AP Chemistry: Iron Complex Solutions
Gabriella Posess
AP Chemistry- Period C
11/24/11
Iron Complex Lab
Purpose: Be able to tell the difference between iron (II) and iron (III) solutions by performing redox reactions between irons oxidation states.
Materials: test tube rack
6 test tubes- 25 x150 mm
25 mL graduated cylinder
6 stoppers
6 mL distilled water
6 mL Iron (III) chloride solution (0.02M FeCl3 x 6H2O)
6 mL Iron (II) sulfate solution (0.02M FeSO4 x 7H2O)
1 drop Potassium ferricyanide solution(0.1M K3Fe(CN)6)
1 drop Potassium ferrocyanide solution(0.1M K4Fe(CN)6 x H2O)
1 drop Potassium thiocyanate solution(0.1M KSCN)
Procedure:
Place 6 test tubes in a test tube rack. Label the tubes 1 through 6. Label tubes 1 through 3 with Fe 2+ and 4 through 6 as Fe 3+. …show more content…
Place the stoppers in the tubes and invert the tubes to mix.
Add 2 mL 0.02 M iron (III) chloride solution and 2 mL of distilled water to test tubes 4 through 6. Place the stoppers in the tubes and invert the tubes to mix.
Add 1/2 drop 0.1 M potassium ferrocyanide solution to tube 1. Both Irons are in the +2 state, so instead of forming a dark blue precipitate they form a light blue precipitate of potassium iron (II) hexacyanoferrate (II).
Add 1/2 drop 0.1 M potassium ferrocyanide solution to tube 4. A deep blue precipitate will form because of the presence of iron (II) and iron (III) ions. This precipitate is iron (III) hexacyanoferrate (II). As the Iron (II) in test tube 1 is slowly oxidized into iron (III), it will begin to turn darker blue.
Add 1/2 drop of 0.1 M potassium ferricyanide solution to tube 2. A deep blue precipitate will form with the iron (III) sulfate. In this reaction, the ferricyanide ions oxidize iron (II) to iron (III), forming ferrocyanide
AP Chemistry- Period C
11/24/11
Iron Complex Lab
Purpose: Be able to tell the difference between iron (II) and iron (III) solutions by performing redox reactions between irons oxidation states.
Materials: test tube rack
6 test tubes- 25 x150 mm
25 mL graduated cylinder
6 stoppers
6 mL distilled water
6 mL Iron (III) chloride solution (0.02M FeCl3 x 6H2O)
6 mL Iron (II) sulfate solution (0.02M FeSO4 x 7H2O)
1 drop Potassium ferricyanide solution(0.1M K3Fe(CN)6)
1 drop Potassium ferrocyanide solution(0.1M K4Fe(CN)6 x H2O)
1 drop Potassium thiocyanate solution(0.1M KSCN)
Procedure:
Place 6 test tubes in a test tube rack. Label the tubes 1 through 6. Label tubes 1 through 3 with Fe 2+ and 4 through 6 as Fe 3+. …show more content…
Place the stoppers in the tubes and invert the tubes to mix.
Add 2 mL 0.02 M iron (III) chloride solution and 2 mL of distilled water to test tubes 4 through 6. Place the stoppers in the tubes and invert the tubes to mix.
Add 1/2 drop 0.1 M potassium ferrocyanide solution to tube 1. Both Irons are in the +2 state, so instead of forming a dark blue precipitate they form a light blue precipitate of potassium iron (II) hexacyanoferrate (II).
Add 1/2 drop 0.1 M potassium ferrocyanide solution to tube 4. A deep blue precipitate will form because of the presence of iron (II) and iron (III) ions. This precipitate is iron (III) hexacyanoferrate (II). As the Iron (II) in test tube 1 is slowly oxidized into iron (III), it will begin to turn darker blue.
Add 1/2 drop of 0.1 M potassium ferricyanide solution to tube 2. A deep blue precipitate will form with the iron (III) sulfate. In this reaction, the ferricyanide ions oxidize iron (II) to iron (III), forming ferrocyanide