Thursday, November 20, 2008

OXIDATION And REDUCTION, balancing redox reaction


What is a chemical reaction?
Examples of chemical changes, physical changes, and some gray areas.
Chemical equations of chemical reactions.
Balancing equations.
Balancing ionic equations with polyatomic ions.
Balancing combustion (burning) equations.
Balancing by overview.
Synthesis reactons.
Decomposition reactions.
Single replacement reactions.
Double replacement reactions.
Some chemical equations for you to balance.


A chemical reaction is material changing from a beginning mass to a resulting substance. The hallmark of a chemical reaction is that new material or materials are made, along with the disappearance of the mass that changed to make the new. This does not mean that new elements have been made. In order to make new elements, the nuclear contents must change. There are magnitudes of difference in the amounts of energy in ordinary chemical reactions compared to nuclear reactions, the rearrangement of the nuclei of atoms to change to new elements is enormous compared to the smaller energies of chemical changes. The alchemists, in their efforts to change less expensive metals to gold, did not have the fundamental understanding of what they were attempting to do to appreciate the difference.

A chemical equation is a way to describe what goes on in a chemical reaction, the actual change in a material. Chemical equations are written with the symbols of materials to include elements, ionic or covalent compounds, aqueous solutions, ions, or particles. There is an arrow pointing to the right that indicates the action of the reaction. The materials to the left of the arrow are the reactants, or materials that are going to react. The materials to the right of the arrow are the products, or materials that have been produced by the reaction. The Law of Conservation of Mass states that in a chemical reaction no mass is lost or gained. The Law of Conservation of Mass applies to individual types of atom. One could say that for any element, there is no loss or gain of that element in a chemical reaction. There are such things as reversible reactions, reactions in which the products reassemble to become the original products. Reversible reactions are symbolized in chemical equations by a double-headed arrow, but the standard remains to call the materials on the left the reactants and the materials on the right the products.

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Chemical reactions, also called chemical changes, are not limited to happening in a chemistry lab. Here are some examples of chemical reactions with the corresponding chemical equations:

A silver spoon tarnishes. The silver reacts with sulfur in the air to make silver sulfide, the black material we call tarnish.
2 Ag + S Ag2S

An iron bar rusts. The iron reacts with oxygen in the air to make rust.
4 Fe + 3 O2 2 Fe2O3

Methane burns. Methane combines with oxygen in the air to make carbon dioxide and water vapor.
CH4 + 2 O2 CO2 + 2 H2O

An antacid (calcium hydroxide) neutralizes stomach acid (hydrochloric acid).
Ca(OH)2 + 2 HCl CaCl2 + 2 H2O

Glucose (simple sugar) ferments to ethyl alcohol and carbon dioxide. The sugar in grapes or from grain ferments with yeast to make the alcohol and carbon dioxide. The carbon dioxide is the gas that bubbles out of beer or champaign.
C6H12O6 (glucose) 2 C2H5OH (ethyl alcohol) + 2 CO2

Alcohol plus oxygen becomes vinegar and a molecule of water. As in the fermentation of glucose, this is a more complex reaction than it appears here because it is a biochemical reaction.
C2H5OH + O2 HC2H3O2 + H2O

As a general rule, biochemical happenings make poor examples of basic chemical reactions because the actual reaction is carried on within living things and under enzyme control.


Here are some examples of changes that are NOT chemical reactions. In each case, the original material or materials may be reclaimed by physical processes.

Water boils out of a kettle or condenses on a cold glass.
An aluminum pot is put on a burner and gets hot.
Dry ice goes from a solid to a gaseous form of carbon dioxide (sublimation).
Gold melts or solidifies.
Sand is mixed in with salt.
A piece of chalk is ground to dust.
Glass breaks.
An iron rod gets magnetized.
A lump of sugar dissolves in water.


Even more telling are the gray areas. Are these changes chemical or physical? Why? (Punch the * discussion link after each one for a discussion on why that example is a gray area.)

Table salt dissolves in water. *
A hydrated crystal, such as blue vitriol, is dried with heat. *
Lightning makes ozone (O3) from oxygen (O2). The ozone then reverts to oxygen. *
Carbon dioxide dissolves in water. *
Ammonia gas dissolves in water. *
With pressure and heat graphite becomes diamond. *
An egg is cooked.*
A tree dies. *

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In order to write the chemical equations, you must first know the formulas for the materials involved. The formulas must be written on the proper side of the arrow - - reactants on the left and products on the right. The order in which the reactants and products are written does not matter, just as long as every material is on the proper side. Once the materials involved in the reaction are written correctly, DON�T TOUCH THEM. If you need to draw a box around each participant in the reaction to keep your grubby paws off the materials, do it.

Very often you will see the descriptions of the materials in the reaction in parentheses after the material. A gas is shown by (g). A solid material is shown by (s). A liquid is shown by (l). A material dissolved in water (an aqueous solution) is shown by (aq). An upwards pointing arrow () indicates a gas being produced, and a downwards pointing arrow indicates a solid precipitate being produced.

