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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.

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SKILLS Project

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  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

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