1. Chapter 4 Chemical Quantities and Aqueous Reactions

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3. Stoichiometry (4.2)

4. Amounts of Reactants and Products
Answer the question: “How much?”
How much product will be formed?
How much reactant is needed?
Stoichiometry - The quantitative study of reactants and products in a chemical reaction.
We need to learn one more “tool.”

5. We will consider the following reaction:
2H2 + O H2O
This reaction says:
2 molec H2 reacts with 1 molec O2 to give 2 molec water
2 moles H2 reacts with 1 mole O2 to give 2 moles water
We want to think of the coefficients in terms of moles.
If 4 moles of H2 reacts with 2 moles O2, how many moles of water would be formed?
4 moles H2O

6. 4 mol H2 X mol H2O mol H2 2 = 4 mol H2O
The coefficients in a balanced equation are always used to convert between moles of substances in a chemical reaction.
Let us see how we use this as a unit factor.
4 mol H2 X
mol H2O
mol H2 2
= 4 mol H2O
numbers come from balanced equation

7. because 6x moles of H2O as C6H12O6, the number makes sense
Practice  How many moles of water are made in the combustion of 0.10 moles of glucose?
Given:
Find:
0.10 moles C6H12O6
moles H2O
Conceptual Plan:
Relationships:
C6H12O O2 → 6 CO2 + 6 H2O 1 mol C6H12O6 : 6 mol H2O
mol H2O
mol C6H12O6
Solution:
Check:
0.6 mol H2O = 0.60 mol H2O
because 6x moles of H2O as C6H12O6, the number makes sense 7

8. Amounts of Reactants and Products
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the number of moles of the sought quantity
Convert moles of sought quantity into desired units

9. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O

10. Predicting Amounts from Stoichiometry
The amounts of any substance in a chemical reaction can be determined from the amount of just one substance
How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2

11. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasolne
Assuming that gasoline is octane, C8H18, the equation for the reaction is
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
The equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the conceptual plan will be
g C8H18
mol CO2
g CO2
mol C8H18

12. 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2
Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasoline
Given:
Find:
3.4 x 1015 g C8H18
g CO2
Conceptual Plan:
Relationships:
1 mol C8H18 = g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2
g C8H18
mol CO2
g CO2
mol C8H18
Solution:
Check:
because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

13. The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags:
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
How many grams of N2 are produced from 6.00 g of NaN3? N2
3.88 g
1.72 g
0.138 g
2.59 g of 30

14. Limiting Reagents and Yield (4.3)

15. The Limiting Reactant
For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others
When this reactant is used up, the reaction stops and no more product is made
The reactant that limits the amount of product is called the limiting reactant
sometimes called the limiting reagent
the limiting reactant gets completely consumed
Reactants not completely consumed are called excess reactants
The amount of product that can be made from the limiting reactant is called the theoretical yield

16. Limiting Reagents 2NO + 2O2 2NO2 NO is the limiting reagent
O2 is the excess reagent

17. Here’s a “non-chemical” example.
We want to build toy cars using wooden blocks and wheels.
Each car has one body (large block) and four wheels.
4 wheels block  1 car
We look in our toy box and find we have 10 large blocks and 24 wheels.
Which is our limiting “reagent”?
The wheels
So, how many cars can we build?
Six

18. The first step in doing a “Limiting Reagent” problem is to identify it as one.
When measured info is given for each of the reactants (usually two,) you have a limiting reactant problem.

19. Procedure for Limiting Reagent Problems
Calculate the moles of product formed for each reactant given.
The reactant which produces the least amount of product is the limiting reagent.
Using the information obtained by the limiting reagent, finish the problem.

20. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent

21. Use limiting reagent (Al) to calculate amount of product that can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3

22. Reaction Yield Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100

23. Nitroglycerin (C3H5N3O9) is a powerful explosive
Nitroglycerin (C3H5N3O9) is a powerful explosive. Its decomposition may be represented by:
4C3H5N3O9  6N CO H2O + O2
What is the maximum amount of O2 in grams that can be obtained from 2.00 x102 g of nitroglycerin? Calculate the percent yield in this reaction if the amount of O2 generated is found to be 6.55 g.
molar mass of C3H5N3O9 = g
32.00 g O2
1 mol O2 x
1 mol C3H5N3O9
227.1 g C3H5N3O9 x
1 mol O2
4 mol C3H5N3O9 x
2.00 x 102 g
= 7.05 g
% Yield =
Actual Yield
Theoretical Yield
x 100
= g/7.05 g x 100
= 92.9%

24. Ammonia is produced using the Haber process:
3 H2 + N2  2 NH3
What percent yield of ammonia produced from 15.0 kg each of H2 and N2, if 13.7 kg of product are recovered? Assume the reaction goes to completion.
7.53 x 10-2 %
1.50 x 10-1 %
75.3%
15.0%
16.2% of 30

25. Solution Concentration (4.4)

26. Solutions
When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous
the salt is still there, as you can tell from the taste, or simply boiling away the water
Homogeneous mixtures are called solutions
The component of the solution that changes state is called the solute
The component that keeps its state is called the solvent
if both components start in the same state, the major component is the solvent

27. Solution Concentration: Molarity
Moles of solute per 1 liter of solution
Used because it describes how many molecules of solute in each liter of solution

28. Preparing 1 L of a 1.00 M NaCl Solution

29. because most solutions are between 0 and 18 M, the answer makes sense
Example 4.5: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution
Given:
Find:
25.5 g KBr, 1.75 L solution
molarity, M
Conceptual Plan:
Relationships:
1 mol KBr = g, M = moles/L
g KBr
mol KBr
L sol’n M
Solution:
Check:
because most solutions are between 0 and 18 M, the answer makes sense

30. Using Molarity in Calculations
Molarity shows the relationship between the moles of solute and liters of solution
If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar
2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
1 L solution : 2 moles sugar

31. Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH?
Given:
Find:
0.125 M NaOH, mol NaOH
liters, L
Conceptual Plan:
Relationships:
0.125 mol NaOH = 1 L solution
mol NaOH
L sol’n
Solution:
Check:
because each L has only mol NaOH, it makes sense that mol should require a little more than 2 L

32. Practice – How would you prepare 250.0 mL of 0.150 M CaCl2?
Given:
Find:
250.0 mL solution
mass CaCl2, g
Conceptual Plan:
Relationships:
1.00 L sol’n = mol; 1 mL = 0.001L; 1 mol = g
mL sol’n
L sol’n
g CaCl2
mol CaCl2
Solution:
Dissolve 4.16 g of CaCl2 in enough water to total mL
Check:
the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter

33. moles solute in solution 1 = moles solute in solution 2
Dilution
Often, solutions are stored as concentrated stock solutions
To make solutions of lower concentrations from these stock solutions, more solvent is added
the amount of solute doesn’t change, just the volume of solution
moles solute in solution 1 = moles solute in solution 2
The concentrations and volumes of the stock and new solutions are inversely proportional
M1∙V1 = M2∙V2

34. Example 4. 7: To what volume should you dilute 0. 200 L of 15
Example 4.7: To what volume should you dilute L of 15.0 M NaOH to make 3.00 M NaOH?
Given:
Find:
V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Conceptual Plan:
Relationships:
M1V1 = M2V2
V1, M1, M2 V2
Solution:
Check:
because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does

35. Sulfuric acid is found in some types of batteries. What volume of 3
Sulfuric acid is found in some types of batteries. What volume of 3.50 M H2SO4 is required to prepare mL of 1.25 M H2SO4?
17.5 mL
700. mL
89.3 mL
109 mL
None of the above of 30

36. Solution Stoichiometry
Because molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

37. Example 4. 8: What volume of 0
Example 4.8: What volume of M KCl is required to completely react with L of M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)?
Given:
Find:
0.150 M KCl, L of M Pb(NO3)2
L KCl
Conceptual Plan:
Relationships:
1 L Pb(NO3)2 = mol, 1 L KCl = mol, 1 mol Pb(NO3)2 : 2 mol KCl
L Pb(NO3)2
mol KCl
L KCl
mol Pb(NO3)2
Solution:
Check:
because you need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x the volume of Pb(NO3)2

