Download 3.9-10 Redox titrations

Redox titrations
Textbook reference: p 218-222
Redox titrations
• Carry out redox titrations and carry out
structured calculations involving MnO4• Understand how to carry out redox titrations
and carry out structured calculations
involving MnO4• Perform non-structured titration calculations,
based on experimental results
• Carry out structured titration calculations
based on experimental results
Redox titrations
• These are based on the same principles as
acid-base titrations but involves the transfer
of electrons rather than the transfer of
protons
• You need an indicator species which is any
chemical that will change colour when all of
the oxidising agent has been used up
Self-indicating reactions
• Some redox titrations do not need an indicator
like acid-base titrations do
• The strong oxidiser potassium manganate (VII)
ions are a good example
MnO4- (aq) 
Mn2+ (aq)
Purple
Almost colourless
Can you think of any other oxidising agents that
change colour as they change oxidation state?
Iron in redox reactions
• Whenever iron is used with an oxidising agent
it will most probably be as the 2+ ion, e.g.
when metallic iron is dissolved in sulfuric acid
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
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
Step 1
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯ ———>
Mn2+
No need to balance Mn;
equal numbers
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
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
Step 1
Step 2
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯ ———>
+7
Mn2+
+2
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]
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
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
Step 1
Step 2
Step 3
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯ ———> Mn2+
+7
+2
MnO4¯ + 5e¯ ———> Mn2+
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
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
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
Step 1
Step 2
Step 3
Step 4
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯ ———> Mn2+
+7
+2
MnO4¯ + 5e¯ ———> Mn2+
MnO4¯ + 5e¯ + 8H+ ———> Mn2+
Total charges on either side are not equal;
LHS = 1- and 5- = 6RHS = 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
BALANCING REDOX HALF EQUATIONS
1
2
3
4
Work out formulae of the species before and after the change; balance if required
Work out oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
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
Step 1
Step 2
Step 3
Step 4
Step 5
MnO4¯ being reduced to Mn2+ in acidic solution
MnO4¯ ———> Mn2+
+7
+2
MnO4¯ + 5e¯ ———> Mn2+
MnO4¯ + 5e¯ + 8H+ ———> Mn2+
MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O
Everything balances apart from oxygen and hydrogen O LHS = 4
H
now balanced
RHS = 0
LHS = 8
You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced
RHS = 0
5.3 Exercise 1
BALANCING REDOX HALF EQUATIONS
Q.
Balance the following half equations...
Na
—>
Na+
Fe2+
—>
Fe3+
I2
—>
I¯
C2O42-
—> CO2
H2O2
1
2
3
4
5
—>
O2
H2O2
—>
H2O
NO3-
—>
NO
NO3-
—>
NO2
SO42-
—>
SO2
REMINDER
Work out the formula of the species before and after the change; balance if required
Work out the oxidation state of the element before and after the change
Add electrons to one side of the equation so that the oxidation states balance
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 charges
If the equation still doesn’t balance, add sufficient water molecules to one side
BALANCING REDOX HALF EQUATIONS
Q.
Balance the following half equations...
Na
—>
Na+
+
e-
Fe2+
—>
Fe3+
+
e-
+
2e-
I2
+
2e-
C2O42H2O2
—>
2I¯
—> 2CO2
—>
O2
+ 2H+ + 2e-
H2O2 + 2H+ + 2e- —>
2H2O
NO3- + 4H+ + 3e- —>
NO
+
2H2O
NO3- + 2H+ + e-
—>
NO2
+
H 2O
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
Step 2
Step 3
Step 4
Write out the two half equations
Multiply the equations so that the number of electrons in each is the same
Add the two equations and cancel out the electrons on either side
If necessary, cancel any other species which appear on both sides
Worked example
MnO4-(aq) + 8H+(aq) + 5eFe2+(aq)


Mn2+(aq) + 4H2O(l)
Fe3+(aq) + e-
1. Construct the fully balanced redox ionic equation for the manganate(VII)
ion oxidising the iron(II) ion.
2. 24.3 cm3 of 0.02 mol dm-3 KMnO4 reacted with 20.0 cm3 of an iron(II)
solution.
3. Calculate the concentration of the iron(II) ion.
4. How do recognise the end-point in the titration?
5. Calculate the percentage of iron in a sample of steel wire if 1.51 g of the
wire was dissolved in excess of dilute sulphuric acid and the solution made
up to 250 cm3 in a standard graduated flask. 25.0 cm3 of this solution was
pipetted into a conical flask and needed 25.45 cm3 of 0.02 mol dm-3
KMnO4 for complete oxidation.
• Calculate the percentage of iron in a
sample of steel wire if 1.51 g of the wire
was dissolved in excess of dilute
sulphuric acid and the solution made up
to 250 cm3 in a standard graduated
flask. 25.0 cm3 of this solution was
pipetted into a conical flask and needed
25.45 cm3 of 0.02 mol dm-3 KMnO4 for
complete oxidation.