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SATII化学专题分析Redox and Electrochemistry

　　SAT考试是美国高中生进入美国大学的标准入学考试，包括SAT1和SAT2。SAT2是专项测验，有数学、物理、化学、生物、外语等，大部分为选择题，是可以选科的。今天我们来学习SATII化学专题分析Redox and Electrochemistry，希望对大家的复习有所帮助。

　　Oxidation-reduction (redox) reactions are another important type of reaction that you will see questions about on the SAT II Chemistry test. The test writers will expect you to be able to identify elements that are oxidized and reduced, know their oxidation numbers, identify half-cells, and balance redox reactions. The following is a brief overview of the basics.

　　Oxidation-Reduction

　　Oxidation-reduction reactions involve the transfer of electrons between substances. They take place simultaneously, which makes sense because if one substance loses electrons, another must gain them. Many of the reactions we’ve encountered thus far fall into this category. For example, all single-replacement reactions are redox reactions. Before we go on, let’s review some important terms you’ll need to be familiar with.

　　Electrochemistry: The study of the interchange of chemical and electrical energy.

　　Oxidation: The loss of electrons. Since electrons are negative, this will appear as an increase in the charge (e.g., Zn loses two electrons; its charge goes from 0 to +2). Metals are oxidized.

　　Oxidizing agent (OA): The species that is reduced and thus causes oxidation.

　　Reduction: The gain of electrons. When an element gains electrons, the charge on the element appears to decrease, so we say it has a reduction of charge (e.g., Cl gains one electron and goes from an oxidation number of 0 to -1). Nonmetals are reduced.

　　Reducing agent (RA): The species that is oxidized and thus causes reduction.

　　Oxidation number: The assigned charge on an atom. You’ve been using these numbers to balance formulas.

　　Half-reaction: An equation that shows either oxidation or reduction alone.

　　Rules for Assigning Oxidation States

　　A reaction is considered a redox reaction if the oxidation numbers of the elements in the reaction change in the course of the reaction. We can determine which elements undergo a change in oxidation state by keeping track of the oxidation numbers as the reaction progresses. You can use the following rules to assign oxidation states to the components of oxidation-reduction reactions:

　　The oxidation state of an element is zero, including all elemental forms of the elements (e.g., N2, P4, S8, O3).

　　The oxidation state of a monatomic ion is the same as its charge.

　　In compounds, fluorine is always assigned an oxidation state of -1.

　　Oxygen is usually assigned an oxidation state of -2 in its covalent compounds. Exceptions to this rule include peroxides (compounds containing the group), where each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H2O2).

　　Hydrogen is assigned an oxidation state of +1. Metal hydrides are an exception: in metal hydrides, H has an oxidation state of -1.

　　The sum of the oxidation states must be zero for an electrically neutral compound.

　　For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.

　　Now try applying these rules to a problem.

　　Example

　　Assign oxidation numbers to each element in the following:

　　H2S

　　MgF2

　　Explanation

　　The sum of the oxidation numbers in this compound must be zero since the compound has no net charge. H has an oxidation state of +1, and since there are two H atoms, +1 times 2 atoms = +2 total charge on H. The sulfur S must have a charge of -2 since there is only one atom of sulfur, and +2 - 2 = 0, which equals no charge.

　　F is assigned an oxidation state of -1 (according to rule 3), and there are two atoms of F, so this gives F a total charge of -2. Mg must have a +2 oxidation state since +2 - 2 = 0 and the compound is electrically neutral.

　　This time the net charge is equal to -3 (the charge of the polyatomic ion—according to rule 7). Oxygen is assigned a -2 oxidation state (rule 4). Multiply the oxidation number by its subscript: -24 = -8. Since there is only 1 phosphorus, just use those algebra skills: P + -8 = -3. Phosphorus must have a +5 charge.

　　Example

　　When powdered zinc metal is mixed with iodine crystals and a drop of water is added, the resulting reaction produces a great deal of energy. The mixture bursts into flames, and a purple smoke made up of I2 vapor is produced from the excess iodine. The equation for the reaction is

　　Zn(s) + I2(s)ZnI2(s) + energy

　　Identify the elements that are oxidized and reduced, and determine the oxidizing and reducing agents.

