Saturday, April 25, 2009 http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method

Redox reactions - in which charge is transferred between reagents - can be balanced by inspection, although it can be extremely difficult. But let's start with a very simple example:

Cu + Fe3+ -> Cu2+ + Fe2+

At first sight the equation seems to be already balanced, but if we check charges - it is not. There is +3 charge on the left, and +4 charge on the right, so we have to do something about it. Using the inspection method we should select the most complicated molecule first. Let's say it is Fe3+:

Cu + 1Fe3+ -> Cu2+ + Fe2+

If so, we can already add 1 in front of Fe2+:

Cu + 1Fe3+ -> Cu2+ + 1Fe2+

Trying to balance copper will lead us nowhere (atoms are balanced, but charges will be left unbalanced as they were from the beginning), so let's look at the charge - it can be balanced just like atoms. We have +3 on the left and +2 on the right - so we need additional +1 on the right side. Do you remember that we can use fractions at this stage? Half Cu2+ will do:

Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+

What is wrong now is the left side of equation - too much copper at the moment. Once again, half will do:

1/2Cu + Fe3+ -> 1/2Cu2+ + 1Fe2+

Final touch - multiply everything by 2 to remove fractions:

Cu + 2Fe3+ -> Cu2+ + 2Fe2+

Atoms - balanced (one of copper and two of iron on both sides). Charge - balanced (+6 on both sides). We have just balanced the redox equation by using the inspection method. But in the case of more difficult ones - like this for example:

FeSO4 + KMnO4 +H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O

this approach is a waste of time. Much easier and faster ways of balancing redox reactions are oxidation numbers method and half reaction method.

AdSense

When balancing redox reactions we have always - apart from all the rules pertaining to balancing chemical equations - additional information about electrons moving. In the case of oxidation numbers method we assume electrons are transferred between atoms (which is only an approximation), in the case of half reactions method we assume there are two systems exchanging electrons (which is much closer to reality, although what we observe may be a multistep process with numerous intermediate reagents).

Let's try to use half reactions method to balance reaction equation already mentioned in the balancing by inspection section:

FeSO4 + KMnO4 +H2SO4 -> Fe2(SO4)3 + MnSO4 + H2O

First of all, let's concentrate on what is really important - on the net ionic reaction:

Fe2+ + MnO4- + H+ -> Fe3+ + Mn2+ + H2O

All other ions (K+ and SO42-) are only spectators and we don't need them to balance the equation. Besides, charges are what is really important in the half reactions method.

At first glance you can see that in the reaction iron gets oxidized and permanganate gets reduced:

Fe2+ -> Fe3+

MnO4- -> Mn2+

We will start equation balancing with balancing these half reactions - using electrons to balance charge. In the case of iron oxidation half reaction atoms are already balanced, but charge is not. To balance charge we will add one electron on the right side:

Fe2+ -> Fe3+ + e-

We can use electrons safely, as the final step of balancing will be electron cancellation.

Now we have to balance permanganate reduction half reaction:

MnO4- -> Mn2+

Before balancing charges we have to balance atoms. What to do with the oxygen? We know that the reaction takes place in acidic conditions (see sulfuric acid in the skeletal reaction at the beginning) - so we can add H+ on the left and H2O on the right:

MnO4- + H+ -> Mn2+ + H2O

Using simple balancing by inspection we will add two coefficients to balance atoms:

MnO4- + 8H+ -> Mn2+ + 4H2O

Once the atoms are balanced it is time to balance charge with electrons - there is +7 charge on the left side of the equation and +2 on the right side. To balance charges we have to add 5 electrons on the left:

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

At the moment we have two balanced half reactions. Here comes the final trick - we add these half reactions, multiplying them by such coefficients that electrons cancel out (easiest approach is to use just numbers of electrons, although if often means you will have to find lowest coefficients later). To have five electrons in both equations we have to multiply first equation by 5:

5Fe2+ -> 5Fe3+ + 5e-

and when we add both half reactions we will get

MnO4- + 8H+ + 5Fe2+ + 5e- -> Mn2+ + 4H2O + 5Fe3+ + 5e-

Canceling out the electrons:

MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 4H2O + 5Fe3+

And the reaction is balanced.

Balancing hydrogen and oxygen in the half reactions method requires knowledge about the conditions in which reaction takes place. To balance oxygen we can add H+ on the side where there is oxygen excess and water on the second, just as we did in the above example. But we can also use OH- and water to do the trick, for example half reaction:

ClO- -> Cl-

is not balanced, but once we add water and OH-:

ClO- + H2O -> Cl- + 2OH-

we have not only balanced atoms but we are also ready to balance charge by adding two electrons on the left:

ClO- + H2O + 2e- -> Cl- + 2OH-

and the half reaction is ready to be used. General rule says that if the reaction takes place in acidic conditions we use water and H+ to balance oxygens, and if the reaction takes place in basic conditions - we use OH- and water. Don't worry if it looks like the reaction produces H+ in the solution that was already acidic - while it will influence reaction equilibrium, we are concentrating on the equation balancing, not on the equilibrium right now.

Also note that in really hard cases, when you have no idea what is really going on in the solution, you may look for the half reactions (and amount of electrons consumed/produced) in the standard potentials tables.

