2012 topic 09 oxidation and reduction reactions
-
Upload
david-young -
Category
Technology
-
view
2.929 -
download
1
Transcript of 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+
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 (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)
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
CALCULATING OXIDATION STATE - 2
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
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
OXIDATION 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
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
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 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.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 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 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
REDOX 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
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
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 = +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]
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2
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
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
Example 2 MnO4¯ being reduced to Mn2+ in acidic solution
Step 1 MnO4¯ ———> Mn2+
Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+
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
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
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+
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+
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
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
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
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
BALANCING 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
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)
COMBINING 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
COMBINING 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
COMBINING 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
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.
COMBINING 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
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 EQUATIONS
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