3 a redox reaction

CHAPTER 3CHAPTER 3

Oxidation Oxidation and and

ReductionReduction

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(A) Redox Reactions, RR(A) Redox Reactions, RRAfter this lesson, you should be able to:• State what oxidation is• State what reduction is• Explain what redox reaction is• State what oxidizing agent is• State what reducing agent is• Calculate the oxidation number of an element in a

compound• Relate the oxidation number of an element to the name of its

compound using the IUPAC nomenclature• Explain with examples oxidation and reduction processes in

terms of the change in oxidation number• Explain with examples oxidation and reduction processes in

terms of electron transfer• Explain with examples oxidizing agents and reducing agents

in redox reactions• Write oxidation and reduction half-equations and ionic

equations.

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What is Redox Reaction?What is Redox Reaction?

• Redox reaction are chemical reactions involving oxidation and reduction occurring simultaneously.

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Explanation of Explanation of Redox Reaction, RRRedox Reaction, RR

Redox reactions can be explained in term of:

– Loss or gain of oxygen

– Loss or gain of hydrogen

– Transfer of electrons

– Changes in oxidation number

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Explanation of RR based on Loss or Gain of Oxygen

• Oxidation is a chemical reaction in which oxygen is added to a substance

• Reduction is defined as the loss of oxygen from a substance

• The substance that causes oxidation is called oxidizing agent (oxidant)

• The substance that causes reduction is called the reducing agent (reductant)

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2CuO(s) + C(s) 2Cu(s) + CO2(g)

• CuO is reduced to Cu

• C is oxidized to CO2

• CuO acts as oxidizing agent (oxidant)

• C acts as reducing agent (reductant)

Gains oxygen (oxidation)

Loses oxygen (reduction)

Example 1:

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Explanation of RR based on Loss or Gain of Hydrogen

• Oxidation is the loss of hydrogen from a substance

• Reduction is the gain of hydrogen from a substance

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H2S(g) + Cl2 (g) S(s) + 2HCl(g)

• H2S is oxidized to S

• Cl2 is reduced to HCl

• Cl2 acts as oxidizing agent (oxidant)

• H2S acts as reducing agent (reductant)

Gains hydrogen (reduction)

Loses hydrogen (oxidation)

Example 2:

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Practice A1:

Study the following equations and identify the oxidized substances, reduced substances, oxidant and reductant.

a)2HBr(aq) + Cl2(l) 2HCl(aq) + Br2(l)

b)Mg(s) + PbO(s) MgO(s) + Pb(s)

c)CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

d)Fe3O4(s) + 4CO(g) 3Fe(s) + 4CO2(g)

e)PbS(s) + 4H2O2(aq) PbSO4(s) + 4H2O(l)

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Explanation of RR based on Transfer of Electrons

• Oxidation is the loss of electrons• Reduction is the gain of electrons

• Oxidizing Agent is electron acceptors• Reducing Agent is electron donors

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

Loss e- Zn(s) Zn2+ (aq) + 2e-

Gain e- Cu2+ (aq) + 2e- Cu(s)

------------------------------------------------------------- Zn(s) + Cu2+ (s) + 2e- Zn2+ (aq) + Cu(s) + 2e-

----------------------------------------------------------------------

Ionic Eq. Zn(s) + Cu2+ (s) Zn2+ (aq) + Cu(s)

=========================================

Gains electron (reduction)

Loses electron (oxidation)

Example 3:

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Practice A2:Study the following redox reactionsa) Cu(s) + 2Ag+(aq) Cu2+ (aq) + 2Ag(s)b) Cl2(g) + 2Br−(aq) 2Cl−(aq) + Br2(l)

c) Ca(s) + 2HCl(aq) CaCl2(aq) + H2(g)

d) 2Na(s) + Cl2(g) 2NaCl(s)

e) 2Fe2+(aq) + Br2(aq) 2Fe3+(aq) + 2Br−(aq)

For each of the above reaction above,i) Write the half equationsii) Identify the

– Oxidized substance– Reduced substance– Oxidizing agent– Reducing agent based on the transfer of electrons

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What is an Oxidation Number?