AdSense

Before we will get to explanation very important disclaimer: oxidation numbers don't exist. They were invented to help in charge accounting needed when balancing redox reaction equations, but they don't refer to any real life chemical concept.

The general idea behind the oxidation numbers (ON) method for balancing chemical equations is that electrons are transferred between charged atoms. These charges - assigned to atoms - are called oxidation numbers, just to remind you that they don't reflect real structure of the reagents.

There are several simple rules used for assigning oxidation numbers to every atom present in any compound:

* First of all, charged mono atomic ion has oxidation number equal to its charge. Thus Na+ has oxidation number +1, Fe3+ has oxidation number +3, F- has oxidation number of -1 and S2- has oxidation number of -2.

* Second rule says that the oxidation number of a free element is always 0. Thus oxidation number of Cu is 0, oxidation number of O in O2 is 0, the same holds for S in S8 and so on.

* Oxygen in almost all compounds has oxidation number -2.

* Hydrogen in almost all compounds has oxidation number +1.

* Some elements usually have the same oxidation number in their compounds:

o alkali metals - Li, Na, K, Rb, Cs - oxidation numbers are +1

o alkaline earth metals - Be, Mg, Ca, Sr, Ba - oxidation numbers are +2

o halogens (except when they form compounds with oxygen or one another) - oxidation numbers are -1 (always true for fluorine)

* Last rule says that the charge of the ion or molecule equals sum of oxidation numbers of all atoms.

There are some exceptions to the rules 2 and 3 - for example oxygen in peroxides has oxidation number of -1, and hydrogen in hydrides has oxidation number -1 too, but these are very rare cases.

Before we will try to balance any equations let's use above rules to assign oxidation numbers to atoms in several substances.

For example - what is oxidation number of sulfur in SO2? Particle is not charged, so oxidation number of sulfur must equal sum of oxidation numbers of oxygens, but with the opposite sign. Oxygen oxidation number is -2, there are two oxygens - that gives -4 together, so sulfur must have ON=+4.

What is oxidation number of atoms in CrO42-? Oxygen is -2 and there are 4 oxygens - that gives overall of -8, ion has charge of -2, so central atom must have ON=+6.

How do we use oxidation numbers for balancing? First of all, we have to understand that oxidation means increase of oxidation number, while reduction means decrease of oxidation number. In both cases change of oxidation number is due to electrons lost (oxidation) or gained (reduction). We calculate oxidation numbers for all atoms present in the reaction equation (note that it is not that hard as it sounds, as for most atoms oxidation numbers will not change) and we look for a ratio that makes the number of electrons lost equal to the number of electrons gained. That gives us additional information needed for reaction balancing.

Let's try with following reaction:

KIO3 + KI + H2SO4 -> K2SO4 + H2O + I2

First of all - we don't need any spectators here, as they are only making things look more difficult then they are in reality. Quick glance tells us that the net ionic reaction is

IO3- + I- + H+ -> H2O + I2

Looks like IO3- is oxidizing agent here and I- is reducting agent. I- has oxidation number of -1, iodine in IO3- has oxidation number of +5. On the right side in I2 both iodine atoms have oxidation number 0. It means that iodine in IO3- must gain 5 electrons. These electrons come from I- - one for every I- ion. Assuming (just like we do in the inspection method) that IO3- is the most complicated molecule and it's coefficient is 1 we will need five I- for the redox process to complete:

1IO3- + 5I- + H+ -> H2O + I2

Now that the ration between oxidizer and reducing agent is known we use simple techniques we know from the inspection method to balance remaining elements. There are six atoms of iodine on the left, so we need three I2 molecules to balance iodine:

1IO3- + 5I- + H+ -> H2O + 3I2

And the final, trivial step is balancing oxygen, hydrogen and water:

IO3- + 5I- + 6H+ -> 3H2O + 3I2

Other case we can try is oxidation of Mn2+ with NaBiO3 in acidic conditions:

Mn2+ + BiO3- + H+ -> MnO4- + Bi3+ + H2O

Using methods for oxidation numbers calculation we can easily check that manganese is oxidized from +2 to +7 (freeing five electrons) and bismuth is reduced from +5 to +3 (accepting two electrons). To balance electrons transferred we can put coefficients 2 and 5 on the left side of reaction equation:

2Mn2+ + 5BiO3- + H+ -> MnO4- + Bi3+ + H2O

Rest can be balanced by inspection and is not difficult to do, yielding:

2Mn2+ + 5BiO3- + 14H+ -> 2MnO4- + 5Bi3+ + 7H2O

Now the same equation can be also easily balanced as a full (non net-ionic) version:

4MnSO4 + 10NaBiO3 + 14H2SO4 -> 4NaMnO4 + 5Bi2(SO4)3 + 14H2O + 3Na2SO4

Apart from the three already described methods, there is also a general method, often less user friendly - but thanks to its systematic approach perfect for use in computer programs. That's the method EBAS - our chemical reaction equation balancer - uses.

Let's try an algebraic method for

H3BO3 -> H4B6O11 + H2O

It can be rather easily balanced by inspection, but let's try a more systematic approach.

