BALANCING EQUATIONS, CHEMICAL REACTION TYPES & REDOX EQUATIONS


Lecture Video Link:

H.W.: 1a - 1f

H.W.: WORKSHEET: BALANCING CHEMICAL EQUATIONS: WORKSHEET - - - ANSWERS TO ODD QUESTIONS


IDENTIFYING CHEMICAL REACTIONS:

Matter can combine or break apart to produce new types of matter with very ________________________________. When this occurs, it is said that matter has undergone a chemical reaction.

EVIDENCE FOR CHEMICAL REACTIONS:

Studying chemistry can be just like solving a mystery. To determine whether or not a chemical reaction has occurred, one needs to look for observable clues. If any one of these four things occur (and more than one can occur at the same instance), a chemical reaction has occurred.

The RELEASE OF ________________________________

A ________________________________ CHANGE

The FORMATION OF A ________________________________

There is a CHANGE IN HEAT and LIGHT (both of these must occur)


WRITING A CHEMICAL EQUATION:

Many chemical reactions are important principally for the ________________________________ they release. The popular fuel, propane, is used for cooking food and heating homes. Propane is in one of a group of compounds called Hydrocarbons.

A hydrocarbon contains carbon and hydrogen atoms (and sometimes oxygen is within the molecule), and they combine with oxygen in chemical reactions to form carbon dioxide and water. The reaction that occurs when propane is burned can be represented by the following word equation.

Propane + Oxygen ® Carbon Dioxide + Water + Energy

A chemical reaction is an equation that shows what happens in a reaction. All chemical reactions are composed of ________________________________ (chemicals that are present before the reaction) and ________________________________ (chemicals that are produced after the reaction takes place.)

Na+1 + Cl-1 ® NaCl

In the chemical equation above, the Na+1 and the Cl-1 are the reactants. The NaCl is the product. The symbol between the Cl-1 and the NaCl, ( ® ), is the YIELDS sign. The arrow points towards the products and is used to show how the reaction takes place.

 Symbol Used  Meaning

+
 Read plus or and. Used between two formulas to indicate reactants combined or products formed

 
Read yields or produces. Used to separate reactants (on the left) from products (on the right). The arrow points in the direction of change (we will always point the arrow toward the RIGHT)

( s )
Read solid. Written after a symbol or formula to indicate that the physical state of the substance is solid.

( l )
Read liquid. Written after a symbol or formula to indicate that the physical state of the substance is liquid.

( g )
Read gas. Written after a symbol or formula to indicate that the physical state of the substance is gaseous.

 ( aq )
Read aqueous. Written after a symbol or formula to indicate that the substance is dissolved in water.

 
Indicates that the reaction is reversible.

 N.R.
Read No Reaction. Indicates that the given reactants do not react with each other.


SYMBOLS THAT INDICATE THE STATE:

Equations can also be written to indicate the physical state of the reactants and products. In fact, sometimes an equation cannot be fully understood unless this information is shown.

The symbols (g), (l), (s), and (aq) indicate whether the substance is a ________________________________, a liquid, a ________________________________, or one dissolved in water. The above graphic shows these other conventions used in writing chemical equations.

REVERSIBLE REACTIONS:

While the arrow in an equation shows the direction of change, it implies that reactions occur in only one direction. This is not always the case, under suitable conditions, many chemical reactions can be reversed.

For example:
H2 (g) + O2 (g)
® H2O (l) + Energy

You have already learned that water can be separated into its elements. The reaction can be written as follows.

Energy + H2O (l) ® H2 (g) + O2 (g)

(please note: these equations (water) are not balanced, yet!)

Notice that the second equation is the reverse of the first. A chemical reaction that is reversible can be described by a single equation in which a double arrow shows that the reaction is possible in both directions.

H2 (g) + O2 (g) ® H2O (l) + Energy

The equation says that hydrogen and oxygen combine to form water, releasing energy in the process. It also says that with the addition of energy to water, under suitable conditions, hydrogen and oxygen will be formed.


BALANCING EQUATIONS:

In a chemical reaction, matter cannot be created ________________________________. For example in the reaction...

H2 + Cl2 ® HCl

There are 2 atoms of hydrogen and 2 atoms of chlorine on the reactants side, but only one atom of hydrogen and chlorine is on the products side. To fix this problem and to follow the rule that matter cannot be created nor destroyed, we must balance the equation.

H2 + Cl2 ® 2 HCl

By balancing the equation, you must have equal numbers of each of the different atoms on both the reactants and the products sides. To balance equations, there are a few rules that must be followed.

First, locate the most complex compound and start balancing each of the different atoms (saving oxygen and hydrogen for last). For example,

NaCl + H2 ® HCl + Na

1. Start with NaCl. (This is the most complex compound because HCl has hydrogen in it and we save that for last.)

2. There is one atom of sodium on each side so move on to the chlorine.

3. There is one atom of chlorine on each side, so move on to the HCl.

4. There is one atom of hydrogen in HCl and two atoms of hydrogen on the reactants side, so put a 1/2 coefficient in front of H2, so there is only one atom of hydrogen on each side.

NaCl + 1/2 H2 ® HCl + Na

5. Since there can NOT be any fractions in the final answer - multiply all the coefficients by 2 as to eliminate the 1/2 coefficient.

2 NaCl + H2 ® 2 HCl + 2 Na

6. Check for the lowest common coefficients.


Another way of working this problem is to take inventory of each atom on the Reactants side and another inventory of each atom on the Products side of the yields sign. Given the following equation below:

KClO3 ® KCl + O2

First, take "Inventory of each atom" by counting the number of atoms for each element on each side of the yields sign.

Reactants:

1 - K

1 - Cl

3 - O

Products:

1 - K

1 - Cl

2 - O

Second, If you have an Even number of one-type of element on one side you must have an Even number of that element on the other side. Therefore, look for the even - odd combinations. Which in this case is oxygen.

For now, the number of potassium atoms, and the number of chlorine atoms are equal - so we will not do anything to those molecules that contain potassium (K) or chlorine (Cl) at this time (in order to change potassium or chloride numbers).

However, there are three Oxygen atoms in Potassium Chlorate (KClO3), and Two Oxygen atoms in oxygen gas (O2). In most cases, begin working with the "odd" number first to get your even number of atoms. So, place a two in front of the Potassium Chlorate molecule:

___2___ KClO3 ® KCl + O2

And take "Inventory" again. Remember, that the Coefficient in front of the molecule means that there are now that many molecules.

Example: ___2___ KClO3 = KClO3 + KClO3 (or two molecules of potassium chlorate). Therefore we must use the coefficient as a multiplier for each element - for example: 2 multiplied by O3 = Six oxygen: 2 x 3 oxygen = 6 oxygen.

Reactants:

2 - K

2 - Cl

6 - O

Products:

1 - K

1 - Cl

2 - O

By placing the two in front of the potassium chlorate molecule - that gives us our Even number of Oxygen atoms (which we want) and that also changes the number of potassium atoms and chlorine atoms. So let us balance the potassium and chlorine atoms on each side of the yields sign - by placing a two in front of the potassium chloride molecule (KCl).

2 KClO3 ® ___2___ KCl + O2

And take "Inventory" again.

Reactants:

2 - K

2 - Cl

6 - O

Products:

2 - K

2 - Cl

2 - O

After taking "Inventory" we notice that we need six oxygen atoms on the products side of the yield sign to balance the equation. So we place a three in front of the oxygen gas molecule.

2 KClO3 ® 2 KCl + ___3___ O2

And take "Inventory" again.

Reactants:

2 - K

2 - Cl

6 - O

Products:

2 - K

2 - Cl

6 - O

Now that all our atoms are equal on the reactants side as well as on the products side - we are finished. Our balanced equation now reads:

2 KClO3 ® 2 KCl + 3 O2

Reading the above equation: Two molecules of Potassium Chlorate yields two molecules of Potassium Chloride and three molecules of Oxygen gas.

Remember, not all equations are this easy to balance, and may require changing the numbers often, but to assist in a few short-cuts - save oxygen and hydrogen for last to balance, and save any "free-standing" elements for last as well. Free-standing means those elements which do not form molecules, or form diatomic molecules (H2 , N2 , O2 , F2 , Cl2 ).


