Chemical Equilibria Essay Research Paper Darryl FaganPd

Chemical Equilibria Essay, Research Paper

Darryl Fagan

Pd. 4 01-20-00

AP Chemistry

An Activity Series

Chemistry Lab 5

Introduction;

In this experiment you will study some metals and some nonmetals to find their relative reactivity. A ranking according to reactivity is called an activity series. For example, an activity series containing the elements calcium, gold and iron would put the reactive calcium at the top, iron in the middle, and the unreactive gold at the bottom. If a piece of iron metal is placed in a solution of gold nitrate, the iron will dissolve forming positive ions while the solid gold metal appears. The more reactive metal will displace ions of the less reactive metal from solution. The less reactive element will appear as the solid element.

Purpose;

To determine an activity series for metals using a microscale technique, and to determine an activity series for halogens using a solvent extraction technique.

Procedure;

a. Determine an activity series for some metals.

Place a 24-well plate so that there are 6 wells across and 4 wells down.

Place 1 dropper full of copper (II) nitrate solution in rows 2 through 4 in the first column. Put one dropper full of magnesium nitrate in wells 1, 3, and 4 of the second column. Place 1 dropper full of lead (II) nitrate in wells 1, 2, and4 of the third column. Put one dropper full of zinc nitrate in wells 1, 2, and 3 of the fourth column, and put 1 dropper full of silver nitrate on each of the 4 wells in the fifth column.

Put a small piece of copper metal in each of the wells containing a solution in the first row. Add magnesium metal to each of the solutions in the second row, lead to the third, and zinc to the fourth row. Allow to stand at least 5 minutes. Determine if a reaction has occurred in each well by observing if a metal precipitate has formed or if the surface of the metal has become coated. If a metal ion is reduced by a metal, than the reserve reaction should not occur. One metal is more reactive than another if the metal will replace the metal ion in its compounds. Record your data.

Dispose of all materials by the suggested manor of Methods #26a and #26b.

b. Determine the activity series for some halogens.

In this part you will test to see if the halogens Cl , Br ,and I ,can be reduced by the halide ion Cl , Br , and I . To determine what products will be formed, you will need to have a test, which will tell which halogen is present, Halogens dissolve in the nonpolar solvent mineral oil forming different color solutions. Mineral oil does not dissolve in water, but when shaken with an aqueous halogen solution, the halogen is extracted from the water into the mineral oil. The color of the mineral oil layer indicates which halogen is present.

First, test to see what color each halogen shows in mineral oil. Place 1 dropper full of chlorine water, 1 dropper of Bromine water, and 1 dropper of iodine water into separate 10mm test tubes. Add one dropper of mineral oil to each, cork the tube and shake it for ten seconds. Let the mineral oil layer separate and record the color that each halogen shows when dissolved in mineral oil.

Test to see if the halide ions give a color to mineral oil. Place 1 dropper of NaCl, KI, and NaBr solutions into different test tubes, add a dropper of mineral oil to each, cork and shake to determine if the halide ions impart a color to the mineral oil layer. Report your data.

React each halogen with the other two halide ions to see if a halide ion can reduce other halogens. Place 1 dropper of NaBr solution into one test tube and 1 dropper of KI solution into a second test tube. Add 1 dropper of chlorine water to each, cork and shake to mix. Now add 1 dropper of mineral oil, cork and shake again. When the mineral oil layer has separated, record its color and determine if a reaction has occurred. Record your data. If chlorine appears no reaction has occurred, if either bromine or iodine appears there was a reaction.

Repeat the test using bromine water mixed with NaCl and KI solutions, and iodine water mixed with CaCl and NaBr solutions.

Materials;

24-well plate

Forceps

Test tubes, 13 X 100mm

Test tube rack

Cork stoppers for test tubes

Copper Copper (II) nitrate, 0.1 M

Zinc Zinc nitrate, 0.1 M

Magnesium Magnesium nitrate, 0.1 M

Lead Lead (II), 0.1 M

Silver nitrate, 0.1 M

Chlorine water Sodium chloride, 0.1 M

Bromine water Sodium bromide, 0.1 M

Iodine water Potassium iodide, 0.1 M

Mineral oil

Method;

In the three parts of the lab, you will be using two methods, for the first section you will be using a microscale technique. For the remaining two experiments you will be using a solvent extraction technique.

