REDOX A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

REDOXREDOXA guide for A level studentsA guide for A level students

KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING2008 2008

SPECIFICATIONSSPECIFICATIONS

INTRODUCTION

This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.

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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

Before you start it would be helpful to…

• Recall the layout of the periodic table

• Be able to balance simple equations

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 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


...







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

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 STATESOXIDATION 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)









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 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

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)


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-



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

• 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



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


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






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



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 C = +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




REDOX REACTIONSREDOX REACTIONS



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

REDUCTION

OXIDATION




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

C2O42- —> CO2

H2O2 —> O2

H2O2 —> H2O

Cr2O72- —> Cr3+

Cr2O72- —> CrO4

2-

SO42- —> SO2


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

1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 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 5 If equation still doesn’t balance, add sufficient water molecules to one side

BALANCING REDOX HALF EQUATIONSBALANCING 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)

As everything balances, there is no need to proceed to Steps 4 and 5




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

No need to balance Mn; equal numbers




Step 1 MnO4¯ ———> Mn2+


Overall charge on MnO4¯ is -1; sum of the OS’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 @ +6 2 @ +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



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 EQUATIONSCOMBINING 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¯








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







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


SUMMARY






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

Mg and H+

Cr2O72- and Fe2+

H2O2 and MnO4¯

C2O42- and MnO4¯

S2O32- and I2

Cr2O72- and I¯

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)

½ I2 + e¯ ——> I¯ (x6)

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

BALANCING

REDOX

EQUATIONS

ANSWERS

REVISION CHECKREVISION CHECK

What should you be able to do?

Recall the definitions for oxidation and reduction in terms of oxygen, hydrogen and electrons

Write balanced equations representing oxidation and reduction

Know the trend in electronegativity across periods

Predict the oxidation state of elements in atoms, simple ions, compounds and complex ions

Recognize, in terms of oxidation state, if oxidation or reduction has taken place

Balance ionic half equations

Combine two ionic half equations to make a balanced redox equation

CAN YOU DO ALL OF THESE? CAN YOU DO ALL OF THESE? YES YES NONO

You need to go over the You need to go over the relevant topic(s) againrelevant topic(s) again

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REDOXREDOXTHE ENDTHE END

© 2008 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2008 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

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REDOX A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS