Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 11: 11 February

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Chemical Reactions, cont.

Ionic reactions - dissolving and precipitates, cont.

• Mix barium chloride and potassium sulfate:

Ba²⁺_(aq) + 2 Cl^-_(aq) + 2 K⁺_(aq) + SO₄^2-  BaSO_4(s) + 2 K⁺_(aq) + 2 Cl^-

Again, we want to write a net ionic equation showing only the ions which reacted:

Ba²⁺ + SO₄^2-  BaSO_4(s)

Notice that net ionic equations are very general expressions. Essentially they are saying that any time we have these species present they will react, regardless of what else happens to be there! (Sometimes folks are confused when they add ions which should react and they don't. This is usually a case where something else reacted first, so the ions of interest really weren't there!).

Since all ions in aqueous reactions are considered to be hydrated, we do not generally include _(aq) as part of the formula. But remember, it is assumed!

It is useful to remember some simple "rules" (really more like guide lines) to help in predicting reactions. For common compounds such as we see in general chemistry we can use the following rules:

1. Nitrates (NO₃^-) are all soluble.
2. Alkali metal (Li⁺, Na⁺, K⁺, Cs⁺, and Rb⁺) and ammonium (NH₄⁺) salts are all soluble, with the exception of a few Lithium salts.
3. Chloride, bromide, and iodide (Cl^-, Br^-, and I^-) salts are generally soluble, except for the salts of silver, lead(II) and mercury(I) (Ag⁺, Pb²⁺ and Hg₂²⁺).
4. Sulfates are soluble, except for the salts of barium {BaSO₄}, lead(II) {PbSO₄}, mercury(II) {HgSO₄}, and calcium {CaSO₄}.
5. Most hydroxides are only slightly soluble (but see rule 2).
6. Sulfides (S^2-), carbonates (CO₃^2-), phosphates (PO₄^3-), and chromates (CrO₄^2-) are only slightly soluble (but see rule 2).

(Additional examples of solving net-ionic equations can be found in the Discussion Module.)

Oxidation/Reduction (Redox) Reactions

In these reactions we see a transfer of electrons from one atom or molecule to another. First let's look at some terms.

• Oxidation refers to taking electrons away from a substance. So to oxidize means to behave like oxygen normally does and "steal" electrons.
• Reduction refers to recieving electrons - it is the opposite of oxidation.
• Note that oxidation and reduction always go together. When oxygen oxidizes it is itself reduced (it gains electrons).
  • Consider the example of burning natural gas:

CH₄ + 2 O₂ CO₂ + 2 H₂O

Notice that the methane is oxidized by the oxygen. We say that the carbon and hydrogen are both oxidized to give the new covalent products, water and carbon dioxide, because the electrons are not evenly shared, they are pulled toward the oxygens in each bond.

Examples:

• A piece of clean copper is placed in a solution of silver nitrate. Assuming the copper goes to Cu(II) write a balanced net ionic equation. Notice that in this case you must balance the charge in order to get the final equation.

Ag⁺ + NO₃^- + Cu⁰ Ag⁰ + NO₃^- + Cu²⁺

Balancing: 2Ag⁺ + Cu⁰ 2Ag⁰ + Cu²⁺ 

• Consider the reaction of calcium metal with hydrochloric acid. This is similar to the reaction of sodium and water, and hydrogen gas is given off. Write a net ionic equation for this reaction:

H⁺ + Cl^- + Ca⁰ H_{2 (g)} + Ca²⁺ + Cl^-

Balancing: 2 H⁺ + Ca⁰ H_{2 (g)} + Ca²⁺

• A piece of clean iron is placed in a solution of copper(II) sulfate. Assuming iron goes to Fe(III) write a balanced equation. Again the problem is to balance the charge.

Cu²⁺ + SO₄^2- + Fe⁰ Cu⁰ + Fe³⁺ + SO₄^2-

Balancing: 3 Cu²⁺ + 2 Fe⁰ 3 Cu⁰ + 2 Fe³⁺

(Additional examples of solving net-ionic equations can be found in the Discussion Module.)

Balancing Redox Equations

There are two common methods for balancing redox reactions: the oxidation number method and the half-reaction method. The half-reaction method works very well for ionic reactions, it is relatively easy to give partial credit, and it is the only method I will use in this class. If you know how to do the other method you are welcome to do so, but be careful to make sure you show your work or I won't be able to give partial credit!

The Half-Reaction Method

In the half-reaction method what we do is first break an equation into two parts and then balance the parts individually. Presented stepwise:

Acid Solution:

Separate the reaction into two half-reactions.

Balance each half-reaction separately:

Add two equations together

Cancel items appearing on both sides.

Example. Balance the following equation as it occurs in acid solution:

MnO₄^- + Cl^- Mn²⁺ + Cl₂

First break the equation into two half reactions, one for Mn and one for Cl

10 e^- + 16 H⁺ + 2 MnO₈^-+ 10 Cl^- 2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 10 e^-

16 H⁺ + 2 MnO₄^- + 10 Cl^- 2 Mn²⁺ + 8 H₂O + 5 Cl₂

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© R A Paselk

Last modified 11 February 2011