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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