2012 topic 09 oxidation and reduction reactions

IB Chemistry Power Points

Topic 09

Oxidation and Reduction

www.pedagogics.ca

REDOXA guide for A level students

KNOCKHARDY PUBLISHING2008

SPECIFICATIONS

Much taken from

OXIDATIONis a GAIN OF OXYGEN

2Mg + O2 ——> 2MgO

magnesium has been oxidised as it has gained oxygen

is the REMOVAL (LOSS) OF HYDROGEN

C2H5OH ——> CH3CHO + H2

ethanol has been oxidised as it has ‘lost’ hydrogen

OXIDATION & REDUCTION – Simplified Definitions

OXIDATION & REDUCTION – Simplified Definitions

REDUCTIONis a GAIN OF HYDROGEN

C2H4 + H2 ——> C2H6

ethene has been reduced as it has gained hydrogen

is the REMOVAL (LOSS) OF OXYGEN

CuO + H2 ——> Cu + H2O

copper(II) oxide has been reduced as it has ‘lost’ oxygen

However as chemistry became more sophisticated, it was realised that another definition was required

...

OXIDATION Removal (loss) of electrons ‘OIL’

species will get less negative or more positive

REDUCTION Gain of electrons ‘RIG’species will become more negative or less

positive

REDOX When reduction and oxidation take place

OXIDATION AND REDUCTION IN TERMS OF ELECTRONS

Oxidation and reduction are not only defined as changes in O

and H

OXIDATION & REDUCTION – Better Definitions

OIL - Oxidation Is the Loss of electrons

RIG - Reduction Is the Gain of electrons

Used to... tell if oxidation or reduction has taken place

work out what has been oxidised and/or reduced

construct half equations and balance redox equations

FOR ATOMS AND SIMPLE IONSThe number of electrons which must be added or removed to become neutral

Atoms Na = 0 neutral already ... no need to add any electrons

Cations Na+ = +1 need to add 1 electron to make Na+ neutral

Anions Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral

OXIDATION NUMBERS (STATES)

Q. What are the oxidation states of the elements in the following?

a) C b) Fe3+ c) Fe2+

d) O2- e) He f) Al3+


a) C b) Fe3+ c) Fe2+

d) O2- e) He f) Al3+


a) C (0) b) Fe3+ (+3) c) Fe2+ (+2)

d) O2- (-2) e) He (0) f) Al3+ (+3)


a) C (0) b) Fe3+ (+3) c) Fe2+ (+2)

d) O2- (-2) e) He (0) f) Al3+ (+3)

OXIDATION STATES

• because CO2 is a neutral molecule, the sum of the oxidation states must be ?

• for this, one element must have a positive OS and the other must be ?

Explanation

MOLECULESThe SUM of the oxidation states adds up to ZERO

ELEMENTS H in H2 = 0 both are the same and must add up to Zero

COMPOUNDS C in CO2 = +4O in CO2 = -2 1 x +4 and 2 x -2 = Zero

HOW DO YOU DETERMINE THE VALUE OF AN ELEMENT’S OXIDATION STATE?• from its position in the periodic table and/or• the other element(s) present in the formula (oxygen is almost always -2 etc)

HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?• the more electronegative species will have the negative value

OXIDATION STATES

in SO42- the oxidation state of S = +6 there is ONE S

O = -2 there are FOUR O’s +6 + 4(-2) = -2 so the ion has a 2- charge

COMPLEX IONSThe SUM of the oxidation states adds up to THE CHARGE

e.g. NO3- sum of the oxidation states = - 1

SO42- sum of the oxidation states = - 2

NH4+ sum of the oxidation states = +1

Example SO42-

OXIDATION STATES

What is the oxidation number of Mn in MnO4¯ ?

• the oxidation state of oxygen in most compounds is - 2• there are 4 O’s so the sum of its oxidation states - 8• overall charge on the ion is - 1• therefore the sum of all the oxidation states must add up to - 1• the oxidation states of Mn four O’s must therefore equal - 1• therefore the oxidation state of Mn in MnO4¯is +7

+7 + 4(-2) = - 1

COMPLEX IONSThe SUM of the oxidation states adds up to THE CHARGE

e.g. NO3- sum of the oxidation states = - 1

SO42- sum of the oxidation states = - 2

NH4+ sum of the oxidation states = +1

Example

HYDROGEN +1 except 0 atom (H) and molecule (H2)

-1 hydride ion, H¯ in sodium hydride NaH

OXYGEN -2 except 0 atom (O) and molecule (O2)-1 in hydrogen peroxide, H2O2

+2 in F2O

HALOGENS -1 except 0 atom (X) and molecule (X2)

OXIDATION STATES

CALCULATING OXIDATION STATE - 1

Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other values

Q. Give the oxidation state of the element other than O, H or F in...SO2 NH3 NO2 NH4

+ IF7 Cl2O7

NO3¯ NO2¯ SO32- S2O3

2- S4O62- MnO4

2-

What is odd about the value of the oxidation state of S in S4O62- ?