38. Types of Aqueous Solutions (4.5)

39. What Happens When a Solute Dissolves?
There are attractive forces between the solute particles holding them together
There are also attractive forces between the solvent molecules
When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules
If the attractions between solute and solvent are strong enough, the solute will dissolve

40. Table Salt Dissolving in Water
Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal
When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions
The result is a solution with free moving charged particles able to conduct electricity

41. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
To conduct electricity, there must be mobile ions. This occurs when a compound breaks up (dissociates or ionizes) into its ions in water.

42. The term “dissociation” is used for the breaking up of a compound into cations and anions.
The term “ionization” is used to describe the separation of acids and bases into ions.

43. Molecular View of Electrolytes and Nonelectrolytes
To conduct electricity, a material must have charged particles that are able to flow
Electrolyte solutions all contain ions dissolved in the water
ionic compounds are electrolytes because they dissociate into their ions when they dissolve
Nonelectrolyte solutions contain whole molecules dissolved in the water
generally, molecular compounds do not ionize when they dissolve in water
the notable exception being molecular acids

44. Salt vs. Sugar Dissolved in Water
ionic compounds dissociate into ions when they dissolve
molecular compounds do not dissociate when they dissolve

45. Conduct electricity in solution?
Cations (+) and Anions (-)
Strong Electrolyte – 100% dissociation
NaCl (s) Na+ (aq) + Cl- (aq)
H2O
Weak Electrolyte – not completely dissociated
CH3COOH CH3COO- (aq) + H+ (aq)

46. Ionization of acetic acid
CH3COOH CH3COO- (aq) + H+ (aq)
A reversible reaction. The reaction can occur in both directions.
Acetic acid is a weak electrolyte because its ionization in water is incomplete.

47. CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water
CaCl2
HNO3
(NH4)2CO3
CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
HNO3(aq)  H+(aq) + NO3−(aq)
(NH4)2CO3(aq)  2 NH4+(aq) + CO32−(aq)

48. Solubility of Ionic Compounds
Some ionic compounds, such as NaCl, dissolve very well in water at room temperature
Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature
Compounds that dissolve in a solvent are said to be soluble, where as those that do not are said to be insoluble
NaCl is soluble in water, AgCl is insoluble in water
the degree of solubility depends on the temperature
even insoluble compounds dissolve, just not enough to be meaningful

49. Solubility Rules Compounds that Are Generally Soluble in Water

50. Solubility Rules Compounds that Are Generally Insoluble in Water

52. Practice – Determine if each of the following is soluble in water
KOH
AgBr
CaCl2
Pb(NO3)2
PbSO4
KOH is soluble because it contains K+
AgBr is insoluble; most bromides are soluble, but AgBr is an exception
CaCl2 is soluble; most chlorides are soluble, and CaCl2 is not an exception
Pb(NO3)2 is soluble because it contains NO3−
PbSO4 is insoluble; most sulfates are soluble, but PbSO4 is an exception

53. Which of the following compounds will be insoluble in water?
Ca(OH)2
Na3PO4
CaS
Hg2Cl2
LiF

54. Types of Reactions in Aqueous Solution
Precipitation Reactions (4.6)
Acid - Base Reactions (4.7)
Oxidation-Reduction Reactions (4.8)

55. Precipitation Reactions
Precipitation reactions are reactions in which a solid forms when we mix two solutions
reactions between aqueous solutions of ionic compounds
produce an ionic compound that is insoluble in water
the insoluble product is called a precipitate
picture from Tro Intro Chem, 2nd Ed.

56. 2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2 KNO3(aq)

57. Predicting a precipitation reaction and writing the chemical equation
Write the reactants for the reaction
Show the reactants in their ionic form, i.e., if they are soluble, write their dissociated form
Determine which cation-anion combination yields an insoluble compound

58. Write the products of the reaction.
The insoluble product is written in its combined form with an (s) after it.
The remaining ions are written as aqueous ions.
This reaction equation is called the ionic equation.
Cancel the “spectator ions,” which are not involved in the overall reaction.
What remains is the “net ionic equation,” which shows only the species that actually take part in the reaction.