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Now comes the fun part, balancing the reaction. The Law of Conservation of Mass states that in a chemical reaction there is no loss of mass. Each type of element will have the same amount before the reaction and after the reaction, or as reactant and product. But you can�t change the materials that participate in the reaction, so you must write an integer coefficient in front of (to the left of) each material in the reaction to make sure every type of atom has the same number on each side of the reaction. Let�s start with the reaction of the Haber process:

Nitrogen gas plus hydrogen gas under pressure and at high temperature turn into ammonia. First write the materials correctly. Nitrogen and hydrogen are diatomic gases. Ammonia is a binary covalent memory item. The nitrogen and hydrogen are the reactants, and the ammonia is the product. Leave room for the coefficients in front of the materials.

_ N2 + _ H2 _ NH3

You can begin with either the nitrogen or the hydrogen. There are two nitrogen atoms on the left and only one on the right. In order to balance the nitrogen atoms, place a �2� in front of the ammonia.

_ N2 + _ H2 2 NH3

There are two hydrogens on the left and six on the right. We balance the hydrogens by placing a �3� in front of the hydrogen gas.

_ N2 + 3 H2 2 NH3

Now go back and check to make sure everything is balanced. There are two nitrogen and six hydrogens on both sides of the reaction. It is balanced. There is no coefficient shown in front of the nitrogen. There is no need to write ones as coefficients. The reaction equation is:

N2 + 3 H2 2 NH3

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Silver nitrate and calcium chloride solutions combined produce a precipitate of silver chloride and leave a solution of calcium nitrate. This time we have ionic compounds in the reaction. Until you are sure of the compounds, you might want to write the ionic materials as the ions, as demonstrated here.

_ Ag+(NO3)- + _ Ca2+Cl-2 _ Ag+Cl- + _ Ca2+ (NO3) -2

Notice that from one side to the other there is no change in the nitrate ion. In this case you can count the nitrate ion as a whole rather than splitting it up into nitrogen and oxygen. Your thoughts might go this way: How many silvers on the right? One. How many silvers on the left? One. They are the same. How many nitrates on the left? One. How many nitrates on the left? One. How many nitrates on the right? Two. We need to put a coefficient of two in front of the silver nitrate.

2 AgNO3 + _ CaCl2 _ AgCl + _ Ca (NO3)2

This changes the balance of silvers, so we have to put a two in front of the silver chloride.

2 AgNO3 + _ CaCl2 2 AgCl + _ Ca (NO3)2

Now let�s check again. Two silvers on each side. Two nitrates on each side. One calcium on each side and two chlorides on both sides. The balanced reaction is:

2 AgNO3 + CaCl2 2 AgCl + Ca (NO3)2

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Sulfuric acid and potassium hydroxide neutralize each other to make water and potassium sulfate. Here is an acid-base neutralization. These make a salt (Not necessarily common table salt.) and water. (Notice the ionic materials are written with the ion notation so they are sure to be right. Water and sulfuric acid are memory items and should not need to be written in ion form, though you could write the ions to make sure they are right.)

_ H2SO4 + _ K+(OH)- _ K+2(SO4)2+ + _ H2O

The water is made from the hydrogen ion of the acid and the hydroxide ion of the base. Notice that it is a lot easier to understand how to balance the reaction if you write the water as if it were an ionic compound.

_ H2(SO4)+ _ K+(OH)- _ K+2(SO4)2+ + _ H+(OH)-

This is easier now because the hydrogen in the acid does not get confused with the hydrogen in the hydroxide of the base. Two hydrogens on each side. One sulfate on both sides. Two potassiums and two hydroxides on each side.

H2(SO4)+ 2 K(OH) K2(SO4) + 2 H(OH)

The reaction is now balanced.

Next is an example of having to go around the equation again. Phosphoric acid and calcium hydroxide react to make water and calcium phosphate.

_ H3PO4 + _ Ca2+(OH)-2 _ H+(OH)- + _ Ca2+3(PO4)3-2

First put a three on the water to balance the hydrogen in the phosphoric acid.

_ H3PO4 + _ Ca(OH)2 3 H(OH) + _ Ca3(PO4)2

Now put a two on the phosphoric acid to balance the phosphate from the calcium phosphate.

2 H3PO4 + _ Ca(OH)2 3 H(OH) + Ca3(PO4)2

We have changed the amount of hydrogen ion, so we will have to change it on the right again.

2 H3PO4 + _ Ca(OH)2 6 H(OH) + Ca3(PO4)2

And change the coefficient in front of the Ca(OH)2 to match the calcium on the right side.

2 H3PO4 + 3 Ca(OH)2 6 H(OH) + Ca3(PO4)2

Only now does the rest of the equation balance with six hydrogens, six hydroxides, two phosphates, and three calciums on each side.

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Most burning reactions are the oxidation of a fuel material with oxygen gas. Complete burning produces carbon dioxide from all the carbon in the fuel, water from the hydrogen in the fuel, and sulfur dioxide from any sulfur in the fuel. Methane burns in air to make carbon dioxide and water.