What does 'balanced' mean? It means that for every element, there are the same number of atoms on both sides of the reaction equation. Our reaction has three coefficients A, B and C:

AH3BO3 -> BH4B6O11 + CH2O

From a mathematical point of view 'balanced' means that for every element we can write an equation, comparing the number of atoms on both sides of the reaction equation. An example for boron:

A = 6×B + 0×C

where coefficients 1, 6 and 0 are taken from the compound formulae. We can write similar equations for all elements - hydrogen:

3×A = 4×B + 2×C

and oxygen:

3×A = 11×B + C

As there are no free terms in this set of equations, it has a trivial solution (A = B = C = 0) which we are not interested in. We have three equations, and three unknowns. Such equation set is not a thing that you may want to solve manually, although when balancing chemical equations in most cases it can be done relatively easy, as most equations don't contain all unknowns. In this case we have very simple equation A = 6×B that we can use to substitute 6×B for A in the second and third equation to get:

18×B = 4×B + 2×C

18×B = 11×B + C

After some rearranging:

7×B = C

7×B = C

Both equations are identical. In algebra it usually means that the set of equations doesn't have a unique solution, but in the case of chemical equations we have one additional information - all coefficients must be integer and they must be the smallest ones. To find them we can assume one of the coefficients to be 1:

B = 1

If so

A = 6

C = 7

and indeed

6H3BO3 -> H4B6O11 + 7H2O

is the balanced reaction equation.

This first example doesn't look convincing - why do we have to solve set of equations when the reaction equation can be easily balanced by other means? Good point - but what if the reaction can be not easily balanced?

P2I4 + P4 + H2O -> PH4I + H3PO4

Try for a moment. Looks easy but soon gets surprisingly hard and the coefficients become pretty high, which makes you wonder if you have not made some mistake (*see some balancing hints at the bottom of that page). What about general, algebraic method?

We need five coefficients, and there are only four equations (one for each element present) - but it shouldn't bother us, as we know that we have additional information that works as an additional equation.

AP2I4 + BP4 + CH2O -> DPH4I + EH3PO4

Setting up equations:

P: 2×A + 4×B = D + E

I: 4×A = D

H: 2×C = 4×D + 3×E

O: C = 4×E

Balances for iodine and oxygen make this set look much easier then expected. C = 4×E and D = 4×A are substitutions that we are about to use to reduce number of unknowns:

1st equation: 2×A + 4×B = 4×A + E

3rd equation: 8×E = 4×D + 3×E

So we have now after some canceling:

4×B = 2×A + E

4×A = D

5×E = 4×D

C = 4×E

For someone fluent in mathematics it is obvious that we have already finished - it is now enough to assume that one of the variables equals 1 to calculate values of all others. Assuming A = 1 and simply substituting calculated values we have:

A = 1

B = 13/10

C = 64/5

D = 4

E = 16/5

These are hardly integer, but all we have to do is to find the smallest common denominator to have a list of integer coefficients in numerators. In this case the smallest common denominator is 10, so if we multiply all numbers by 10 we get:

A = 10

B = 13

C = 128

D = 40

E = 32

And you may check that these are the correct coefficients. Imagine finding them by inspection method!

It's also easy to use the algebraic method to balance redox reaction with charged species. Let's try it for

ACr2O72- + BH+ + CFe2+ -> DCr3+ + EH2O + FFe3+

What equations do we have? Four balances of atoms:

Cr: 2A = D

O: 7A = E

H: B = 2E

Fe: C = F

But that's not enough to balance equation - we have six coefficients and four equations. For algebraic method we can have one equation less than variables - so we need a fifth equation. So far we have not accounted for charge balance, and that will be our last equation needed to balance reaction:

-2A + B + 2C = 3D + 3F

(note that sign of coefficients in the last equation depends on the charge sign). We are ready to solve. First of all, we know that

E = 7A

so

B = 14A

Let's get rid of B and D in the charge balance equation:

-2A + 14A + 2C = 6A + 3F

Sorting, and using C=F:

C = 6A

That's almost ready. Let's put A = 1 and calculate all other coefficients simply using already known values:

C = 6

D = 2

E = 7

B = 14

F = 6

and balanced equation takes form:

Cr2O72- + 14H+ + 6Fe2+ -> 2Cr3+ + 7H2O + 6Fe3+

That's all. Not that hard.

Probably the most important characteristics of the algebraic method is that - contrary to the inspection method - is guaranteed to give you an answer. If the reaction can be balanced, you will find coefficients. If the reaction can't be balanced - you will find it out seeing that there are more unknowns than independent equations (remember - one more is not a problem), or that equations are contradictory. With inspection method you will never have a proof that the equation can't be balanced.

There is an additional reason for this method to be important. Computers are very effective in solving sets of simultaneous equations, for example using method called Gauss elimination. What is difficult for humans is a perfect task for the number cruncher built in every processor. That's why EBAS is able to balance the so-called Blakley equation (20 unknowns in 19 equations) in a blink of an eye on a 10 year old PC.

Note that there are - although rare - cases, when reaction occurring in reality cannot be balanced with the algebraic approach. Some of these cases (together with explanation) are described in "failures" section.

