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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 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-

 

Pick the spontaneous reaction

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

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 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

You must look at the possible oxidation of:

SO42-

H2O Cu

Strongest Reducing Agent

-

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

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.