topic 9 oxidation and reduction. 9.1.1 define oxidation and reduction in terms of electron loss and...

Topic 9

Oxidation and Reduction

9.1.1 Define oxidation and reduction in terms of electron loss and gain.

Oxidation: The loss of electronsFe2+(aq) → Fe3+(aq) + e-

Reduction: The gain of electrons2H+(aq) + 2e- → H2(g)

9.1.1 Define oxidation and reduction in terms of electron loss and gain.

Helpful Mnemonic

This is Leo the Lion

LEO goes GER

Loss of Electrons is Oxidation

Gain of Electrons is Reduction

9.1.2 Deduce the oxidation number of an element in a compound.

A species is oxidized when it loses electrons.– Here, zinc loses two electrons to go from neutral

zinc metal to the Zn2+ ion.


A species is reduced when it gains electrons.– Here, each of the H+ gains an electron and they

combine to form H2.


• It may be easier to find what is being reduced and oxidized by splitting the equation into “half equations”.

• For example, with Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

It can be split up as:Zn(s) → Zn2+(aq) + 2e-

and 2H+(aq) + 2e- → H2(g)


• It is not always easy to split equations into half equations.

• Consider the following reaction:

Can you tell which is being oxidized? If not, then we need to use oxidation numbers.

N2(g) + 3H2(g) 2NH3(g)


Oxidation NumberThe charge that an atom would have if all

covalent bonds were broken so that the more electronegative element kept all the electrons.


Oxidation Number Rules– Elements in elemental state = 0– F = -1 (always)– O = -2 (except in H2O2 where its +1)– H = +1 (except in hydrides H-)– Halides = -1 except when bonded to oxygen or other halides

higher in the group (more reactive one will be -1)

The sum of the oxidation numbers in a neutral compound is 0.The sum of the oxidation numbers in a polyatomic ion is the

charge on the ion.


Find the oxidation number for the following:

Nitrogen in N2 =

Carbon in CH4 =

Sulfur in H2SO4 =

Phosphorous in PCl4+ =

Iodine in IO4- =

Answers: 0, -4, +6, +5, +7

– Elements in elemental state = 0

– F = -1 (always)

– O = -2 (except in H2O2 where its +1)

– H = +1 (except in hydrides H-)

– Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1)

9.1.3 State the names of compounds using oxidation numbers.

For elements that have a variable oxidation number, the oxidation state is signified by Roman numerals.

Example: Fe+3 would be written as Iron(III)How would you write the following?FeCl2 FeCl3 MnO4

- Cr2O3

Answers: iron(II) chloride, iron(III) chloride, permanganate (VII), chromium(III) oxide

Challenge: How would you write the formula for ammonium dichromate?

Answer: (NH4)2Cr2O7

9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.

Let’s go back to the equation:

What is the oxidation number for nitrogen on both sides?

Has it been oxidized or reduced?

Answer: Oxidation number goes from 0 to -3. It has gained electrons, therefore it has been reduced.

N2(g) + 3H2(g) 2NH3(g)


Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)


MnO4− + C2O4

2- Mn2+ + CO2

First, assign oxidation numbers.

+7 +3 +4+2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.

Next, find out if carbon and manganese are being oxidized or reduced.

9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.

• Let’s look at an equation that we worked with before….

• What is wrong with this equation?• Answer: It is not balanced!• We have worked with half equations before

(zinc and hydrogen). Now we’ll dig deeper.

MnO4− + C2O4

2- Mn2+ + CO2


General rules for balancing half equations

– 1) Balance atoms being oxidized or reduced– 2) Add H20 to balance Oxygen atoms

– 3) Add H+(aq) to balance Hydrogen atoms

– 4) Add e- to balance charge


Oxidation Half-Reaction

C2O42− CO2

To balance the carbon, we add a coefficient of 2:

C2O42− 2 CO2


Oxidation Half-ReactionC2O4

2− 2 CO2

The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.

C2O42− 2 CO2 + 2 e−


Reduction Half-Reaction

MnO4− Mn2+

The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.

MnO4− Mn2+ + 4 H2O



MnO4− Mn2+ + 4 H2O

To balance the hydrogen, we add 8 H+ to the left side.

8 H+ + MnO4− Mn2+ + 4 H2O



8 H+ + MnO4− Mn2+ + 4 H2O

To balance the charge, we add 5 e− to the left side.

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

9.2.2 Deduce redox equations using half-equations.

Combining the Half-ReactionsNow we evaluate the two half-reactions

together:C2O4

2− 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.


Combining the Half-Reactions5 C2O4

2− 10 CO2 + 10 e−

10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O

When we add these together, we get:

10 e− + 16 H+ + 2 MnO4− + 5 C2O4

2−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−


Combining the Half-Reactions10 e− + 16 H+ + 2 MnO4

− + 5 C2O42−

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

The only thing that appears on both sides are the electrons. Subtracting them, we are left with:

16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2



PracticeGiven two half-equations:

Cr2O72-

(aq) → Cr3+(aq)

Fe2+ → Fe3+

Deduce the half-equations for each, then deduce the redox equation.