EXAMPLE:

Write and balance the equation for the burning, or combustion of ethylene, C2H4.

- Since ethylene is a hydrocarbon like propane, the equation for the combustion of propane can serve as a model.

The equation for the combustion of ethylene are these.

C2H4 + O2 ® CO2 + H2O

- Since both Carbon and Hydrogen occur only once on each side of the arrow, you can begin with either element. If you start with hydrogen, the equation can be balanced by placing a 2 in front of the water on the right.

C2H4 + O2 ® CO2 + 2 H2O

Now the number of hydrogen atoms is balanced, but the number of carbon and oxygen atoms is not. Balance the carbon next. Notice that if you change the coefficient of C2H4, you also change the number of hydrogens.

C2H4 + O2 ® 2 CO2 + 2 H2O

Now both the carbon and hydrogen are balanced, but there are six oxygen atoms indicated on the right and only two on the left. Placing a three in front of the oxygen in the equation will complete the process.

C2H4 + 3 O2 ® 2 CO2 + 2 H2O

PRACTICE PROBLEMS:

1. Write and balance the equation for the reaction of sodium and water to produce sodium hydroxide and hydrogen gas.

 

 

 

2. Write and balance the equation for the formation of magnesium nitride from its elements.

 

 

 

Answers:


HOMEWORK PROBLEMS:

Write the balanced equations for each of the following reactions:

1a. Cu + H20 ® CuO + H2

 

1b. Al(NO3)3 + NaOH ® Al(OH)3 + NaNO3

 

1c. KNO3 ® KNO2 + O2

 

1d. Fe + H2SO4 ® Fe2(SO4)3 + H2

 

1e. O2 + CS2 ® CO2 + SO2

 

1f. Mg + N2 ® Mg3N2

 

1g. When copper(II) carbonate is heated, it forms copper(II) oxide and carbon dioxide gas.

 

1h. Sodium reacts with water to produce sodium hydroxide and hydrogen gas.

 

1i. Copper combines with sulfur to form copper(I) sulfide.

 

1j. Silver nitrate reacts with sulfuric acid to produce silver sulfate and nitric acid (HNO3)

Answers:


Lecture Video Link:

H.W.: 1g - 1j

H.W.: WORKSHEET: WORD EQUATIONS - BALANCING WORKSHEET - - - ANSWERS TO ODD QUESTIONS


Lecture Video Link:

H.W.: COMBUSTION, SYNTHESIS, & DECOMPOSITION WORKSHEET - - - ANSWERS TO ODD QUESTIONS


BALANCING COMBUSTION REACTIONS

Combustion is an exothermic reaction in which a substance combines with oxygen forming products in which all elements are combined with oxygen. It is a process we commonly call burning. Usually energy is released in the form of heat and light. The general form of combustion equations for hydrocarbons is:

CxHy + O2 ® CO2 + H2O

Most combustion reactions are the oxidation of a fuel material with oxygen gas. A complete combustion 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

First, place 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

or

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

2 C3H7OH + 13 O2 ® 6 CO2 + 8 H2O

FOR HONORS CHEMISTRY - ALWAYS REDUCE THE COEFFICIENTS TO THE LOWEST WHOLE NUMBER RATIO !!!


SYNTHESIS REACTIONS

ALSO CALLED COMBINATION, CONSTRUCTION, OR COMPOSITION REACTIONS

The title of this section contains four names for the same type of reaction. Your text may use any of these. I prefer 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

Predicting synthesis reactions: What would you expect for a product if aluminum metal reacts with chlorine gas? Based on the observed regularity, the following reaction seems likely.

2 Al (s) + 3 Cl2 (g) ® 2 AlCl3 (s)

Keep in mind that this is not a prediction that aluminum metal will react with chlorine. Whether a reaction occurs depends on many factors. At the moment, you have no basis for concluding that a reaction between aluminum and chlorine will occur.

