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 Oxidation - Reduction Reactions

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كاتب الموضوعرسالة
امير الجوالة
عضو هام
عضو هام
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عدد المساهمات : 113
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تاريخ التسجيل : 17/11/2009

مُساهمةموضوع: Oxidation - Reduction Reactions   الأحد مارس 14, 2010 11:28 am

The Process of Discovery: Oxidation and Reduction

The first step toward a theory of chemical reactions was taken by Georg Ernst Stahl in 1697 when he proposed the phlogiston theory, which was based on the following observations.
Metals have many properties in common.
Metals often produce a "calx" when heated. (The term calx is defined as the crumbly residue left after a mineral or metal is roasted.)
These calxes are not as dense as the metals from which they are produced.
Some of these calxes form metals when heated with charcoal.
With only a few exceptions, the calx is found in nature, not the metal.
These observations led Stahl to the following conclusions.
Phlogiston (from the Greek phlogistos, "to burn") is given off whenever something burns.
Wood and charcoal are particularly rich in phlogiston because they leave very little ash when they burn. (Candles must be almost pure phlogiston because they leave no ash.)
Because they are found in nature, calxes must be simpler than metals.
Metals form a calx by giving off phlogiston.
Metal <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> calx + phlogiston
Metals can be made by adding phlogiston to the calx.
Calx + phlogiston <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> metal
Because charcoal is rich in phlogiston, heating calxes in the presence of charcoal sometimes produces metals.
This model was remarkably successful. It explained why metals have similar properties they all contained phlogiston. It explained the relationship between metals and their calxesthey were related by the gain or loss of phlogiston. It even explained why a candle goes out when placed in a bell jar the air eventually becomes saturated with phlogiston.
There was only one problem with the phlogiston theory. As early as 1630, Jean Rey noted that tin gains weight when it forms a calx. (The calx is about 25% heavier than the metal.) From our point of view, this seems to be a fatal flaw: If phlogiston is given off when a metal forms a calx, why does the calx weigh more than the metal? This observation didn't bother proponents of the phlogiston theory. Stahl explained it by suggesting that the weight increased because air entered the metal to fill the vacuum left after the phlogiston escaped.
The phlogiston theory was the basis for research in chemistry for most of the 18th century. It was not until 1772 that Antoine Lavoisier noted that nonmetals gain large amounts of weight when burned in air. (The weight of phosphorus, for example, increases by a factor of about 2.3.) The magnitude of this change led Lavoisier to conclude that phosphorus must combine with something in air when it burns. This conclusion was reinforced by the observation that the volume of air decreases by a factor of 1/5th when phosphorus burns in a limited amount of air.
Lavoisier proposed the name oxygene (literally, "acid-former") for the substance absorbed from air when a compound burns. He chose this name because the products of the combustion of nonmetals such as phosphorus are acids when they dissolve in water.

P4(s) + 5 O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> P4O10(s) P4O10(s) + 6 H2O(l) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 4 H3PO4(aq)
Lavoisier's oxygen theory of combustion was eventually accepted and chemists began to describe any reaction between an element or compound and oxygen as oxidation. The reaction between magnesium metal and oxygen, for example, involves the oxidation of magnesium.

2 Mg(s) + O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 MgO(s)
By the turn of the 20th century, it seemed that all oxidation reactions had one thing in common oxidation always seemed to involve the loss of electrons. Chemists therefore developed a model for these reactions that focused on the transfer of electrons. Magnesium metal, for example, was thought to lose electrons to form Mg2+ ions when it reacted with oxygen. By convention, the element or compound that gained these electrons was said to undergo reduction. In this case, O2 molecules were said to be reduced to form O2- ions.


A classic demonstration of oxidation-reduction reactions involves placing a piece of copper wire into an aqueous solution of the Ag+ ion. The reaction involves the net transfer of electrons from copper metal to Ag+ ions to produce whiskers of silver metal that grow out from the copper wire and Cu2+ ions.

