AN INTRODUCTION TO SYNTHETIC ORGANIC CHEMISTRY KNOCKHARDY PUBLISHING.
REDOX A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS.
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Transcript of REDOX A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS.
![Page 1: REDOX A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS.](https://reader031.fdocuments.net/reader031/viewer/2022031909/56649f505503460f94c725c4/html5/thumbnails/1.jpg)
REDOXREDOXA guide for A level studentsA guide for A level students
KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING2008 2008
SPECIFICATIONSSPECIFICATIONS
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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.
Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.
Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...
www.knockhardy.org.uk/sci.htm
Navigation is achieved by...
either clicking on the grey arrows at the foot of each page
or using the left and right arrow keys on the keyboard
KNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHING
REDOXREDOX
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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
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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
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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
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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
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...
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 - OXIDATION & REDUCTION - DefinitionsDefinitions
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...
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 - OXIDATION & REDUCTION - DefinitionsDefinitions
OIL - Oxidation Is the Loss of electrons
RIG - Reduction Is the Gain of electrons
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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
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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+
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
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
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OXIDATION STATESOXIDATION STATES
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
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
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OXIDATION STATESOXIDATION STATES
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
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• 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 STATESOXIDATION STATES
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
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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 STATESOXIDATION STATES
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
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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 STATESOXIDATION STATES
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
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OXIDATION STATESOXIDATION 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
Examples
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OXIDATION STATESOXIDATION STATES
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
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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
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
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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 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
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OXIDATION STATESOXIDATION 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 (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)
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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 STATESOXIDATION STATES
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
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OXIDATION STATESOXIDATION 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
What will be the usual and the maximum oxidation state in compounds of?
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
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
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OXIDATION STATESOXIDATION STATES
CALCULATING OXIDATION STATE - 2
The position of an element in the periodic table can act as a guide
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
What will be the usual and the maximum oxidation state in compounds of?
Li Br Sr O B NUSUAL +1 -1 +2 -2 +3 -3 or +5MAXIMUM +1 +7 +2 +6 +3 +5
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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
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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 C = +4 O = -2
SOCl2 S = +4 Cl = -1 O = -2
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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
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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
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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
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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
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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
C2O42- —> CO2
H2O2 —> O2
H2O2 —> H2O
Cr2O72- —> Cr3+
Cr2O72- —> CrO4
2-
SO42- —> SO2
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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
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
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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
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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
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
BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
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No need to balance Mn; equal numbers
BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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]
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
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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
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
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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
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
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+
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Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+
Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O now balanced
BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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
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
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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
BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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BALANCING REDOX HALF EQUATIONSBALANCING 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
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BALANCING REDOX HALF EQUATIONSBALANCING REDOX HALF EQUATIONS
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
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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
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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)
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
The reaction between manganate(VII) and iron(II)
Step 1 Fe2+ ——> Fe3+ + e¯ OxidationMnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
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
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
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
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
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
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¯
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
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COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
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+
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
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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+
COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
SUMMARY
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
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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
COMBINING HALF EQUATIONSCOMBINING HALF EQUATIONS
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¯
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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
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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
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You need to go over the You need to go over the relevant topic(s) againrelevant topic(s) again
Click on the button toClick on the button toreturn to the menureturn to the menu
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WELL DONE!WELL DONE!Try some past paper questionsTry some past paper questions
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REDOXREDOXTHE ENDTHE END
© 2008 JONATHAN HOPTON & KNOCKHARDY PUBLISHING© 2008 JONATHAN HOPTON & KNOCKHARDY PUBLISHING