9.1
OXIDATION AND REDUCTION

Redox (reduction-oxidation) reactions play a key role in many chemical and biochemical processes.
NATURE OF SCIENCE (1.9)
How evidence is used – changes in the definition of oxidation and reduction from one involving specific elements (oxygen and hydrogen), to one involving electron transfer, to one invoking oxidation numbers is a good example of the way that scientists broaden similarities to general principles.

Access to a supply of clean drinking water has been recognized by the United
Nations as a fundamental human right, yet it is estimated that over one billion people lack this provision. Disinfection of water supplies commonly uses oxidizing agents such as chlorine or ozone to kill microbial pathogens.

Chemistry has developed a systematic language that has resulted in older names becoming obsolete. What has been lost and gained in this process?
Oxidation states are useful when explaining redox reactions. Are artificial conversions a useful or valid way of clarifying knowledge?

UNDERSTANDING/KEY IDEA
9.1.A
Oxidation and reduction can be considered in terms of oxygen gain/hydrogen loss, electron transfer or change in oxidation number.

 Early definitions for oxidation and reduction were based upon observations of the gain and loss of oxygen and hydrogen during chemical change.
 Oxidation: gain of oxygen or loss of hydrogen
 Reduction: loss of oxygen or gain of hydrogen
 It is now recognized that oxidation/reduction occur whenever there is a shift in electron density from one atom to another, whether complete or partial.

 Oxidation is the loss of electrons.
 Atom gets more positive.
 Reduction is the gain of electrons.
 Atom gets more negative.
 Reactions that involve oxidation and reduction are called redox reactions.
 You can never have one without the other.

UNDERSTANDING/KEY IDEA
9.1.B
An oxidizing agent is reduced and a reducing agent is oxidized.

 Redox reactions ALWAYS involve the simultaneous oxidation of one reactant with the reduction of another through the transfer of electrons.
 The reactant causing the oxidation of the other reactant is called the oxidizing agent. (It is the one that was reduced.)
 The reactant causing the reduction of the other reactant is called the reducing agent. (It is the one that is oxidized.)

APPLICATION/SKILLS
Be able to identify the species oxidized and reduced and the oxidizing and reducing agents, in redox reactions.

 If you can determine which one was reduced (got more negative), then the other three are easy to find.
 Remember that if it got reduced, then it is the oxidizing agent and vice versa.

 Identify the oxidizing and reducing agents in the following equation.
2Al + 3PbCl
2
→ 2AlCl
3
+ 3Pb
First write down the oxidation numbers.
0 +2 -1 +3 -1 0
2Al + 3PbCl
2
→ 2AlCl
3
+ 3Pb
Al went from 0 to +3 – oxidized (red agent)
Pb went from +2 to 0 – reduced (ox agent)
*Note the oxidizing agent is the whole compound (PbCl
2
), not just the element.*

Identify the substances being oxidized and reduced.
Cu + 2Ag + → 2Ag + Cu 2+
Copper goes from a zero to a +2.
It got more positive so it was oxidized.
Silver goes from +1 to zero.
It got more negative so it was reduced.

APPLICATION/SKILLS
Be able to deduce redox reactions using half-equations in acidic or neutral solutions.

 Since oxidation cannot occur without reduction, you can identify each type of reaction within a redox reaction.

These are called the ½ equations.
 Include the electrons in your equations.


Deduce the two ½ equations for the following reaction:
Zn + Cu 2+ → Zn 2+ + Cu
Zinc goes from 0 to 2+ (more pos – oxidized)
Cu goes from 2+ to 0 (more neg – red)
Oxidation: Zn → Zn 2+ + 2e -
Reduction: Cu 2+ + 2e → Cu
Disproportionation – when the same element is both oxidized and reduced.
















WRITE DOWN THE ½ REACTIONS.
BALANCE ALL ELEMENTS BESIDES OXYGEN AND HYDROGEN.
BALANCE OXYGEN WITH WATER.
BALANCE HYDROGENS WITH H +
DETERMINE OVERALL CHARGES ON BOTH SIDES.
BALANCE THE CHARGES ON EACH SIDE WITH e -
MULTIPLY BY A FACTOR IF NEEDED SO THE e CANCEL.
ADD EQUATIONS AND CANCEL OUT COMMON FACTORS ON
BOTH SIDES.