Cu(s) + 2 Ag+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> Cu2+(aq) + 2 Ag(s)
The Cu2+ ions formed in this reaction are responsible for the light-blue color of the solution. Their presence can be confirmed by adding ammonia to this solution to form the deep-blue Cu(NH3)42+ complex ion.
Chemists eventually recognized that oxidation-reduction reactions don't always involve the transfer of electrons. There is no change in the number of valence electrons on any of the atoms when CO2 reacts with H2, for example,

CO2(g) + H2(g) CO(g) + H2O(g)
as shown by the following Lewis structures:

Chemists therefore developed the concept of oxidation number to extend the idea of oxidation and reduction to reactions in which electrons are not really gained or lost. The most powerful model of oxidation-reduction reactions is based on the following definitions.
Oxidation involves an increase in the oxidation number of an atom.
Reduction occurs when the oxidation number of an atom decreases.
According to this model, CO2 is reduced when it reacts with hydrogen because the oxidation number of the carbon decreases from +4 to +2. Hydrogen is oxidized in this reaction because its oxidation number increases from 0 to +1.




Oxidation-Reduction Reactions
We find examples of oxidation-reduction or redox reactions almost every time we analyze the reactions used as sources of either heat or work. When natural gas burns, for example, an oxidation-reduction reaction occurs that releases more than 800 kJ/mol of energy.

CH4(g) + 2 O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> CO2(g) + 2 H2O(g)
Within our bodies, a sequence of oxidation-reduction reactions are used to burn sugars, such as glucose (C6H12O6) and the fatty acids in the fats we eat.

C6H12O6(aq) + 6 O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>6 CO2(g) + 6 H2O(l) CH3(CH2)16CO2H(aq) + 26 O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 18 CO2(g) + 18 H2O(l)
We don't have to restrict ourselves to reactions that can be used as a source of energy, however, to find examples of oxidation-reduction reactions. Silver metal, for example, is oxidized when it comes in contact with trace quantities of H2S or SO2 in the atmosphere or foods, such as eggs, that are rich in sulfur compounds.

4 Ag(s) + 2 H2S(g) + O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Ag2S(s) + 2 H2O(g)
Fortunately, the film of Ag2S that collects on the metal surface forms a protective coating that slows down further oxidation of the silver metal.
The tarnishing of silver is just one example of a broad class of oxidation-reduction reactions that fall under the general heading of corrosion. Another example is the series of reactions that occur when iron or steel rusts. When heated, iron reacts with oxygen to form a mixture of iron(II) and iron(III) oxides.

2 Fe(s) + O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 FeO(s) 2 Fe(s) + 3 O2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Fe2O3(s)
Molten iron even reacts with water to form an aqueous solution of Fe2+ ions and H2 gas.

Fe(l) + 2 H2O(l) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> Fe2+(aq) + 2 OH-(aq) + H2(g)
At room temperature, however, all three of these reactions are so slow they can be ignored.
Iron only corrodes at room temperature in the presence of both oxygen and water. In the course of this reaction, the iron is oxidized to give a hydrated form of iron(II) oxide.

2 Fe(s) + O2(aq) + 2 H2O(l) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 FeO H2O(s)
Because this compound has the same empirical formula as Fe(OH)2, it is often mistakenly called iron(II), or ferrous, hydroxide. The FeO H2O formed in this reaction is further oxidized by O2 dissolved in water to give a hydrated form of iron(III), or ferric, oxide.

4 FeO H2O(s) + O2(aq) + 2 H2O(l) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Fe2O3 3 H2O(s)
To further complicate matters, FeO H2O formed at the metal surface combines with Fe2O3 3 H2O to give a hydrated form of magnetic iron oxide (Fe3O4).

FeO H2O(s) + Fe2O3 <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 3 H2O(s) Fe3O4 n H2O(s)
Because these reactions only occur in the presence of both water and oxygen, cars tend to rust where water collects. Furthermore, because the simplest way of preventing iron from rusting is to coat the metal so that it doesn't come in contact with water, cars were originally painted for only one reason to slow down the formation of rust.



Assigning Oxidation Numbers
The key to identifying oxidation-reduction reactions is recognizing when a chemical reaction leads to a change in the oxidation number of one or more atoms. It is therefore a good idea to take another look at the rules for assigning oxidation numbers. By definition, the oxidation number of an atom is equal to the charge that would be present on the atom if the compound was composed of ions. If we assume that CH4 contains C4- and H+ ions, for example, the oxidation numbers of the carbon and hydrogen atoms would be -4 and +1.
Note that it doesn't matter whether the compound actually contains ions. The oxidation number is the charge an atom would have if the compound was ionic. The concept of oxidation number is nothing more than a bookkeeping system used to keep track of electrons in chemical reactions. This system is based on a series of rules, summarized in the table below.