59. Writing Net Ionic Equations -Example
Write the balanced molecular equation.
Write the ionic equation showing the strong electrolytes completely dissociated into cations and anions.
Cancel the spectator ions on both sides of the ionic equation
Check that charges and number of atoms are balanced in the net ionic equation
Write the net ionic equation for the reaction of silver
nitrate with sodium chloride.
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
Ag+ + NO3- + Na+ + Cl AgCl (s) + Na+ + NO3-
Ag+ + Cl AgCl (s)

60. Practice – Write the ionic and net ionic equation for each
K2SO4(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2SO4(s)
2K+(aq) + SO42−(aq) + 2Ag+(aq) + 2NO3−(aq)
® 2K+(aq) + 2NO3−(aq) + Ag2SO4(s)
2 Ag+(aq) + SO42−(aq) ® Ag2SO4(s)
Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l)
2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq)
® 2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq) ® CO2(g) + H2O(l)
K2SO4(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2SO4(s)
Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l)

61. Types of Reactions in Aqueous Solution
Acid - Base Reactions (4.7)

62. Acid-Base Reactions 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
Also called neutralization reactions because the acid and base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
The net ionic equation for an acid-base reaction is
H+(aq) + OH(aq)  H2O(l)
as long as the salt that forms is soluble in water

63. Acids and Bases in Solution
Acids ionize in water to form H+ ions
more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+
most chemists use H+ and H3O+ interchangeably
Bases dissociate in water to form OH ions
bases, such as NH3, that do not contain OH ions, produce OH by pulling H off water molecules
In the reaction of an acid with a base, the H+ from the acid combines with the OH from the base to make water
The cation from the base combines with the anion from the acid to make the salt
acid + base salt + water

64. Some Additional Examples
HF + NaOH 
2HNO3 + Ba(OH)2 
H2SO4 + 2LiOH 
NaF + H2O
Ba(NO3) H2O
Li2SO4 + 2H2O

65. Titration
Often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration
In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio.
the unknown solution is added slowly from an instrument called a burette

66. Acid-Base Titrations
The difficulty is determining when there has been just enough titrant added to complete the reaction
the titrant is the solution in the burette
In acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity
the chemical is called an indicator
At the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH
also known as the equivalence point

67. Titration The titrant is the base solution in the burette.
As the titrant is added to
the flask, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change.
At the titration’s endpoint,
just enough base has been added to neutralize all the acid. At this point the indicator changes color.

68. Color Change of Phenolphthalein in a Titration

69. 12.54 mL of 0.100 M NaOH Given: 10.00 mL HCl
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Write down the given quantity and its units
Given: mL HCl
12.54 mL of M NaOH

70. Write down the quantity to find, and/or its units
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Write down the quantity to find, and/or its units
Find: concentration HCl, M

71. Collect needed equations and conversion factors
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Collect needed equations and conversion factors
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH  1 L sol’n

72. Write a conceptual plan
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Write a conceptual plan mL
NaOH L
NaOH
mol
NaOH
mol
HCl mL
HCl L
HCl

73. Apply the conceptual plan
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the conceptual plan
= 1.25 x 103 mol HCl

74. Apply the conceptual plan
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the conceptual plan

75. Check the solution HCl solution = 0.125 M
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Check the solution
HCl solution = M
The units of the answer, M, are correct.
The magnitude of the answer makes sense because
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.

