L8_2015.3.26_Oxidation and Reduction, Electrolysis

Lecture 8 Oxidation and Reduction,

Electrolysis

Human Biology and Organic Chemistry I HS 37-006-31 (41)

Dr. Thomas Lui ([email protected])

1. Redox reaction

2. Electrochemical cell

3. Electrolysis

1

Reduction-Oxidation (Redox)

Reaction involves oxidation and reduction is called redox reaction Oxidation and reduction must occur together

Mg(s) + H2O(g) MgO(s) + H2(g)

Definition In terms of oxygen addition/removal

Oxidation : Gain of oxygen Mg is oxidized Reduction : Loss of oxygen H2O is reduced

In terms of electron transfer

Oxidation : Loss of electron Mg Mg2+ + 2e-

Reduction : Gain of electron H2O + 2e- H2 + O

2-

Oxidation

Reduction

! Mg(OH)2 is produced if H2O(l) is used

2

Redox Reaction

However, some redox reactions may not involve oxygen

Mg + 2HCl MgCl2 + H2

The definition of redox reaction in terms of oxygen addition or removal is

not the best

To determine the redox reaction, oxidation number (or oxidation state) is

always applied

Oxidation

Reduction

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Oxidation Number (O.N.)

A useful tool to determine whether a substance has been oxidized or reduced

Oxidation number of an atom in an element is 0 e.g. O.N. of Cl2 is 0

For a simple ionic compound, O.N. of the element is the same as the

charge carried by the ion

Example Oxidation Number

Cation Anion

NaCl Na+ : +1 Cl- : -1

MgO Mg2+ : +2 O2- : -2

ZnI2 Zn2+ : +2 I- : -1

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Oxidation Number (O.N.)

For a covalent compound or polyatomic ion, the O.N. can be determined using the concept of electronegativity. Assuming that the compound is ionic in nature. The element with higher electronegativity value is considered as anion, while element with lower electronegativity value is considered as cation

Example Oxidation Number

NO2 N : +4 O : -2

NO3- N : +5 O : -2

NH4+ N : -3 H : +1

SO42- S : +6 O : -2

5

Redox Reaction Changing in Oxidation Number

Mg + 2HCl MgCl2 + H2

Oxidation

Mg Mg2+ + 2e- Oxidation : in O.N.

Reduction

2H+ + 2e- H2 Reduction : in O.N.

In the above example, Mg is described as reducing agent (R.A.) (the agent that reduces others and oxidizes itself), while HCl is described as oxidizing agent (O.A.) (the agent that oxidizes others and reduces itself)

Oxidation

Reduction

0 +2

+1 0

6

Classwork

Specify which of the following equations represent redox reactions, and indicate the oxidizing and reducing agents, if any.

a) CH4 + H2O CO + 3H2

b) 2AgNO3 + Cu Cu(NO3)2 + 2Ag

c) Zn + 2HCl ZnCl2 + H2

d) Fe + 2HCl FeCl2 + H2

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Will the Reaction Occur? The Reactivity Series

Left : Cu + AgNO3 Right : Ag + Cu(NO3)2

Why does the reaction occur in the left picture, but not in the right picture?

8


Any element with higher reactivity will react with the ion of any element with lower reactivity

Elements at the top of the table readily lose electron, and hence are strong R.A.

Elements at the bottom of the table are less willing to lose electron, and hence are weak R.A.

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As Cu is more willing to lose electron compared with Ag, it transfers electrons to Ag+, so that Cu is oxidized while Ag+ is reduced.