• The oxidation number or oxidation state of an element is the charge that the atom of the element would have if complete transfer of electron occurs.

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Rules in assigning Oxidation Number

Rule 1:Rule 1:

• The oxidation number of an atom in its elemental state is zero.

For example:

The oxidation number of each atom in

Mg, Cu, Na, H2, O2, Cl2 and P4 is zero.

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Rule 2:Rule 2:•The oxidation number of monoatomic ion is equal to its charge

For example:

Ion Na+ Mg2+ Al3+ Br− S2− N3−

Oxidation Number

+1 +2 +3 −1 −2 −3

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Rule 3:Rule 3:•The oxidation number of hydrogen in a compound is always +1 except in metal hydrides, where it is −1.

For example:

The oxidation number of H in H2O and NH3 is +1. However, the oxidation number of H in sodium hydride, NaH is −1

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Rule 4:Rule 4:•The oxidation number of oxygen in a compound is always −2 except in peroxides.

For example: The oxidation number of O in H2O and MgO is −2 . However, the oxidation number of O in hydrogen peroxide, H2O2 is −1

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Rule 5:Rule 5:

•The oxidation number of fluorine in all its compound is −1.

•The oxidation number of other halogens

(Cl, Br, I) in their compounds is −1 except when they combine with more electronegative elements such as oxygen or nitrogen.

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Rule 6:Rule 6:

•The sum of the oxidation numbers of all the elements in the formula of a compoundformula of a compound must be zero.

•The sum of the oxidation numbers of all the elements in the formula of a polyatomic ionformula of a polyatomic ion must be equal to the charge of the ion.

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Calculation of Oxidation Number, ON

Example 1:Determine the oxidation number of nitrogen in NH3.

Assume that the oxidation number of nitrogen is XThe ON of H in NH3 is +1 (rule 3)

The sum of ON of all atoms = 0 (rule 6)Thus , x + 3(+1) = 0 x + 3 = 0 x = −3

NH3

x +1

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Example 2:

Determine the oxidation number of copper in Cu2O

Assume that the oxidation number of copper is X

The ON of O in Cu2O is −2 (rule 4)

The sum of ON of all atoms = 0 (rule 6)

Thus , 2x + (−2) = 0

2x −2 = 0

x = +1

Cu2O

x −2

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Example 3:

Determine the oxidation number of sulphur in SO42−

Assume that the oxidation number of sulphur is X

The ON of O in SO42− is −2 (rule 4)

The sum of ON of all atoms = −2 (rule 6)

Thus , x + 4(−2) = −2

x −8 = −2

x = +6

SO4 2−

x −2

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Example 4:Determine the oxidation number of manganese in MnO4

−

Assume that the oxidation number of manganese is XThe ON of O in MnO4

− is −2 (rule 4)The sum of ON of all atoms = −1 (rule 6)Thus , x + 4(−2) = −1 x −8 = −1 x = +7

MnO4−

x −2

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Practice A3:

Determine the oxidation number of the underlined elements in the following compound

a) CO2 b)MgF2 c) H3PO4 d) V2O5

e) CO f) NH4+ g) SO3h) ClO4

-

i) N2O j) H2O2 k) S2O32− l) CrO4

2−

m) Cr2O72− n) Al2O3 o) BrO3

− p)VO2

q) PbO22− r) NO3

− s) NO2− t) CO3

2−

u) HCl v) HClO w) HClO2 x)ClO2

y) HClO3 z)HClO4

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Oxidation Number and IUPAC Nomenclature

