Electrochem review session

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Chemistry 30

Electrochemistry

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Electrochemistry Unit – Curricular Alignment/Table of Contents

REDUCTION OXIDATION: AN INTRODUCTION TO TERMS ...3

Half Reactions:...4

Oxidizing and Reducing Agents: ...4

DIPLOMA EXAMINATION PRACTICE...5

ASSIGNING OXIDATION NUMBERS:...8

OXIDATION NUMBER CHANGES ...8

PRACTICE ASSIGNING OXIDATION NUMBERS...10

WILL A REDOX REACTION BE SPONTANEOUS? ...11

PRACTICE MAKING REDUCTION HALF REACTION TABLES...12

PREDICTING REDOX REACTIONS...14

CHEMISTRY 30: PREDICTING A REDOX REACTION...16

BALANCING REDOX EQUATIONS ...22

METHOD 1 THE OXIDATION NUMBER CHANGE METHOD: ...22

METHOD 2 BALANCING USING HALF REACTIONS...24

BALANCING REDOX REACTIONS USING OXIDATION NUMBERS OR HALF REACTIONS...29

DISPROPORTIONATION/AUTOXIDATION: A WORKSHEET...31

TITRATIONS IN REDOX REACTIONS ...36

REDOX TITRATIONS AND THEIR SOLUTIONS USING STOICHIOMETRY...37

ELECTROCHEMISTRY AND ELECTROCHEMICAL CELLS ...41

VOLTAIC CELLS...41

HALF CELLS AND CALCULATING ELECTRIC POTENTIAL...48

Hydrogen as the standard to calculate cell potentials...48

Calculating Cell Potentials ...49

CHEMISTRY 30: VOLTAIC CELLS PREDICTIONS...59

BATTERIES, WET CELLS AND DRY CELLS...67

FUEL CELLS...70

CORROSION ...76

PREVENTION:...77

ELECTROLYTIC CELLS...80

PREDICTING THE ANODE AND CATHODE HALF REACTIONS IN ELECTROLYSIS:...80

Electrolysis of water...81

Electrolysis of molten sodium chloride...81

Electrolysis of brine (Chlor-Alkali Process):...82

ELECTROPLATING: ...83

ELECTROLYTIC CELLS WORKSHEET...85

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Reduction Oxidation: An Introduction to Terms

Reduction oxidation reactions are chemical changes that occur when electrons are transferred between reactants.

• Redox reactions for short.

• Burning of wood in a fireplace, or burning of food in our bodies are

examples of oxidation reactions.

• All oxidation reactions are accompanied by reduction reactions.

Oxidation originally meant the combination of an element with oxygen to give oxides.

e.g. The rusting of iron to form iron(III) oxide.

Fe(s) + 3O2(g)  2Fe2O3(s)

Compounds can also be oxidized. Methane gas burns in oxygen:

CH4(g) + 2O2(g)  CO2(g) + 2 H2O(l)

Reduction has meant the loss of oxygen from a compound: e.g. The reduction of iron ore to iron metal by heating with coke (Charcoal)

2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g)

Iron oxide is reduced to metal iron – the volume decreases or is reduced.

Remember: Redox reactions happen at the same time…iron is reduced, carbon is oxidized in the above example.

Today: Oxidation is the complete or partial loss of electrons, or gain of oxygen.

Reduction is the complete or partial gain of electrons or loss of oxygen.

Two ways we can remember this is:

Lose Electrons Oxidation – Gain Electrons Reduction

LEO the Lion Says GER

Oxidation Is Losingelectrons – Reduction Is Gainingelectrons

OIL RIG

How This Relates to the Curriculum:

30–B1.1k define oxidation and reduction operationally and theoretically

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Identification of Oxidation and Reduction

Clues for Oxidation:

• A metal atom becomes a positive ion. • A non-metal ion becomes an element.

• All or part of a chemical reacts to give products containing more oxygen.

• A chemical reacts directly with oxygen.

Clues for Reduction:

• A positive metal ion becomes an atom. • A non-metal atom becomes a negative ion.

• All or part of a chemical reacts to give products containing less oxygen.

Examples of Redox Reactions:

Reactions between metals and non-metals:

Mg + S  Mg2+_{+ S}

2-• The net reaction is a transfer of two electrons from a magnesium atom

to a sulfur atom.

• Mgloses 2 e- and is oxidized, the Sgains 2 e- and is reduced.

Half Reactions:

• To see this transfer more clearly, we could separate this net redox

reaction into its two separate oxidation and reduction reactions (called half reactions because they are half of the net reaction).

• Half reactions help us to see where the electrons are coming from and

going to.

Oxidation Half reaction:

Mg  Mg2+_{+ 2e} -Reduction Half reaction:

S + 2e-__S2-

Oxidizing and Reducing Agents:

The substance that donates the electrons is a reducing agent (RA).

• Mg reduces sulphur.

The substance that accepts electrons is an oxidizing agent (OA).

• S oxidizes Mg.

Mg(s) + S(s)  MgS(s)

Example Problem

What is oxidized and what is reduced in this single replacement reaction? 2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2 Ag(s)

Solution:

• Write out the equation and show ions

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1. Br2(l) + 2Fe2+(aq)→ 2Br-(aq) + 2Fe3+(aq)

2. 2Fe3+(aq) + 2I-(aq)→ I2(s) + 2Fe2+(aq)

3. Cl2(g) + 2Br-(aq)→ 2Cl-(aq) + Br2(l)

4. Hg(l) + Br2(l)→ 2Br-(aq) + Hg2+(aq)

5. Hg2+

(aq) + 2Fe2+(aq)→ 2Fe3+(aq) + Hg(l)

• Write the 2 half reaction equations for the above single replacement reactions.

• Indicate which half reaction is a reduction half reaction and which is an oxidation half reaction

Mass and Molar Relationships in Redox Reactions:

When Zn(s) is placed in CuSO4(aq) a redox reaction occurs

Zn(s)→ Zn2+(aq) + 2e- LEO (loss of electrons oxidation )

Cu2+

(aq) + 2e-→ Cu(s) GER (gaining electrons reduction)

OVERALL Zn(s) + Cu2+(aq)→ Zn2+(aq) + Cu(s)

• Mass lost by Zn ≠ mass gained by Cu2+ because they have different

molar masses.

• # e- lost by Zn = # e- gained by Cu2+(aq)

• Moles Zn used = moles Cu formed because both half reactions

involve the transfer of 2 moles of electrons for every mole of metal/metal ion. This is not always the case in other redox reactions (e.g. when copper metal reacts with silver ions.)

Cu(s) + 2Ag+(aq)→ 2Ag(s) + Cu2+(aq)

Oxidation ½ reaction equation Cu(s)→ Cu2+(aq) + 2e

-Reduction ½ reaction equation Ag+(aq) + e-→ Ag(s)

Two moles of silver react with one mole of copper, because in order for one mole of copper to be oxidized, two moles of electrons are released. Therefore two moles of silver must be reduced to accept the two moles of electrons from the oxidizing copper.

Reducing agent = Cu(s)

Oxidizing agent = Ag+

(aq)

Remember:

• An oxidizing agent removes e- and is itself reduced in the process • A reducing agent gives up e- and in the process is itself oxidized

Zn(s) is oxidized and is the reducing agent

Cu2+

(aq) is reduced and is the oxidizing agent

Diploma Examination Practice

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1. Match the equations, as numbered above, with the corresponding descriptions listed below.

