Part C: Single Displacement Reactions

1. Activity series (or reduction potential) of some common metals. Sometimes, when you put a metal into a solution containing the ions of another metal, the two metals change places. The metal ions in solution precipitate out to form the elemental form of that metal (a reduction occurs), and the elemental metal becomes ionized and enters the solution (an oxidation occurs).

   We say that a metal is "reduced" when its charge decreases, in this case from a positively-charged ion to the elemental metal which has a zero charge. Conversely, a metal is oxidized when its charge increases, in this case from the zero charge of an elemental metal to the positive charge of the metal ion.

   The metal that is being reduced (gaining electrons) is called an oxidizing agent because it is taking the electrons away from the other metal, thus forcing the other metal to be oxidized.

   The metal that is being oxidized (losing electrons) is called a reducing agent because it is giving its electrons to the other metal, thus forcing the other metal to be reduced.

   Metals that are "more easily reduced" will more likely be converted from the ion form to the elemental metal. The ions are more likely to gain electrons and are said to have a "higher reduction potential." The "activity" of a metal is its tendency to become an ion (to be ionized). So the metals with higher reduction potentials (more easily reduced) are less active.

   The more active metals are more likely to be oxidized from the elemental form to the ion when in the presence of a less active metal. If it's already in ionic form, then it won't change because that's how it would rather exist, as an ion.

   The metal that ends up as an element is less active, because it was reduced (gaining electrons) by the metal that was being oxidized (losing electrons). Confusing? It really isn't. Remember, one of the metals is gaining electrons (being reduced) and the other loses them, becoming the ion (being oxidized). The metal that ends up with the electrons is the less active metal because it has a higher reduction potential (more easily reduced).

   IN YOUR REPORT, PLEASE WRITE THE CHEMICAL EQUATION WHETHER IT OCCURS OR NOT. WRITE "N.R." IN PARENTHESES NEXT TO THE EQUATION IF THE REACTION DOES NOT OCCUR.

   For more information on the activity series of metals, look at Table 8.2 and Section 12.5 in your textbook.

2. Reactivity of halogens. This part of the lab is often the hardest for students to understand. It involves the oxidation and reduction of three halogens (Cl, Br, and I). Among the nonmetals, it is a general tendency for the elements closer to fluorine to attract electrons more strongly than elements that are farther away on the periodic table. Therefore, if a halide ion (X^-) and halogen (X₂) are put into an aqueous solution together, a transfer of electrons could take place from the halide to the halogen, thus oxidizing the halide ion to a halogen and reducing the halogen to a halide ion. This will only happen if the element closer to fluorine on the periodic table starts out as the halogen; it prefers to exist as the halide ion and will capture the electrons from the other element. If the element that is closer to fluorine in the periodic table starts out as a halide (X^-), then it will be happy to stay that way, keeping its extra electrons.

   You'll be able to tell if the reaction happens or not by an ingenious method. You'll add an organic solvent into which only the halogens (X₂) can dissolve. The organic solvent will form a separate layer and float on top of the aqueous layer. When you mix the two layers, the halogen will dissolve into the organic layer and the halide ion (X^-) will stay behind in the aqueous layer. The key is that each halogen (Cl₂, Br₂, and I₂) will show up as a distinct color in the organic layer.

   At first, you'll mix a known solution of a single halogen with the organic solvent and find out what color corresponds to each halogen. Then you'll mix various halogens an halides and determine which element exists as the halogen (X₂) by observing the color of the organic layer. The other element will exist as the halide ion (X^-) and will not show up in the organic layer. See if the theory holds true: that the halogen closest to fluorine in the periodic table would prefer to exist as the halide ion, forcing the lower elements to exist as the neutral halogen (X₂) molecules.

   IN YOUR REPORT, PLEASE WRITE THE CHEMICAL EQUATION WHETHER IT OCCURS OR NOT. WRITE "N.R." IN PARENTHESES NEXT TO THE EQUATION IF THE REACTION DOES NOT OCCUR.

   (Note: On page 68 of your lab manual, please change the word "reduce" to "oxidize.")

   For more information on the oxidation and reduction of halogens, refer to Table 8.2 and example (d) under Single Displacement Reactions in your book.

Part D: Double Displacement Reactions

Solubility.This part of the experiment is fun. You get to mix various ionic compounds or acids and watch to see if a precipitate (solid) is formed by the combinations of ions in the mixture. You can look in solubility tables to find out whether or not you expect a precipitate to form. Each reaction will be a double displacement reaction, in which the anions will change from one cation to the other.

IN YOUR REPORT, PLEASE WRITE THE CHEMICAL EQUATION WHETHER IT OCCURS OR NOT. WRITE "N.R." IN PARENTHESES NEXT TO THE EQUATION IF THE REACTION DOES NOT OCCUR.

For more information on solubility, refer to Appendix IV, Section 16.12, and Table 16.4 in your textbook.