Rules for Assigning Oxidation Numbers
The oxidation number of an atom is zero in a neutral substance that contains atoms of only one element. Thus, the atoms in O2, O3, P4, S8, and aluminum metal all have an oxidation number of 0.
The oxidation number of monatomic ions is equal to the charge on the ion. The oxidation number of sodium in the Na+ ion is +1, for example, and the oxidation number of chlorine in the Cl- ion is -1.
The oxidation number of hydrogen is +1 when it is combined with a nonmetal. Hydrogen is therefore in the +1 oxidation state in CH4, NH3, H2O, and HCl.
The oxidation number of hydrogen is -1 when it is combined with a metal. Hydrogen is therefore in the -1 oxidation state in LiH, NaH, CaH2, and LiAlH4.
The metals in Group IA form compounds (such as Li3N and Na2S) in which the metal atom is in the +1 oxidation state.
The elements in Group IIA form compounds (such as Mg3N2 and CaCO3) in which the metal atom is in the +2 oxidation state.
Oxygen usually has an oxidation number of -2. Exceptions include molecules and polyatomic ions that contain O-O bonds, such as O2, O3, H2O2, and the O22- ion.
The nonmetals in Group VIIA often form compounds (such as AlF3, HCl, and ZnBr2) in which the nonmetal is in the -1 oxidation state.
The sum of the oxidation numbers of the atoms in a molecule is equal to the charge on the molecule.
The most electronegative element in a compound has a negative oxidation number.
Any set of rules, no matter how good, will only get you so far. You then have to rely on a combination of common sense and prior knowledge. Questions to keep in mind while assigning oxidation numbers include the following: Are there any recognizable ions hidden in the molecule? Does the oxidation number make sense in terms of the known electron configuration of the atom?



Recognizing Oxidation-Reduction Reactions
Chemical reactions are often divided into two categories: oxidation-reduction or metathesis reactions. Metathesis reactions include acid-base reactions that involve the transfer of an H+ ion from a Brnsted acid to a Brnsted base.

CH3CO2H(aq) +OH-(aq) CH3CO2-(aq)+H2O(l) Brnsted acid Brnsted base Brnsted base Brnsted acid
They can also involve the sharing of a pair of electrons by an electron-pair donor (Lewis base) and an electron-pair acceptor (Lewis acid).

Co3+(aq)+6 NO2-(aq) Co(NO2)63-(aq) Lewis acid Lewis base
Oxidation-reduction reactions or redox reactions can involve the transfer of one or more electrons.

Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)
They can also occur by the transfer of oxygen, hydrogen, or halogen atoms.

CO2(g) + H2(g) CO(g) + H2O(g) SF4(g) + F2(g) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> SF6(g)
Fortunately, there is an almost foolproof method of distinguishing between metathesis and redox reactions. Reactions in which none of the atoms undergoes a change in oxidation number are metathesis reactions. There is no change in the oxidation number of any atom in either of the metathesis reactions, for example.


The word metathesis literally means "interchange" or "transposition," and it is used to describe changes that occur in the order of letters or sounds in a word as a language develops. Metathesis occurred, for example, when the Old English word brid became bird. In chemistry, metathesis is used to describe reactions that interchange atoms or groups of atoms between molecules.
When at least one atom undergoes a change in oxidation state, the reaction is an oxidation-reduction reaction. Each of the reactions in the figure below is therefore an example of an oxidation-reduction reaction.





Practice Problem :
Classify each of the following as either a metathesis or an oxidation-reduction reaction. Note that mercury usually exists in one of three oxidation states: mercury metal, Hg22+ ions, or Hg2+ ions.
(a) Hg22+(aq) + 2 OH-(aq) <IMG height=9 alt="---->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow.gif" width=17> Hg2O(s) + H2O(l)
(b) Hg22+(aq) + Sn2+(aq) <IMG height=9 alt="---->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow.gif" width=17> 2 Hg(l) + Sn4+(aq)
(c) Hg22+(aq) + H2S(aq) <IMG height=9 alt="---->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow.gif" width=17> Hg(l) + HgS(s) + 2 H+(aq)
(d) Hg2CrO4(s) + 2 OH-(aq) <IMG height=9 alt="---->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow.gif" width=17> Hg2O(s) + CrO42-(aq) + H2O(l)




Balancing Oxidation-Reduction Equations
A trial-and-error approach to balancing chemical equations involves playing with the equationadjusting the ratio of the reactants and productsuntil the following goals have been achieved.

