before you start it would be helpful to… recall the layout of the periodic table be able to...
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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
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 & REDUCTION - OXIDATION & REDUCTION - DefinitionsDefinitions
...
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 & REDUCTION - OXIDATION & REDUCTION - DefinitionsDefinitions
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+
Q. What are the oxidation states of the elements in the following?
a) C b) Fe3+ c) Fe2+
d) O2- e) He f) Al3+
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 OXIDATION NUMBERNUMBER
Q. What are the oxidation states of the elements in the following?
a) C (0) b) Fe3+ (+3) c) Fe2+ (+2)
d) O2- (-2) e) He (0) f) Al3+ (+3)
Q. What are the oxidation states of the elements in the following?
a) C (0) b) Fe3+ (+3) c) Fe2+ (+2)
d) O2- (-2) e) He (0) f) Al3+ (+3)
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 OXIDATION NUMBERNUMBER
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
OXIDATION OXIDATION NUMBERNUMBER
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 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
OXIDATION OXIDATION NUMBERNUMBER
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 OFAN ELEMENT’S OXIDATION STATE?
• from its position in the periodic table and/or• the other element(s) present in the formula
OXIDATION OXIDATION NUMBERNUMBER
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
OXIDATION OXIDATION NUMBERNUMBER
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
OXIDATION OXIDATION NUMBERNUMBER
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)
OXIDATION OXIDATION NUMBERNUMBER
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-
What is odd about the value of the oxidation state of S in S4O62- ?
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
OXIDATION NUMBEROXIDATION NUMBER
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
What is odd about the value of the oxidation state of S in S4O62- ?
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
OXIDATION NUMBEROXIDATION NUMBER
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
OXIDATION STATESOXIDATION STATES
CALCULATING OXIDATION STATE - 2
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 STATESOXIDATION STATES
CALCULATING OXIDATION STATE - 2
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
OXIDATION STATESOXIDATION 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 STATESOXIDATION 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
REDOX When reduction and oxidation take place
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
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
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
REDOX When reduction and oxidation take place
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)
REDOX REACTIONSREDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
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 REACTIONSREDOX REACTIONS
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
Q. State if the changes involve oxidation (O) or reduction (R) or neither (N)
Fe2+ —> Fe3+
I2 —> I¯
F2 —> F2O
REDOX REACTIONSREDOX REACTIONS
REDUCTION in O.S. INCREASE in O.S.Species has been REDUCED Species has been OXIDISED
OXIDATION AND REDUCTION IN TERMS OF ELECTRONS
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
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.
Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)
Redox reactions of metals with acidsRedox reactions of metals with acids
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.
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