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 half-equation method.
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In this method we use the half-equation for the oxidising agent and the half-equation for the reducing agent then add them together.
Iron(III) salts are reduced to iron(II) salts.
The half equation is
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+
To obtain a balanced equation 2 electrons muct be inserted on the LHS
Cl2 + 2e- 2Cl-
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
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-
4 2Fe3+ + 2e- 2Fe2+
3 2Cl- Cl2 + 2e-
2Fe3+ + 2Cl- 2Fe2+ +Cl2
There must be the same no of electrons on each side which then cancel.
Ethanedioate ion is C2O42- and the halfequation is
C2O42- 2CO2 + 2e-
2MnO4- + 16H+ + 5C2O42- 2Mn2+ + 8H2O + 10CO2
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?
0.200 x 25 = 0.005000 mols
Ratio of ethanedioate to manganate(VII) = 2:5
no of mols of manganate(VII) in 17.2ml =
0.005000 x 2 = 0.002000 mols
conc of manganate(VII) = 0.002000 x 1000
= 0.116 mol/L