Goals for Balancing Chemical Equations
1. The number of atoms of each element on both sides of the equation is the same and therefore mass is conserved.
2. The sum of the positive and negative charges is the same on both sides of the equation and therefore charge is conserved. (Charge is conserved because electrons are neither created nor destroyed in a chemical reaction.)

There are two situations in which relying on trial and error can get you into trouble. Sometimes the equation is too complex to be solved by trial and error within a reasonable amount of time. Consider the following reaction, for example.

3 Cu(s) + 8 HNO3(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 3 Cu2+(aq) + 2 NO(g) + 6 NO3-(aq) + 4 H2O(l)
Other times, more than one equation can be written that seems to be balanced. The following are just a few of the balanced equations that can be written for the reaction between the permanganate ion and hydrogen peroxide, for example.

2 MnO4-(aq) + H2O2(aq) + 6 H+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Mn2+(aq) + 3 O2(g) + 4 H2O(l) 2 MnO4-(aq) + 3 H2O2(aq) + 6 H+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Mn2+(aq) + 4 O2(g) + 6 H2O(l) 2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l) 2 MnO4-(aq) + 7 H2O2(aq) + 6 H+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 2 Mn2+(aq) + 6 O2(g) + 10 H2O(l)
Equations such as these have to be balanced by a more systematic approach than trial and error.



The Half-Reaction Method of Balancing Redox Equations
A powerful technique for balancing oxidation-reduction equations involves dividing these reactions into separate oxidation and reduction half-reactions. We then balance the half-reactions, one at a time, and combine them so that electrons are neither created nor destroyed in the reaction.
The steps involved in the half-reaction method for balancing equations can be illustrated by considering the reaction used to determine the amount of the triiodide ion (I3-) in a solution by titration with the thiosulfate (S2O32-) ion.
STEP 1: Write a skeleton equation for the reaction. The skeleton equation for the reaction on which this titration is based can be written as follows.

I3- + S2O32- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> I- + S4O62-
STEP 2: Assign oxidation numbers to atoms on both sides of the equation. The negative charge in the I3- ion is formally distributed over the three iodine atoms, which means that the average oxidation state of the iodine atoms in this ion is -1/3. In the S4O62- ion, the total oxidation state of the sulfur atoms is +10. The average oxidation state of the sulfur atoms is therefore +21/2.

I3- + S2O32- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> I-+S4O62--1/3+2 -2-1+21/2 -2
STEP 3: Determine which atoms are oxidized and which are reduced.
STEP 4: Divide the reaction into oxidation and reduction half-reactions and balance these half-reactions one at a time. This reaction can be arbitrarily divided into two half-reactions. One half-reaction describes what happens during oxidation.

Oxidation: S2O32- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>S4O62- +2 +21/2
The other describes the reduction half of the reaction.

Reduction:I3-<IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>I--1/3-1
It doesn't matter which half-reaction we balance first, so let's start with the reduction half-reaction. Our goal is to balance this half-reaction in terms of both charge and mass. It seems reasonable to start by balancing the number of iodine atoms on both sides of the equation.

Reduction:I3- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 3 I-
We then balance the charge by noting that two electrons must be added to an I3- ion to produce 3 I- ions,

Reduction:I3- +2 e- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54> 3 I-
as can be seen from the Lewis structures of these ions shown in the figure below.


We now turn to the oxidation half-reaction. The Lewis structures of the starting material and the product of this half-reaction suggest that we can get an S4O62- ion by removing two electrons from a pair of S2O32- ions, as shown in the figure below.

Oxidation: 2 S2O32- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>S4O62-+2 e-
STEP 5: Combine these half-reactions so that electrons are neither created nor destroyed. Two electrons are given off in the oxidation half-reaction and two electrons are picked up in the reduction half-reaction. We can therefore obtain a balanced chemical equation by simply combining these half-reactions.