A biological redox reaction carried out in a plant cell but not in an animal cell __________

A biological redox reaction carried out in both animal and plant cells __________

A spontaneous, non-_{biological redox reaction}

__________

A non-_{spontaneous, non}-_{biological redox reaction}

__________

2. Oxidation–reduction reactions occur in biological systems. A net oxidation–reduction reaction that occurs in the body is

A. Mg2+_{(aq) + 2OH}–_(aq)_→_Mg(OH) 2(s)

B. HCO3–(aq) + H3O+(aq) → H2CO3(aq) + H2O(l)

C. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

D. C6H12O6(aq) + 6O2(g) → 6 CO2(g) + 6H2O(l)

Use the following equations to answer the next question.

3. The equations that represent oxidation–reduction reactions, listed in any order, are

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4. If you had a low level of the biological catalyst, alcohol dehydrogenase,

A. you would feel hotter than normal

B. the concentration of NADH(aq) would increase

C. your blood alcohol level would decrease at a faster rate than normal

D. your blood alcohol level would remain high for a longer period than normal

Use the following information to answer the next question.

5. Which of the following observations could have been made during this experiment?

A. There was no reaction.

B. A thick white precipitate formed.

C. A colourless gas was produced and the test tube cooled off. D. A colourless gas was produced and the test tube warmed up.

6. As the H2S(g) forms S(l), the sulphur atoms

A. gain 2 e–_{and are oxidized}

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Assigning Oxidation Numbers:

Up to now we have defined the charges of ions, and have used these charges to determine the gain or loss of electrons or reduction and oxidation. What happens when the species are molecules or polyatomic ions!?

An oxidation number is a positive or negative number assigned to an atom in a molecule or polyatomic ion based primarily on electronegativities (virtual charge).

We use these to balance redox reactions and identify whether species are getting oxidized or reduced.

Generally: an oxidation number is the charge that an atom would have if the electrons in each bond were assigned to the atom of the more electronegative element.

• In ionic compounds they are equal to the ionic

charges.

e.g. CaCl2 or NaCl

• In simple molecular compounds like water the

numbers are assigned as if the partial transfer of electrons were a complete transfer.

e.g. H2O O = -2, and each H = +1

How to Assign Oxidation Numbers: The Fundamental Rules

1. The oxidation number of any pure element is zero (including diatomic

and polyatomic elements). Thus the oxidation number of H in H

2 is

zero.

2. The oxidation number of a monatomic ion is equal to its charge. Thus the oxidation number of Cl in the Cl- ion is -1, that for Mg in the Mg+2_{ion is +2, and that for oxygen in O}2-_{ion is -2.}

3. The sum of the oxidation numbers in a compound is zero if neutral, or

equal to the charge if an ion.

4. The oxidation number of alkali metals in compounds is +1, and that of alkaline earths in compounds is +2. The oxidation number of F is -1 in all its compounds (this includes most halogens as well).

5. The oxidation number of H is +1 in most compounds. Exceptions is the ionic hydrides, such as NaH (where H = -1).

6. The oxidation number of oxygen (O) is -2 in most compounds. Exception is peroxides, such as H2O2 or Na2O2, where O = -1.

a. A peroxide is a chemical substance that contains a peroxo

unit, one that has a chemical formula of O22-. The most

familiar example of a peroxide is hydrogen peroxide, shown here on the right with the peroxo unit in red. In lab slang the term "peroxide" is sometimes used for hydrogen peroxide.

Example What is the oxidation number of each element in the following?

+4 -2 +4 -2 +1 +6 -2 SO2 CO32- K2SO4

Oxidation Number Changes

How This Relates to the Curriculum:

30–B1.2k define oxidizing agent, reducing agent, oxidation number, half-reaction, disproportionation

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+1 +5 –2 0 +2 +5 –2 0

2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2 Ag(s)

• silver is reduced as it decreases from +1 to 0 • copper is oxidized as it increases from 0 to +2

Practice: Use the change is oxidation number to identify which elements are oxidized and reduced in each of the following: Cl2(g) + 2HBr(aq)  2HCl(aq) + Br2(l)

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Practice Assigning Oxidation Numbers

1. Determine the Oxidation Numbers as indicated in the following examples:

Entity

H

2

S S

8

S

2

Cl

2

SO

2

SO

32

-

SO

42

-

HSO

4

- S

4

O

62

-

ON

Sulphur

Entity

H

3

P

4

P

2

Cl

4

PF

3

Na

3

P

PO

33

-

PO

43

-

HPO

42

-

ON

Phosphorus

Entity

Mn

3

P

2

MnO

2

MnCl

2

Mn

3

O

4

MnO

42

-

MnO

4

-

MnSO

4

ON

Manganese

Entity

HCl

Cl

2

OCl

2

PCl

3

MgCl

2

ClO-

ClO

3

-

ClF

3

ON

Chlorine

Entity

K

3

N

2

N

2

H

4

NH

3

N

2

O

NO

2

NH

4

NO

3

NO

2

-

NO

3

-ON

Nitrogen

Entity

Cr

CrCl

₃

PbCrO

₄

Cr

₂

O

₃

CrO

₄₂

-

Cr

₂

O

₇₂

-

ON

Chromium

Entity

O

3

H

+

H

2

O

2

LiH

CaO

2

PbO

2

OCl

2

ON

Oxygen/Hydrogen

Entity

CH

4

CaC

2

CO

2

C

2

H

5

OH

CH

3

CO

2

-

ON

Carbon

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Will a REDOX Reaction be Spontaneous?

We can determine whether a redox reaction will be spontaneous or not by looking at our table of reduction half reactions on our data booklet page 7.

Notice that these reactions have the reduced species on the product side.

The strongest Oxidizing Agent is the species at the top left hand side of the page.

The strongest Reducing Agent is the species at the bottom right hand side of the page.

A spontaneous redox reaction occurs only if the oxidizing agent is above the reducing agent in a table of redox half reactions.

To set up a redox table:

e.g. Given the following reactions, set up a reduction half reaction table:

1 3Co2+

(aq) + 2In(s)  2In3+(aq) + 3Co(s)

2 Cu2+(aq) + Co(s)  Co2+(aq) + Cu(s)

3 Cu2+

(aq) + Pd(s)  no evidence of reaction

Take one reaction at a time. Identify the oxidizing agent and reducing agents

Reaction 1: Co2+_{is the oxidizing agent and In is the reducing agent.}

Reaction is spontaneous – the oxidizing agent must be above the reducing agent on the Redox list.

Reaction 2: Cu2+ is the oxidizing agent and Co is the reducing agent. Reaction is spontaneous, so Cu2+_{is above Co on the list.}

Reaction 3: Not spontaneous, but if a reaction did occur – Cu2+ would be the oxidizing agent, and

Pd(s) would be the reducing agent. Because it is not spontaneous, then Cu2+ must be below Pd on the list.

Pd Cu Co In

Now you can create your list of reduction half reactions

Stongest Oxidizing agent Pd2+(aq) + 2e- ↔ Pd(s)

Cu2+

(aq) + 2e- ↔ Cu(s)

Co2+

(aq) + 2e- ↔ Co(s)

In3+(aq) + 3e- ↔ In(s) Strongest reducing agent

How This Relates to the Curriculum:

30–B1.6k predict the spontaneity of a redox reaction, based on standard reduction potentials, and compare their predictions to experimental results

30–B1.3s analyze data and apply mathematical and conceptual models to develop and assess possible solutions

• evaluate data from an experiment to

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Practice Making Reduction Half Reaction Tables

Use the following results tables in 1 - 3 to construct

mini reduction

half reaction tables

for the entities involved (• = spontaneous

reaction).