*Balancing hints for P2I4 + P4 + H2O -> PH4I + H3PO4 reaction:

Contrary to what balancing by inspection rules say, start with balancing oxygen from phosphoric acid, then balance hydrogen, iodine, and finally phosphorus. Leaving oxygen and hydrogen to the end is asking for troubles.

hello goodbye; 5:34 AM

redox

oxidation–reduction reaction

chemical reactionalso called redox reaction

Main

any chemical reaction in which the oxidation number of a participating chemical species changes. The term covers a large and diverse body of processes. Many oxidation–reduction reactions are as common and familiar as fire, the rusting and dissolution of metals, the browning of fruit, and respiration and photosynthesis—basic life functions.

General considerations » Major classifications

Most oxidation–reduction (redox) processes involve the transfer of oxygen atoms, hydrogen atoms, or electrons, with all three processes sharing two important characteristics: (1) they are coupled—i.e., in any oxidation reaction a reciprocal reduction occurs, and (2) they involve a characteristic net chemical change—i.e., an atom or electron goes from one unit of matter to another. Both reciprocity and net change are illustrated below in examples of the three most common types of oxidation–reduction reactions.

General considerations » Major classifications » Oxygen-atom transfer

Carbon reacts with mercury(II) oxide (a compound in which mercury has a bonding capacity expressed as +2; see below Oxidation-state change) to produce carbon dioxide and mercury metal. This reaction can be written in equation form:

Carbon, receiving oxygen, is oxidized; mercury(II) oxide, losing oxygen, undergoes the complementary reduction; and the net change is the transfer of two oxygen atoms from mercury(II) oxide units to a carbon atom.

General considerations » Major classifications » Hydrogen-atom transfer

Hydrogen atoms are transferred from hydrazine, a compound of nitrogen and hydrogen, to oxygen in the following reaction:

Hydrazine, losing hydrogen, is oxidized to molecular nitrogen, while oxygen, gaining hydrogen, is reduced to water.

General considerations » Major classifications » Electron transfer

Zinc metal and copper(II) ion react in water solution, producing copper metal and an aqueous (denoted by aq) zinc ion according to the equation

With the transfer of two of its electrons, the zinc metal is oxidized, becoming an aqueous zinc ion, while the copper(II) ion, gaining electrons, is reduced to copper metal. Net change is the transfer of two electrons, lost by zinc and acquired by copper.

Because of their complementary nature, the oxidation and reduction processes together are referred to as redox reactions. The reactant that brings about the oxidation is called the oxidizing agent, and that reagent is itself reduced by the reducing agent. In the examples given above, mercury(II) oxide, oxygen, and the copper(II) ion are oxidizing agents, and carbon, hydrazine, and zinc are the reducing agents.

General considerations » General theory » Stoichiometric basis

Describing the redox processes as above conveys no information about the mechanism by which change takes place. A complete description of the net chemical change for a process is known as the stoichiometry of the reaction, which provides the characteristic combining proportions of elements and compounds. Reactions are classified as redox and nonredox on the basis of stoichiometry; oxygen-atom, hydrogen-atom, and electron transfer are stoichiometric categories.

General considerations » General theory » Oxidation-state change

Comprehensive definitions of oxidation and reduction have been made possible by modern molecular structure theory. Every atom consists of a positive nucleus, surrounded by negative electrons, which determine the bonding characteristics of each element. In forming chemical bonds, atoms donate, acquire, or share electrons. This makes it possible to assign every atom an oxidation number, which specifies the number of its electrons that can be involved in forming bonds with other atoms. From the particular atoms in a molecule and their known bonding capacities, the bonding pattern within a molecule is determined, and each atom is regarded as being in a specific oxidation state, expressed by an oxidation number.

Redox processes are defined as reactions accompanied by oxidation-state changes: an increase in an atom’s oxidation number corresponds to an oxidation; a decrease, to a reduction. In this generalized theory, three examples of ways in which oxidation-state changes can occur are by oxygen-atom (gain, oxidation; loss, reduction), hydrogen-atom (loss, oxidation; gain, reduction), and electron (loss, oxidation; gain, reduction) transfer. The oxidation-state change definition is usually compatible with the above rules for applying the oxygen-atom-transfer and hydrogen-atom-transfer criteria and always compatible with the electron-transfer criterion when it is applicable. The oxidation state of any atom is indicated by a roman numeral following the name or symbol for the element. Thus, iron(III), or Fe(III), means iron in an oxidation state of +3. The uncombined Fe(III) ion is simply Fe³⁺.

General considerations » Historical origins of the redox concept

Of the chemical processes now regarded as redox reactions, combustion was the earliest focus of philosophical and scientific attention. The Greek scientific philosopher Empedocles listed fire as one of the four elements of matter. In more modern times the phlogiston theory enjoyed scientific popularity. This theory was first articulated in 1697 by G.E. Stahl of Germany. As noted earlier, it asserted that matter releases an elementary constituent, phlogiston, during combustion. Thus, the burning of charcoal was interpreted as the loss of phlogiston from carbon to the air. The theory was also applied to processes other than combustion; in the recovery of a metal from its oxide by heating with charcoal, for example, phlogiston was regarded as being transferred from carbon to the oxide.

Phlogiston saturation was believed to be responsible for the limited ability of air in a closed container to support combustion. A notable consequence of the phlogiston theory was the notion that an oxide of a metal, such as mercury(II) oxide (HgO), was a chemically simpler substance than the metal itself: the metal could be obtained from the oxide only by the addition of phlogiston. The phlogiston theory, however, could provide no acceptable explanation of the gain in weight when an oxide is formed from a metal.