Formula of Compound Oxidation Number of Underlined Metal

IUPAC name

FeCl2 +2 Iron(II) chloride

FeCl3 +3 Iron(III) chloride

CuCl +1 Copper(I) chloride

CuSO4 +2 Copper(II) sulphate

Mn(NO3)2 +2 Manganese(II) nitrate

MnO2 +4 Manganese(IV) oxide

K4Fe(CN)6 +2Potassium hexacyanoferrate (II)

K3Fe(CN)6 +3 Potassium hexacyanoferrate (III)

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Common Names and IUPAC Names for some compound

Molecular Formula

ON of metal

Common Name IUPAC name

Na2SO3 +4 Sodium sulphite Sodium sulphate(IV)

Na2SO4 +6 Sodium sulphate Sodium sulphate(VI)

NaNO2 +3 Sodium nitrite Sodium nitrate(III)

NaNO3 +5 Sodium nitrate Sodium nitrate(V)

HNO2 +3 Nitrous acid Nitric(III) acid

HNO3 +5 Nitric acid Nitric (V) acid

H2SO4 +6 Sulphuric acid Sulphuric (VI) acid

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Explanation of RR based on The Changes in Oxidation Number

• An increase in oxidation number indicates Oxidation

• A decrease in oxidation number indicates Reduction

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2Mg(s) + O2 (s) 2MgO(s) Oxidation 0 0 +2 −2

Number

Decrease in oxidation number

Increase in oxidation number

Example 4:

0

+2

-2

Mg2+

MgO2

O2-

• Increase in Oxidation Number

• (Oxidation)• decrease in Oxidation Number

• (Reduction)

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Explanation

• The oxidation number of Magnesium increases from 0 to +2.

• Magnesium undergoes oxidation to magnesium ion

• The oxidation number of Oxygen decrease from 0 to −2

• Oxygen undergoes reduction to oxide ion

• Magnesium acts as reducing agent

• Oxygen acts as oxidizing agent

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Practice A4a) 2H2 + O2 2H2Ob) 2Na + Br2 2NaBrc) Pb + 2Ag+ Pb2+ + 2Agd) Zn + 2HCl ZnCl2 + H2

Explain the above redox reactions based on the changes in oxidation number. Your explanation should includes:

i) oxidized and reduced substance in each reaction. Give reason for your answer.

ii) oxidizing agent and reducing agent in each reaction. State what happens to them

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Summary on the Definition of Oxidation and Reduction

OxidationOxidation ReductionReduction

Gain of Oxygen Loss of Oxygen

Loss of Hydrogen Gain of Hydrogen

Loss of Electrons Gain of Electron

Increase in Oxidation Number

Decrease in Oxidation Number

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Examples of Redox Reaction

1. Combustion2. Extraction of Metals3. Corrosion of Metals4. Electrochemistry ( Reaction happen in

Electrolytic Cell and Voltaic Cell)5. Change of Fe2+ to Fe3+ and vice versa6. Displacement of Metal from its salt solution7. Displacement of halogen from its halide solution8. Transfer of electrons at a distance

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Examples of Non Redox Reaction

• Neutralization

• Precipitation Reaction

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Change of Fe2+ to Fe3+ & vice versa

• Iron metal (Fe) exhibits two oxidation numbers, i.e. +2 and +3

• Fe2+ ion can be easily converted into Fe3+ ion.• Fe3+ ion can also be easily converted into Fe2+ ion.

Fe2+ Fe3+

Loss of electron

Gain of electron

oxidation

reduction

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Oxidation of Fe2+ to Fe3+

Procedure:1. Pour 2cm3 of freshly prepared iron(II) sulphate,

FeSO4 solution into a test tube.

2. Using a dropper, add bromine water drop by drop until no further changes are observed.

3. Warm the test tube gently4. Add NaOH solution slowly into the test tube

until it excess.5. Record the observation.

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Solution Used Observations

FeSO4

+ Br2

Reddish brown bromine water was decolourized.

•Green FeSO4 solution turn brown.