_ CH4 + _ O2 _ H2O + _ CO2

Easy. Put a two in front of the water to take care of all the hydrogens and a two in front of the oxygen. Anything you have to gather (any atom that comes from two or more sources in the reactants or gets distributed to two or more products) should be considered last.

CH4 + _ O2 2 H2O + CO2

CH4 + 2 O2 2 H2O + CO2

What if the oxygen does not come out right? Let�s consider the equation for the burning of butane, C4H10.

_ C4H10 + _ O2 _ CO2 + _ H2O

Insert the coefficients for carbon dioxide and water.

_ C4H10 + _ O2 4 CO2 + 5 H2O

We now have two oxygens on the left and thirteen oxygens on the right. The real problem is that we must write the oxygen as a diatomic gas. The chemical equation is not any different from an algebraic equation in that you can multiply both sides by the same thing and not change the equation. Multiply both sides by two to get the following.

2 C4H10 + _ O2 8 CO2 + 10 H2O

Now the oxygens are easy to balance. There are twenty-six oxygens on the right, so the coefficient for the oxygen gas on the left must be thirteen.

2 C4H10 + 13 O2 8 CO2 + 10 H2O

Now it is correctly balanced. What if you finally balanced the same equation with:

4 C4H10 + 26 O2 16 CO2 + 20 H2O


6 C4H10 + 39 O2 24 CO2 + 30 H2O

Either equation is balanced, but not to the lowest integer. Algebraically you can divide these equations by two or three to get the lowest integer coefficients in front of all of the materials in the equation.

Now that we are complete pyromaniacs, let�s try burning isopropyl alcohol, C3H7OH.

_ C3H7OH + _ O2 _ CO2 + _ H2O

First take care of the carbon and hydrogen.

_ C3H7OH + _ O2 3 CO2 + 4 H2O

But again we come up with an oxygen problem. The same process works here. Multiply the whole equation (except oxygen) by two.

2 C3H7OH + _ O2 6 CO2 + 8 H2O

Now the number nine fits in the oxygen coefficient. (Do you understand why?) The equation is balanced with six carbons, sixteen hydrogens, and twenty oxygens on each side.

2 C3H7OH + 9 O2 6 CO2 + 8 H2O

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Some equations are just mean, nasty, and rotten and defy your efforts to balance them. For some of these equations, a process I call overview is useful. Take as an example the smelting of magnetite, an iron ore.

_ Fe3O4 + _ CO _ CO2 + _ Fe

Unless you just happen to hit it right, you are unlikely to balance this equation with the trial method. (Go ahead and try it before you read further.)

The process overview shows that for each oxygen that the magnetite has, one carbon monoxide must turn to carbon dioxide. The carbon monoxide and carbon dioxide must have a coefficient that is four times the coefficient of the magnetite. Leave the magnetite coefficient and put a '4' in front of the carbon monoxide and carbon dioxide.

_ Fe3O4 + 4 CO 4 CO2 + _ Fe

The carbon and oxygen is balanced, leaving only the iron to be balanced.

Fe3O4 + 4 CO 4 CO2 + 3 Fe


The balancing of equations involving a reduction and oxidation will be considered in the chapter on redox ( reduction and oxidation reactions).

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The title of this section contains four names for the same type of reaction. Your text may use any of these. Chemtutor prefers the first of the names and will use �synthesis� where your text may use one of the other words. The hallmark of a synthesis reaction is a single product. A synthesis reaction might be symbolized by:

A + B AB

Two materials, elements or compounds, come together to make a single product. Some examples of synthesis reactions are: Hydrogen gas and oxygen gas burn to produce water.

2 H2 + O2 2 H2O and

sulfur trioxide reacts with water to make sulfuric acid.

H2O + SO3 H2SO4

What would you see in a �test tube� if you were witness to a synthesis reaction? You would see two different materials combine. A single new material appears.

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Of the names for this type of reaction, Chemtutor again prefers the first. Mozart composed until age 35. After that, he decomposed. Yes, a decomposition is a coming apart. A single reactant comes apart into two or more products, symbolized by:

XZ X + Z

Some examples of decomposition reactions are: potassium chlorate when heated comes apart into oxygen gas and potassium chloride

2 KClO3 2 KCl + 3 O2

and heating sodium bicarbonate releases water and carbon dioxide and sodium carbonate.

6 NaHCO3 3 Na2CO3 + 3 H2O + 3 CO2

In a �test tube� you would see a single material coming apart into more than one new material.

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Here is an example of a single replacement reaction: silver nitrate solution has a piece of copper placed into it. The solution begins to turn blue and the copper seems to disappear. Instead, a silvery-white material appears.