Q. Give the oxidation state of the element other than O, H or F in...SO2 NH3 NO2 NH4

+ IF7 Cl2O7

NO3¯ NO2¯ SO32- S2O3

2- S4O62- MnO4

2-

What is odd about the value of the oxidation state of S in S4O62- ?

OXIDATION STATES

A. The oxidation states of the elements other than O, H or F are

SO2 O = -2 2 x -2 = - 4 overall neutral S = +4

NH3 H = +1 3 x +1 = +3 overall neutral N = - 3

NO2 O = -2 2 x -2 = - 4 overall neutral N = +4

NH4+ H = +1 4 x +1 = +4 overall +1 N = - 3

IF7 F = -1 7 x -1 = - 7 overall neutral I = +7

Cl2O7 O = -2 7 x -2 = -14 overall neutral Cl = +7

NO3¯ O = -2 3 x -2 = - 6 overall -1 N = +5

NO2¯ O = -2 2 x -2 = - 4 overall -1 N = +3

SO32- O = -2 3 x -2 = - 6 overall -2 S = +4

S2O32- O = -2 3 x -2 = - 6 overall -2 S = +2

MnO42- O = -2 4 x -2 = - 8 overall -2 Mn = +6

METALS • have positive values in compounds

• value is usually that of the Group Number Al is +3

• where there are several possibilities the values go no higher than the Group No. Sn can be +2 or +4

Mn can be +2,+4,+6,+7

NON-METALS • mostly negative based on their usual ion Cl usually -1 • can have values up to their Group No. Cl +1 +3 +5 or +7

OXIDATION STATES


The position of an element in the periodic table can act as a guide

OXIDATION STATES

Q. What is the theoretical maximum oxidation state of the following elements?

Na P Ba Pb S Mn Cr

What will be the usual and the maximum oxidation state in compounds of?

Li Br Sr O B N +1

Q. What is the theoretical maximum oxidation state of the following elements?

Na P Ba Pb S Mn Cr


Li Br Sr O B N +1

METALS • have positive values in compounds

• value is usually that of the Group Number Al is +3

• where there are several possibilities the values go no higher than the Group No. Sn can be +2 or +4

Mn can be +2,+4,+6,+7

NON-METALS • mostly negative based on their usual ion Cl usually -1 • can have values up to their Group No. Cl +1 +3 +5 or +7



OXIDATION STATES



A. What is the theoretical maximum oxidation state of the following elements?

Na P Ba Pb S Mn Cr+1 +5 +2 +4 +6 +7 +6


Li Br Sr O B NUSUAL +1 -1 +2 -2 +3 -3 or +5MAXIMUM +1 +7 +2 +6 +3 +5

manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2

sulphur(VI) oxide for SO3 S is in the +6 oxidation state

dichromate(VI) for Cr2O72- Cr is in the +6 oxidation state

phosphorus(V) chloride for PCl5 P is in the +5 oxidation state

phosphorus(III) chloride for PCl3 P is in the +3 oxidation state

OXIDATION STATES

THE ROLE OF OXIDATION STATE IN NAMING SPECIES

To avoid ambiguity, the oxidation state is often included in the name of a species

Q. Name the following... PbO2

SnCl2

SbCl3

TiCl4

BrF5

OXIDATION STATES

Q. Name the following... PbO2 lead(IV) oxide

SnCl2 tin(II) chloride

SbCl3 antimony(III) chloride

TiCl4 titanium(IV) chloride

BrF5 bromine(V) fluoride

manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2

sulphur(VI) oxide for SO3 S is in the +6 oxidation state

dichromate(VI) for Cr2O72- Cr is in the +6 oxidation state

phosphorus(V) chloride for PCl5 P is in the +5 oxidation state

phosphorus(III) chloride for PCl3 P is in the +3 oxidation state

THE ROLE OF OXIDATION STATE IN NAMING SPECIES

To avoid ambiguity, the oxidation state is often included in the name of a species


OXIDATION Removal (loss) of electrons ‘OIL’species will get less negative or more positive

REDUCTION Gain of electrons ‘RIG’species will become more negative or less positive

