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Redox Titrations A redox titration is the same as an acid-base titration except it involves a redox reaction and generally does not require an indicator. Reagents are chosen so that the reaction is spontaneous and one of the half reactions has a natural colour change . Redox Titrations - PowerPoint PPT Presentation
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Redox Titrations
A redox titration is the same as an acid-base titration except it involves a redox reaction and generally does not require an indicator. Reagents are chosen so that the reaction is spontaneous and one of the half reactions has a natural colour change.
Redox Titrations
A redox titration is the same as an acid-base titration except it involves a redox reaction and generally does not require an indicator. Reagents are chosen so that the reaction is spontaneous and one of the half reactions has a natural colour change.
Pick a suitable reagent for redox titration involving IO3- in acid
solution. F- I- SO42- Cl-
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
[Fe2+] =
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
[Fe2+] =
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
0.00675 L MnO4- x 0.100 mole
[Fe2+] = 1 L
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
1MnO4- + 8H+ +5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
0.00675 L MnO4- x 0.100 mole x 5 moles Fe2+
[Fe2+] = 1 L 1 mole MnO4-
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
0.00675 L MnO4- x 0.100 mole x 5 moles Fe2+
[Fe2+] = 1 L 1 mole MnO4-
0.0250 L
6.75 mL of 0.100 M KMnO4 is required to titrate 25.0 mL of FeCl2. Calculate the [Fe2+].
MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+
0.00675 L 0.0250L
0.100 M ? M
0.00675 L MnO4- x 0.100 mole x 5 moles Fe2+
[Fe2+] = 1 L 1 mole MnO4-
0.0250 L
= 0.135 M
Write the anode and cathode reactions.
Pt Pt
H2O2(aq)MnO4- in acid
NaNO3aq)
Inert electrodes- look at the solution for the reactions
voltmeter
Cathode: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Anode: H2O2 → O2 + 2H+ + 2e-
What happens to the mass of the cathode?Constant
What happens to the mass of the anode?Constant
What happens to the pH of the cathode?Increases
What happens to the pH of the anode?Decreases
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu
2K+
SO42-
H2O
-
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu
2K+
SO42-
H2O
-
Cathode
Reduction
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu
2K+
SO42-
H2O
-
Cathode
Reduction
2H2O + 2e-→H2 +2OH-
-0.41 v
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu
2K+
SO42-
H2O
-
Cathode
Reduction
2H2O + 2e-→H2 +2OH-
-0.41 v
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu
Cu might oxidize2K+
SO42-
H2O
-
Cathode
Reduction
2H2O+2e-→H2(g)+ 2OH-
-0.41 v
+
AnodeOxidation
Cu(s) → Cu2++2e-
-0.34 v
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu
2K+
SO42-
H2O
2H2O + Cu(s) → Cu2++ H2 +2OH- E0 = -0.75 v MTV = +0.75 v
Non-Inert ElectrodesA non-inert Anode might oxidize.The Cathode will stay inert.
DC Power - +
K2SO4(aq)
CuCu-
Cathode
Reduction
2H2O+2e-→H2(g)+ 2OH-
-0.41 v
+
AnodeOxidation
Cu(s) → Cu2++2e-
-0.34 v
2K+
SO42-
H2O