2 AgNO3 + Cu Cu(NO3)2 + 2 Ag

A solution of an ionic compound has available an element. The element replaces one of the ions in the solution and a new element appears from the ion in solution. This type of reaction is called a replacement because a free element replaces one of the ions in a compound. There are two types of single replacement reactions, anionic and cationic. A cationic single replacement is what happened in the case of the silver being replaced by the copper in the above reaction because both the silver and the copper are only likely to make cations. An anionic single replacement is also possible. Into a potassium iodide solution chlorine gas is bubbled. The chlorine is used up and the solution turns purple-brown from the iodine. This is an example of an anionic single replacement reaction.

2 KI + Cl2 2 KCl + I2

Could you start with copper II nitrate and silver metal and get silver nitrate and copper metal, or could you start with potassium chloride and iodine and get potassium iodide and chlorine? No. The reactions don�t work that way. You can arrange cations or anions in a list of which ion will replace the next. This type of list is an activity series. The activity series of cation elements (metals) shows that gold is the least active metal. That should not be surprising, because gold does not tarnish. If we were to consider the Group 1 elements only on the activity list, lithium is the least active and francium is the most active, with each larger element being more active than the smaller one above it on the Periodic Chart. On the other side of the chart we could consider an activity series for anions. Taking just the halogens, the smallest halogen, fluorine is the most active. As the size of the halogen increases down the chart, the activity decreases. If an element is more active than the element of the same sign in an ionic solution, the more active element will replace it.

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Some texts refer to single and double replacement reactions as solution reactions or ion reactions. That is understandable, considering these are mostly done in solutions in which the major materials we would be considering are in ion form. Chemtutor thinks that there is some good reason to call double replacement reactions de-ionizing reactions because a pair of ions are taken from the solution in these reactions. Let�s take an example.

AgNO3 + KCl AgCl(s) + KNO3

Above is the way the reaction might be published in a book, but the equation does not tell the whole story. Dissolved silver nitrate becomes a solution of silver ions and nitrate ions. Potassium chloride ionizes the same way. When the two solutions are added together, the silver ions and chloride ions find each other and become a solid precipitate. (They �rain� or drop out of the solution, this time as a solid.) Since silver chloride is insoluble in water, the ions take each other out of the solution.

Ag+ + (NO3)- + K+ +Cl- AgCl + K+ + (NO3)-

Here is another way to take the ions out of solution. Hydrochloric acid and sodium hydroxide (acid and base) neutralize each other to make water and a salt. Again the solution of hydrochloric acid is a solution of hydrogen (hydronium ions in the acid and base section) and chloride ions. The other solution to add to it, sodium hydroxide, has sodium ions and hydroxide ions. The hydrogen and hydroxide ions take each other out of the solution by making a covalent compound (water).

HCl + NaOH HOH + NaCl or

H+ + Cl- + Na+ + (OH)- HOH + Na+ + Cl-

One more way for the ions to be taken out of the water is for some of the ions to escape as a gas.

CaCO3 + 2 HCl CaCl2 + H2O + CO2

Ca2+ + (CO3)2- + 2 H+ + 2 Cl- Ca2+ + 2 Cl- + H2O + CO2

The carbonate and hydrogen ions became water and carbon dioxide. The carbon dioxide is lost as a gas to the ionic solution, so the equation can not go back.

One way to consider double replacement reactions is as follows: Two solutions of ionic compounds are really just sets of dissolved ions, each solution with a positive and a negative ion material. The two are added together, forming a mixture of four ions. If two of the ions can form (1) an insoluble material, (2) a covalent material such as water, or (2) a gas that can escape, it qualifies as a reaction. Not all of the ions are really involved in the reaction. Those ions that remain in solution after the reaction has completed are called spectator ions, that is, they are not involved in the reaction. There is some question as to whether they can see the action of the other ions, but that is what they are called.

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1. sulfur trioxide and water combine to make sulfuric acid.

2. lead II nitrate and sodium iodide react to make lead iodide and sodium nitrate.

3. calcium fluoride and sulfuric acid make calcium sulfate and hydrogen fluoride (Hydrofluoric acid)

4. calcium carbonate will come apart when you heat it to leave calcium oxide and carbon dioxide.

5. ammonia gas when it is pressed into water will make ammonium hydroxide.

6. sodium hydroxide neutralizes carbonic acid

7. zinc sulfide and oxygen become zinc oxide and sulfur.

8. lithium oxide and water make lithium hydroxide

9. aluminum hydroxide and sulfuric acid neutralize to make water and aluminum sulfate.

10. sulfur burns in oxygen to make sulfur dioxide.

11. barium hydroxide and sulfuric acid make water and barium sulfate.

12. aluminum sulfate and calcium hydroxide become aluminum hydroxide and calcium sulfate.

13. copper metal and silver nitrate react to form silver metal and copper II nitrate.

14. sodium metal and chlorine react to make sodium chloride.

15. calcium phosphate and sulfuric acid make calcium sulfate and phosphoric acid.

16. phosphoric acid plus sodium hydroxide.

17. propane burns (with oxygen)

18. zinc and copper II sulfate yield zinc sulfate and copper metal

19. sulfuric acid reacts with zinc

20. acetic acid ionizes.