(2 S2O32-<IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>S4O62- + 2 e-)+ (I3- + 2 e- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>3 I-)I3- + 2 S2O32- <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>3 I- + S4O62-
STEP 6: Balance the remainder of the equation by inspection, if necessary. Since the overall equation is already balanced in terms of both charge and mass, we simply introduce the symbols describing the states of the reactants and products.

I3-(aq) + 2 S2O32-(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>3 I-(aq) + S4O62-(aq)



Redox Reactions In Acidic Solutions
Some might argue that we don't need to use half-reactions to balance equations because they can be balanced by trial and error. The half-reaction technique becomes indispensable, however, in balancing reactions such as the oxidation of sulfur dioxide by the dichromate ion in acidic solution.

H+ SO2(aq)+Cr2O72-(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>SO42-(aq)+Cr3+(aq)
The reason why this equation is inherently more difficult to balance has nothing to do with the ratio of moles of SO2 to moles of Cr2O72-; it results from the fact that the solvent takes an active role in both half-reactions.



The reaction between oxalic acid and potassium permanganate in acidic solution is a classical technique for standardizing solutions of the MnO4- ion. These solutions need to be standardized before they can be used because it is difficult to obtain pure potassium permanganate. There are three sources of error.
Samples of KMnO4 are usually contaminated by MnO2.
Some of the KMnO4 reacts with trace contaminants when it dissolves in water, even when distilled water is used as the solvent.
The presence of traces of MnO2 in this system catalyzes the decomposition of MnO4- ion on standing.
Solutions of this ion therefore have to be standardized by titration just before they are used. A sample of reagent grade sodium oxalate (Na2C2O4) is weighed out, dissolved in distilled water, acidified with sulfuric acid, and then stirred until the oxalate dissolves. The resulting oxalic acid solution is then used to titrate MnO4- to the endpoint of the titration, which is the point at which the last drop of MnO4- ion is decolorized and a faint pink color persists for 30 seconds.




Solutions of the MnO4- ion that have been standardized against oxalic acid, using the equation balanced in the previous practice problem, can be used to determine the concentration of aqueous solutions of hydrogen peroxide, using the equation balanced in the following practice problem.

Practice Problem 5: An endless number of balanced equations can be written for the reaction between the permanganate ion and hydrogen peroxide in acidic solution to form the manganese (II) ion and oxygen:

MnO4-(aq)+H2O2(aq)Mn2+(aq)+O2(g)
Use the half-reaction method to determine the correct stoichiometry for this reaction.



Answer
2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq)2 Mn2+(aq) + 5 O2(g) + 8 H2O(l)

Redox Reactions in Basic Solutions
Half-reactions are also valuable for balancing equations in basic solutions. The key to success with these reactions is recognizing that basic solutions contain H2O molecules and OH- ions. We can therefore add water molecules or hydroxide ions to either side of the equation, as needed.
The following equation describes the reaction between the permanganate ion and hydrogen peroxide in an acidic solution.

2 MnO4-(aq) + 5 H2O2(aq) + 6 H+(aq) <IMG height=12 alt="------->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>2 Mn2+(aq) + 5 O2(g) + 8 H2O(l)
It might be interesting to see what happens when this reaction occurs in a basic solution.
Practice Problem 6: Write a balanced equation for the reaction between the permanganate ion and hydrogen peroxide in a basic solution to form manganese dioxide and oxygen.

MnO4-(aq) + H2O2(aq)MnO2(s) + O2(g)
Answer

2 MnO4-(aq) + 3 H2O2(aq) 2 MnO2(s) + 3 O2(g) + 2 OH-(aq) + 2 H2O(l)


Reactions in which a single reagent undergoes both oxidation and reduction are called disproportionation reactions. Bromine, for example, disproportionates to form bromide and bromate ions when a strong base is added to an aqueous bromine solution.

OH- Br2Br - + BrO3-


Practice Problem 7: Write a balanced equation for the disproportionation of bromine in the presence of a strong base.

Answer
3 Br2(aq) + 6 OH-(aq) 5 Br-(aq) + BrO3-(aq) + 3 H2O(l)




Molecular Redox Reactions
Lewis structures can play a vital role in understanding oxidation-reduction reactions with complex molecules. Consider the following reaction, for example, which is used in the Breathalyzer to determine the amount of ethyl alcohol or ethanol on the breath of individuals who are suspected of driving while under the influence.

