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20.2 Oxidation Numbers >20.2 Oxidation Numbers >

1 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 20Oxidation-Reduction Reactions

20.1 The Meaning of Oxidation and Reduction

20.2 Oxidation Numbers

20.3 Describing Redox Equations



Why does a sparkler have such a bright light?

CHEMISTRY & YOUCHEMISTRY & YOU

If you have ever seen or held a sparkler, then you know that sparklers give off very bright light. They are like handheld fireworks.



Assigning Oxidation NumbersAssigning Oxidation Numbers

Assigning Oxidation Numbers

What is the general rule for assigning oxidation numbers?




An oxidation number is a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction.




As a general rule, a bonded atom’s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element.




In binary ionic compounds, the oxidation numbers of the atoms equal their ionic charges.

• The compound sodium chloride is composed of sodium ions (Na1+) and chloride ions (Cl1–).

• The oxidation number of sodium is +1.

• That of chlorine is –1.– Notice that the sign is put before the oxidation

number.




Because water is a molecular compound, no ionic charges are associated with its atoms.• Oxygen is reduced in the formation of water.

• Oxygen is more electronegative than hydrogen.

• The two shared electrons in the H–O bond are shifted toward oxygen and away from hydrogen.

– The oxidation number of oxygen is –2.

– The oxidation number of each hydrogen is +1.




Oxidation numbers are often written above the chemical symbols in a formula.

H2O+1 –2




Rules for Assigning Oxidation Numbers

1. The oxidation number of a monatomic ion is equal in magnitude and sign to its ionic charge. For example, the oxidation number of the bromide ion (Br1–) is –1; that of the Fe3+ ion is +3.

2. The oxidation number of hydrogen in a compound is +1, except in metal hydrides, such as NaH, where it is –1.

3. The oxidation number of oxygen in a compound is –2, except in peroxides, such as H2O2, where it is –1, and in compounds with the more electronegative fluorine, where it is positive.

4. The oxidation number of an atom in uncombined (elemental) form is 0. For example, the oxidation number of the potassium atoms in potassium metal (K) or of the nitrogen atoms in nitrogen gas (N2) is 0.

5. For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0.

6. For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.



Sample Problem 20.2Sample Problem 20.2

Assigning Oxidation Numbers to Atoms

What is the oxidation number of each kind of atom in the following ions and compounds?

a. SO2 c. Na2SO4

b. CO32– d. (NH4)2S



Use the set of rules you just learned to assign and calculate oxidation numbers.

Analyze Identify the relevant concepts.1




Solve Apply concepts to this situation. 2


SO2

+4 –2

a. There are two oxygen atoms, and the oxidation number of each oxygen is –2 (Rule 3).

The sum of the oxidation numbers for the neutral compound must be 0 (Rule 5).

Therefore, the oxidation number of sulfur is +4, because +4 + (2 × (–2)) = 0.





b. The oxidation number of each oxygen is –2 (Rule 3).

CO32–

? –2





CO32–

+4 –2

b. The sum of the oxidation numbers of the carbon and oxygen atoms must equal the ionic charge, –2 (Rule 6).

The oxidation number of carbon must be +4, because +4 + (3 × (–2)) = –2.





c. The oxidation number of each sodium ion, Na+, is the same as its ionic charge, +1 (Rule 1).

The oxidation number of oxygen is –2 (Rule 3).

Na2SO4

+1 ? –2





c. For the sum of the oxidation numbers in the compound to be 0 (Rule 5), the oxidation number of sulfur must be +6, because (2 × (+1)) + (+6) + (4 × (–2)) = 0.

Na2SO4

+1 +6 –2





d. Ammonium ions, NH4+, have an ionic charge

of +1, so the sum of the oxidation numbers of the atoms in the ammonium ion must be +1.

The oxidation number of hydrogen is +1 in this ion.

So, the oxidation number of nitrogen must be –3.

NH4+

? +1

? + 4(+1) = +1–3 + 4(+1) = +1





d. Two ammonium ions have a total charge of +2.

Since the compound (NH4)2S is neutral, sulfur must have a balancing oxidation number of –2.

(NH4)2S –3 +1 –2



• The results are consistent with the rules for determining oxidation numbers.

• Also, addition of the oxidation numbers correctly gives the final overall charge for the ion and the three neutral compounds.

Evaluate Do the results make sense?3




Chromium in its uncombined state is a dull silvery color. Orange potassium dichromate (K2Cr2O7) and purple chromium(III) potassium sulfate (CrK(SO4)2·12H2O) are both compounds of chromium. What is the oxidation number of chromium in each compound?



Chromium in its uncombined state is a dull silvery color. Orange potassium dichromate (K2Cr2O7) and purple chromium(III) potassium sulfate (CrK(SO4)2·12H2O) are both compounds of chromium. What is the oxidation number of chromium in each compound?

K2Cr2O7

+1 +6 –2

CrK(SO4)2·12H2O+3 +1 +6 –2



Oxidation-Number Changes in Oxidation-Number Changes in Chemical ReactionsChemical Reactions

Oxidation-Number Changes in Chemical Reactions

How are oxidation and reduction defined in terms of a change in oxidation number?




The figure below shows what happens when copper wire is placed in a solution of silver nitrate.