 FOR BASIC SOLUTIONS, ADD OH TO BOTH SIDES TO GET RID
OF H + ’S.

Balance the following redox reaction:
NO
3
+ Cu → NO + Cu
Step 1: Write the ½ reactions
2+
Ox: Cu → Cu 2+
Red: NO
3
→ NO
Step 2: Balance all elements besides H and O
Cu and N are balanced already.
Step 3: Balance oxygen with water
Ox: Cu → Cu 2+
Red: NO
3
→ NO + 2H
2
O

Step 4: Balance hydrogens with H +
Ox: Cu → Cu 2+
Red: 4H + + NO
3
→ NO + 2H
2
O
Step 5: Determine overall charges on both sides
Ox: (zero) Cu → Cu 2+ (+2)
Red: (+3) 4H + + NO
3
→ NO + 2H
2
O (zero)
Step 6: Balance the charges on each side with e-
Ox:
Red: 3e + 4H +
Cu → Cu 2+ + 2e -
+ NO
3
→ NO + 2H
2
O

Step 7: Multiply by a factor if needed to cancel out the e-
3
(Cu → Cu 2+ + 2e )
2 (3e + 4H + + NO
3
→ NO + 2H
2
O)
3Cu → 3Cu 2+ + 6e -
6e + 8H + + 2NO
3
→ 2NO + 4H
2
O
Step 8: Add both equations and cancel out anything common on both sides.
3Cu → 3Cu 2+ + 6e -
8H +
6e + 8H + + 2NO
3
+ 3Cu + 2NO
3
-
→ 2NO + 4H
2
O
→ 2NO + 3Cu 2+ + 4H
2
O

UNDERSTANDING/KEY IDEA
9.1.C
Variable oxidation numbers exist for transition metals and for most main-group non-metals.

APPLICATION/SKILLS
Be able to deduce the oxidation states of an atom in an ion or a compound.

GUIDANCE
Oxidation states should be represented with the sign before the given number, not after like as in ions.

 Oxidation numbers are shown with the
+or- in front of the number such as +7.
 Ions with their charges are shown with the number first followed by the charge such as 2+.

 The concept of oxidation numbers provides a way to keep track of electrons in redox reactions.
 They are not really charges, but we will use them to assign numbers in covalent compounds to show that the overall charge in a compound is zero.
 It is essentially a convention to assign which element has electron control.

 1. Elements by themselves are zero.
 2. In simple ions, the oxidation number is the same as its charge.
 3. Oxidation numbers in a neutral compound must add up to zero.
 4. Oxidation numbers in a polyatomic ion must add up to the charge of the ion.
 You can predict many oxidation numbers from the periodic table.

GUIDANCE
Know that the oxidation state of hydrogen can be -1 in metal hydrides and oxygen can be -1 in peroxides.

 Fluorine is always (-1).
 Oxygen is usually (-2) except:
 Peroxides (-1)
 OF
2
(+2)
 Hydrogen is (+1) except:
 Metal hydrides NaH (-1)
 Chlorine is (-1) except:
 When combined with O or F, then it is (+1).

 Find the oxidation states for the elements in H
2
SO
4 and Na
2
C
2
O
4.
H
2
SO
4
: H +1 (2 of them for a total of +2)
O -2 (4 of them for a total of -8)
S +6 (make the overall charge 0)
Na
2
C
2
O
4
: Na +1 (2 of them for a total of +2)
O -2 (4 of them for a total of -8)
C +3 (2 of them to equal +6)

APPLICATION/SKILLS
Be able to deduce the name of a transition metal compound from a given formula, applying oxidation numbers represented by Roman numerals.

GUIDANCE
Oxidation number and oxidation state are often used interchangeably, though IUPAC does formally distinguish between the two terms. Oxidation numbers are represented by Roman numerals according to IUPAC.

 You know how to do this – Yeah!!!
 Use Roman numerals when you have more than one choice of ion.


You may see covalent compounds using
Roman numerals.
NO – nitrogen monoxide or
 NO
2 nitrogen II oxide
– nitrogen dioxide or nitrogen IV oxide

UNDERSTANDING/KEY IDEA
9.1.D
The activity series ranks metals according to the ease with which they undergo oxidation.

 Not all oxidizing and reducing agents are the same strength.
 Their strength depends upon how easily they lose or gain electrons.