You can predict an equation for the reaction between aluminum and chloride because of the observed regularity that elements react to form compounds. You can write the formula for the product because of regularities concerning chemical formulas. You learned that aluminum always has a +3 charge in compounds, and that chlorine always has a -1 charge when it forms a binary compound with a metal.

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.


DECOMPOSITION REACTIONS

ALSO CALLED DESYNTHESIS, DECOMBINATION, OR DECONSTRUCTION

Of the names for this type of reaction, I prefer the name decomposition. 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:

AB ® A + B

A decomposition reaction is opposite of a synthesis reaction. In a decomposition reaction, a compound breaks down to form two or more simpler substances.

What is the equation for the decomposition of water, H2O? Since water contains only two elements, the decomposition products can be predicted as the individual elements.

2 H2O(l) ® 2 H2 (g) + O2 (g)


Decomposition versus Dissociation:

1. 2 NaCl(l) ® 2 Na (s) + Cl2 (g) Electrolysis Reaction - Decomposition Reaction

2. NaCl(s) ® Na+1(aq) + Cl-1(aq) Dissolved in Water - Dissociation Reaction

Reaction "A", a Decomposition Reaction, there is a _____________________ CHANGE from the sodium chloride molecule which produces sodium metal and chlorine gas, substances with properties very different from those of salt (NaCl) when an electric current is passed through it. Most decomposition reactions form ELEMENTAL SUBSTANCES.

The change in Reaction "B" produces _____________________ substances. The sodium ions and chloride ions are present in sodium chloride. The ions (sodium and chloride) have very different properties from those of neutral elements shown in equation "A". Reactions similar to "B" are considered a Dissociation not a Decomposition. Other Dissociations are described by the following equations:

KBr (s) ® K+1(aq) + Br-1(aq)

and

CaI2 (s) ® Ca +2(aq) + 2 I -1(aq)


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.


Problem: Write an equation for the decomposition of lithium chloride, LiCl, and an equation for its dissociation.

Answer:

2 LiCl(s) ® 2 Li(s) + Cl2 (g) Decomposition

LiCl(s) ® Li+1(aq) + Cl-1(aq) Dissociation


Lecture Video Link:

H.W.: SINGLE & DOUBLE REPLACEMENT WORKSHEET - - - ANSWERS TO ODD QUESTIONS


SINGLE REPLACEMENT REACTIONS

ALSO CALLED SINGLE DISPLACEMENT, SINGLE SUBSTITUTION, OR ACTIVITY REPLACEMENT

Synthesis reactions occur between two or more different elements. But elements may also react with compounds. A reaction in which one element takes the place of another element as part of a compound, is called a single replacement reaction. In this type of reaction, a metal always replaces another metal and a nonmetal always replaces another nonmetal. The general equation for a single replacement reaction is:

A + BC ® AC + B

Notice that element "A" replaces "C" in the compound "BC." Is the product, "C", an element or a compound?

Consider the following reaction:

Cl2 (g) + 2 KBr (aq) ® 2 KCl (aq) + Br2 (aq)

If chlorine gas is bubbled through a solution of potassium bromide, chlorine replaces bromine in the compound and elemental bromine is produced.

Notice that before reacting, chlorine is uncombined. It is a free (uncombined) element. Bromine, however, is combined with potassium in the compound potassium bromide. After reacting, the opposite is true. The chlorine is now combined with potassium in the compound potassium chloride, and bromine exists as a free element. The reaction is usually described with the words, chlorine has replaced bromine, and the reaction is called a single replacement reaction.

Replacement reactions are not reversible. In other words, the following reaction will NOT take place!

2 KCl (aq) + Br2 (aq) ® Cl2 (aq) + 2 KBr (aq)

Therefore, the reaction will not happen so the equation is written as:

KCl (aq) + Br2 (aq) ® N.R.

Predicting if a reaction will occur:

Activity Series: Halogens:

There is an interesting regularity observed in replacement reactions involving the halogens: Fluorine, Chlorine, Bromine, and Iodine. Each halogen will react to replace any of the halogens below it in the periodic table, but will not replace those above. For example, chlorine will replace Bromine and Iodine, but it will not replace Fluorine.