1. Pb

2+

_(aq)

V

2+

_{(aq) Cd}

2+

_{(aq) Zn}

2+

_(aq)

Zn(s)

• -

V(s)

• -

•

• Cd(s)

• -

-

Pb(s)

-

2. Cd

2+

_{(aq) V}

2+

_{(aq) Ra}

2+

_{(aq) Be}

2+

_(aq)

Be(s)

•

• -

-

Ra(s)

•

• -

• V(s)

• -

-

Cd(s)

-

3. Ga

3+

_{(aq) In}

3+

_{(aq) Al}

3+

_(aq)

Tl

+

_(aq)

Ga(s)

-

• -

• Al(s)

•

• -

• In(s)

-

• Tl(s)

-

4. Construct a table of reduction half reactions from the

following experimental evidence:

Co

2+

_{(aq) + Zn(s) ---> Co(s) + Zn}

2+

_(aq)

Mg

2+

_{(aq) + Zn(s) --->}

no evidence of reaction

5. What is the redox half reaction table for manganese, gallium, cadmium and cerium

based on the following reaction evidence?

2Ga(s)

+

3Cd

2+

_(aq)

---> 2Ga

3+

_{(aq) +}

_3Cd(s)

Ga(s)

+

Mn

2+

_(aq)

_--->

no evidence of reaction

3Mn

2+

_{(aq) +}

_2Ce(s)

_--->

_3Mn(s)

+

2Ce

3+

_(aq)

6. What is the redox half reaction table for strontium, nickel, platinum, hydrogen and

cerium? The results series of reactions are reported below.

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7. Construct a table of reduction half reactions from the following evidence.

Be(s)

+

Cd

2+

_(aq)

_--->

Be

2+

_(aq)

+

_Cd(s)

Cd(s)

+

2H

+

_(aq)

_--->

Cd

2+

_(aq)

+

H

₂

_(g)

Ca

2+

_{(aq) +}

_Be(s)

_--->

no evidence of reaction

Cu(s)

+

2H

+

_(aq)

_--->

no evidence of reaction

8. Construct a table of reduction half reactions from the following evidence.

Pd(s)

+

Cu

2+

_(aq)

_--->

no evidence of reaction

3Co

2+

_{(aq) +}

_2In(s)

_--->

_3Co(s)

+

2In

3+

_(aq)

Cu

2+

_{(aq) +}

_Co(s)

_--->

_{Cu(s) +}

Co

2+

_(aq)

9. Construct a table of reduction half reactions from the following evidence.

Ag(s)

+

I

₂

(s)

--->

no evidence of reaction

Ag(s)

+

Br

₂

_(l)

_--->

_2AgBr(s)

Cu(s)

+

I

2

_(s)

_--->

CuI

₂

_(s)

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Predicting REDOX Reactions

You may be given a mixture of aqueous substances (ions and molecules), and then be asked to predict which species will react in a redox reaction.

In order to predict redox reactions we:

1. First have to identify the types of atoms and ions in our mixture. 2. Then we go to our table of half reactions and label each species as

either an oxidizing agent (found on the left side) or a reducing agent (found on the right side). Be sure to combine ions if they are required for reactions (e.g. MnO4-(aq) is a stong oxidizing agent only in the

presence of acids – H+ (aq)).

3. The reaction that will occur is between the strongest oxidizing agent (SOA), and the strongest reducing agent (SRA). Identify these species from your list.

4. Write down each of their half reactions – reduction reaction for oxidizing agent, and oxidation reaction for the reducing agent (remember to switch around the reducing agent reaction). 5. Even up the electrons on either side by multiplying reactions by

lowest common denominator.

6. You can write “spont.” Over the reaction arrow to indicate that this reaction will occur spontaneously as long as the oxidizing agent is found above the reducing agent.

e.g. Aqueous potassium permanganate is mixed with an acidic iron (II) sulphate solution.

Does a redox reaction take place?

1. List species present

– K+(aq) MnO4-(aq) H+(aq) Fe2+(aq) SO42-(aq) H2O(l)

2. Identify oxidizing agents and reducing agents

K+(aq) Fe2+(aq)

MnO4-(aq) H+(aq) H2O(l)

H+(aq)

H2O(l)

Fe2+(aq)

SO42-(aq)

3. Strongest Oxidizing agent – MnO4-(aq), H+(aq) Stongest reducing agent - Fe2+(aq)

4. Half reactions:

MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)

Fe2+

(aq)  Fe3+(aq) + e- 5. Even out electrons

MnO4-(aq) + 8H+(aq) + 5e-  Mn2+(aq) + 4H2O(l)

5Fe2+

(aq)  5Fe3+(aq) + 5e-

How This Relates to the Curriculum:

30–B1.5k compare the relative strengths of oxidizing and reducing agents, using empirical data

30–B1.7k write and balance equations for redox reactions in acidic and neutral solutions by

• using half-reaction equations obtained

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Chemistry 30: Predicting a Redox Reaction

For the following questions, apply the rules for predicting Redox reactions to determine the strongest oxidizing agent and strongest reducing agent, and identify if the reaction will be spontaneous or non-spontaneous. Please answer all questions in the example format provided. Use the example from the previous page as a guide.

1. Copper is dipped in hydrochloric acid.

2. Copper is dipped in nitric acid.

3. Iron is exposed to moist air.

4. An aqueous solution of gold (III) chloride is mixed with

hydrogen peroxide.

5. Sodium is added to water.

6. Aluminum is added to water.

7. Hydrogen gas is bubbled through a solution of iron(III)

chloride.

8. Chlorine is bubbled through an aqueous solution of sodium

bromide.

9. Bromine is bubbled through an aqueous solution of sodium

chloride.

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Diploma Examination Practice

7. The oxidation number of Mo in CaMoO4(s) is

A. +2 B. +4 C. +6 D. –2

8. Which of the following substances is the strongest reducing agent? A. V2+_(aq)

B. V3+_(aq)

C. VO2+(aq)

D. VO2+_(aq)

9. The oxidizing agents above, listed from strongest to weakest, are A. U3+_(aq)_{, La}3+_(aq)_{, Y}3+_(aq)

B. U3+(aq), Y3+(aq), La3+(aq)

C. Y3+(aq), U3+(aq), La3+(aq)

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10. Match each of the chemical changes listed above with the reaction species given below.

tin __________ nitrate ion __________ nickel(II) ion __________ hydrogen ion __________

11. The oxidation number of nitrogen in each compound listed above is, respectively, _____ , _____ , _____ , and _____ .

12. Electrolysis of MgCl2(aq) will not produce magnesium metal because

A. Cl–(aq) is a stronger oxidizing agent than Mg2+(aq)

B. H2O(l) is a stronger reducing agent than Mg2+(aq)

C. H2O(l) is a stronger oxidizing agent than Mg2+(aq)

D. Cl–(aq) is a stronger reducing agent than Mg2+(aq)

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14. In the process of being oxidized, the zinc A. gained electrons to produce more Zn(s) B. lost electrons and became Zn2+(aq) C. gained protons to produce Zn2+(aq) D. lost protons and became Zn(s)

