

SKILLS Project. Oxidation Numbers, Reducing and Oxidizing Agents. Oxidation Numbers. Remember, oxidation numbers are a rough measure of the charge of individual elements within a chemical compound.

jill
Télécharger la présentation

SKILLS Project

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. SKILLS Project Oxidation Numbers, Reducing and Oxidizing Agents

  2. Oxidation Numbers • Remember, oxidation numbers are a rough measure of the charge of individual elements within a chemical compound. • We can use these values to predict the quantity and flow of electrons within RedOx reactions.

  3. Tips and hints: • Elemental substances, diatomics: 0 • Ex: Cu(s), O2, H2, Cl2 • Non-metals receive charges based on their electronegativity and place on the periodic table. • The most electronegative element is given its charge first. • F > O > Cl ……. • Group 1 (Alkali) metals are always going to be (1+), unless they are in their elemental (solid) forms.

  4. Tips and Hints, cont’d • You can usually make the assumption that group 2 (Alkaline Earth) metals will be (2+) unless in their elemental forms. • Monatomic ions have an oxidation number equal to their visible charge. • Ex: Cu2+ = 2+, Pb4+ = 4+, etc • Hydrogen is (1+) with non-metals, (1-) with metals, and (0) in H2 (elemental). • The individual charges within a compound will add up to a compound’s overall charge.

  5. Example #1: Water, H2O • Oxygen receives a (2-) charge first, being the most electronegative element. 1+` x 2- H2O 2x - 2 = 0 2(x) + (-2) = 0 • Solving for (x), we find that a single hydrogen in water has a (1+) charge. • As a result, hydrogen is our “unknown” element. Keep in mind, we expect it to be (1+). • We can set up a simple algebraic equation from this information. • Our two unknowns (2x) from the hydrogens plus the (-2) from the oxygen gives the overall “visible” neutral charge of 0.

  6. Example #2: Carbonate, CO32- • Once again, oxygen receives a (2-) charge first, being the most electronegative element. This charge will be multiplied by 3 as there are 3 O’s. x 4+ 2- CO32- x - 6 = -2 x + 3(-2) = -2 • These charges add up to a total overall visible charge of (-2). • Set up your equation….. • This means that our single carbon atom will be our unknown, with a charge of (x). • Note: Neutral carbon has 4 valence electrons. The three bonded oxygens are more than enough to “steal” all of these, but not to break carbon’s octet. • Solve for (x). We find that the oxidation number of carbon in the carbonate polyatomic is (4+)

  7. Example #3: Ammonium, NH4+ x 3- 1+ NH4+ x + 4(+1) = +1 x + 4 = +1 • As a result, our single nitrogen atom is our unknown, (x). • To start off, we know that hydrogen is always (1+) in the presence of other non-metals such as nitrogen. Note that this applies to each of the 4 hydrogen atoms. • From this, we can set up our equation. The sum of all the individual charges should be (+1) from the overall charge of ammonium ion. • Solving the equation, we find the nitrogen has an overall charge of (3-). This is consistent with the fact that nitrogen needs 3 electrons to complete an octet.

  8. Example #4: KMnO4 1+ 7+ x 2- KMnO4 x + 4(-2) = 0 (+1) 1 + x – 8 = 0 + • As a result, manganese (Mn) is our unknown charge. Note: transition metals almost always act as unknowns in any compound. • Potassium permanganate contains 3 elements: potassium, manganese, and oxygen. Oxygen is the most electronegative and receives a charge of (2-). • Set up the equation, taking the subscripts in the compound into account. The overall charge of the compound is 0. • Potassium is a group 1 metal and receives a charge of (1+). • Solving the equation, we find that the charge on manganese is (7+).

  9. Example #5: NaC2H3O2 1+ x 0 1+ 2- NaC2H3O2 1 + 2x + 3 - 4 = 0 2x + 3(+1) = 0 (+1) + + 2(-2) • We can do the same for hydrogen in the presence of other non-metals. • Like transition metals, carbon is often going to be an unknown when determining oxidation numbers. Note, there are two carbons in this compound. • With all our charges in place, we can set up the equation…. • Solving for (x), we find that the oxidation number of a single carbon in sodium acetate is (0). • Now, we can assign a (1+) charge to sodium. Remember, we can do this automatically for any group 1 metal. • Once again, we can use oxygen as our starting point in this compound with an assigned charge of (2-).