Activity Series: Metals:

Metals also undergo replacement reactions, and regularities similar to those described for the halogens are observed. Metals can be listed in a series in which each metal will replace all the metals below it on the list, but none of the metals above it. Such list is commonly called an activity series. The activity series for metals is determined by experiments in which pairs of metals are compared for reactivity.


The MORE active elements are closer to the TOP of the list, while the LESS active elements are closer to the BOTTOM of the list.

Use the Activity Series to predict whether the following reaction can occur under normal conditions.

Mg (s) + CuSO4 (aq) ® Cu (s) + MgSO4 (aq)

Answer:

Since Magnesium is above copper on the activity series, magnesium is more active and will replace copper in the compound CuSO4. This reaction will occur.


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. How could you predict that this reaction would take place without stepping into the lab?

Cu + 2 AgNO3 ® Cu(NO3)2 + 2 Ag

Answer: Again, you need to look at the activity series and locate copper and silver. Since copper is higher on the list (more active) than silver (less active), it is safe to assume this reaction will occur. Confirmation, again, can only take place in the lab.


DOUBLE REPLACEMENT REACTIONS

ALSO CALLED DOUBLE DISPLACEMENT OR METATHESIS

Double replacement reactions or metathesis reactions (metathesis is a Greek term meaning "changing partners" and accurately describes what happens.) The general equation for a double replacement reaction is:

AB + XY ® AY + XB

Predicting Double Replacement Reactions:

Deciding whether a double replacement reaction will occur is a matter of predicting whether an insoluble product can form. If sodium nitrate is substituted for lead nitrate, you will see no reaction when the solutions in the two test tubes are mixed. Why Not? If you write out all possible combinations of metal and nonmetal ions and check the table below, you will see that none of them is insoluble. The combinations of a number of different positive and negative ions to form precipitates and soluble compounds.



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. I think 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 (See Chart Above) is insoluble in water, the ions take each other out of the solution.

Ag+1 + (NO3)-1 + K+1 +Cl-1 ® AgCl + K+1 + (NO3)-1


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+1 + Cl-1 + Na+1 + (OH)-1 ® HOH + Na+1 + Cl-1


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

Ca+2 + (CO3)-2 + 2 H+1 + 2 Cl-1 ® Ca+2 + 2 Cl-1 + 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.


Lecture Video Link:

H.W.: 2a-2d

H.W.: REDOX WORKSHEET

MORE PRACTICE FOR THE QUIZ


OXIDATION and REDUCTION:

Many familiar chemical processes belong to a class of reactions called Oxidation-Reduction, or Redox, reactions.

Every minute, redox reactions are taking place in your body and all around you.

Reactions in batteries, burning of wood in a campfire, corrosion of metals, ripening of fruit, and combustion of gasoline to name a few examples.


You have already seen some examples of oxidation and reduction reactions, when you classified reactions earlier this chapter. Recall that in a combustion reaction, such as the combustion of methane or the rusting of iron, oxygen is a reactant.

CH4 (g) + 2 O2 (g) ® CO2 (g) + 2 H2O (g)

4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)

Hopefully, you learned that these reactions can be classified as synthesis or combustion.

The term oxidation also seems reasonable for describing these reactions because, in both cases, a reactant combines with oxygen.

The term reduction is used to describe the reverse process. An example of reduction is the decomposition of water.

2 H2O (g) ® 2 H2 (g) + O2 (g)

However, not all oxidation reactions involve oxygen. There are many similar reactions between metals and nonmetals that are classified as oxidation or reduction reactions.


REDOX REACTIONS:

When a piece of copper is placed in a colorless silver nitrate solution, you can tell that a chemical reaction occurs because the solution turns blue over a period of time. You also notice a silvery coating that forms on the piece of copper.

You know that copper(II) ions in solution are blue, so copper(II) ions must be forming.

The silver nitrate solution contains silver ions; these ions must be coming out of solution to form solid silver-colored material.

The unbalanced equation for the reaction between solid copper and silver ions is this:

Cu(s) + Ag+1(aq) ® Cu+2(aq) + Ag(s)

The equation tells you that solid copper atoms are changing to copper ions at the same time that silver ions are changing to solid silver atoms.