Answer

Cr2O72-

(aq) + 14H+(aq) + 6Fe2+

(aq) →

2Cr3+(aq) + 7H2O(l) + 6Fe3+

(aq)

9.2.3 Define the terms oxidizing agent and reducing agent.

Oxidizing agent: Substance that is reduced and causes the oxidation of another substance in a redox reaction.

Reducing agent: Substance that is oxidized and causes the reduction of another substance in a redox reaction.

9.2.4 Identify the oxidizing and reducing agents in redox equations.

Identify the oxidizing and reducing agents in the following equations:

Sn2+(aq) + 2Fe3+

(aq) → Sn4+(aq) Fe2+

(aq)

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

9.2.4 Identify the oxidizing and reducing agents in redox equations.

Deduce the following half equations, deduce the redox equation, and identify the oxidizing agent and the reducing agent.

• MnO4-(aq) → Mn-2

(aq)

• SO2(aq) → SO42-

(aq)

9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents.

• Recall in acids and bases that a strong acid had a weak conjugate base.

• Same in redox reactions. The conjugate of a powerful oxidizing agent is a weak reducing agent.

F2 + 2e- ↔ 2F-Strong oxidizing

agentWeak reducing

agent


• Mr. F can really attract the electrons (more electronegative).

• When Mr. F has the electrons, he doesn’t want to let them go.

• So although he is a good oxidizing agent, he is a poor reducing agent. (He doesn’t like to reduce the number of his electrons!)

9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing

agents.

• Think back to Topic 3 on Periodicity.• What are the trends in electronegativity?


Compare

What exception do you see?

Hydrogen (Lithium is anotherexception)

9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-

(aq)

Feasible? A: Yes

I2(aq) + 2Cl-(aq) → Cl2(aq) + 2I-

(aq)

Feasible?

A: No

Chlorine attracts electrons more strongly than iodine, so chlorine is a better oxidizing agent.

http://www.youtube.com/watch?v=SQit4Yxl4HU&list=EC5C3F606F1DFA4133

9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.

Zn(s) + Cu2+(aq) → Cu(s) + Zn2+

(aq)

Feasible?A: Yes

Cu(s) + Zn2+(aq) → Zn(s) + Cu2+

(aq)

Feasible?A: No

These examples are all displacement reactions, because they involve a more reactive metal or non-

metal displacing the reactive one from its salt.

9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell.

• Anode(-): Oxidation, forms a negative charge• Cathode(+): Reduction, forms a positive charge

2+2+e

-e-e-

e-

e-e-

2+

2+

e-e-


• Lets harness some Energy!!

2+2+

2+

2+

2+

Zn(s) Cu(s)

2+2+

e-e-

2+

2+2+

e-e-

Problem, the highly negative charge on electrode causes (+) ions to be attracted

back

Solution

Balance (-) charge by replacing it

with some more negative ions


• Lets harness some Energy!!

2+

2+2+

2+

2+

Zn(s) Cu(s)

2+2+

e-e-

2+

2+2+

http://www.dynamicscience.com.au/tester/solutions/chemistry/redox/galvan5.swf

+-

+-

+

-

+

-+

-

e-e-

2+





9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive

electrode (cathode).• ANOX and RED CAT• The oxidation occurs

at the anode.• The reduction occurs

at the cathode.• Which of the metals is

being reduced?• So which is the

cathode?

9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).

• Lead and zinc are set up in a voltaic cell.• Which one would be oxidized? Which one is

being reduced?• A: Zinc is being oxidized. Lead is being

reduced. • Which one would be the cathode and which

would be the anode?• Zinc would be the anode, lead is the cathode.

9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.

Need to have a liquid containing ions, which is called an electrolyte. (http://www.youtube.com/watch?v=Pu9XZyQPsy0)

9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode).

• The anode attracts anions.• When the anions reach they anode, they lose

electrons.• So are they oxidized or reduced?• A: oxidized• When cations reach the cathode they gain

electrons and they are reduced.

9.5.3 Describe how current is conducted in an electrolytic cell.

• Electricity is supplied from an external source and used to create a non-spontaneous reaction.

• Electrolyte solution can conduct electricity because the ions move towards oppositely charged electrodes.

• Electrons flow From Anode To CAThode (FATCAT)

9.5.4 Deduce the products of the electrolysis of a molten salt.

Sodium chloride• Negative chloride ions are attracted to the

positive ions. There they lose electrons and are oxidized to chlorine gas:

2Cl-(l) → Cl2(g) + 2e-

• Positive sodium ions are attracted to the negative cathode. They gain electrons and are reduced to sodium metal:

Na+(l) + e- → Na(l)

9.5.4 Deduce the products of the electrolysis of a molten salt.

QuestionFor every 2 mol of electrons that flow through

the circuit, how many mol of chlorine gas and sodium metal will be produced?

A: 1 mol of chlorine gas and 2 mol of sodium.

topic 9 oxidation and reduction. 9.1.1 define oxidation and reduction in terms of electron loss and...

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