15. In the balanced redox reaction equation

3Cu(s) + 2NO3–(aq) + 8H+(aq) →3 Cu2+(aq) + 2NO(g) + 4H2O(l),

the oxidation number of nitrogen A. decreases by 3

B. increases by 3 C. increases by 2 D. decreases by 6

16. In a reaction, Sn2+_(aq)

A. will undergo oxidation when combined with Pb(NO3)2(aq)

B. act as a reducing agent when combined with Ni(s) C. always act as an oxidizing agent

D. act as an oxidizing agent when combined with Cd(s)

17. A redox reaction occurs when an iron nail is placed in a solution of copper (II) sulphate. Elemental copper begins to form, and the colour of the solution changes. In this reaction, the reducing agent is

A. Fe(s) B. Cu(s) C. Fe2+(aq) D. Cu2+_(aq)

18. Four metals represented by the symbols R, S, T, and V and their ions combine with each other in the following manner:

When the oxidizing agents are arranged from strongest to weakest, the order is

A. S2+_{(aq), T}+_{(aq), R}3+_{(aq), V}2+_(aq)

B. V2+_{(aq), R}3+_{(aq), T}+_{(aq), S}2+_(aq)

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19. In this reaction, the oxidation state of iron A. changes from 0 to +3

B. changes from +2 to +3 C. changes from +2 to 0 D. does not change

20. The order of oxidizing agents, from strongest to weakest, is A. X2+_(aq)_{, Z}3+_(aq)_{, A}2+_(aq)_{, D}+_(aq)

B. A2+_(aq)_{, Z}3+_(aq)_{, D}+_(aq)_{, X}2+_(aq)

C. Z3+_(aq)_{, X}2+_(aq)_{, A}2+_(aq)_{, D}+_(aq)

D. X2+_(aq)_{, D}+_(aq)_{, Z}3+_(aq)_{, A}2+_(aq)

21. Which of the following reducing agents is the strongest? A. I –_(aq)

B. Br2(l)

C. H2O(l)

D. Al(s)

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23. The net ionic equation for the conversion of copper(II) oxide in copper ore is

CuO(s) + 2H+(aq) →Cu2+(aq) + H2O(l)

The copper in the copper (II) oxide is A. reduced

B. oxidized

C. the oxidizing agent

D. neither oxidized nor reduced

24. A student has one coin made of copper and one coin made of nickel. Which of the following solutions could the student use to demonstrate which of these metals is the stronger reducing agent?

A. Hg2+(aq)

B. Fe3+(aq)

C. Fe2+(aq)

D. Sn4+(aq)

25. In the stomach, the reaction between hydrochloric acid and iron occurs because the

A. iron donates protons to the acid B. acid donates electrons to the iron C. iron accepts protons from the acid D. acid accepts electrons from the iron

Use the following information to answer the next two questions.

26. In the decomposition equation, the product species that would have an oxidation state of zero are

A. hydrogen and nitrogen B. carbon and hydrogen C. nitrogen and oxygen D. carbon and oxygen

27. Reactions producing carbon dioxide cause concern among environmentalists because CO2(g) is

A. a poisonous gas

B. a major greenhouse gas

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Balancing Redox Equations

Some redox reactions are too complex to try the TLAR method (from Chemistry 20) for balancing.

We use the basis that the total number of electrons gained in reduction must equal the total number of electrons lost in oxidation.

Method 1

The Oxidation Number Change Method:

Balance the equation by comparing the increases and decreases in oxidation numbers.

Balancing equations by using oxidation numbers

1. Assign oxidation numbers to each element and identify the elements whose oxidation number has changed.

2. On the reactant side, use arrows to show this change as a gain or loss of e- per atom / per formula unit

3. Balance the e-_{exchange between the oxidizing agent and the reducing}

agent with equation coefficients

4. Balance all atoms but oxygen and hydrogen 5. Balance O with H2O

6. Balance H atoms with H+_{in acid solutions or with H2O and add an}

equal number of OH-_{to the other side in basic solutions.}

7. Check atoms, check charges

Example: Cr2O72-(aq) + I-(aq)→ I2(aq) + Cr3+(aq)

1. Assign oxidation #

Cr2O72-(aq) + I-(aq)→ I2(aq) + Cr3+(aq)

+6 -2 -1 0 +3

The elements that changed are I and Cr

2. The Cr atom has changed from 6 to 3 which is a gain of 3 e- per atom, but Cr2O72- has 2 atoms of Cr in the formula, so this is a gain of 6 e- per formula

unit. The I atom has changed from –1 to 0, which is a loss of 1 e-_{per atom,}

and since I-_{has only 1 atom in the formula, it is also 1e}-_{per formula unit (f.u.)}

Cr2O72-(aq) + I-(aq)→ I2(aq) + Cr3+(aq)

+6 -2 -1 0 +3

↑ ↓

3e-/atom 1e-/atom 6e-/f.u. 1e-/f.u.

How This Relates to the Curriculum:

30–B1.7k write and balance equations for redox reactions in acidic and neutral solutions by

• using half-reaction equations obtained

from a standard reduction potential table

• developing simple half-reaction

equations from information provided about redox changes

• assigning oxidation numbers, where

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Cr2O72- + 6I-→ I2 + Cr3+

4. Balance all atoms except H and O Cr2O72- + 6I-→ 3I2 + 2Cr3+

5. Balance O atoms first with H2O, then H atoms with H+for acid solution 14H+_{+ Cr}

2O72- + 6I-→ 3I2 + 2Cr3+ + 7H2O

6. Check : atoms are balanced and so are the charges (+6 net )

Balance the following equations using oxidation numbers

1. I2(s) + NO3-(aq)→ IO3-(aq) + NO(g) (acid solution)

2. ClO3-(aq) + C2H4(aq)→ CO2(g) + Cl-(aq) (basic solution)

3. C3H7OH(aq) + Cr2O72-(aq)→ C2H5COOH(aq) + Cr3+(aq) (acid solution)

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Method 2

Balancing Using Half Reactions

Last year we balanced chemical reactions by making sure we had the same number and types of atoms on one side of the arrow as were on the other side. For example, H2 + O2→ H2O is not a balanced reaction. However, if we react

two moles of hydrogen gas with one mole of oxygen gas to produce two moles of water, we now have a balanced reaction: 2 H2 + O2→ 2 H2O.

Some people call this the TLAR method, or “That Looks About Right”.

Oxidation-reduction, or Redox reactions involve a transfer of electrons among reactants, and we must worry about keeping track of charge, as well as

keeping track of mass (the number and types of atoms on each side of the reaction). Because of the added task of accounting for charge, Redox reactions are very difficult to balance using the TLAR method. Fortunately we have a systematic approach we can use to balance Redox reactions.

Balancing Redox Reactions

1. Break down the Redox reaction into its separate half reactions. Perform Steps 2-5 on EACH half reaction.

2. Balance all elements except H and O.

3. Balance oxygen atoms by adding an equal number of H2O

molecules to the other side of the reaction (remember the arrow is an equals sign).

4. Balance hydrogen atoms by adding an equal number of H+_{ions to} the other side of the reaction.

5. Balance the charge by adding electrons.

6. Make the numbers of electrons in the half reactions the same, and recombine the half reactions. Cancel any identical species on opposite sides.

7. For acidic solutions, go to Step 8. For basic solutions:

a. Add the same number of OH-_{ions to BOTH sides, as}

there are H+_.

b. Make H2O out of the OH-_{and H}+_{and cancel H2O}

molecules appearing on both sides.

8. Check the mass (number and types of atoms) and charge balance of your reaction.

That is a pretty complicated checklist, so let’s do a couple of examples and explain the checklist steps as we do them.