21. steam methane to get hydrogen and carbon dioxide

22. calcium oxide and aluminum make aluminum oxide and calcium

23. chlorine gas and sodium bromide yield sodium chloride and bromine


1. SO3 + H2O H2SO4

2. Pb(NO3)2 + 2NaI PbI2 + 2NaNO3
DOUBLE REPLACEMENT (lead II iodide precipitates)

3. CaF2 + H2SO4 CaSO4 + 2 HF
DOUBLE REPLACEMENT (calcium sulfate precipitates)

4. CaCO3 CaO + CO2

5. NH3 + H2O NH4OH

6. 2 NaOH + H2CO3 Na2CO3 + 2 H2O

7. 2 ZnS + O2 2 ZnO + 2 S

8. Li2O + H2O 2 LiOH

9. 2 Al(OH)3 + 3 H2SO4 6 H2O + Al2(SO4)3

10. S + O2 SO2

11. Ba(OH)2 + H2SO4 2 H2O + BaSO4

12. Al2(SO4)3 + 3 Ca(OH)2 2 Al(OH)3 + 3 CaSO4
(BOTH calcium sulfate and aluminum hydroxide are precipitates.)

13. Cu + 2AgNO3 2Ag + Cu(NO3)2

14. 2Na + Cl2 2 NaCl

15. Ca3(PO4)2 + 3 H2SO4 3 CaSO4 + 2 H3PO4

16. H3(PO4) + 3 NaOH Na3PO4 + 3 H2O

17. C3H8 + 5 O2 4 H2O + 3 CO2

18. Zn + CuSO4 ZnSO4 + Cu

19. H2SO4 + Zn ZnSO4 + H2

20. HC2H3O2 H+ + (C2H3O2)-

21. 2 H2O + CH4 4 H2 + CO2

22. 3 CaO + 2 Al Al2O3 + 3 Ca

23. Cl2 + 2 NaBr 2 NaCl + Br2


How are redox reactions different?

Oxidation states.

Is it a redox reaction?

Half reactions.

Reduction or oxidation?

Practice with assigning oxidation states.

Balancing redox reactions.


Redox is the term used to label reactions in which the acceptance of an electron (reduction) by a material is matched with the donation of an electron (oxidation). A large number of the reactions already mentioned in the Reactions chapter are redox reactions.

Synthesis reactions are also redox reactions if there is an exchange of electrons to make an ionic bond. If chlorine gas is added to sodium metal to make sodium chloride, the sodium has donated an electron and the chlorine has accepted an electron to become a chloride ion or an attached chlorine.

If a compound divides into elements in a decomposition, a decomposition reaction could be a redox reaction. The electrolysis of water is a redox reaction. With a direct electric current through it, water can be separated into oxygen and hydrogen. H2O ==> H2 + O2 The oxygen and hydrogen in the water are attached by a covalent bond that breaks to make the element oxygen and the element hydrogen. Learning more about the conditions for redox reactions will show that the electrolysis of water is a redox reaction.

A single replacement reaction is always a redox reaction because it involves an element that becomes incorporated into a compound and an element in the compound being released as a free element.

A double replacement reaction usually is not a redox reaction.

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Before we go any further into redox, we must understand oxidation states. The idea of oxidation state began with whether or not a metal was attached to an oxygen. Unattached (free) atoms have an oxidation state of zero. Since oxygen almost always takes in two electrons when it is not a free element, the combined form of oxygen (oxide) has an oxidation state of minus two. The exception to a combined oxygen taking two electrons is the peroxide configuration. Peroxide can be represented by -O-O- where the each dash is a covalent bond and each �O� is an oxygen atom. Peroxide can be written as a symbol, (O2)2-. The over-simplified way of showing this is that each oxygen atom has a negative one oxidation state, but that is not really so because the peroxides do not come in individual oxygen atoms. Peroxides are not as stable as oxides, and there are very many fewer peroxides in nature than oxides. H2O2 is hydrogen peroxide.

Hydrogen in compound always an oxidation state of plus one except as a hydride. A hydride is a compound of a metal and hydrogen. Hydrides react with water, so there are no hydrides found in nature. The formula XH or XH2 or XH3 or even XH4 where X is a metal is the general chemical formula for hydride.

The rules for oxidation state are in some ways arbitrary and unnatural, but here they are:

1. Any free (unattached) element with no charge has the oxidation state of zero. Diatomic gases such as O2 and H2 are also in this category.

2. All compounds have a net oxidation state of zero. The oxidation state of all of the atoms add up to zero.

3. Any ion has the oxidation state that is the charge of that ion. Polyatomic ions (radicals) have an oxidation state for the whole ion that is the charge on that ion. The ions of elements in Group I, II, and VII (halogens) and some other elements only have one likely oxidation state.

4. Oxygen in compound has an oxidation state of minus two, except for oxygen as peroxide, which is minus one.

5. Hydrogen in compound has an oxidation state of plus one, except for hydrogen as hydride, which is minus one.

6. In radicals or small covalent molecules, the element with the greatest electronegativity has its natural ion charge as its oxidation state.