3 CH3CH2OH(g) + 2 Cr2O72-(aq) + 16 H+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l)
We could balance the oxidation half-reaction in terms of the molecular formulas of the starting material and the product of this half-reaction.

Oxidation: C2H6O C2H4O2
It is easier to understand what happens in this reaction, however, if we assign oxidation numbers to each of the carbon atoms in the Lewis structures of the components of this reaction, as shown in the figure below.


The carbon atom in the CH3 group in ethanol is assigned an oxidation state of -3 so that it can balance the oxidation states of the three H atoms it carries. Applying the same technique to the CH2OH group in the starting material gives an oxidation state of -1.
The carbon in the CH3 group in the acetic acid formed in this reaction has the same oxidation state as it did in the starting material: -3. There is a change in the oxidation number of the other carbon atom, however, from -1 to +3. The oxidation half-reaction therefore formally corresponds to the loss of four electrons by one of the carbon atoms.

Oxidation:CH3CH2OHCH3CO2H +4 e-
Because this reaction is run in acidic solution, we can add H+ and H2O molecules as needed to balance the equation.

Oxidation:CH3CH2OH+ H2OCH3CO2H +4 e-+4 H+
The other half of this reaction involves a six-electron reduction of the Cr2O72- ion in acidic solution to form a pair of Cr3+ ions.

Reduction: Cr2O72-+6 e- 2 Cr3+
Adding H+ ions and H2O molecules as needed gives the following balanced equation for this half-reaction.

Reduction:Cr2O72-+14 H++6 e-2 Cr3++7 H2O
We are now ready to combine the two half-reactions by assuming that electrons are neither created nor destroyed in this reaction.

3(CH3CH2OH + H2O CH3CO2H + 4 e- + 4 H+) 2(Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O) 3 CH3CH2OH + 2 Cr2O72- + 28 H+ + 3 H2O 3 CH3CO2H + 4 Cr3+ + 12 H+ + 14 H2O
Simplifying this equation by removing 3 H2O molecules and 12 H+ ions from both sides of the equation gives the balanced equation for this reaction.

3 CH3CH2OH(g) + 2 Cr2O72-(aq) + 16 H+(aq) 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l)




Common Oxidizing Agents and Reducing Agents
In looking at oxidation-reduction reactions, we can focus on the role played by a particular reactant in a chemical reaction. What is the role of the permanganate ion in the following reaction, for example?

2 MnO4-(aq) + 5 H2C2O4(aq) + 6 H+(aq) <IMG height=12 alt="----->" src="http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/graphics/rarrow2.gif" width=54>10 CO2(g) + 2 Mn2+(aq) + 8 H2O(l)
Oxalic acid is oxidized to carbon dioxide in this reaction and the permanganate ion is reduced to the Mn2+ ion.

Oxidation: H2C2O4CO2+3+4 Reduction: MnO4- Mn2++7+2
The permanganate ion removes electrons from oxalic acid molecules and thereby oxidizes the oxalic acid. Thus, the MnO4- ion acts as an oxidizing agent in this reaction. Oxalic acid, on the other hand, is a reducing agent in this reaction. By giving up electrons, it reduces the MnO4- ion to Mn2+.
Atoms, ions, and molecules that have an unusually large affinity for electrons tend to be good oxidizing agents. Elemental fluorine, for example, is the strongest common oxidizing agent. F2 is such a good oxidizing agent that metals, quartz, asbestos, and even water burst into flame in its presence. Other good oxidizing agents include O2, O3, and Cl2, which are the elemental forms of the second and third most electronegative elements, respectively.
Another place to look for good oxidizing agents is among compounds with unusually large oxidation states, such as the permanganate (MnO4-), chromate (CrO42-), and dichromate (Cr2O72-) ions, as well as nitric acid (HNO3), perchloric acid (HClO4), and sulfuric acid (H2SO4). These compounds are strong oxidizing agents because elements become more electronegative as the oxidation states of their atoms increase.
Good reducing agents include the active metals, such as sodium, magnesium, aluminum, and zinc, which have relatively small ionization energies and low electro-negativities. Metal hydrides, such as NaH, CaH2, and LiAlH4, which formally contain the H- ion, are also good reducing agents.
Some compounds can act as either oxidizing agents or reducing agents. One example is hydrogen gas, which acts as an oxidizing agent when it combines with metals and as a reducing agent when it reacts with nonmetals.