•When NaOH solution was added, a brown precipitate formed.•The precipitate is insoluble in excess NaOH

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Half Equation:

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

Reduction : Br2(aq) + 2e- 2Br−(aq)

------------------------------------------------------------------------------------

Ionic Equation :

2Fe2+ (aq) + Br2 (aq) 2Fe3+(aq) + 2Br− (aq)

===========================================================


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Reduction of Fe3+ to Fe2+

Procedure:1. Pour 2cm3 of iron(III) sulphate, Fe2(SO4)3

solution into a test tube.2. Add half a spatula of zinc powder to the

solution. Shake the mixture until no further changes are observed.

3. Filter the mixture.4. Add NaOH solution slowly into the filtrate until

in excess.5. Record the observation.

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Solution Used Observations

Fe2(SO4)3

+ Zn

Part of Zn powder dissolved.Brown Fe2(SO4)3 solution turn green. When NaOH solution was added to the filtrate, a green precipitate was formed.The precipitate is insoluble in water.

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Half Equation:

Oxidation : Zn(s) Zn2+ (aq) + 2e-Reduction : Fe3+ (aq) + e- Fe2+(aq)--------------------------------------------------------------------------Ionic Equation :

Zn(s) + 2Fe3+ (aq) Zn2+ (aq) + 2Fe2+(aq)

============================================

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Displacement of Metals

• A metal displacement reaction involves a metal and the salt solution of another metal.

• Displacement Reaction took place if any of these observation is obtained: – a deposition of solid occurs at the bottom of the test

tube.– a change in colour of the salt solution– a decrease in the amount or size of the metal used– Test tube becomes hotter

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Displacement of Metals• A more electropositive

metal can displace a less electropositive metal from its aqueous salt solution.

• A less electropositive metal cannot displace a more electropositive metal from its aqueous salt solution

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A more electropositive metal is located at higher position in the electrochemical series, ES.

A less electropositive metal is located at lower position in the electrochemical series, ES.

KK

NaNa

CaCa

MgMg

AlAl

ZnZn

FeFe

SnSn

PbPb

CuCu

HgHg

AgAg

More electropositive

Less

electropositive

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Displacement of MetalsExamples:

Fe(s) + CuSO4(aq) Cu(s) + FeSO4(aq)

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

Mg(s) + FeSO4(aq) Fe(s) + MgSO4(aq)

Cu(s) + FeSO4(aq) No reaction

Zn(s) + MgSO4(aq) No reaction

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Zinc displaces copper metal from copper(II) sulphate solutionZn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq)

Half Equation: Zn(s) Zn2+(aq) + 2e- (Oxidation) Cu2+(aq) + 2e- Cu(s) (Reduction)

Overall ionic equation:

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

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Observation:

(a) Brown copper metal deposited

(b) The colour of the solution changes from blue to colourless.

(c) The temperature of the mixture increases.

( all displacement reaction are exothermic)

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Displacement of Halogen • A more reactive halogen can displace a less

reactive halogen from its aqueous halide salt solution.

• A less reactive halogen cannot displace a more reactive halogen from its aqueous halide salt solution.

Cl, Br, I

Less reactiveMore reactive

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Colour of halogen

HalogenColour in

aqueous solutionColour in

1,1,1-trichlorethane

Chlorine Pale yellow Colourless

Bromine Brown Brown

Iodine Brown Purple

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Displacement of halogen

Procedure:1. 1cm3 of aqueous potassium iodide solution, 1cm3 of

bromine water and 1cm3 of 1,1,1-trichloroethane are added into a test tube, labelled A. The mixture is shaken.

2. Step 1 is repeated by adding 1cm3 of aqueous potassium bromide solution, 1cm3 of chlorine water and 1cm3 of 1,1,1-trichloroethane are added into a test tube, labelled B. The mixture is shaken.