REDOX REACTIONS


Oxidation and reduction are not only defined as changes in O and H

+7+6+5+4+3+2+1 0-1-2-3-4

REDUCTION

OXIDATION


OXIDATION Removal (loss) of electrons ‘OIL’species will get less negative or more positive

REDUCTION Gain of electrons ‘RIG’species will become more negative or less positive

REDUCTION in O.N. Species has been REDUCED

e.g. Cl is reduced to Cl¯ (0 to -1)

INCREASE in O.N. Species has been OXIDISED

e.g. Na is oxidised to Na+ (0 to +1)

REDOX REACTIONS


Oxidation and reduction are not only defined as changes in O and H

+7+6+5+4+3+2+1 0-1-2-3-4

REDUCTION

OXIDATION

REDUCTION in O.S. INCREASE in O.S.Species has been REDUCED Species has been OXIDISED

REDOX REACTIONS


Q. State if the changes involve oxidation (O) or reduction (R) or neither (N)

Fe2+ —> Fe3+

I2 —> I¯

F2 —> F2O

C2O42- —> CO2

H2O2 —> O2

H2O2 —> H2O

Cr2O72- —> Cr3+

Cr2O72- —> CrO4

2-

SO42- —> SO2

REDOX REACTIONS

REDUCTION in O.S. INCREASE in O.S.Species has been REDUCED Species has been OXIDISED


Q. State if the changes involve oxidation (O) or reduction (R) or neither (N)

Fe2+ —> Fe3+ O +2 to +3 I2 —> I¯ R 0 to -1

F2 —> F2O R 0 to -1

C2O42- —> CO2 O +3 to +4

H2O2 —> O2 O -1 to 0

H2O2 —> H2O R -1 to -2

Cr2O72- —> Cr3+ R +6 to +3

Cr2O72- —> CrO4

2- N +6 to +6

SO42- —> SO2 R +6 to +4

OXIDATION STATES - Review

CALCULATING OXIDATION STATE – MOST IMPORTANT

Q. What is the oxidation state of each element in the following compounds/ions ?

CH4

PCl3

NCl3

CS2

ICl5

BrF3

PCl4+

H3PO4

NH4Cl

H2SO4

MgCO3

SOCl2

OXIDATION STATES


Q. What is the oxidation state of each element in the following compounds/ions ?

CH4 C = - 4 H = +1

PCl3 P = +3 Cl = -1

NCl3 N = +3 Cl = -1

CS2 C = +4 S = -2

ICl5 I = +5 Cl = -1

BrF3 Br = +3 F = -1

PCl4+ P = +5 Cl = -1

H3PO4 P = +5 H = +1 O = -2

NH4Cl N = -3 H = +1 Cl = -1

H2SO4 S = +6 H = +1 O = -2

MgCO3 Mg = +2 C = +4 O = -2

SOCl2 S = +4 Cl = -1 O = -2

END OF PART ONE

WRITING & BALANCING REDOX HALF EQUATIONS

Example 1 Iron(II) being oxidised to iron(III)

Step 1 Fe2+ ——> Fe3+

Step 2 +2 +3 Step 3 Fe2+ ——> Fe3+ + e¯ now balanced

An electron (charge -1) is added to the RHS of the equation...this balances the oxidation state change i.e. (+2) ——> (+3) + (-1)

WRITING & BALANCING REDOX HALF EQUATIONS

No need to balance Mn; equal numbers

In the case of reactions occurring in acidic solutions, more effort is required to write the half reactions.

If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges

If equation still doesn’t balance, add sufficient water molecules to one side

Example 2 MnO4¯ being reduced to Mn2+ in acidic solution

Step 1 MnO4¯ ———> Mn2+

Overall charge on MnO4¯ is -1; sum of the ON’s of all atoms must add up to -1

Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8

To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]



Step 2 +7 +2




The oxidation states on either side are different; +7 —> +2 (REDUCTION)

To balance; add 5 negative charges to the LHS [+7 + (5 x -1) = +2]

You must ADD 5 ELECTRONS to the LHS of the equation



Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+




Total charges on either side are not equal; LHS = 1- and 5- = 6-RHS = 2+

Balance them by adding 8 positive charges to the LHS [ 6- + (8 x 1+) = 2+ ]

You must ADD 8 PROTONS (H+ ions) to the LHS of the equation



Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+

Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+






Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+

Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+

Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O now balanced

Everything balances apart from oxygen and hydrogen O LHS = 4 RHS = 0

H LHS = 8 RHS = 0

You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced




Watch out for cases when the species is present in different amounts oneither side of the equation ... IT MUST BE BALANCED FIRST

