Transcript Redox Titrations - Dr Ashby's Chemistry Pages

Redox Titrations
Oxidation-reduction reactions involve a transfer of
electrons. The oxidising agent accepts electrons and the
reducing agent gives electrons. In working out the
equation for a redox reaction we can use the halfequation method.
In this method we use the half-equation for the oxidising
agent and the half-equation for the reducing agent then
add them together.
Examples of half-reaction equations
Iron(III) salts are reduced to iron(II) salts.
The half equation is
Fe3+  Fe2+
For the equation to balance the charge on the RHS must
equal the charge on the LHS. This can be accomplished
by inserting an electron on the LHS
Fe3+ + e-
 Fe2+
When chlorine acts as an oxidising agent it is reduced to
Cl- ions
Cl2

2Cl-
To obtain a balanced equation 2 electrons muct be inserted
on the LHS
Cl2 + 2e- 
2Cl-
Potassium mangante(VII) is an oxidising agent. In acid
solutions it is reduced to a manganese(II) salt
MnO4- +
8H+ + 5e- 
Mn2+ + 4H2O
Potassium dichromate(VI) is also an oxidising agent. It is
reduced to a chromium(III) salt
Cr2O7- +
14H+ +
6e-  2Cr3+
+ 7H2O
Using half equations to obtain the equation for the reaction
Find the equation for the reaction of iron(III) with chloride
ions
The 2 half reactions needed are
1
Fe3+ + e-  Fe2+
2
Cl2 + 2e- 
2Cl-
We need iron(III) and chloride as reactants so equation 2
must be reversed
3
2Cl-  Cl2 + 2e-
Now we must balance the electrons, then add the half
equations together to obtain the full equation. To
balance the electrons we multiply equation 1 by 2
1
Fe3+ + e-
4
2Fe3+ + 2e-  2Fe2+
2Cl-  Cl2 + 2e2Fe3+ + 2Cl-  2Fe2+ + Cl2
3
 Fe2+ x2
There must be the same no of electrons on each side
which then cancel.
Construct an equation for the reaction of potassium
manganate(VII) with ethanedioate ion
Ethanedioate ion is C2O42- and the half equation is
C2O42-  2CO2 + 2e-
2MnO4- +
16H+ +
5C2O42-

2Mn2+ + 8H2O + 10CO2
Now we can do a titration calculation using this equation!
A 25.0ml portion of sodium ethanedioate solution of
concentration 0.200 mol/L is warmed and titrated against
a solution of potassium manganate(VII). If 17.2ml of
potassium manganate(VII) is required what is its
concentration?
No of moles ethanedioate in 25.0ml of solution of conc
0.200 mol/L =
0.200 x 25 = 0.005000 mols
1000
Ratio of ethanedioate to manganate(VII) = 2:5
 no of mols of manganate(VII) in 17.2ml =
0.005000 x 2
5
=
0.002000 mols
 conc of manganate(VII) = 0.002000 x 1000
17.2
= 0.116 mol/L