Before you start it would be helpful to… Recall the layout of the periodic table Be able to balance simple equations REDOX.

Before you start it would be helpful to…

• Recall the layout of the periodic table

• Be able to balance simple equations

REDOXREDOX

CONTENTS

• Definitions of oxidation and reduction

• Calculating oxidation state

• Use of H, O and F in calculating oxidation state

• Naming compounds

• Redox reactions

• Balancing ionic half equations

• Combining half equations to form a redox equation

• Revision check list

REDOXREDOX

OXIDATIONGAIN OF OXYGEN

2Mg + O2 ——> 2MgO

magnesium has been oxidised as it has gained oxygen

REMOVAL (LOSS) OF HYDROGEN

C2H5OH ——> CH3CHO + H2

ethanol has been oxidised as it has ‘lost’ hydrogen

OXIDATION & REDUCTION - OXIDATION & REDUCTION - DefinitionsDefinitions


REDUCTIONGAIN OF HYDROGEN

C2H4 + H2 ——> C2H6

ethene has been reduced as it has gained hydrogen

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 electronsspecies will get less negative or more positive

REDUCTION Gain of electronsspecies 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 Removal (loss) of electronsspecies will get less negative or more positive

REDUCTION Gain of electronsspecies will become more negative or less positive





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

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

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

cations Na in Na+ = +1 need to add 1 electron to make Na+ neutral

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

OXIDATION NUMBEROXIDATION NUMBER

OXIDATION OXIDATION NUMBERNUMBER

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)








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

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

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


Explanation



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

HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?

• the more electronegative species will have the negative value• electronegativity increases across a period and decreases down a group• O is further to the right than C in the periodic table so it has the negative value




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

HOW DO YOU DETERMINE THE VALUE OFAN ELEMENT’S OXIDATION STATE?

• from its position in the periodic table and/or• the other element(s) present in the formula




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


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

Examples


What is the oxidation state (OS) 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

Examples

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

FLUORINE -1 except 0 atom (F) and molecule (F2)


CALCULATING OXIDATION NUMBER - 1

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

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

FLUORINE -1 except 0 atom (F) and molecule (F2)

OXIDATION STATESOXIDATION STATES

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-


CALCULATING OXIDATION NUMBER - 1

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


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 (14/2)

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 (4/2)

S4O62- O = -2 6 x -2 = -12 overall -2 S = +2½ ! (10/4)

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


An oxidation state must be a whole number (+2½ is the average value)

METALS • have positive values in compounds

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

NON-METALS • mostly negative based on their usual ion Cl usually -1


CALCULATING OXIDATION STATE - 2

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



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



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 = +4 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 H = +4 O = -2

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

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

Q. Name the following... PbO2

SnCl2

SbCl3

TiCl4

BrF5


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 REACTIONSREDOX REACTIONS



+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.S. Species has been REDUCED

e.g. Cl is reduced to Cl¯ (0 to -1)

INCREASE in O.S. Species has been OXIDISED

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




+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



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

Fe2+ —> Fe3+

I2 —> I¯

F2 —> F2O


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

Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)

Redox reactions of metals with acidsRedox reactions of metals with acids



We can assign oxidaton numbers to each atom in any equation in We can assign oxidaton numbers to each atom in any equation in order to –order to –

Identify whether a redox reaction has taken placeIdentify whether a redox reaction has taken place

Work out what has been oxidised and what has been reduced.Work out what has been oxidised and what has been reduced.



The metal is oxidised, forming positive metal ionsThe metal is oxidised, forming positive metal ions

The hydrogen ion in the acid is reduced, forming the element The hydrogen ion in the acid is reduced, forming the element hydrogen, as a gashydrogen, as a gas

We can write the above equation to show the role of the hydrogen We can write the above equation to show the role of the hydrogen ion, Hion, H++

Mg (s) + 2HH++ (aq)(aq) Mg Mg2+2+ +H +H 2 (g)2 (g)

We can assign oxidaton numbers to each atom in any equation in We can assign oxidaton numbers to each atom in any equation in order to –order to –Identify whether a redox reaction has taken placeIdentify whether a redox reaction has taken placeWork out what has been oxidised and what has been reduced.Work out what has been oxidised and what has been reduced.

Before you start it would be helpful to… Recall the layout of the periodic table Be able to balance simple equations REDOX.

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