2 Na(s) + H2(g) 2 NaH(s) H2(g) + Cl2(g) 2 HCl(g)
Another example is hydrogen peroxide, in which the oxygen atom is in the -1 oxidation state. Because this oxidation state lies between the extremes of the more common 0 and -2 oxidation states of oxygen, H2O2 can act as either an oxidizing agent or a reducing agent.



The Relative Strengths of Oxidizing and Reducing Agents
Spontaneous oxidation-reduction reactions convert the stronger of a pair of oxidizing agents and the stronger of a pair of reducing agents into a weaker oxidizing agent and a weaker reducing agent. The fact that the following reaction occurs, for example, suggests that copper metal is a stronger reducing agent than silver metal and that the Ag+ ion is a stronger oxidizing agent than the Cu2+ ion.

Cu(s)+2 Ag+(aq) Cu2+(aq)+2 Ag(s) stronger
reducing
agent stronger
oxidizing
agentweaker
oxidizing
agent weaker
reducing
agent
On the basis of many such experiments, the common oxidation-reduction half-reactions have been organized into a table in which the strongest reducing agents are at one end and the strongest oxidizing agents are at the other, as shown in the table below. By convention, all of the half-reactions are written in the direction of reduction. Furthermore, by convention, the strongest reducing agents are usually found at the top of the table.

The Relative Strengths of Common Oxidizing Agents and Reducing Agents
K+ + e- K Best Ba2+ + 2 e- Ba reducingCa2+ + 2 e- Ca agentsNa+ + e- Na Mg2+ + 2 e- MgH2 + 2 e- 2 H- Al3+ + 3 e- AlMn2+ + 2 e- MnZn2+ + 2 e- Zn Cr3+ + 3 e- Cr S + 2 e- S2- 2 CO2 + 2 H+ + 2 e- H2C2O4Cr3+ + e- Cr2+ Fe2+ + 2 e- FeCo2+ + 2 e- CoNi2+ + 2 e- NiSn2+ + 2 e- SnPb2+ + 2 e- Pb Fe3+ + 3 e- Fe 2 H+ + 2 e- H2 S4O62- + 2 e- 2 S2O32-Sn4+ + 2 e- Sn2+ Cu2+ + e- Cu+ O2 + 2 H2O + 4 e- 4 OH-Cu+ + e- CuI2 + 2 e- 2 I-oxidizing MnO4- + 2 H2O + 3 e- MnO2 + 4 OH-powerO2 + 2 H+ + 2 e- H2O2 Reducing increasesFe3+ + e- Fe2+ powerHg22+ + 2 e- 2 Hg increasesAg+ + e- Ag Hg2+ + 2 e- Hg H2O2 + 2 e- 2 OH- HNO3 + 3 H+ + 3 e- NO + 2 H2O Br2(aq) + 2 e- 2 Br- 2 IO3- + 12 H+ + 10 e- I2 + 6 H2O CrO42- + 8 H+ + 3 e- Cr3+ + 4 H2O Pt2+ + 2 e- Pt MnO2 + 4 H+ + 2 e- Mn2+ + 2 H2O O2 + 4 H+ + 4 e- 2 H2O Cr2O72- + 14 H+ + 6 e- 2 Cr3+ + 7 H2O Cl2(g) + 2 e- 2 Cl- PbO2 + 4 H+ + 2 e- Pb2+ + 2 H2O MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O Au+ + e- Au H2O2 + 2 H+ + 2 e- 2 H2O Co3+ + e- Co2+ Best S2O82- + 2 e- 2 SO42- oxidizingO3(g) + 2 H+ + 2 e- O2(g) + H2O agents F2(g) + 2 H+ + 2 e- 2 HF(aq)
Fortunately, you don't have to memorize these conventions. All you have to do is remember that the active metals, such as sodium and potassium, are excellent reducing agents and look for these entries in the table. The strongest reducing agents will be found at the corner of the table where sodium and potassium metal are listed
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