Example 3 Cr2O72- being reduced to Cr3+ in acidic solution

Step 1 Cr2O72- ———> Cr3+ there are two Cr’s on LHS

Cr2O72- ———> 2Cr3+ both sides now have 2

Step 2 2 Cr @ +6 2 Cr @ +3 both Cr’s are reduced

Step 3 Cr2O72- + 6e¯ ——> 2Cr3+ each Cr needs 3 electrons

Step 4 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+

Step 5 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+ + 7H2O now balanced

BALANCING REDOX HALF EQUATIONS


REMINDER1 Work out the formula of the species before and after the change; balance if required2 Work out the oxidation state of the element before and after the change3 Add electrons to one side of the equation so that the oxidation states balance4 If the charges on all the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges5 If the equation still doesn’t balance, add sufficient water molecules to one side

Q. Balance the following half equations...

Na —> Na+

Fe2+ —> Fe3+

I2 —> I¯

C2O42- —> CO2

H2O2 —> O2

H2O2 —> H2O

NO3- —> NO

NO3- —> NO2

SO42- —> SO2


Q. Balance the following half equations...

Na —> Na+ + e-

Fe2+ —> Fe3+ + e-

I2 + 2e- —> 2I¯

C2O42- —> 2CO2 + 2e-

H2O2 —> O2 + 2H+ + 2e-

H2O2 + 2H+ + 2e- —> 2H2O

NO3- + 4H+ + 3e- —> NO + 2H2O

NO3- + 2H+ + e- —> NO2

+ H2O

SO42- + 4H+ + 2e- —> SO2 + 2H2O

COMBINING HALF EQUATIONS

A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...

Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides

The reaction between manganate(VII) and iron(II)



Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction






Step 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1







Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯



OXIDIZING AGENT AND REDUCING AGENT

In the reaction between manganate(VII) and iron(II)

MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯

The manganate (VII) ion is reduced (gains electrons) and is called the oxidizing agent.

The iron(II) ion is oxidized (loses electrons) and is called the reducing agent.





Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯

Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+






Q. Construct balanced redox equations for the reactions between...

Mg (OX) and H+ (RED)

Fe2+ (OX) and Cr2O72- (RED)

MnO4¯ (RED) and H2O2 (OX)

MnO4¯ (RED) and C2O42- (OX)

S2O32- (OX) and I2 (RED)

Cr2O72- (RED) and I- (OX)

Mg ——> Mg2+ + 2e¯ (x1)H+ + e¯ ——> ½ H2 (x2)

Mg + 2H+ ——> Mg2+ + H2

Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)

Fe2+ ——> Fe3+ + e¯ (x6)

Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2O

MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2)

H2O2 ——> O2 + 2H+ + 2e¯ (x5)

2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O

MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2)

C2O42- ——> 2CO2 + 2e¯ (x5)

2MnO4¯ + 5C2O42- + 16H+ ——> 2Mn2+ + 10CO2 + 8H2O

2S2O32- ——> S4O6

2- + 2e¯ (x1)

½ I2 + e¯ ——> I¯ (x2)

2S2O32- + I2 ——> S4O6

2- + 2I¯

Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1)

I¯ ——> ½ I2 + e¯ (x6)

Cr2O72- + 14H+ + 6I ¯ ——> 2Cr3+ + 3I2 + 7H2O

BALANCING

REDOX

EQUATIONS

ANSWERS

END OF PART TWO

Reactivity Series

Displacement reactions

Mg

Magnesium

SO4Cu

Copper sulphate

The magnesium DISPLACES the copper from copper sulphate

SO4Mg

Magnesium sulphate

Cu

Copper

A displacement reaction is one where a MORE REACTIVE metal will DISPLACE a LESS REACTIVE metal from a compound.

We have looked at several reactions:Fe + Cu(NO3)2 Cu + Fe(NO3)2

Li + H2O LiOH + H2

Such experiments reveal trends. The activity series ranks the relative reactivity of metals.It allows us to predict if certain chemicals will undergo single displacement reactions when mixed: metals near the top are most reactive and will displace metals near the bottom.

Q: Which of these reactions occur?Fe + CuSO4 Ni + NaCl Li + ZnCO3 Al + CuCl2

Cu + Fe2(SO4)3

Yes, Fe is above Cu

NR (no reaction)

No, Ni is below Na

Zn + Li2CO3Cu + AlCl3

Yes, Al is above Cu

KNaLiCaMgAlZnFeNiSnPbH

CuHgAgAu

Yes, Li is above Zn

Trends in Oxidation and Reduction

Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents

Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

2012 topic 09 oxidation and reduction reactions

Technology

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