Data tables;

Cu (aq) Mg (aq) Pb (aq) Zn (aq) Ag (aq)

Cu (s) X No No No Yes

Mg (s) Yes X Yes Yes Yes

Pb (s) Yes Yes X Yes Yes

Zn (s) Yes No Yes X Yes

Cl (aq) Br (aq) I (aq)

Cl (aq) X Clear-Yellow Clear

Br (aq) Clear-Yellow X Clear-Yellow

I (aq) Clear Pink-Clear X

Mineral Oil Yellow-Clear Clear Pink-Purple

NaCl KI NaBr

Br (aq) Clear Clear X

I (aq) Clear X Clear

Conclusion;

I believe that this experiment though very rough in nature has completed the purpose of administering the methods of determining activity series in metals through a microscale technique, and determining an activity series for halogens using a solvent extraction technique. All reactions occurred as predetermined by their placement on a textbook reactivity series, and no errors occurred in the collection of data. The experiment contained no possibility of error, the reactions either work or do not work.

Discussion;

1.) Write balanced net ionic equations for all reactions that occurred with metals.

Cu (s) + 2Ag (aq)  Cu (aq) + 2Ag (s)

Mg (s) + Cu (aq)  Mg (aq) + Cu (s)

Mg (s) + Pb (aq)  Mg (aq) + Pb (s)

Mg (s) + Zn (aq)  Mg (aq) + Zn (s)

Mg (s) + 2Ag (aq)  Mg (aq) + 2Ag (s)

Pb (s) + Cu (aq)  Pb (aq) + Cu (s)

Pb (s) + Mg (aq)  Pb (aq) + Mg (s)

Pb (s) + Zn (aq)  Pb (aq) + Zn (s)

Pb (s) + 2Ag (aq)  Pb (aq) + 2Ag (s)

Zn (s) + Cu (aq)  Zn (aq) + Cu (s)

Zn (s) + Pb (aq)  Zn (aq) + Pb (s)

Zn (s) + 2Ag (aq)  Zn (aq) + 2Ag (s)

2.) List the metals in order of decreasing ease of oxidation. Compare your list with the activity series found in the textbook. How do the two lists correlate?

Test Results Text Book

Magnesium Magnesium

Lead Lead

Zinc Zinc

Copper Copper

Silver Silver

 The two results correlate precisely as they should.

3.) Write half-reactions for each of the meal ions. Arrange the reaction list in order of decreasing ease of reduction. Compare your listing with a listing found in a table of standard reduction potentials. How do the two lists correlate?

Cu + 2Ag  Cu + 2Ag

Cu  Cu + 2e- 2Ag + 2e-  2Ag

Mg + Cu  Mg + Cu

Mg  Mg + 2e- Cu + 2e-  Cu

Mg + Pb  Mg + Pb

Mg  Mg +2e- Pb + 2e-  Pb

Mg + Zn  Mg + Zn

Mg  Mg + 2e- Zn + 2e-  Zn

Mg + 2Ag  Mg + 2Ag

Mg  Mg + 2e- 2Ag + 2e-  2Ag

Pb + Cu  Pb + Cu

Pb  Pb + 2e- Cu +2e-  Cu

Pb + Mg  Pb + Mg

Pb  Pb + 2e- Mg + 2e-  Mg

Pb + Zn  Pb + Zn

Pb  Pb + 2e- Zn + 2e-  Zn

Pb + 2Ag  Pb + 2Ag

Pb  Pb + 2e- 2Ag + 2e-  2Ag

Zn + Cu  Zn + Cu

Zn  Zn + 2e- Cu + 2e-  Cu

Zn + Pb  Zn + Pb

Zn  Zn + 2e- Pb + 2e-  Pb

Zn + 2Ag  Zn + 2Ag

Zn  Zn + 2e- 2Ag + 2e-  2Ag

A; order of occurrence according to data

B ; order of occurrence according to textbook

4.) Explain what is meant by solvent extraction.

Solvent extraction is the process in which two substances integrated with different substances that can not combine are introduced and the ensuing reaction occurs with no interference from the two solvents. In this case water and mineral oil were the two substances, these substances readily separate when introduced and therefore are ideal candidates for this technique.

5.) How can you tell if a reaction occurs in the halogen experiment?

When a reaction occurs in the halogen experiment there are loose ions floating in the solution. Different substances give off different color ions, and the base solvent do not have an interchange of ions, therefore if there is no color to the solution, no reaction may have occurred because of the lack of change in color. If the fluid mixture however does change in color, there was an exchange of electrons, and the now colorful ions are present in the solution.

6.) Why should you not expect the halide ion to dissolve in mineral oil?

Mineral oil is a highly neutral substance and reacts with nothing, it will also not mix with other fluids making it ideal for the solvent extraction technique used in determining the activity series for the halogens.

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