Now would be a good time to try the oxidation state problems beginning the practice page at the end of this chapter. Problems 1-30 are good examples for practice of assigning oxidation states.

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A redox reaction will have at least one type of atom releasing electrons and another type of atom accepting electrons. How can you most easily tell if a reaction is redox? Label every atom on both the reactant and product side of the equation with its oxidation number. If there is a change in oxidation number from one side of the equation to the other of the same species of atom, it is a redox reaction. Each complete equation must have at least one atom species losing electrons and at least one atom species gaining electrons. The loss and gain of electrons will be reflected in the changes of oxidation number.

Let�s take the following equation: K2(Cr2O7) + KOH ==> 2 K2(CrO4) + H2O Is it a redox equation or not? Potassium dichromate and potassium hydroxide make potassium chromate and water. Some of the atoms are easy. All of the oxygens in compound have an oxidation state of minus two. All of the hydrogens have an oxidation state of plus one. Potassium is a group one element, so it should have an oxidation state of plus one in the compounds. That seems to make sense because dichromate and chromate ions have a charge of minus two and there are two potassium atoms in each compound. Hydroxide ion has a charge of minus one and it has one potassium. But what about the chromium atoms? We can do a little primitive math on the material either from the starting point of the compound or the ion to find the oxidation state of chromium in that compound. The entire compound must have a net oxidation state of zero, so the oxidation numbers of two potassiums one chromium and four oxygens must equal to zero. 2 K + Cr + 4 O = 0 We know the oxidation state of everything else but the chromium. 2(+1) + Cr + 4 (-2) = 0 and Cr = +6. Or we could do it from the point of view of the chromate ion. Cr + 4 O = -2 The oxygens are minus two each. Cr + 4 (-2) = -2 Either way Cr = +6. Now the dichromate; 2 K + 2 Cr + 7 O = 0 and 2 (+1) + 2 Cr + 7 (-2) = 0. Then 2 Cr = +12 and Cr = +6. You can do the math for the dichromate ion to see for yourself that the chromium does not change from one side of this equation to the other. As suspicious-appearing as the equation might have seemed to you, it is not a redox reaction.

Consider copper metal in silver nitrate solution becomes silver metal and copper II nitrate. The oxygens do not change. Oxygen in compound is negative two on both sides. The nitrogen can not change. It does not move out of the nitrate ion where it has an oxidation state of plus five. (Is that right?) The other two have to change because they both are elements with a zero oxidation state on one side and in compound on the other. Silver goes from plus one to zero and Copper goes from zero to plus two.
AgNO3 + Cu ==> Cu(NO3)2 + Ag (not a balanced reaction)

Think of this on a number line. The copper is oxidized because its oxidation number goes up from zero to plus two. The silver is reduced because its oxidation number reduces from plus one to zero.

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Consider the reaction: AgNO3 + Cu ==> Cu(NO3)2 + Ag (not a balanced reaction)

Half reactions are either an oxidation or a reduction. Only the species of atom that is involved in a change is in a half reaction. In the above reaction, silver goes from plus one to zero oxidation state, but to account for everything, the electrons must be placed into the half reaction. e- + Ag+ ==> Ag0 (Reduction) Notice that the half reaction must be balanced in charge also and that the only way to balance it is to add electrons to the more positive side. The other half reaction is that of copper. Cu0 ==> Cu+2 + 2 e- (Oxidation)

This time the material is oxidized and the electrons must appear on the product side. We must double the silver half reaction to cancel out the electrons from right to left. The two half reactions can be added together to make one reaction, thus.
2( e- + Ag+ ==> Ag0)
Cu0 ==> Cu+2 + 2 e- and the total reaction is:
Cu0 + 2Ag+ ==> Cu+2 + 2Ag0

In the complete reaction the number of electrons lost must equal the number of electrons gained. The number of electrons used in the reduction half reaction must equal the number of electrons produced in the oxidation half reaction. The entire half reactions must be multiplied by numbers that will equalize the numbers of electrons, and the final complete balanced chemical reaction must show these number relationships.

One of the important bits of information from adding the half reactions in this case is that the entire chemical equation will have to have two silver atoms for every copper atom in the reaction for the reaction to balance electrically. This type of information from the half reactions is sometimes the easiest or only way to balance a chemical equation. The redox balancing problems beginning with number 31 at the end of the chapter are good help for your further understanding.

From doing this math on a number of materials, you will find that it is possible to get some strange-looking oxidation states, to include some fractional ones. The oxidation state math works on fractional oxidation states also, even though fractional charges are not possible.

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A reduction of a material is the gain of electrons. An oxidation of a material is the loss of electrons. This system comes from the observation that materials combine with oxygen in varying amounts. For instance, an iron bar oxidizes (combines with oxygen) to become rust. We say that the iron has oxidized. The iron has gone from an oxidation state of zero to (usually) either iron II or iron III. This may be difficult to remember. The easier way to tell if a half reaction is a reduction or oxidation is to plot the changing ion into the number line. If the oxidation state of the ion goes up the number line, it is an oxidation. If it goes down the number line, it is a reduction. Based on the KIS principle (Keep It Simple), remember only one rule for this.