  10. Practice on Your Own: 0 1+ 2- 2+ 1- • MgH2 • C6H12O6 2+ 4+ 2- 5+ 1+ 2- • CaCO3 • H3PO4 1+ 7+ 2- 5+ 1- • HClO4 • BrF5 2- 1+ 3+ 2- • Fe2O3 • C2H2 4+ 2- 2- 2+ • SO2 • NO

  11. So, how do we use oxidation numbers? • Oxidation-reduction equations consist of two separate halves, an oxidation and a reduction. • Changes in oxidation numbers indicate which elements are being oxidized or reduced. • As a result, you can identify oxidizing and reducing agents.

  12. A few definitions: • Oxidation: loss of electrons. • The loss of electrons produces an oxidation number that is more positive (or less negative). • Reduction: gain of electrons. • The gain of electrons produces an oxidation number that is more negative (or less positive).

  13. Definitions, cont’d: • Oxidizing Agent: • Substance which oxidizes another substance. The oxidizing agent is reduced as a result. • Reducing Agent: • Substance which reduces another substance. The reducing agent is oxidized as a result.

  14. Example 1: Oxidizing/Reducing Agents Ni(s) + Cu2+(aq)  Ni2+(aq) + Cu(s) 0 2+ 2+ 0 RA OA • To determine the oxidizing and reducing agents in this problem, we’ll need to discover the oxidation numbers of each element. • We can start by assigning a value of “0” to Ni(s) and Cu(s). These are elements in their standard states. • Cu2+ and Ni2+ are single ions with visible charges. Their oxidation numbers are equal to their visible charges. • Now, use the oxidation numbers to determine who has been oxidized and who has been reduced. • According to the equation, nickel goes from an oxidation number of 0  2+. Nickel was itself oxidized and functions as the reducing agent. • Copper, Cu, does the exact opposite, going from 2+  0. Cu2+ is being reduced and acts as the oxidizing agent. • Altogether: • Ni(s) is oxidized Reducing Agent • Cu2+(aq) is reduced  Oxidizing Agent

  15. Example 2: Oxidizing/Reducing Agents 2MnO4- + 5C2O42- 10CO2 + 2Mn2+ 7+ 2- 3+ 2- 4+ 2- 2+ OA RA • Next, we find that carbon, C, in C2O42- has also changed its oxidation number from • 3+  4+. C2O42- is being oxidized and is therefore the reducing agent. • This time around, we’ll need to determine each of the oxidation numbers of the elements in each compound to determine who is oxidized and who is reduced. • Starting with MnO4-, we solve for the individual oxidation numbers of each element. • Note, we are only concerned with the charges within each compound. Coefficients have no influence here, so ignore them! • In other words, 5MnO4- would produce the same oxidation numbers as 2MnO4-. • Now that we’ve assigned all oxidation numbers, we can begin searching for the elements that have been oxidized or reduced. • First off, we notice that manganese, Mn, went from 7+  2+ . This means MnO4- was reduced and is the oxidizing agent. • Altogether, • MnO4- is reduced Oxidizing Agent • C2O42- is oxidized  Reducing Agent

  16. Example 3: Oxidizing/Reducing Agents 2Fe2+ + H2O2 + 2H+  2Fe3+ + 2H2O 2+ 1+ 1- 1+ 3+ 1+ 2- RA OA • Once again, we’ll need to determine the oxidation number of each individual element. • Remember, coefficients do not factor into the assignment of oxidation numbers. • A close look reveals that iron, Fe, was oxidized from 2+  3+. This would make Fe2+ the reducing agent. • We can ignore hydrogen as its oxidation number remains 1+ throughout the problem. • Oxygen, on the other hand, goes from 1-  2-, indicating that it has been reduced. As a result, H2O2 is the oxidizing agent. • Overall: • Fe2+ is oxidized reducing agent • H2O2 is reduced  oxidizing agent

  17. Practice on Your Own: • H2(g) + F2(g)  2HF(g) • C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) • Fe2O3(s) + 2Al(s)  Al2O3(s) + 2Fe(s) • Pb(NO3)2(aq) + 2I2(aq)  PbI4(s) + 2NO3-(aq) RA OA RA OA RA OA OA RA

More Related