Cu + 2AgNO3 Cu(NO3)2 + 2Ag

Since NO3

- is not involved in the reaction, the equation is simply written as:

Cu + 2Ag+ Cu2+ + 2Ag

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Balancing of Redox Reaction

Use half equation method

Write the R.A. or O.A. and their corresponding products for each half reaction

Multiply each balanced equation by a number, so that the number of electrons lose in oxidation is identical to that gain in reduction

Combine the half equation, and eliminate the electrons

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Example 1 Chlorine reacts with iron (II) sulphate solution to give chloride ions and iron (III) sulphate

Oxidation : Fe2+ Fe3+ + e- (1)

Reduction : Cl2 + 2e- 2Cl- (2)

Equation (1) 2 : 2Fe2+ 2Fe3+ + 2e- (3) Equation (2) + (3) :

2Fe2+ + Cl2 2Fe3+ + 2Cl-

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Example 2 Acidified potassium permanganate solution reacts with potassium iodide solution to form a brown solution of iodine

Oxidation : 2I- I2 + 2e

- (1)

Reduction : MnO4- Mn2+

Since the reaction takes place in acidified solution, H+ and H2O is used to balance the reduction half equation

MnO4- + 8H+ + 5e- Mn2+ + 4H2O (2)

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Example 2 Acidified potassium permanganate solution reacts with potassium iodide solution to form a brown solution of iodine

Equation (1) 5 : 10I- 5I2 + 10e

- (3) Equation (2) 2 :

2MnO4- + 16H+ + 10e- 2Mn2+ + 8H2O (4)

Equation (3) + (4) :

2MnO4- + 10I- + 16H+ 2Mn2+ + 5I2 + 8H2O

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Example 3 Sodium sulphite solution turns acidified potassium dichromate from orange to green

Since the reaction takes place in acidified solution, H+ and H2O is used to balance the reduction half equation

Oxidation :

SO32- + H2O SO4

2- + 2H+ + 2e- (1) Reduction :

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O (2)

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Example 3 Sodium sulphite solution turns acidified potassium dichromate from orange to green

Equation (1) 3 :

3SO32- + 3H2O 3SO4

2- + 6H+ + 6e- (3) Equation (2) + (3) :

Cr2O72- + 14H+ + 3SO3

2- + 3H2O 2Cr3+ + 7H2O + 3SO4

2- + 6H+

8 4

Simplify : Cr2O7

2- + 8H+ + 3SO32- 2Cr3+ + 4H2O + 3SO4

2-

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Common Changes of Some Common Oxidizing Agents

Oxidizing agent Change

Oxygen O2 (Colorless) 2O

2- (Colorless)

O2 (Colorless) OH- (Colorless)

Halogen X2 2X-

Acidified potassium permanganate

MnO4- (Purple) Mn2+ (Pink)

Alkalified potassium permanganate

MnO4- (Purple) MnO2 (Black)

Acidified managanese (IV) oxide

MnO2 (Black) Mn2+ (Pink)

Acidified potassium dichromate

Cr2O72- (Orange) 2Cr3+ (Green)

conc. nitric acid NO3- (Colorless) NO2 (Brown)

dil. nitric acid NO3- (Colorless) NO (Colorless)

conc. sulphuric acid SO42- (Colorless) SO2 (Colorless)

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Common Changes of Some Common Reducing Agents

Reducing agent Change

Hydrogen H2 (Colorless) 2H+ (Colorless)

Carbon C (Black) CO2 (Colorless) C (Black) CO (Colorless)

Carbon monoxide CO (Colorless) CO2 (Colorless)

Sulphur dioxide SO2 (Colorless) SO42- (Colorless)

Sulphite SO32- (Colorless) SO4

2- (Colorless)

Sulphide S2- (Colorless) S (Pale yellow)

Iron (II) salt Fe2+ (Pale green) Fe3+ (Yellow)

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Classwork

The following reactions are carried out in an acidic medium. Balance each of the following equations. a) Cr + NO3

- Cr3+ + NO

b) Al + MnO4- Al3+ + Mn2+

The following reactions are carried out in an alkaline medium. Balance each of the following equations. a) PO3

3- + MnO4- PO4

3- + MnO2

b) Mg + OCl- Mg(OH)2 + Cl-

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Electrochemical Cell

Electrochemical cell is a device capable of deriving electrical energy from chemical reactions