76. Gas-Evolving Reactions
Some reactions form a gas directly from the ion exchange
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water
K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq)
H2SO3  H2O(l) + SO2(g)

77. Types of Reactions in Aqueous Solution
Oxidation-Reduction Reactions (4.8)

78. Other Patterns in Reactions
The precipitation, acid-base, and gas-evolving reactions all involve exchanging the ions in the solution
Other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions
also known as redox reactions
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)

79. Oxidation-Reduction Reactions
Also called redox reactions
electron-transfer reactions
2Na(s) + Cl2(g)  2NaCl(s)
Na becomes Na+ ; it loses an electron.
2Na  2Na+ + 2e-
Cl2 becomes 2Cl- ; it gains electrons.
Cl2 + 2e-  2Cl-
These are called half reactions. One cannot occur without the other.
This is the oxidation half reaction: loss of electrons.
This is the reduction half reaction: gain of electrons.

80. Redox Reactions Loss of electrons – oxidation (LEO)
Gain of electrons – reduction (GER)
Reduction is easily remembered because it is accompanied by a reduction in the charge.
Oxidizing agent - the substance being reduced (causes oxidation)
Reducing agent - the substance being oxidized (causes reduction)

81. Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)
Zn Zn2+ + 2e-
Zn is oxidized
Zn is the reducing agent
Cu2+ + 2e Cu
Cu2+ is reduced
Cu2+ is the oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)
Cu Cu2+ + 2e-
Ag+ + 1e Ag
Ag+ is reduced
Ag+ is the oxidizing agent

82. Which of the following is a redox reaction?
HBr (aq) + KOH (aq) → H2O (l) + KBr (aq)
SO3 (g) + H2O (l) → H2SO4 (aq)
HBr (aq) + Na2S (aq) → NaBr (aq) + H2S (g)
NH4+ (aq) + 2 O2 (g) → H2O (l) + NO3– (aq) + 2 H+ (aq)
None of the above

83. Oxidation Number
To understand more complex redox reactions, we must know the oxidation number to determine what is oxidized and what is reduced.
Oxidation number - the number of charges an atom would have if electrons were transferred completely.

84. Rules for assigning oxidation numbers to atoms
In free elements, each atom has an oxidation number of zero.
H2, Br2 S8, K, Al ON = 0
For monatomic ions, the oxidation number is the same as the charge of the ion.
Li+ ON = +1 S2- ON = -2

85. Rules (continued)
The ON of oxygen is usually -2. A common exception is in hydrogen peroxide H2O2 and other peroxides (O22-) in which the is ON is -1.
The ON of hydrogen is usually +1, except when it is bonded to metals in binary compounds, in which case it is -1.
H2O ON of H = +1
LiH ON of H = -1

86. Rules (continued)
Fluorine has an ON of -1. Other halogens can be -1, if they are the halide anions, or they can have various positive ONs, depending on what they are combined with.
NaF ON of F = -1
NaCl ON of Cl = -1
HClO4 ON of Cl  -1 revisit later

87. Rules (continued)
In a neutral molecule, the sum of the ONs of all the atoms must be zero. In a polyatomic ion, the sum of the ONs of all the elements must equal the net charge of the ion.
Oxidation numbers do not have to be integers. e.g., O2- ON = -½

88. The oxidation numbers of elements in their compounds

89. Oxidation numbers of all the elements in the following ?
IF7
Oxidation numbers of all the elements in the following ?
F = -1
7x(-1) + ? = 0
I = +7
K2Cr2O7
NaIO3
Na = +1
O = -2
O = -2
K = +1
3x(-2) ? = 0
7x(-2) + 2x(+1) + 2x(?) = 0
I = +5
Cr = +6

90. Determine the oxidation number of the red element in each of the following compounds:
H2PO4– SO32– N2O4

91. Oxidation and Reduction: Another Definition
Oxidation occurs when an atom’s oxidation state increases during a reaction
Reduction occurs when an atom’s oxidation state decreases during a reaction
CH O2 → CO2 + 2 H2O
− – −2
oxidation
reduction

92. Sn4+ + Ca → Sn2+ + Ca2+ F2 + S → SF4
Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions:
Sn4+ + Ca → Sn Ca2+
F2 + S → SF4
Ca is oxidized, Sn4+ is reduced
Ca is the reducing agent, Sn4+ is the oxidizing agent −1
S is oxidized, F is reduced
S is the reducing agent, F2 is the oxidizing agent