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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

SKILLS Project

Oxidation Numbers, Reducing and Oxidizing Agents


Oxidation numbers

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.


Tips and hints

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.


Tips and hints cont d

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.


Example 1 water h 2 o

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.


Example 2 carbonate co 3 2

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+)


Example 3 ammonium nh 4

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.


Example 4 kmno 4

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+).


Example 5 nac 2 h 3 o 2

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-).


Practice on your own

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


So how do we use oxidation numbers

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.


A few definitions

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).


Definitions cont d

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.


Example 1 oxidizing reducing agents

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


Example 2 oxidizing reducing agents

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


Example 3 oxidizing reducing agents

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


Practice on your own1

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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