General considerations » Historical origins of the redox concept » Combustion and oxide formation

Late in the 18th century, the interrelated work of Joseph Priestley and Antoine-Laurent Lavoisier led to the overthrow of the phlogiston theory. Lavoisier saw Priestley’s discovery of oxygen in 1774 as the key to the weight gains known to accompany the burning of sulfur and phosphorus and the calcination of metals (oxide formation). In his Traité élémentaire de chimie, he clearly established that combustion consists of a chemical combination between oxygen from the atmosphere and combustible matter (see below Combustion and flame). By the end of the century, his ideas were widely accepted and had been successfully applied to the more complex processes of respiration and photosynthesis. Reactions in which oxygen was consumed were classified as oxidations, while those in which oxygen was lost were termed reductions.

General considerations » Historical origins of the redox concept » Electrochemical reactions

During the 19th century, the evolving field of electrochemistry led to a broadened view of oxidation. It was possible, for instance, to produce the ferric, or iron(III), ion from the ferrous, or iron(II), ion at the anode (positive electrode, where electrons are absorbed from solution) of an electrochemical cell (a device in which chemical energy is converted to electrical energy), according to the equation:

Molecular oxygen could effect a similar transformation, according to the equation:

The similarity of the two processes led to a precursor of the electron-transfer explanation for redox reactions. After the discovery of the electron, the conviction that oxidation and reduction are accomplished through electron loss and gain became firmly entrenched. Thus, early in the 20th century chemists tended to attribute all redox reactions to the transfer of electrons. Later work on chemical bonding, however, demonstrated the incorrectness of that description. An electronegativity scale (listing of elements in descending order of their tendency to attract and hold bonding electrons) provided a firm basis for the oxidation-state assignments on which oxidation–reduction definitions have become based.

General considerations » Examples of oxidation–reduction reactions

Molecular oxygen is a conspicuously important oxidizing agent. It will directly oxidize all but a few of the metals and most of the nonmetals as well. Often these direct oxidations lead to normal oxides, such as those of lithium (Li), zinc (Zn), phosphorus (P), and sulfur (S).

Organic foodstuffs are oxidized to carbon dioxide and water in respiration. The reaction stoichiometry can be illustrated for glucose, a simple sugar:

Although the oxygen–glucose reaction is slow at ambient temperatures outside the living cell, it proceeds quickly under the influence of enzymatic catalysis within the body. Essentially all organic compounds react with oxygen under appropriate conditions, but the reaction rates at ordinary temperatures and pressures vary greatly.

Many other oxidizing agents serve as oxygen-atom sources. Hydrogen peroxide (H₂O₂), acid chromate ion (HCrO₄⁻), and hypochlorous acid (HClO) are reagents often used in oxygen-atom-transfer reactions—for example, in the following reactions:

In the simplest hydrogen-atom transfers, molecular hydrogen serves as the hydrogen-atom source. The hydrogenations of ethylene and of molecular nitrogen are illustrative in the following equations:

Reactions of molecular hydrogen are characteristically slow at ordinary temperatures. The hydrogenation of molecular nitrogen and of olefins such as ethylene (an olefin is an unsaturated hydrocarbon compound; it has at least two adjacent carbon atoms joined by a double bond to which other atoms or groups of atoms can be joined directly) is a process of extraordinary commercial importance and requires catalysts to occur at useful rates.

Hydrogen-atom transfer from an organic molecule to a suitable acceptor is a common mode of organic oxidation. The oxidation of formic acid by permanganate and that of ethanol by acid chromate share stoichiometry that features hydrogen-atom loss by the organic species, as shown in the following equations:

The oxidizing agents permanganate and acid chromate, typical of many hydrogen-atom acceptors, undergo complicated changes rather than simple hydrogen-atom addition.

Electron-transfer stoichiometry is usually associated with metal ions in aqueous solution, as shown in the following equations:

Many positively charged metal ions have been shown to be bonded to water molecules, so that their electron-transfer reaction occurs between rather complex molecular groups. The iron ion formulas above, for example, are more properly written as [Fe(H₂O)₆]²⁺ and [Fe(H₂O)₆]³⁺ to reflect the presence of six water molecules bonded to the metal ion. Simple electron transfer between free ions is known only in the gas phase, as in this argon–sodium reaction:

Several other types of redox reactions do not fall in the oxygen-atom, hydrogen-atom, or electron-transfer categories. Among these are reactions of fluorine, chlorine, bromine, and iodine. These four elements, known as the halogens, form diatomic molecules, which are versatile oxidizing agents. The following examples are typical:

Such reactions often qualify as redox processes only in the broad sense that oxidation-state changes occur. The oxidation-state characterization extends oxidation–reduction chemistry to include examples from the reactions of all the chemical elements.

General considerations » Significance of redox reactions

Oxidation–reduction reactions have vast importance not only in chemistry but in geology and biology as well. The surface of the Earth is a redox boundary between the planet’s reduced metallic core and an oxidizing atmosphere. The Earth’s crust is largely composed of metal oxides, and the oceans are filled with water, an oxide of hydrogen. The tendency of nearly all surface materials to be oxidized by the atmosphere is reversed by the life process of photosynthesis. Because they are constantly renewed by the photosynthetic reduction of carbon dioxide, life’s complex compounds can continue to exist on the Earth’s surface.

