SAT Chemistry Oxidation Reduction - Oxidation States
SAT Chemistry Oxidation Reduction - Oxidation StatesBecause the transfer of electrons in chemical reactions is so prevalent, chemists have developed a manner to keep track of their movement. It’s important to note that for processes involving ionic species, of the type we have looked at so far, it is somewhat straightforward to distinguish redox from non-redox: The sign and magnitude of the charge on the ions involved allows for an easy view of what is taking place.
Using the spontaneous reaction between zinc and iron(III) chloride referenced in the previous section, the zinc started as an atom with no net charge and turned into an ion with a 2+ charge. During chemical reactions, in order for the charge on particles to increase, negative electrons have to be lost and the process of oxidation occurs. A similar analysis of the iron in the reaction shows that the iron began as a 3+ ion (ferric) and ended as a neutral iron atom in the products. In other words, the iron went down in charge. Particles that go down in charge do so by gaining negative electrons. This description gives insight to why the process is referred to as reduction despite the fact that electrons are being gained. Reduction refers to the change in charge involving ionic species, not to the change in the number of electrons possessed by a particle as a result of the process taking place. Since reduction involves a decrease in charge, oxidation must involve an increase in charge (when ionic species are involved). Using the term oxidation may seem like a strange way to describe this process, but its use will be better understood from a historical context when combustion reactions are looked at as redox processes in an upcoming section of this chapter.
As discussed, reactions involving ions are relatively simple to recognize as redox (or not) because it is straightforward to identify changes in charges if they occur. It’s not quite so simple when redox reactions involve molecular species. For these types of reactions, as well as those involving ionic species, a related but noteworthily different term is used to describe the responsibility particles have for the electrons around them as a chemical reaction unfolds. To keep track of the transfer of electrons in all formulas (ionic or molecular), chemists have devised a system of electron bookkeeping called oxidation states (or oxidation numbers.) In this system, an oxidation state is assigned to each member of a formula using rules that recognize the degree to which electrons practically belong to a particular element in a given substance from an ionic bonding perspective. Basically, elements with high electronegativities are given responsibility for the electrons in a bond, ionic or covalent, and the change in this responsibility will be noted by changes in their oxidation states. In this regard, the oxidation states system assumes an ionic perspective for all bonding, i.e., electrons are not shared but belong to one element ar the other in a chemical bond. Although we know that electrons are shared in a large number of chemical bonds, particularly those described as covalent (or molecular), assigning responsibility for the electrons in this way allows for easy recognition of how the accountability for electrons changes during all chemical processes.
Oxidation states are designated by a small number superscript preceded by a plus or minus sign. This is not to be confused with the ionic charges we have been using thus far that are shown as plus or minus signs to
the right of the magnitude of ionic charge as a superscript.
The Rules for Assigning an Oxidation State
The basic rules for assigning an oxidation state to an element in a substance’s formula are given below. By applying these simple rules, you can assign oxidation states to elements in practically all substances you may encounter as a beginning chemistry student. To apply these rules, remember that 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 be equal to the charge on the ion.
- The oxidation state of an element of an atom in an element is zero. Examples: O for Na(s), O2(g), and Hg(l)
- The oxidation state of a monatomic ion is the same as its charge. Examples: +1 for Na+ and -1 for Cl-.
- The oxidation state for fluorine is -1 in its compounds. Examples: HF, hydrogen fluoride, has one H at +1 and one F at -1. PF3 has one P at +3 and three F’s at -1 each. (Note that in each compound the sum of the oxidation states is equal to zero.)
- The oxidation state of oxygen is usually -2 in its compounds. Example: H2O has two H’s at +1 each and one O at -2. (Exceptions occur when the oxygen is bonded to fluorine and the oxidation state of fluorine takes precedence, and in peroxide compounds where the oxidation state is assigned the value of -1.)
- The oxidation state of hydrogen in most compounds is +1. Examples: H2O, HCl, and NH3. (In hydrides, where hydrogen acts like an anion compounded with a metal, there is an exception, however. In this case, hydrogen is assigned the value of -1. Examples: LiH and KH.)
Example 1 __________________________________________
In Na2SO4, what is the oxidation state of sulfur?
The first thing to recognize is that this compound is an ionic substance. Ionic substances have two parts—the cation and the anion. In this case, the cation is monatomic and the anion is polyatomic. The cation is Na+ and the anion is SO042- (sulfate.) By Rule #2, the oxidation state of the sodium is +1 because the oxidation states of monatomic ions are the same as their charges. By Rule #4, the oxidation state of the oxygen in the sulfate is -2. Now you can look at the complete formula and calculate the oxidation state of the sulfur.
Since the positive sum and the negative sum must equal 0,
(+2) + x + (-8) = 0
The sulfur must have a +6 oxidation state.
Example 2 _______________________________________________
What is the oxidation state of carbon in the molecule CO2?
By Rule #4, the oxidation state of oxygen is -2, and since there are two oxygen atoms in this formula the total negative sum is -4. Since the positive and negative sums must add to zero, the oxidation state of carbon is +4 in this compound. Keep in mind that oxidation states are not “real" charges and carbon dioxide is not an ionic substance. In this case, the oxidation state of +4 for carbon indicates that for electron bookkeeping purposes, carbon will not be responsible for the electrons it is sharing in the bond between the carbon and oxygen atoms and the responsibility will lie with the oxygen. When CO2 is involved in a chemical process and carbon’s responsibility for electrons changes (i.e., its oxidation state changes), a redox reaction will be recognized.
Example 3 _______________________________________________
In the polyatomic ion dichromate ('Cr2O72-), what is the oxidation state of chromium?
In a polyatomic ion, the algebraic sum of the positive and negative oxidation states of all the elements must equal the charge on the ion.