　　Explanation

　　Assign oxidation numbers to each species. Zn and I2 are both assigned values of 0 (rule 1). For zinc iodide, I has an oxidation number of -1 (group 7A—most common charge), which means that for zinc, the oxidation number is +2.

　　Evaluate the changes that are taking place. Zn goes from 0 to +2 (electrons are lost and Zn is oxidized). The half-reaction would look like this:

　　Zn0Zn2+ + 2e-

　　And I2 goes from 0 to -1 (it gains electrons and so is reduced). This half-reaction would look like this:

　　Here, zinc metal is the reducing agent—it causes the reduction to take place by donating electrons—while iodine solid is the oxidizing agent; iodine solid accepts electrons.

　　Voltaic (or Galvanic) Cells

　　Redox reactions release energy, and this energy can be used to do work if the reactions take place in a voltaic cell. In a voltaic cell (sometimes called a galvanic cell), the transfer of electrons occurs through an external pathway instead of directly between the two elements. The figure below shows a typical voltaic cell (this one contains the redox reaction between zinc and copper):

　　As you can see, the anode is the electrode at which oxidation occurs; you can remember this if you remember the phrase “an ox”—“oxidation occurs at the anode.” Reduction takes place at the cathode, and you can remember this with the phrase “red cat”—“reduction occurs at the cathode.” An important component of the voltaic cell is the salt bridge, which is a device used to maintain electrical neutrality; it may be filled with agar, which contains a neutral salt, or be replaced with a porous cup. Remember that electron flow always occurs from anode to cathode, through the wire that connects the two half-cells, and a voltmeter is used to measure the cell potential in volts.

　　Batteries are cells that are connected in series; the potentials add to give a total voltage. One common example is the lead storage battery (car battery), which has a Pb anode, a PbO2 cathode, and H2SO4 electrolyte is their salt bridge.

　　Standard Reduction Potentials

　　The potential of a voltaic cell as a whole will depend on the half-cells that are involved. Each half-cell has a known potential, called its standard reduction potential (Eº). The cell potential is a measure of the difference between the two electrode potentials, and the potential at each electrode is calculated as the potential for reduction at the electrode. That’s why they’re standard reduction potentials, not standard oxidation potentials. Here is the chart:

　　On this reduction potential chart, the elements that have the most positive reduction potentials are easily reduced and would be good oxidizing agents (in general, the nonmetals), while the elements that have the least positive reduction potentials are easily oxidized and would be good reducing agents (in general, metals). Let’s try a quick problem.

　　Example

　　Which of the following elements would be most easily oxidized: Ca, Cu, Fe, Li, or Au?

　　Explanation

　　Use the reduction potential chart: nonmetals are at the top and are most easily reduced. Metals are at the bottom and are most easily oxidized. Lithium is at the bottom of the chart—it’s the most easily oxidized of all. So the order, from most easily oxidized to least easily oxidized, is Au, Fe, Cu, Ca, Li.

　　Example

　　Which one of the following would be the best oxidizing agent: Ba, Na, Cl, F, or Br?

　　Explanation

　　Using the reduction potential chart and the fact that oxidizing agents are the elements that are most easily reduced, we determine fluorine is the best oxidizing agent.

　　Electrolytic Cells

　　While voltaic cells harness the energy from redox reactions, electrolytic cells can be used to drive nonspontaneous redox reactions, which are also called electrolysis reactions. Electrolytic cells are used to produce pure forms of an element; for example, they’re used to separate ores, in electroplating metals (such as applying gold to a less expensive metal), and to charge batteries (such as car batteries). These types of cells rely on a battery or any DC source—in other words, whereas the voltaic cell is a battery, the electrolytic cell needs a battery. Also unlike voltaic cells, which are made up of two containers, electrolytic cells have just one container. However, like in voltaic cells, in electrolytic cells electrons still flow from the anode to the cathode. An electrolytic cell is shown below.