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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
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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 spontaneousand 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 spontaneousand 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 MnO4- in acid H2O2(aq) voltmeter NaNO3aq) Inert electrodes- look at the solution for the reactions
Cathode Anode
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 Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu K2SO4(aq) 2K+ SO42- H2O
Non-Inert Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu 2K+ SO42- H2O K2SO4(aq) -
Non-Inert Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu K2SO4(aq) - Cathode Reduction 2K+ SO42- H2O
Non-Inert Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu K2SO4(aq) - Cathode Reduction 2H2O + 2e-→H2 +2OH- -0.41 v 2K+ SO42- H2O
Non-Inert Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu K2SO4(aq) - Cathode Reduction 2H2O + 2e-→H2 +2OH- -0.41 v Cu might oxidize 2K+ SO42- H2O
You must look at the possible oxidation of: SO42- H2O Cu Strongest Reducing Agent
Non-Inert Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu K2SO4(aq) - Cathode Reduction 2H2O+2e-→H2(g)+ 2OH- -0.41 v + Anode Oxidation Cu(s)→ Cu2++2e- -0.34 v 2K+ SO42- H2O
Non-Inert Electrodes A non-inertAnodemight oxidize. The Cathode will stay inert. DC Power - + Cu Cu K2SO4(aq) - Cathode Reduction 2H2O+2e-→H2(g)+ 2OH- -0.41 v + Anode Oxidation Cu(s)→ Cu2++2e- -0.34 v 2K+ SO42- H2O 2H2O + Cu(s)→ Cu2++H2 +2OH- E0 = -0.75 v MTV = +0.75 v
Is Al a reactive or non-reactive metal? Look on page 8 Reactive as Al is a relatively strong reducing agent. Why is Al used for boats, patio furniture, swing sets, and trucks boxes? Al makes a clear transparent Al2O3 paint like coating that prevents further oxidation.