In the first example, we will balance the following Redox reaction in acidic solution:

CrO42-(aq) + Cl-(aq)→ Cr3+(aq) + ClO2-(aq)

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Step l. Break down the Redox reaction into its separate half reactions.

How do we know what the half reactions are? One species is going to be oxidized and lose electrons, and the other will be reduced, gaining electrons. We must use our knowledge of assigning oxidation numbers to find out which atoms lose electrons and which atoms gain. Chlorine has a –1 oxidation number on the reactants side, and a +3 oxidation number on the products side. It loses electrons and is therefore oxidized. Chromium goes from a +6

oxidation number to a +3, so it is reduced. The half reactions are: Cl-(aq)→ ClO2-(aq) (Oxidation)

CrO42-(aq)→ Cr3+(aq) (Reduction)

We need to perform the next three steps on each half reaction individually.

Step 2. Balance all elements except H and O.

In this reaction, the chlorine and chromium atoms are already balanced.

Step 3. Balance oxygen atoms by adding an equal number of H2O molecules to the other side of the reaction.

We can add as many water molecules as we want because all these reactions are taking place in aqueous solution.

2H2O(l) + Cl

-(aq)→ ClO2-(aq)

CrO42-(aq)→ Cr3+(aq) + 4H2O(l)

Step 4. Balance hydrogen atoms by adding an equal number of H+_{ions to}

the other side of the reaction.

If the solution is acidic, we also have excess H+_{ions that we may add as we}

see fit. If the solution is basic, we start by adding H+, and then take care of them in Step 7.

2H2O(l) + Cl-(aq)→ ClO2-(aq) +4H+(aq) 8H+(aq) + CrO42-(aq)→ Cr3+(aq) + 4H2O(l) Step 5. Balance the charge by adding electrons.

You almost always end up adding electrons to the same side as you added H+ ions. The whole purpose of balancing Redox reactions is to cancel out electrons. Make sure that you are adding electrons to opposite sides of thereaction arrow for your oxidation and reduction half reactions. This is a good way to make sure that you are on the right track.

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Step 6. Make the numbers of electrons in the half reactions the same, and recombine the half reactions. Cancel any identical species on opposite sides.

In order to make the number of electrons equal so they can cancel out, we need to multiply the entire top reaction by three. We multiply the bottom reaction by four. In the process of balancing Redox reactions, we will also be able to tell the number of electrons transferred during the reaction. This knowledge will help us find ∆G for the reaction, or its equilibrium constant, K.

6H2O(l) + 3Cl-(aq) → 3ClO2-(aq) + 12H+(aq) + 12e

-12e-_{+ 32H}+

(aq) + 4CrO42-(aq)→ 4Cr3+(aq) + 16H2O(l)

Now we add the two half reactions back together and cancel out anything wecan. We must be sure we cancel out the electrons!

6H2O(l) + 3Cl-(aq) + 12e- + 32H+(aq) + 4CrO42-(aq)→ 3ClO2-(aq) + 12H+(aq)

+ 12e-_{+ 4Cr}3+

(aq) + 16H2O(l)

The final equation is 3Cl

-(aq) + 4CrO42-(aq) + 20H+(aq)→ 3ClO2-(aq) + 4Cr3+(aq) +10H2O(l) Step 8. Check the mass and charge balance of your reaction.

This reaction occurs in acidic solution, so we skip Step 7 and make sure that the mass and charges are balanced in our reaction.

Atoms (left side) Atoms (right side) 3 Cl 3 Cl

4 Cr 4 Cr

16 O 6 O + 10 O = 16 O 20 H 20 H

Mass checks!

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The second example will be to balance the following reaction in basic solution:

BrO3-(aq) + F2(g)→ BrO4-(aq) + F-(aq)

Step l. Break down the Redox reaction into its separate half reactions.

Bromine’s oxidation number changes from +5 to +7 so it is oxidized. Fluorine’s oxidation number changes from 0 to –1, so it is reduced.

BrO3-(aq)→ BrO4-(aq) Oxidation

F2(g)→ F-(aq) Reduction

Step 2. Balance all elements except H and O.

Bromine is balanced in the first half reaction, but fluorine is not balanced in the second. You must place a 2 in front of F- to balance the total number of atoms/ions.

BrO3-(aq)→ BrO4-(aq)

F2(g)→2F-(aq)

Step 3. Balance oxygen atoms by adding an equal number of H2O molecules

to the other side of the reaction.

H2O(l) + BrO3-(aq)→ BrO4-(aq)

F2(g)→ 2F-(aq)

Step 4. Balance hydrogen atoms by adding an equal number of H+ ions to the other side of the reaction.

H2O(l) + BrO3-(aq)→ BrO4-(aq) + 2H+(aq)

F2(g)→ 2F-(aq)

Step 5. Balance the charge by adding electrons.

Once again, here is a good place to make sure that each half reaction gets electrons on different sides of the arrow. We also see from this step that two electrons are transferred in this reaction.

H2O(l) + BrO3-(aq)→ BrO4-(aq) + 2H+(aq) + 2e -2e-+ F2(g)→ 2F-(aq)

Step 6. Make the numbers of electrons in the half reactions the same, and recombine the half reactions. Cancel any identical species on opposite sides.

There is no need to multiply the reactions by a certain number in order to get the electrons to cancel out. We may combine the half reactions and cancel directly. In this case, the only species that cancels is the electrons, and the resulting equation is

H2O(l) + BrO3-(aq) + F2(g)→ BrO4-(aq) + 2H+(aq) + 2F-(aq) Step 7. For basic solutions:

a. Add the same number of OH-_{ions to both sides, as there are H}+_.

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2OH-(aq) + H2O(l) + BrO3-(aq) + F2(g)→ BrO4-(aq) + 2H+(aq) + 2F-(aq) + 2OH-(aq)

b. Make H2O out of the OH- and H+ and cancel H2O molecules appearing on both sides.

The 2H+ and 2OH- ions on the right side of the equation combine to form two water molecules. The one water molecule on the left side cancels, leaving one water molecule on the right side. The reaction is now

2OH

-(aq) + BrO3-(aq) + F2(g)→ BrO4-(aq) + 2F-(aq) + H2O(l) Step 8. Check the mass and charge balance of your reaction.

Even if your given equation is not balanced, your finished product will be balanced for both mass and charge.

Atoms (left side) Atoms (right side) 2 O + 3 O = 5 O 4 O + 1 O = 5 O

2 H 2 H

1 Br 1 Br

2 F 2 F

Mass checks!

Charge: (-2) + (-1) + 0 = (-1) + (-2) + 0 -3 = -3

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Balancing Redox Reactions Using Oxidation Numbers or Half

Reactions

Balance the following equations using oxidation numbers or Half Reactions as outlined in your rule sheets.