• The oxidation number of silver decreases from +1 to 0 as each silver ion (Ag1+) gains an electron and is reduced to silver metal (Ag0).





• Copper’s oxidation number increases from 0 to +2 as each atom of copper metal (Cu0) loses two electrons and is oxidized to copper(II) ion (Cu2+).





2AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Ag(s) +1 +5 –2 0 +2 +5 –2 0




This figure illustrates a redox reaction that shows what occurs when a shiny iron nail is dipped into a solution of copper(II) sulfate.

• The iron reduces Cu2+ ions in solution and is simultaneously oxidized to Fe2+.

• The iron becomes coated with metallic copper.




You can define oxidation and reduction in terms of a change in oxidation number.

An increase in the oxidation number of an atom or ion indicates oxidation.

A decrease in the oxidation number of an atom or ion indicates reduction.



What happens to the oxidation numbers of metals as they burn in a sparkler?




What happens to the oxidation numbers of metals as they burn in a sparkler?


As the metals burn, they gain oxygen or undergo oxidation. A substance that undergoes oxidation has an increase in oxidation number. Therefore, as metals burn, their oxidation numbers increase.




Identifying Oxidized and Reduced Atoms

Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reactions. Also identify the oxidizing agent and the reducing agent.

a. Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l)

b. C(s) + O2(g) → CO2(g)



• An increase in oxidation number indicates oxidation.

• A decrease in oxidation number indicates reduction.

• The substance that is oxidized in a redox reaction is the reducing agent.

• The substance that is reduced is the oxidizing agent.







a. Use the rules to assign oxidation numbers to each atom in the equation.

0 +1 –1 +1 –1 0

Cl2(g) + 2HBr(aq) → 2HCl(aq) + Br2(l)

The oxidation number of each chlorine in Cl2 is 0 because of Rule 4.





a. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced.

• The element chlorine is reduced because its oxidation number decreases (0 to –1).

• The bromide ion from HBr(aq) is oxidized because its oxidation number increases (–1 to 0).

0 +1 –1 +1 –1 0






a. Finally, identify the oxidizing and reducing agents.

• Chlorine is reduced, so Cl2 is the oxidizing agent.

• The bromide ion from HBr(aq) is oxidized, so Br– is the reducing agent.

0 +1 –1 +1 –1 0






b. Use the rules to assign oxidation numbers to each atom in the equation.

0 0 +4 –2

C(s) + O2(g) → CO2(g)





b. Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced.

• The element carbon is oxidized because its oxidation number increases (0 to +4).

• The element oxygen is reduced because its oxidation number decreases (0 to –2).

0 0 +4 –2

C(s) + O2(g) → CO2(g)





b. Finally, identify the oxidizing and reducing agents.

• Carbon is oxidized, so C is the reducing agent.

• Oxygen is reduced, so O2 is the oxidizing agent.

0 0 +4 –2

C(s) + O2(g) → CO2(g)



• It makes sense that what is oxidized in a chemical reaction is the reducing agent because it loses electrons—it becomes the agent by which the atom that is reduced gains electrons.

• Conversely, it makes sense that what is reduced in a chemical reaction is the oxidizing agent because it gains electrons—it is the agent by which the atom that is oxidized loses electrons.

Evaluate Do the results make sense?3





Identifying Oxidized and Reduced AtomsUse changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction. Also identify the oxidizing agent and the reducing agent.

Zn(s) + 2MnO2(s) + 2NH4Cl(aq) → ZnCl2(aq) + Mn2O3(s) + 2NH3(g) + H2O(l)



• An increase in oxidation number indicates oxidation.

• A decrease in oxidation number indicates reduction.

• The substance that is oxidized in a redox reaction is the reducing agent.

• The substance that is reduced is the oxidizing agent.







Use the rules to assign oxidation numbers to each atom in the equation.

0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2






Then use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced.

• The element zinc is oxidized because its oxidation number increases (0 to +2).

• The manganese ion is reduced because its oxidation number decreases (+4 to +3).

0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2






Finally, identify the oxidizing and reducing agents.

• Zinc is oxidized, so Zn is the reducing agent.

• Manganese (in MnO2) is reduced, so Mn4+ is the oxidizing agent.

0 +4 –2 –3 +1 –1 +2 –1 +3 –2 –3 +1 +1 –2




Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction.

2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s)



Use changes in oxidation number to identify which atoms are oxidized and which are reduced in the following reaction.

2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s)+1 +5 –2 +1 –2 +2 –2 +1 –2 0

Sulfur is oxidized because its oxidation number increases (–2 to 0). Nitrogen is reduced because its oxidation number decreases (+5 to +2).

2HNO3(aq) + 3H2S(g) → 2NO(g) + 4H2O(l) + 3S(s)



Key Concepts Key Concepts

As a general rule, a bonded atom’s oxidation number is the charge that it would have if the electrons in the bond were assigned to the atom of the more electronegative element.

An increase in the oxidation number of an atom or ion indicates oxidation. A decrease in the oxidation number of an atom or ion indicates reduction.



Glossary TermGlossary Term

oxidation number: a positive or negative number assigned to an atom to indicate its degree of oxidation or reduction; the oxidation number of an uncombined element is zero



Redox reactions are identified by changes in oxidation number.

BIG IDEABIG IDEA

Reactions



END OF 20.2END OF 20.2

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