Consists of 2 half cells connected by a salt bridge. Each half cell consists of an electrode dipping into an electrolyte

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Electrochemical Cell

Oxidation occurs at anode while reduction occurs at cathode

Electrolyte Compound that ionizes when dissolved in solvents

Produce free mobile ions to complete the circuit Normally ionic compound

Strong electrolyte Complete ionization when dissolved in water (e.g. NaCl)

Weak electrolyte Dissociate to a small extent (e.g. CH3COOH)

Salt bridge

To complete the circuit Provide cations and anions to replace those consumed at the electrodes,

and hence maintain a balance in charge between 2 half cells

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Electrochemical Cell Cell Diagram

A shorthand way of describing a electrochemical cell

ZnZn2+Cu2+Cu anode cathode

phase boundary

salt bridge

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Alkaline Battery An Example of Electrochemical Cell

Anode (Oxidation) : Zn2+ + 2OH- ZnO + H2O + 2e

-

Cathode (Reduction) :

2MnO2 + H2O + 2e- Mn2O3 + 2OH

-

Overall :

Zn2+ + 2MnO2 ZnO + Mn2O3

Membrane separator functions as salt bridge

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Rusting of Iron A Redox Reaction in Daily Life

Anode (Oxidation) : Fe Fe2+ + 2e-

Cathode (Reduction) : O2 + 2H2O + 4e- 4OH-

Fe2+ and OH- are reacted to form Fe(OH)2, which was further oxidized to form Fe3+

2Fe2+ + 4OH- 2Fe(OH)2 2Fe(OH)2 + O2 + (x-2)H2O Fe2O3xH2O Rust

Rusting process can be speeded up by the presence of an acid (speeds up the dissolution of iron) or a dissolved salt in water ( the electrical conductivity of water)

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Corrosion Protection

Sacrificial protection Coating a layer of metal which is more reactive than iron (such as zinc) on

the surface of iron The metal, rather than iron, is oxidized, while iron remains intact

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Classwork e-

Substances: Zn Mg ZnSO4 MgSO4 Salt bridge

(i)

(ii)

(iii)

(iv)

(v)

a) Match the numbers with the substances

b) Write the ionic equation occurred at each half cell, and hence give the equation for the overall reaction of the cell

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Electrolysis

Using a direct electric current to drive a non-spontaneous chemical reaction

Electrolysis Electrochemical cell

Anode (Oxidation) : 2H2O O2 + 4H+ + 4e-

Cathode (Reduction) : 2H+ + 2e- H2 Overall 2H2O 2H2 + O2

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Electrolysis

Anode (Oxidation) : 2Br- Br2 + 2e-

Cathode (Reduction) : Pb2+ + 2e- Pb Overall PbBr2 Pb + Br2

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Use of Electrolysis

Electroplating Using electrical current to reduce the dissolved metal cation so that they

form a thin layer of metal coating on an electrode

Anode (Oxidation) : Ag Ag+ + e-

Cathode (Reduction) : Ag+ + e- Ag

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Use of Electrolysis

Refining of copper Remove the impurity found in cupper ore Anode : Impure copper

Only copper and impurities more easily oxidized than Cu (e.g. Zn and Fe) dissolve at anode

Cathode : Pure copper Cu2+ ions, but not Zn2+ and Fe2+ ions, are reduced

Anode (Oxidation) : Cu Cu2+ + 2e-

Cathode (Reduction) : Cu+ + 2e- Cu

Amount of Cu dissolved at anode = Amount of Cu precipitate at cathode

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Reference

1. Bettelheim FA, Brown WH, Campbell MK & Farrell SO (2010). Introduction to general, organic & biochemistry, 9th edition: Thomson Brooks/Cole.

2. McMurry J, Castellion ME, Ballantine DS, Hoeger CA & Peterson VE (2010). Fundamentals of general, organic & biological chemistry, 6th edition: Pearson Prentice Hall.

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L8_2015.3.26_Oxidation and Reduction, Electrolysis