1. AuI₃_{(aq) + SO}₂_{(g) ---> H}₂SO₄_{(aq) + Au(}s) + HI(aq) (acidic)

2. SeO₃2-_{(aq) + I}-_{(aq) ---> Se(}s) + I₂(s) (acidic)

3. CH₃_OH(l) + MnO₄-(aq) ---> CO₃2-(aq) + MnO₄2-(aq) (basic)

4. C₂H₆_O(l) + BrO₃-(aq) ---> CO₂(g) + Br-(aq) (acidic)

5. NO₃-_(aq)+_P

4(s) ---> H3PO4(aq) + NO2(g) (acidic)

6. NO₃-_{(aq) + Zn(}s) ---> ZnO₂2-(aq) + NH₃(aq) (basic)

7. SO₃2-_{(aq) + Cr}₂O₇2-_{(aq) ---> SO}₄2-_{(aq) + Cr}3+_(aq) (acidic)

8. Cl-_{(aq) + VO}₄3-_{(aq) + I}-_{(aq) ---> ICl(aq) + VO}2+_(aq) (acidic)

9. AuCl₄-_{(aq) + C}₂O₄2-_{(aq) ---> Au(}s) + Cl-(aq) + CO₂(g)

10. H₂PO₂-_{(aq) + TeO}₄2-_{(aq) ---> PO}₄3-_{(aq) + Te(}s) (acidic)

11. _CdS(s) + NO₃-(aq) ---> Cd2+(aq) + S₈(s) + NO(g) (acidic)

12. As₄₍s) + OCl-(aq) ---> H₃AsO₄(aq) + Cl-(aq) (acidic)

13. S2-_{(aq) + NO}₃-_{(aq) ---> S}₈₍s) + NO(g) (acidic)

14. Cr₂O₇2-_{(aq) + HCHO(aq) ---> Cr}3+_{(aq) + HCOOH(aq)} (acidic)

15. CH₃NO₂_{(aq) + Ti}3+_{(aq) ---> CH}₃NH₂_{(aq) + Ti}4+_(aq) (acidic)

16. MnO4-_{(aq) + Sb}

2O3(aq) ---> Mn 2+

(aq) + Sb2O5(aq) (acidic)

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Answers - 1 - 18 for Balancing Worksheet

1. 2AuI₃_{(aq) + 3SO}₂_{(g) + 6H}₂_O(l) ----> 3H₂SO₄(aq) + 2Au(s) + 6HI(aq)

2. SeO₃2-_{(aq) + 4I}-_{(aq) + 6H}+_{(aq) ---> Se(}s) + 2I₂(s) + 3H₂O(l)

3. CH₃_OH(l) + 6MnO₄-(aq) + 8 OH-(aq)----> CO₃2-(aq) + 6MnO₄2-(aq) + 6H₂O(l)

4. C₂H₆_O(l) + 2BrO₃-(aq) ---> 2CO₂(g) + 2Br-(aq) + 3H₂O(l)

5. 20 NO₃-_(aq)+_P

4(s) + 20 H +

(aq) ----> 4H3PO4(aq) + 20 NO2(g) + 4H2O(l)

6. NO₃-_{(aq) + 4 Zn(}s) + 7OH-(aq) ---> 4ZnO₂2-(aq) + NH₃(aq) + 2H₂O(l)

7. 3SO₃2-_{(aq) + Cr}₂O₇2-_{(aq) + 8H}+_{(aq) ---> 3SO}₄2-_{(aq) + 2Cr}3+_{(aq) + 4H}₂_O(l)

8. Cl-_{(aq) + 2VO}₄3-_{(aq) + I}-_{(aq + 12H}+_{(aq) ---> ICl(aq) + 2 VO}2+_{(aq) + 6H}₂_O(l)

9. 2AuCl₄-_{(aq) + 3C}₂O₄2-_{(aq) ---> 2Au(}s) + 8Cl-(aq) + 6CO₂(g)

10. 3H₂PO₂-_{(aq) + 2TeO}₄2-_{(aq) ---> 3PO}₄3-_{(aq) + 2Te(}s) + 2H+(aq) + 2H₂O(l)

11. 24CdS(s) + 16NO₃-(aq) + 64H+(aq)--> 24Cd2+(aq) + 3S₈(s) + 16NO(g) + 32H₂O(l)

12. As₄₍s) + 10 OCl-(aq) + 6H₂O(l) ---> 4H₃AsO₄(aq) + 10 Cl-(aq)

13. 24S

2-(aq) + 16NO3

-(aq) + 64H+(aq) ---> 3S8(s) + 16NO(g) + 32H2O(l)

14. Cr₂O₇2-_{(aq) + 3HCHO(aq) + 8H}+_{(aq) ---> 2Cr}3+_{(aq) + 3HCOOH(aq) + 4H}2O(l)

15. CH₃NO₂_{(aq) + 6Ti}3+_{(aq) + 6H}+_{(aq) ---> CH}₃NH₂_{(aq) + 6 Ti}4+_{(aq) + 2H}₂_O(l

16. 4MnO₄-_{(aq) + 5Sb}₂O₃_{(aq) +12H}+_{(aq) ----> 4Mn}2+_{(aq) + 5Sb}₂O₅_{(aq) + 6H}₂_O(l)

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Disproportionation/Autoxidation: a Worksheet

In a select number of redox reactions a single species is both oxidized and reduced. This process is called disproportionation. You may also see the term autoxidation, because this single chemical species oxidizes itself - hence the term autoxidation.

During disproportionation, atoms contained within a molecule or polyatomic ion increase in oxidation number (the reducing form of the entity); while other atoms of the same element within that molecule or polyatomic ion decrease in oxidation number (the oxidizing form of said entity).

Balancing by inspection (TLAR method) is always possible in simpler disproportionation.

In order to balance using Redox methods, you have to find the gain and loss of electrons between on e compound and two different species on the other side of the reaction.

Try to identify the element that disproportionates, and then balance the reaction.

1. NH₄NO₃(s) ---> H₂O(g) + N₂O(g)

2. H₂O₂₍l) ---> H₂O(g) + O₂(g)

3. Cl₂_{(g) + H}₂_O(l) ---> HOCl(aq) + H +

(aq) + Cl-(aq)

4. ClO₃-₍s) ---> O₂(g) + Cl

-(aq)

5. Cl₂_{(g) + OH}-_{(aq) --->OCl}-_{(aq) + H}₂_O(l) + Cl

-(aq)

How This Relates to the Curriculum:

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Diploma Examination Practice

28. The balanced oxidation half-reaction for this change is A. H2O(l) + S2O32–(aq) →SO42–(aq) + 4 e– + 2H+(aq)

B. Cl2(g) + 2 e–→2 Cl–(aq)

C. 5H2O(l) + S2O32–(aq) →2 SO42–(aq) + 10H+(aq) + 8 e–

D. 5H2O(l) + S2O32–(aq) + 4 e–→2 SO42–(aq) + 10H+(aq)

29. When the equation V2O5(s) + Mn(s) →VO(s) + MnO2(s) is balanced

using the lowest whole number coefficients, the coefficient of V2O5(s) is __________

Mn(s) is __________ VO(s) is __________ MnO2(s) is __________

30. The balanced net equation and the predicted energy released per mole of propane consumed for this fuel cell are, respectively,

A. C3H8(g) + 5O2(g) →3 CO2(g) + 4H2O(l) ΔH = –2 219.9 kJ

B. C3H8(g) + 5O2(g) →3 CO2(g) + 4H2O(l) ΔH = –103.8 kJ

C. C3H8(g) + 5O2(g) →3 CO2(g) + 4H2O(l) ΔH = –2 043.9 kJ

D. C3H8(g) + O2(g) + 4H2O(l) →3 CO2(g) + 16H+(aq) + 16 e– ΔH = +66.5 kJ

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32. The chemical reaction in which a single species is both oxidized and reduced is known as disproportionation. An example of this type of reaction is

A. 2NH3(aq) + NaOCl(aq) →N2H4(aq) + NaCl(aq) + H2O(l)

B. Cl2(aq) + H2O(l) →HOCl(aq) + H+(aq) + Cl–(aq)

C. 2F2(g) + O2(g) →2OF2(g)

D. 2Na(s) + I2(s) →2NaI(s)

33. The atom that undergoes reduction in this reaction is A. Au

B. H C. N D. Cl

34. When this equation is balanced using lowest whole number coefficients, the coefficient for nitric acid is

A. 2 B. 3 C. 4 D. 5

35. When the redox reaction

is balanced using lowest whole number coefficients, the coefficient of H2O(l) is __________

NO2–(aq) is __________

Al(s) is __________

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Use the following information to answer the next three questions.