For similar reasons, much of chemical technology hinges on the reduction of materials to oxidation states lower than those that occur in nature. Such basic chemical products as ammonia, hydrogen, and nearly all the metals are produced by reductive industrial processes. When not used as structural materials, these products are reoxidized in their commercial applications. The weathering of materials, including wood, metals, and plastics, is oxidative, since, as the products of technological or photosynthetic reductions, they are in oxidation states lower than those stable in the atmosphere.

Solar radiation is converted to useful energy by a redox cycle that operates continually on a global scale. Photosynthesis converts radiant energy into chemical potential energy by reducing carbon compounds to low oxidation states, and this chemical energy is recovered either through enzymatic oxidations at ambient temperatures or during combustion at elevated temperatures.

Theoretical aspects » Oxidation states

The idea of assigning an oxidation state to each of the atoms in a molecule evolved from the electron-pair concept of the chemical bond. Atoms within a molecule are held together by the force of attraction that the nuclei of two or more of them exert on electrons in the space between them. In many cases this sharing of electrons can be regarded as involving electron-pair bonds between adjacent nuclei. Electron-pair bonding is often diagrammed so as to show all the bonding and nonbonding valence electrons; e.g., the structures of atomic hydrogen, atomic chlorine, and hydrogen chloride shown below (each dot represents one valence electron):

The hydrogen chloride diagram reflects the presence, in the internuclear region, of two electrons that are under the mutual attractive influence of both the hydrogen and chlorine nuclei. Oxidation states for the hydrogen and chlorine in HCl are assigned according to the net charges that remain on H and Cl when the shared electrons are assigned to the atom that has the greater attraction for them. Through physical measurements on isolated atoms and simple molecules, these relative attractive powers have been determined. The Table lists the electronegativity values for some important elements.

Pauling electronegativities of selected elements
fluorine	4.0
oxygen	3.5
nitrogen	3.0
chlorine	3.0
bromine	2.8
sulfur	2.5
iodine	2.5
carbon	2.5
hydrogen	2.1
phosphorus	2.1
iron	1.8
sodium	0.9
Source: L. Pauling, The Nature of the Chemical Bond.

In the hydrogen chloride molecule the chlorine is more electronegative than hydrogen and is, therefore, assigned both shared electrons. Chlorine has seven valence electrons in its neutral state. Having acquired an eighth electron in its reaction with hydrogen, chlorine is considered to have an oxidation state of −1. Hydrogen, on the other hand, is assigned +1, having lost the single valence electron that it has in its neutral state. Charges arrived at in this way are the basis for oxidation-state assignments, conventionally represented by roman numerals, such as in H(I) and Cl(−I) for the constituents of HCl. Because determination of oxidation states is simply a method of conceptually distributing shared electrons to individual atoms, the same number of electrons must be accounted for, before and after such assignment. The Table includes examples of molecules that have multiple bonds. The oxidation states of the atoms involved are added up algebraically in the table, and their sum must always equal the net charge on the molecule. There is no physical reality to oxidation states; they simply represent the results of calculations based on a formal rule.

Oxidation states can be assigned for most common molecules with the help of a few guidelines. First, electrons shared by two atoms of the same element are divided equally; accordingly, elements are always in oxidation state of 0, regardless of their allotropic form (allotropic refers to the phenomenon of an element’s having two or more forms; e.g., carbon can exist as diamond or graphite and in both cases is in the 0 oxidation state). Second, only fluorine is more electronegative than oxygen. Therefore, except in compounds containing oxygen–oxygen or oxygen–fluorine bonds, oxygen can be reliably assigned the oxidation state −2. Similarly, hydrogen is less electronegative than fluorine, oxygen, nitrogen, chlorine, sulfur, and carbon (F, O, N, Cl, S, and C), so it is in the +1 oxidation state in its combinations with those elements. For many common compounds containing only hydrogen, oxygen, and a third element, the third element’s oxidation state can be calculated, assuming oxidation numbers of +1 for hydrogen and −2 for oxygen. When bonds are present between two elements that differ little in electronegativity, however, oxidation-state assignments become doubtful, and the distinction between redox and nonredox processes is not evident.

There is a general reluctance, particularly regarding organic systems, to assume oxidation-state changes when the reaction results can be accounted for by the transfer or addition of water (H₂O), ammonia (NH₃), the hydroxide ion (OH⁻), or the ions of hydrogen (H⁺), chlorine (Cl⁻), bromine (Br⁻), or iodine (I⁻), or combinations of these species; e.g., the ammonium ion (NH₄⁺), hydrogen chloride (HCl). The reason is that, in these molecules and ions, the elements are present in their most typical oxidation states: hydrogen(I), chlorine(−I), oxygen(−II), bromine (−I), iodine(−I), and nitrogen(−III).