HALF-REACTIONS:

Consider what is happening to the two reactants (Cu and Ag). Each copper atom loses two electrons to form a copper(II) ion:

Cu(s) ® Cu+2(aq) + 2e-

Each neutral copper atom acquired a +2 charge by losing two electrons. Whenever an atom or ion becomes more positively charged (positive charge increases) in a chemical reaction, the process is called ________________________________.

As you know, electrons do not exist alone. In this reaction, electrons are transferred to the silver ions in solution:

Ag+1(aq) + e- ® Ag(s)

When a silver ion acquires one electron, it loses its +1 charge and becomes a neutral atom. Whenever an atom or ion becomes less positively charged or more negative (positive charge is decreased) in a chemical reaction, the process is called ________________________________.

LEO says GER

(Lose Electrons = Oxidation) says (Gaining Electrons = Reduction)

Each of these equations describes only half of what takes place when solid copper reacts with silver ions.

Reactions that show just half a process are called Half-Reactions.

You always need two half-reactions - one for oxidation and one for reduction - to describe any Redox reaction.

It is impossible for oxidation to occur by itself. The electrons given up by oxidation cannot exist alone; they must be used by a reduction reaction. Thus, there will always be an oxidation half-reaction whenever there is a reduction half-reaction, and vice-versa.


NET EQUATION FOR A REDOX REACTION:

You might simply add the equations for the half-reaction, but doing this may not give you a balanced equation.

Balancing the net equation for a redox reaction also requires consideration of the ________________________________ in each of the half-reaction equations.

If electrons appear in the net equation for the redox reaction, you know you have made a mistake, because electrons cannot exist by themselves.

The overall equation is balanced only when the number of electrons lost in one half-reaction equals the electrons gained in one half-reaction.

Example: Balance the following equation:

Cu(s) + Ag+1(aq) ® Cu+2(aq) + Ag(s)

The equations for the half-reactions can be written like this:

Balance the number of electrons in the two half-reactions by multiplying the second equation by two:

The number of electrons in both equations is now equal. Add the equations for the two half-reactions to give the balanced redox equation:


Oxidation and reduction occur together. The solid copper (from the example) is called the ________________________________, because it brings about the reduction of silver ions.

If the silver ions were not present, the copper would not oxidize.

The substance that contains the silver ion(s) is the ________________________________, because it causes the oxidation of copper atoms.

Notice that the substance that is reduced is the oxidizing agent and the substance that is oxidized is the reducing agent.


And then there are some REDOX equations that need a POSITIVE Charge added. Since we use the electron as our negative symbol we want to use a proton as our positive symbol. The proton can be represented as p+1, but a better way is to use H +1. That way we can add water molecules to the problem to equal the number of "extra" hydrogen. Let's try the question below...


Practice Problem:

Balance the following equation for a redox reaction.

Li(s) + Pb+2(aq) ® Li+1(aq) + Pb(s)

Answer:


HOMEWORK QUESTIONS:

2a. Balance the following equation for a redox reaction.

Cr+3(aq) + Zn(s) ® Cr(s) + Zn+2(aq)

 

 

 

2b. Write a balanced equation for the reduction of iron(III) ions to iron atoms by the oxidation of nickel atoms to nickel(II) ions in an aqueous solution.

 

 

 

2c. In the reaction Fe+2 + Mg ® Fe + Mg+2, which reactant is the oxidizing agent? and reducing agent?

 

 

 

 

2d, Balance the half reaction for: Br-1 + MnO4-1 ® Br2 + Mn+2

 

 

Answers:


OXIDATION NUMBERS:

To keep track of electron transfers in redox reactions more easily, oxidation numbers have been assigned to all atoms and ions. An Oxidation number is the real or ________________________________ on an atom or ion has when all bonds are assumed to be ionic


Practice Problem:

What is the oxidation number for each element in the compound of Na3PO4?

Na =

P =

O =

Answer:


Determine the oxidation number for each element in the compound:

3a. CuSO4

 

3b. Ni3(PO4)2

 

3c. AgIO3

 

3d. Cr(CO3)3

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