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Displacement of halogenResult:

Test tubeColour of CH3CCl3

Inference

A PurpleIodine displaced

B BrownBromine displaced

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DiscussionTest Tube A

Cl2(aq) + 2KBr(aq) Br2(aq) + 2KCl(aq)

Cl2(aq) + 2Br− (aq) Br2(aq) + 2Cl− (aq)ON : 0 −1 0 −1

• Bromine, Br2 dissolves in CH3CCl3 to give a brown colour

• Chlorine, Cl2 is reduced. Reducing agent are the Br− ions.

• Bromide ions are oxidized. Oxidizing agent is Cl2.

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DiscussionTest Tube B

Br2(aq) + 2KI(aq) I2(aq) + 2KBr(aq)

Br2(aq) + 2I− (aq) I2(aq) + 2Br− (aq)ON : 0 −1 0 −1

• Iodine, I2 dissolves in CH3CCl3 to give a purple colour

• Bromine, Br2 is reduced. Reducing agent are the I− ions.

• Iodide ions are oxidized. Oxidizing agent is Br2.

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Transfer of Electron at a Distance I

At electrode X

Iodide ions lose electron and are oxidized to brown iodine.

2I− (aq) I2(aq) + 2e-

ON: -1 0 (oxidation)

Dilute sulphuric acid

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The colour of the solution changes from colourless to brown.

The e- released by the iodide ion flow from electrode X to electrode Y along the connecting wires.

At electrode Y

The bromine molecules surrounding the electrode Y accept the e- and are reduced to bromide ions.

Br2(aq) + 2e- 2Br−(aq)

ON: 0 -1 (Reduction)

The colour of the solution changes from brown to colourless.


Br2(aq) + 2I−(aq) 2Br−(aq) + I2(aq)

oxidant reductant

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Transfer of Electron at a Distance II

Dilute sulphuric acid

• At electrode X• Each Iron(II) ion loses an electron and is oxidized to

brown iron(III) ion.• Fe2+ (aq) Fe3+ (aq) + e-• ON: +2 +3 (oxidation)

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The colour of the solution changes from green to brown.

The e- released by the iron(II) ion flow from electrode X to electrode Y along the connecting wires.

At electrode Y

The bromine molecules surrounding the electrode Y accept the e- and are reduced to bromide ions.

Br2(aq) + 2e- 2Br−(aq)

ON: 0 -1 (Reduction)

The colour of the solution changes from brown to colourless.


Br2(aq) + Fe2+ (aq) 2Br−(aq) + Fe3+ (aq) oxidant reductant

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Transfer of Electron at a Distance III

• At electrode X• Each Iron(II) ion loses an electron and is oxidized to

brown iron(III) ion.• Fe2+ (aq) Fe3+ (aq) + e-• ON: +2 +3 (oxidation)

XX YY

58

The colour of the solution changes from green to yellow/ brown.

The e- released by the iron(II) ion flow from electrode X to electrode Y along the connecting wires.

At electrode Y

The manganate(VII) ion, MnO4− gathered at the electrode Y

accept the e- and are reduced to manganese(II) ion, Mn2+.

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

ON: +7 +2 (Reduction)

The colour of the solution changes from purple to colourless.

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anode 5[Fe2+ (aq) Fe3+ (aq) + e-]

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

……………………………………………………………………… 5Fe2+ (aq) + MnO4

− (aq) + 8H+ (aq) 5Fe3+ (aq)+ Mn2+ (aq) + 4H2O(l)

Substance oxidized : iron(II) ion, Fe2+

Substance reduced : manganese(VII) ion, MnO4−

Oxidizing agent : manganese(VII) ion, MnO4−

Reducing agent : iron(II) ion, Fe2+

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Practice A5

The figure above shows an experiment on the transfer of electron at a distancea) Identify the i) oxidizing agent ii) reducing agent iii) positive electrode iv) negative electrodeb) Explain the changes at the i) negative electrode ii) positive electrode

A B

K2Cr2O7 (aq)

+

H2SO4(aq)

FeSO4(aq)

H2SO4(aq)

3 a redox reaction

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