Cr = 2 x (x) = 2x
O = 7 x (-2) = -14
Since the sum of these values must equal -2 (the charge on the polyatomic ion)
2x + (-14) = -2
x = +6
The oxidation state of chromium in dicnromate is +6.
Using Oxidation States to Recognize Redox Reactions
Once oxidation states can be assigned to elements in the substances involved in a chemical reaction, recognition of the process as being redox or not is straightforward. If the oxidation states change, then a transfer of electrons is taking place. If the oxidation states remain the same, then a redox reaction is not occurring.
Consider these two reactions:
Na2S(aq) + CuSO4 (aq) → 4 Na2SO4(aq) + CuS(s)
2 NaCO + Cl2(g) → 2NaCl(s)
The oxidation states of each element in all the substances can be determined as shown above each of them here:
Because the oxidation states of the elements in the first reaction don’t change, this reaction is not a redox process. You should recognize this as a precipitation reaction, as described in Chapter 8. The second reaction does exhibit a change in oxidation states and should be viewed as a redox reaction. In this reaction the sodium is being oxidized and the chlorine is being reduced. The chlorine could not be reduced (i.e., gain electrons) if the sodium wasn’t being oxidized (i.e., losing electrons.) In one regard, the sodium is acting as a facilitator for the reduction of the chlorine. As such, it is referred to as a reducing agent. Likewise, the sodium would not lose its electron if it had nowhere to go, so the chlorine is referred to as the oxidizing agent in this reaction. Oxidizing agents, then, contain elements that are capable of being reduced by other substances (reducing agents) that contain elements that are capable of being oxidized. Sometimes the terms oxidizer and reducer are used as labels on the bottles of substances with the tendency to act as oxidizing and reducing agents, respectively.
Sample Problem 1 __________________________________________
Identify the elements that are being oxidized and reduced in the following reaction. Also, name the oxidizing and reducing agents.
2H2O(ℓ ) + 2MnO4-(aq) + I-(aq) → 2MnO2(aq) + IO3-(aq) + 2OH-(aq)
The oxidation states of the hydrogen and oxygen are not changing in the reaction. The oxidation states of the manganese and the iodine are changing, as shown below.
Since the manganese is changing from an oxidation state of +7 to that of +4, the manganese is being reduced. The iodine is being oxidized from a value of -1 to a value of +5. Manganese is gaining electron responsibility while iodine is losing it. In that light, then, the permanganate ion (MnO4-), which contains the manganese, is facilitating the process in which iodine is being oxidized and so is referred to as the oxidizing agent. A typical source of the permanganate ion in chemical reactions like this one is from the compound potassium permanganate, KMn04, whose bottle is generally labeled with the term “oxidizer.” Along a line of similar thinking, the iodide ion, I-, can be called the reducing agent in this reaction.Sample Problem 2 ____________________________________________
Completely analyze the following reaction from a redox perspective.
3CO(g) + Fe2O3(s) → 3CO2(g) + 2Fe(s)
The change in the oxidation states of the elements is shown.
Since the oxidation state of carbon is increasing from +2 to +4, it is being oxidized, its responsibility for electrons is decreasing, and the compound of which it is a part, carbon monoxide, is called the reducing agent. Carbon monoxide is a common reducer because of its chemical desire to turn into carbon dioxide when oxygen becomes available from other substances. On the other hand, the oxidation state of iron is decreasing from +3 to 0, it is being reduced, its responsibility for electrons is going up, and the compound of which it is a part, iron(III) oxide, is called the oxidizing agent.
Combustion reactions are those chemical processes in which substances (called fuels) are rapidly oxidized, accompanied by the release of heat and usually light. Combustion is also referred to as burning. Typical, common, and historically well-known combustion reactions involve using oxygen as the oxidizing agent—hence, the name for the process of losing electrons has became known as oxidation. When a combustion reaction is complete, the elements in the burning fuel form compounds with the oxidizing agent and the responsibility for electrons changes.
The reaction between magnesium and oxygen is a common example of combustion. When the reaction occurs
2Mg(s) + O2(g) → 2MgO(s)
the oxidation states of magnesium and oxygen change from 0 each to +2 and -2, respectively. A blinding light and large quantity of heat is released by the system as the reaction unfolds. The amount of heat released when 1 mole of a fuel burns is referred to as its heat of combustion and is symbolized ΔHc. The ΔHc for Mg is 602 kj/mol. The change in enthalpy associated with the burning of 1 mole of carbon
C(s) + O2(g) → CO2(g) + 393.5 kj
is -393.5 kj, also written as ΔHc = -393.5 kj, and it represents the heat released in another combustion/redox reaction. The heats of combustion for many common substances can easily be found in reference tables like the kind found in Appendix B of this book.
Hydrocarbon Fuel Combustion Reactions
Another common combustion reaction involves the burning of hydrocarbon fuels. Methane, CH4, is a typical hydrocarbon fuel. It is the main component of natural gas. Analysis of the reaction
shows that carbon is being oxidized and the oxygen is being reduced. This fuel is used to heat homes and cook food and should be familiar to all chemistry students as the gas used in their laboratory burners.
Example 1 _______________________________________________
Describe the reaction that occurs when propane, C3H8 (the hydrocarbon fuel used in backyard barbecues) combusts, releasing 2,221 kj when one mole is burned.
It should be noted that when hydrocarbon fuels combust completely in the presence of oxygen, the products of the reaction are carbon dioxide and water. Therefore, the reaction is:
Once again, the carbon is oxidized from the fraction -8/3 to +4 (increasing from negative to positive) and the oxygen is reduced from 0 to -2. Oxygen is the oxidizing agent and propane is the reducing agent. When this reaction occurs light and heat are released and the heat of combustion, ΔHc, is -2221 kj.
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