Someone, in a fit of perversity, decided that we needed more description for the process. A material that becomes oxidized is a reducing agent, and a material that becomes reduced is an oxidizing agent.


Water, aluminum, copper

chromium, gold,

carbon zinc , etc electrical potential and voltages

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For each element in the following materials list the number of the rule you use to assign oxidation state of that element and list the oxidation state you have found it to be.

Summary of Oxidation State Rules
1. Free element O.S. = 0 2. Compound total O.S. = 0 3. Ion O.S. = charge
4. Oxygen O.S. = -2 5. Hydrogen O.S. = +1 6. Electronegativity rules

1. NaCl Na 3; Cl 3 Na = +1 , Cl = -1
2. KMnO4 K 3; O 4; Mn 3,4 or 2,3,4 K = +1, O = -2, Mn = +7
3. diamond C 1 C = 0
4. CO2 O 4; C 4,2 C = +4, O = -2
5. CO O 4; C 4,2 C = +2, O = -2
6. KCN K 3; N 6; C 3,6, 2 K = +1, C = +2, N=-3
7. Na4Fe(CN)6 Na 3; N 6; C 6,3; Fe 3,6,2 Na= +1, N= -3, C= +2, Fe= +2
8. Fe2O3 O 4; Fe 4,2 O = -2, Fe = +3
9. Fe3O4 O 4; Fe 4,2 O = -2, Fe = +8/3
10. (ClO4)- O 4; Cl 4,3 O = -2, Cl = +7
11. (ClO3)- O 4; Cl 4,3 O = -2, Cl = +5
12. (ClO2)- O 4; Cl 4,3 O = -2, Cl = +3
13. (ClO)- O 4; Cl 4,3 O = -2, Cl = +1
14. Cl- Cl 3 Cl = -1
15. Cl2 Cl 1 Cl = 0
16. P2O5 O 4; P 4,2 O = -2, P = +5
17. P4O6 O 4; P 4,2 O = -2, P = +3
18. H3PO4 H 5; O 4; P 5,4,2 or 4,3 H = +1, O = -2, P = +5
19. Mg3N2 Mg 3; N 3,2 Mg = +2, N = -3
20. MgH2 Mg 3; H 3,2 Mg = +2, H = -1 (hydride!)
21. NH3 H 5; N 5,2 H = +1, N = -3
22. N2H4 H 5; N 5,2 H = +1, N = -2
23. (NH4)+ H 5; N 5,2 H = +1, N = -3
24. N2 N 1 N = 0
25. (NO3)- O 4; N 4,3 O = -2, N = +5
26. (NO2)- O 4; N 4,3 O = -2, N = +3
27. NO2 O 4; N 4,2 O = -2, N = +4
28. NO O 4; N 4,2 O = -2, N = +2
29. N2O O 4; N 4,2 O = -2, N = +1
30. Na2O2 Na 3; O 2 Na = +1, O = -1 (peroxide!)

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B. For the each word reaction, write the chemical equation without balancing it, write the oxidation state of each element above that element, and write the two half reactions, labeling which is oxidation and which is reduction. You can check your work by balancing the complete reaction using the numbers from the half reaction addition. If you have a problem with an example, check first with the completed balanced equation in the answer section.

31. Hydrogen gas burns in oxygen to make water.

32. Mercuric oxide, a red powder, is put into a test tube and warmed. Liquid mercury forms on the sides and in the bottom of the tube and oxygen gas escapes from the test tube.

33. Potassium chlorate is heated in a test tube. Oxygen gas is made and potassium chloride is left in the bottom of the tube.

34. Hydrochloric acid is poured onto zinc metal to make zinc chloride and hydrogen gas.

35. A copper wire is put into silver nitrate. Silver metal appears and the solution turns blue from copper II nitrate.

36. Magnetite, an ore of iron, is smelted in large hot furnaces by blowing carbon monoxide gas through the ore. The result is liquid (molten) iron and carbon dioxide bubbles.

37. Lead metal and lead IV oxide in sulfuric acid produce lead II sulfate and water. This is the reaction in a common lead-acid car battery.

38. Methane gas burns in oxygen to make water vapor and carbon dioxide.

39. Octane burns with oxygen to make carbon dioxide and water.

40. Concentrated nitric acid is put on copper wire. Water and copper II nitrate in the water solution is produced, along with a brownish gas, nitrogen monoxide or nitric oxide, NO.

41. Potassium dichromate and hydrochloric acid in solution will make chlorine gas, water, chromium III chloride and potassium chloride. (The soluble salts, of course, remain in the water solution.)

42. Potassium permanganate solution when added to potassium cyanide in water solution will make managanese IV oxide and potassium hydroxide and water and potassium cyanate (KOCN).

43. In a sulfuric acid solution potassium permanganate will titrate with oxalic acid to produce manganese II sulfate, carbon dioxide, water, and potassium sulfate in solution.