36. The type of reaction that this equation represents is A. a Brønsted–Lowry acid–base reaction B. an oxidation–reduction reaction C. a formation reaction

D. a combustion reaction

37. When the above equation is balanced, the equation is

A. CH3OH(l) + Cr2O72–(aq) + 14H +(aq) →CH2O(aq) + 2Cr3+(aq) + 7H2O(l)

B. 3CH3OH(l) + Cr2O72–(aq) + 14H +(aq) →3CH2O(aq) + 2Cr3+(aq) + 7H2O(l)

C. 3CH3OH(l) + Cr2O72–(aq) + 8H +(aq) →3CH2O(aq) + 2Cr3+(aq) + 7H2O(l)

D. 3CH3OH(l) + Cr2O72–(aq) + 8H +(aq) →3CH2O(aq) + 2Cr3+(aq) + 8H2O(l)

Use your recorded answer from the previous Multiple Choice to answer the next Numerical Response.

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39. When reaction equation I is balanced with lowest whole number coefficients, the coefficient of

H+(aq) is __________ BrO2–(aq) is __________

BrO3–(aq) is __________

BrO2(aq) is __________

40. In reaction III, the bromine in BrO2–(aq)

A. undergoes oxidation only B. undergoes reduction only C. both loses and gains protons D. both loses and gains electrons

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Titrations in Redox Reactions

There are many industrial processes that require knowledge of

concentrations of ions/metals.

Examples include:

Is there enough silver in silver ore to justify a mine?

OR

What is the concentration of bleach, or hydrogen peroxide in a

pharmaceutical product?

We can use titrations to determine concentrations of substances. ( it is a

very similar process to the acid base titrations we performed in

Chemistry 20).

Unlike acid base titrations, we can’t use indicators to help us identify

when a titration is complete. Visual indications are often the easiest

method to determine when a redox titration is complete. For a visual

indicator, we have to use oxidizing agents that change colour as they

get reduced.

Permanganate ion (MnO

4-

(aq)) and dichromate ion (Cr

2

O

72-

(aq)) are

used – because they are both strong oxidizing agents under acidic

conditions and they change colour when reduced.

Permanganate changes from a pink to a colourless solution as it reduces

from MnO

4-

to Mn

2+

; and dichromate changes from orange to green as

it reduces from Cr

2

O

72-

to Cr

3+

.

The titrant used must be accurately known. A primary standard is a

chemical that is used to find the concentration of a standard solution.

For example it is hard to know the concentration of MnO

4-

ions in

solution, because under acidic conditions, MnO

4-

reacts easily with

water and will change in concentration all the time. We use tin(II)

chloride to determine the concentration of MnO

4-

. Tin(II) is the

primary standard. Tin(II) is acidified because we don’t want to oxidize

water, only tin(II).

When we use permanganate as the oxidizing agent, we know the

primary standard has been oxidized when we see a slightly pink

solution form (it will only last a few seconds because water will oxidize

and cause any pink permanganate to form manganese ion).

How This Relates to the Curriculum:

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Redox Titrations and their Solutions Using Stoichiometry

Example:

1. In a Chemistry 30 titration experiment 25.0 mL of an acidic 0.100 mol/L tin (II) chloride solution required an average of 12.1 mL of aqueous potassium dichromate to effect complete reaction. What is the concentration of the oxidizing agent?

List the ions in solution (include water!)

H

+

(aq), Sn

2+

(aq), Cl

-

(aq), H

2

O(l), K

+

(aq), Cr

2

O

72-

(aq)

Identify the SOA and SRA:

SOA Cr

2

O

72-

(aq) and H

+

(aq)

SRA Sn

2+

(aq)

Write out the reduction and oxidation half reactions:

Reduction:

Cr

2

O

72-

(aq) + 14H

+

(aq) + 6e

-



2Cr

3+

(aq) + 7H

2

O(l)

Oxidation:

Sn

2+

_(aq)

_

_Sn

4+

_{(aq) + 2e}

-Balance out the electrons gained and lost using lowest common

denominator:

Reduction:

Cr

2

O

72-

(aq) + 14H

+

(aq) + 6e

-



2Cr

3+

(aq) + 7H

2

O(l)

Oxidation:

[Sn

2+

(aq)



Sn

4+

(aq) + 2e

-

]x3

3Sn

2+

(aq)



3Sn

4+

(aq) + 6e

-Write out the balanced equation:

And from the question write out the values given in the

question under the corresponding species.

Cr

2

O

72-

(aq) + 14H

+

(aq) + 3Sn

2+

(aq)



2Cr

3+

(aq) + 7H

2

O(l) + 3Sn

4+

(aq)

V = 12.1 mL

V = 25.0 mL

C = 0.100 mol/L

Solve the question using stoichiometry:

2 2

2 2 7

2 2 2

2 7 1

0.100 1

0.025

1 3 0.0121

molCr O molSn

LSn

LSn molSn LCr O

!

+ +

+ + !

" " "

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Problems for You to Try:

2. The following evidence was obtained when 10.00 mL samples of an acidified iron (II) solution were titrated with a 8.08 mmol/L solution of potassium permanganate.

Volumes of KMnO₄_{(aq) added to 10.00 mL samples of acidified Fe}2+_{(aq) solution}

Trial # 1 2 3 4

Final Buret Volume (mL) 17.9 34.3 16.5 33.0

Initial Buret Volume (mL) 1.1 17.9 0.0 16.5

Calculate the concentration of the reducing agent. (66.5 mmol/L)

3. The same potassium permanganate solution from question 2. was added to 10.00 mL samples of a dilute acidified solution of hydrogen peroxide. If an average 13.1 mL of the titrant were required to completely oxidize the reducing agent what is the concentration of the hydrogen peroxide? (26.5 mmol/L)

4. Freshly prepared aqueous potassium permanganate can be

standardized by titrating it against acidified aqueous tin (II) chloride. The following data was obtained when 10.00 mL samples of acidified tin (II) solution were reacted with fresh KMnO4(aq).

Volumes of KMnO4(aq) added 10.00 mL samples of acidified 0.0500 mol/L SnCl₂(aq)

Trial # 1 2 3 4

Final Buret Volume (mL) 18.5 35.4 18.3 35.1

Initial Buret Volume (mL) 1.1 18.5 1.6 18.3

Calculate the concentration of the oxidizing agent. (11.9 mmol/L)

5. A redox titration was completed by titrating 10.0 mL of aqueous tin (II) nitrate with acidified 0.0955 mol/L potassium dichromate solution. If, on average, 12.4 mL of potassium dichromate solution were required for complete reaction what is the molar concentration of the tin (II) nitrate solution? (0.355 mol/L)

6. Iron metal dissolves in sulphuric acid to give a solution of iron (II) ions. A 1.20 g sample of impure iron was dissolved in acid and diluted to a volume of 250 mL. When 10.0 mL samples of this iron (II) solution were titrated with 0.0114 mol/L KMnO4(aq) an average of 13.6 mL of the titrant

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Diploma Examination Practice

42. In an experiment, a student used 11.33 mL of H2O2(aq) to titrate a

17.00 mL sample of acidified

8.0 ×10–3 mol/L KMnO4(aq). If Mn2+(aq) is one of the products, then

the concentration of the H2O2(aq) is

A. 1.2 ×10–2 mol/L

B. 1.5 ×10–2 mol/L

C. 3.0 ×10–2 mol/L

D. 6.0 ×10–2 mol/L

43. According to the student’s data, the concentration of Fe2+(aq) is A. 0.206 mol/L

B. 0.218 mol/L C. 0.213 mol/L D. 0.223 mol/L

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Use the following information to answer the next three questions.