Theoretical aspects » Oxygen-atom transfer reactions

The oxidation-state concept clarifies the relationship between oxygen-atom, hydrogen-atom, and electron transfer. The oxygen- and hydrogen-transfer criteria apply only when oxygen and hydrogen occur in their typical oxidation states. An example of an appropriate reaction involving oxygen-atom transfer is the reduction of ferrous oxide by carbon monoxide:

In terms of oxidation-state changes, this oxygen-atom transfer is equivalent to the two-electron reduction of iron and complementary two-electron oxidation of carbon:

Oxygen, which occurs in the oxidation state −2 in both reactants and products in the first equation, is not shown in the second. In transferring, the oxygen atom leaves two electrons behind, causing the reduction of iron, and acquires two electrons from the carbon atom, oxidizing the carbon.

In a similar way, the hydrogenation of ethylene corresponds to a two-electron reduction of the two-carbon skeleton:

In this example also, the second equation includes only the atoms that change oxidation states: the four hydrogen atoms initially present in ethylene are in the +1 oxidation state in both reactants and products and are therefore omitted. Each of the two neutral hydrogen atoms can be regarded as giving up an electron to, and thereby reducing, one of the carbon atoms. This example also demonstrates clearly that the oxidation that complements the reduction of ethylene is that of the two hydrogen atoms in H₂—namely, from the 0 to the +1 oxidation state. General application of the oxidation-state concept leads to a formal viewpoint toward all redox reactions as electron-transfer reactions.

Theoretical aspects » Half reactions

One of the basic reasons that the concept of oxidation–reduction reactions helps to correlate chemical knowledge is that a particular oxidation or reduction can often be carried out by a wide variety of oxidizing or reducing agents. Reduction of the iron(III) ion to the iron(II) ion by four different reducing agents provides an example:

Production of the same change in the aqueous iron(III) ion by different reductants emphasizes the fact that the reduction is a characteristic reaction of the iron system itself, and, therefore, the process may be written without specifying the identity of the reducing agent in the following way:

Hypothetical equations of this type are known as half reactions. The symbol e⁻, which stands for an electron, serves as a reminder that an unspecified reducing agent is required to bring about the change. Half reactions can be written, equally, for the reducing agents in the four reactions with ferric ion:

Although hypothetical, half reactions are properly balanced chemical processes. Since V²⁺(aq) increases its oxidation number by one, from +2 to +3, in the first half reaction, an electron is shown as a product of the change. Similarly, two electrons are produced when the oxidation number of zinc increases from 0 to +2 in the second half reaction. When half reactions for hypothetical isolated oxidations and reductions are combined, the electrons must cancel if the equation for a possible overall chemical reaction is to result.

The use of half reactions is a natural outgrowth of the application of the electron-transfer concept to redox reactions. Since the oxidation-state principle allows any redox reaction to be analyzed in terms of electron transfer, it follows that all redox reactions can be broken down into a complementary pair of hypothetical half reactions. Electrochemical cells (in which chemical energy can be converted to electrical energy, and vice versa) provide some physical reality to the half-reaction idea. Oxidation and reduction half reactions can be carried out in separate compartments of electrochemical cells, with the electrons flowing through a connecting wire and the circuit completed by some arrangement for ion migration between the two compartments (but the migration need not involve any of the materials of the oxidation–reduction reactions themselves).

Theoretical aspects » Redox potentials for common half reactions

The analysis of the electrical potential, or voltage, developed by pairing various half reactions in electrochemical cells has led to the determination of redox potentials for a substantial number of common half reactions. While a detailed description of redox potentials requires the methods of thermodynamics (the branch of physics concerned with the role played by heat in the transformation of matter or energy), a great deal of useful information can be obtained from redox potentials with minimal recourse to formal theory. Basically, a table of half-cell potentials is a summary of the relative tendencies of different oxidations and reductions to occur. The Table lists selected half reactions and their corresponding reduction potentials (which are symbolized by E°).

The physical significance of the values is directly linked to several agreements about their use. First, the greater the value of E° (the reduction potential), the greater the tendency of a half reaction to proceed from left to right (as written). The half reactions in the

Table are listed from top to bottom in order of decreasing E°: the higher a reaction’s position on the list, the greater the tendency of the reactants to accept electrons. In other words, reagents high on the list, such as fluorine gas (F₂) and permanganate ion (MnO₄⁻), are strong oxidizing agents. Second, the reduction of hydrogen ions (H⁺) to hydrogen gas (H₂) is arbitrarily assigned the value 0 volts. Half cells with positive reduction potentials involve reactants that are more readily reduced than H⁺; conversely, those with negative potentials involve reactants that are more difficult to reduce than hydrogen ions.

With the aid of reduction potentials, it is possible to predict whether a particular oxidation–reduction reaction can occur. The predictions require breaking down the overall reaction into two half reactions of known reduction potentials. For example, if a strip of zinc metal is dipped into a solution containing copper(II) ion, the possibility exists for a redox process, which can be regarded as the sum of the half reactions aqueous zinc ion (Zn²⁺[aq]) to zinc metal (Zn[s]) and aqueous copper ion (Cu²⁺[aq]) to copper metal (Cu[s]), as follows:

Combining these two half reactions requires writing the zinc ion to zinc metal half reaction the reverse of the way it appears in the Table. When the direction of a half reaction is reversed, so that it can be added to another half reaction, the sign of its redox potential is also reversed (in this case, from negative to positive), and the two reduction potential values are then added.