31. 0 0 +1 -2 2( H0 ==>H+1 +e-) Oxidation
H2 + O2 ==> H2O 2e- + O0 ==> O-2 Reduction

Balanced equation 2 H2 + O2 ==> 2 H2O

32. +2 -2 0 0 2e- + Hg+2 ==> Hg0 Reduction
HgO ==> O2 + Hg O-2 ==> O0 + 2e- Oxidation

Balanced equation 2 HgO ==> O2 + 2 Hg

33. +1 +5-2 +1 -1 0 6e- + Cl+5 ==> Cl-1 Reduction
KClO3 ==> KCl + O2 3( O-2 ==> O0 + 2e-) Oxidation

Balanced equation 2 KClO3 ==> 2 KCl + 3 O2

34. +1 -1 0 +2 -1 0 2( e- + H+1 ==> H0) Reduction
H Cl + Zn ==> ZnCl2 + H2 Zn0 ==> Zn+2 + e- Oxidation

Balanced equation 2 HCl + Zn ==> ZnCl2 + H2

35. +1 +5 - 2 0 +2 +5 -2 0 2( e- + Ag+1 ==> Ag0 Reduction
Ag NO3 + Cu ==> Cu(NO3)2 + Ag Cu0 ==> Cu+2 + 2 e- Oxidation

Balanced chemical equation 2 AgNO3 + Cu ==> Cu(NO3)2 + 2 Ag

36. +8/3 -2 +2 -2 0 +4 -2 4 ( C+2 ==> C+4 + 2 e-) Oxidation
Fe3O4 + CO ==> Fe + CO2 3 ( 8/3 e- + Fe+8/3 ==> Fe0) Reduction

Balanced chemical reaction 3 Fe3O4 + 4 CO ==> 3 Fe + 4 CO2

37. 0 +4 -2 +1 +6 -2 +2 +6 -2 +1 -2 Pb0 ==> Pb+2 + 2e- Oxidation
Pb + PbO2 + H2SO4 ==> PbSO4 + H2O 2e- + Pb+4 ==> Pb+2 Reduction

Balanced equation Pb + PbO2 + 2 H2SO4 ==> 2 PbSO4 + 2 H2O - lead oxidizes and reduces.

38. -4 +1 0 +1 -2 +4 -2 C-4 ==> C+4 +8e- Oxidation
CH4 + O2 ==> H2O + CO2 4( 2e- + O0 ==> O-2 ) Reduction

Balanced equation CH4 + 2 O2 ==> 2 H2O + CO2

39. -9/4 +1 0 +4 -2 +1 -2 25 ( 2e- + O0 ==> O-2) Reduction
C8H18 + O2 ==> CO2 + H2O 2 ( C-9/4 ==> C+4 + 25/4 e-) Oxidation

Balanced equation 2 C8H18 + 25 O2 ==> 16 CO2 + 18 H2O

40. 0 +1 +5 -2 +2 -2 +2 +5 -2 +1 -2 2( 3e- + N+5 ==> N+2) Reduction
Cu + H(NO3) ==> NO + Cu(NO3)2 + H2O 3( Cu0 ==>Cu+2 + 2e-) Oxidation

Balanced equation 3 Cu + 8 HNO3 ==> 4 H2O + 2 NO + 3 Cu(NO3)2

41. +1 +6 -2 +1 -1 0 +3 -1 +1 -2 +1 -1 3( Cl-1 ==> Cl0 + 1e-) Oxidation
K2(Cr2O7) + HCl ==> Cl2 + CrCl3 + H2O + KCl 3 e- + Cr+6 ==> Cr+3 Reduction

Balanced equation K2(Cr2O7) + 14 HCl ==> 3 Cl2 + 7 H2O + 2 CrCl3 + 2 KCl

42. +1 +7 -2 +1 +2 -3 +1 -2 +4 -2 +1 -2 +1 +1 -2 +4 -3
KMnO4 + K(CN) + H2O ==> MnO2 + K(OH) + K(OCN)

3( C+2 ==> C+4 +2e-) Oxidation 2(3e- + Mn+7 ==> Mn+4) Reduction

Balanced equation 2 KMnO4 + 3 KCN + H2O ==> 2 MnO2 + 2 KOH + 3 K(OCN)

43. +1 +7 -2 +1 +3 -2 +1 +6 -2 +2 +6 -2 +4-2 +1 -2 +1 +6-2
KMnO4 + H2C2O2 + H2SO4 ==> MnSO4 + CO2 + H2O + K2SO4

5e- + Mn+7 ==> Mn+2 Reduction 5( C+3 ==> C+4 + e-) Oxidation

Balanced 2 KMnO4 + 5 H2C2O4 + 3 H2SO4 ==> 2 MnSO4 + 10 CO2 + 8 H2O + K2SO4

1 comment:

Jon snow said...

Excellent blog to understand the redox reaction.I just want to ask that when nitrogen is oxidize then nitrous oxide formed so is it a redox reaction.
Oxidation Reduction Reactions