45. If 15.0 mL of oxalic acid solution is completely reacted with 20.0 mL of 0.0015 mol/L acidified permanganate solution, then the oxalic acid concentration will be

A. 8.0 x 10–4 mol/L B. 2.4 x 10–3_mol/L

C. 5.0 x 10–3_mol/L

D. 6.0 x 10–3 mol/L

46. A technician reacting oxalic acid with acidified potassium permanganate is not likely to observe

A. an increase in electrical conductivity B. a visible colour change

C. a slight increase in pH D. the formation of a gas

47. Acidic permanganate solutions and acidic dichromate solutions are often used in redox titrations because they are strong

A. reducing agents that change colour when they are oxidized B. oxidizing agents that change colour when they are reduced C. reducing agents that change colour when the acid is

neutralized

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Electrochemistry and Electrochemical Cells

The conversion of chemical energy into electrical energy and the conversion of electrical energy into chemical energy are electrochemical processes.

• All these processes are redox reactions

Example:

Dip zinc into copper solution and zinc becomes copper plated.

The net ionic equation is:

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

You will see the zinc dissolve and copper solid appear. Half reactions:

Oxidation Zn(s)  Zn2+(aq) + 2e

-Reduction Cu2+(aq) + 2e-  Cu(s)

With minor modifications, this reaction can produce an electric current (a flow of electrons).

If a zinc rod is dipped into a copper(II) sulfate solution, electrons are transferred from Zinc metal to Copper(II) ions.

We need to separate these redox half reactions or else electrons get transferred internally, one electrode becomes plated, and then will not function.

Voltaic Cells

• Voltaic cells are electrochemical cells that are used to convert chemical energy into electrical energy.

• Half cells are one part of a voltaic cell in which either oxidation or

reduction occurs.

• Usually it is a metal rod with a solution of one of its ions.

• The two half cells are separated by a porous partition. Which can

either be a salt bridge (solution containing a conducting solution) or porous cup can be used.

Electrochemical Cell with a Salt Bridge Electrochemical Cell with a Porous Cup

How This Relates to the Curriculum:

30–B2.1k define anode, cathode, anion, cation, salt bridge/porous cup, electrolyte, external circuit, power supply, voltaic cell and electrolytic cell

30–B2.2k identify the similarities and

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• These dividers allow the passage of ions from one compartment to the

other but prevent the solutions from mixing completely.

• A wire carries the electrons in the external circuit from one solution

to the other.

• The driving force of a voltaic cell is the spontaneous redox reaction

between a metal and metal ions in solution.

An electrode is a conductor in a circuit that carries electrons to or from a substance other than a metal.

• The electrode where oxidation occurs is called the anode. Electrons

are produced here. It is the negative electrode.

• The electrode at which reduction occurs is called the cathode.

Electrons are consumed here. It is labeled the positive electrode.

Note: Neither electrode is really charged. All parts of the cell are balanced at all times.

Electrolyte is the conducting solution that contains (usually) the same ions as the electrode in solution (allows electrons to move freely)

For this course, most of our electrochemical measurements will take place with standard voltaic cells:

• A standard voltaic cell operates at 25 °C with electrolytes at 1.0 mol/l

concentration and all gaseous components at standard pressures.

• A standard voltaic cell comprises two standard half cells/electrodes. • A standard half cell contains a redox couple.

• A redox couple comprises the reduced and oxidized forms of a

chemical entity.

• A half cell must also contain a solid electrode - usually one half of

the redox couple.

What follows are several diagrammatic and written means of communicating half cells.

A standard silver half cell A standard lead half cell

A g(s)

A g+(aq)

P b(s)

P b2 +(aq)

_{Ag(s) | Ag+(aq)} _{Pb2+(aq) | Pb(s)}

A standard copper half cell A standard zinc half cell C u(s)

C u2 +(aq)

Z n(s)

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If the appropriate redox couple for a half cell contains no (usually metallic) solid capable of acting as an electrode, one is still needed. The electrode must not take part in the electrochemistry - it must be an inert (unreactive)

electrode. The choices of inert electrodes are carbon (cheap but no good where oxygen is a product) and platinum (expensive inert to everything). Look at the following examples -

A standard permanganate half cell A standard dichromate half cell C(s)

MnO4-(aq) H+(aq)

M n2 +(aq)

P t(s)

C r2O72 -(aq) H+(aq)

C r3 +(aq)

_{C(s) | H+(aq), MnO4-(aq), Mn2+(aq)} _{Pt(s) | H+(aq), Cr2O72-(aq), Cr3+(aq)}

Where one of the components of the redox couple is a gas, a gas electrode is required. This type of half cell has a platinum electrode dipped in the electrolyte with the appropriate gas flowing over it at standard pressure.

A standard hydrogen electrode A standard chlorine electrode

P t(s) H+(aq) H2(g)

P t(s) C l-(aq) C l2(g)

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Example – the zinc-copper voltaic cell

A short form for writing voltaic cells is

Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)

Anode salt bridge Cathode

• The anode is always written first, and then the cathode (alphabetical) • The | lines indicate boundaries of phases that are in contact.

• The || lines indicates the salt bridge or porous membrane.

Solution concentration will affect voltage, as will different combinations of metals.

Electrons are produced at the zinc rod in an oxidation half reaction Electrons are consumed

at the copper rod in a reduction half reaction

Because it is oxidized the zinc rod is the anode(negative) Because it is reduced the copper

rod is the cathode(positive)

The electrons move up through the external wire to the copper rod.

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Question One

A voltaic cell is assembled using a copper electrode in a copper (II) nitrate solution, a silver electrode in a silver nitrate solution, a salt bridge containing potassium nitrate and a number of connecting wires attached to a voltmeter.

1. Identify the electrode where the oxidation occurs

2. Describe the oxidation half reaction at the anode

3. Identify the electrode where the reduction occurs

4. Describe the reduction half reaction at the cathode.

4. Describe the direction of the electron flow between the two electrodes

AgNO₃ (aq)

silver

_copper

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Question Two

A voltaic cell is assembled using a gold Au (s) electrode in a gold nitrate Au(NO3) 3 (aq)

solution, a lead (Pb)electrode in a lead nitrate solution Pb(NO3) 2 (aq), a salt bridge containing potassium nitrate and a number of connecting wires attached to a voltmeter.

5. Describe the direction of the electron flow between the two electrodes.

6. Label the diagram.

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Question Three

A voltaic cell is assembled using a zinc electrode in a zinc nitrate solution, a nickel electrode in a nickel nitrate solution, a salt bridge containing potassium nitrate and a number of connecting wires attached to a voltmeter.

5. Describe the direction of the electron flow between the two electrodes

1. Label the diagram