The resulting E° value for the net reaction, +1.10 volts, measures the tendency of the net reaction to occur. If E^o for a particular net reaction is positive, the process may be expected to occur spontaneously when the reactants are mixed at specified concentrations (one mole per litre; see below Oxidation–reduction equilibria). Therefore, it is predicted that copper metal should be deposited on a strip of zinc metal when the latter is immersed in a solution of a copper(II) salt. This reaction is, in fact, readily observed in the laboratory. A more specific physical interpretation of the +1.10 volt value is that it represents the voltage that would be produced by an ideal electrochemical cell based on the copper(II) ion to copper metal and zinc(II) ion to zinc metal half reactions with all the reagents at specified concentrations.

When the same two half cells are combined, with both their directions (and therefore the signs of their redox potentials) reversed, it is predicted that the reverse reaction, the depositing of zinc metal from a zinc(II) ion solution onto a copper strip, will not occur spontaneously. As in the case of E° values for half reactions, those for net redox reactions also change sign when the direction of the reaction is reversed.

The results of the copper–zinc system can be applied more generally to the half reactions in the Table. For example, copper(II) ion in water (Cu²⁺[aq]) is an oxidant strong enough to force a half reaction lower on the table to proceed spontaneously in the opposite direction of that written. Therefore, not only is copper(II) ion expected to oxidize zinc metal (Zn[s]) to zinc(II) ion (Zn²⁺[aq]); it is also predicted to oxidize hydrogen gas (H₂[g]) to hydrogen ion (H⁺) and sodium metal (Na[s]) to sodium ion (Na⁺). Similarly, fluorine gas (F₂[g]), the strongest oxidant listed in the Table, is predicted to oxidize spontaneously the products of all the other half reactions in the table. In contrast, the strongest reducing agent is solid sodium metal (Na[s]), and it is expected spontaneously to reduce the reactants of all the other half cells.

Theoretical aspects » Oxidation–reduction equilibria

In practice many chemical reactions can be carried out in either direction, depending on the conditions. The spontaneous direction predicted for a particular redox reaction by half-cell potentials is appropriate to a standard set of reaction conditions. Specifically, the temperature is assumed to be 25° C with reagents at specified concentrations. Gases are present at one atmosphere pressure and solutes at one mole per litre (one molecular weight in grams dissolved in one litre of solution) concentration (1M). Solids are assumed to be in contact with the reaction solution in their normal stable forms, and water is always taken to be present as the solvent. Many practical problems can be solved directly with standard reduction potentials.

The usefulness of reduction potentials is greatly extended, however, by a thermodynamic relationship known as the Nernst equation, which makes it possible to calculate changes in half-cell potentials that will be produced by deviations from standard concentration conditions. In the reaction between zinc metal and copper(II) ion, standard conditions for zinc and copper metal require simply that both solids be present in contact with the solution; the E° values are not affected by either the total or proportionate amounts of the two metals. The calculation that the overall reaction is spontaneous by +1.10 volts is based on standard one mole per litre (1M) concentrations for aqueous zinc(II) ion (Zn²⁺[aq]) and aqueous copper(II) ion (Cu²⁺[aq]). Using the Nernst equation it is found that E° for the overall reaction will be +1.10 volts as long as both ions are present in equal concentrations, regardless of the concentration level.

On the other hand, if the ratio of the zinc(II) to copper(II) ion concentrations is increased, the reduction potential (E°) falls until, at a very high preponderance of zinc ion, E° becomes 0 volt. At this point, there is no net tendency for the reaction to proceed spontaneously in either direction. If the zinc(II) to copper(II) ion ratio is increased further, the direction of spontaneity reverses, and zinc ion spontaneously oxidizes copper metal. In practice, such high zinc(II) to copper(II) ion concentration ratios are unattainable, which means that the reaction can only be carried out spontaneously with copper(II) ion oxidizing zinc metal. Many reactions with E° values smaller than +1.10 volts under standard conditions can be carried out in either direction by adjusting the ratio of product and reactant concentrations. The point at which E° = 0 volt represents a state of chemical equilibrium. When chemical reactions are at equilibrium, the concentrations of the reagents do not change with time, since net reaction is not spontaneous in either direction. Measurements of half-cell potentials combined with Nernst-equation calculations are a powerful technique for determining the concentration conditions that correspond to chemical equilibrium.

Theoretical aspects » Reaction rates » Predictability

There are practical limitations on predictions of the direction of spontaneity for a chemical reaction, the most important arising from the problem of reaction rates. An analogy can be made with the simple physical system of a block on a sloping plane. Because of the favourable energy change, the block tends spontaneously to slide down, rather than up, the slope, and, at mechanical equilibrium, it will be at the bottom of the slope, since that is the position of lowest gravitational energy. How rapidly the block slides down is a more complex question, since it depends on the amount and kind of friction present. The direction of spontaneity for a chemical reaction is analogous to the downhill direction for a sliding block, and chemical equilibrium is analogous to the position at the bottom of the slope; the rate at which equilibrium is approached depends on the efficiency of the available reaction processes. Between zinc metal and aqueous copper(II) ion, the reaction proceeds without observable delay, but various other spontaneous redox processes proceed at imperceptibly slow rates under ordinary conditions.

http://www.britannica.com/EBchecked/topic/436636/oxidation-reduction-reaction/49298/Redox-potentials-for-common-half-reactions

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