 Metals tend to give up electrons forming positive ions so they cause other elements to become more negative or to be reduced.
 This is why metals are commonly reducing agents.
 More reactive metals lose their electrons more readily so they are stronger reducing agents.

A sample reactivity series
Mg strongest reducing agent
Al (most readily oxidized)
Zn
Fe
Pb
Cu
Ag weakest reducing agent
(least readily oxidized)

 Nonmetals tend to gain electrons forming negative ions so they cause other elements to become more positive or to be oxidized.
 This is why nonmetals are commonly oxidizing agents.
 More reactive nonmetals gain their electrons more readily so they are stronger oxidizing agents.

A sample reactivity series
F
2
Cl
2
Br
2
I
2 strongest oxidizing agent
(most readily reduced) weakest oxidizing agent
(least readily reduced)

 You do not have to memorize the activity series, but you will have to interpret information from one.
 Remember that any metal or nonmetal above another will cause a displacement reaction.
 If it is not higher on the activity series, the reaction will not occur.
 If you were given a series of viable reactions, you should be able to determine the activity series.

 Redox titrations are commonly used in the food and beverage industry.
 They are very similar to acid-base titrations, but sometimes do not need an indicator as a color change can naturally occur at the equivalence point.

 5Fe 2+ + MnO
4
+ 8H + → 5Fe 3+ + Mn 2+ + 4H
2
O
 This reaction used potassium permanganate in an acidic solution as the oxidizing agent, which oxidizes Fe 2+ ions to Fe 3+ ions. The manganese is reduced from Mn 7+ to Mn 2+ .
 The reaction does not need an indicator as it goes from a deep purple to colorless at equivalence.
 Work the example problem on page 420.






Several different redox titrations use an oxidizing agent to react with excess iodine ions to form iodine.
2I + oxidizing agent → I
2
+ reduced product
Examples of oxidizing agents are: KMnO
K
2
Cr
2
O
7 and NaOCl.
4
, KIO
3
,
The I
2 is then titrated with sodium thiosulfate using starch as an indicator.
The starch indicator is added during the titration
(not at the start) and forms a deep blue color by forming a complex with the free I
2
. As the I
2 reduced to I ions, the blue color disappears is marking equivalence.

 Redox equations: oxidation: 2S
2
O
3
2reduction: I
2
→ S
+ 2e → 2I -
4
O
6
2+ 2e -
 Overall equation:
2S
2
O
3
2 + I
2
→ 2I + S
4
O
6
2-
 Work the example problem on page 422.

UNDERSTANDING/KEY IDEA
9.1.E
The Winkler Method can be used to measure biochemical oxygen demand (BOD), used as a measure of the degree of pollution in a water sample.

 The dissolved oxygen content of water is one of the most important indicators of its quality.
 As pollution increases, the dissolved oxygen content decreases as the oxygen is used by bacteria in decomposition reactions.
 The BOD (biological oxygen demand) is used as a means of measuring the degree of pollution.
 BOD is defined as the amount of oxygen used to decompose the organic matter in a sample of water over a specified time period, usually 5 days at a specified temperature.

APPLICATION/SKILLS
Be able to apply the Winkler
Method to calculate BOD.

 The Winkler Method uses redox titrations to measure the dissolved oxygen in water to calculate the BOD.
1.
The dissolved oxygen in the water is “fixed” by the addition of a manganese II salt such as
MnSO
4
.
2.
Reaction of this salt with oxygen in basic solution causes oxidation of Mn(II) to higher oxidation states such as Mn(IV).
2Mn 2+ + O
2
+ 4OH → 2MnO
2
+ 2H
2
O

3.
Acidified iodide ions are added to the solution and are oxidized by the Mn(IV) to I2.
MnO
2
+ 2I + 4H + → Mn 2+ + I
2
+ 2H
2
O
4.
The iodine produced is then titrated with sodium thiosulfate as described earlier.
2S
2
O
3
2 + I
2
→ 2I + S
4
O
6
2-
So we can see that for every 1 mole of O
2 water, 4 moles of S
2
O
3
2are used.
in the
 Work the sample problem page 423.

International Baccalaureate Organization. Chemistry Guide,
First assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford.
Higher Level Chemistry . 2nd ed. N.p.: Pearson Baccalaureate, 2014. Print.
Most of the information found in this power point comes directly from this textbook.
The power point has been made to directly complement the
Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.