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Oxidation-Reduction Reactions

Oxidation is a chemical change in which electrons are lost by an atom or group of atoms, and reduction is a chemical change in which electrons are gained by an atom or group of atoms.

As a simple example of an oxidation-reduction consider what happens when you dip a strip of metallic zinc into a blue solution of copper(II) sulfate.  The strip of zinc becomes coated with a reddish-brown layer of metallic copper.  The molecular equation for this reaction is         

                        Zn(s) + CuSO4(aq) ==> FeSO4(aq) + Cu(s)

In this simple example we can say that zinc has been oxidized and copper has been reduced.  For convenience, we can think of this reaction as two separate parts or half reactions one involving the loss of two electrons by an atom of zinc and the other being the gain of two electrons by the copper(II) ion.  The two half reactions are

Zn(s) ==> Zn2+(aq) + 2 e-        oxidized

Cu2+(aq) + 2 e- ==> Cu(s)        reduced

Two terms used in conection with oxidation-reduction reactions are oxidizing agent, a species that oxidizes another species and is itself reduced, and reducing agent, a species that reduces another species and is itself oxidized.  In our example above, the copper(II) ion is the oxidizing agent, and zinc metal is the reducing agent.  It is not always immediately apparent from the ionic charge alone whether a compound is undergoing oxidation or reduction.  For example, MnO2 reacts with hydrochloric acid to produce, chlorine gas and the Mn2+ ion.  It is obvious that neutral chlorine is produced by oxidation of Cl-.   We assume, therefore, that MnO2 is undergoing reduction, in spite of the fact that a neutral substance, MnO2 is converted to a cationic substance, Mn2+.  The concept of oxidation numbers was developed as a simple way of keeping track of electrons in a reaction.  By using oxidation numbers, you can determine whether electrons have been transferred and thereby whether an oxidation-reduction reaction has occurred.

Oxidation Numbers    

The oxidation number (sometimes called the oxidation state) of an atom in a substance is defined as the charge of the atom if it existed as a monoatomic ion, or a hypothetical charge assigned to the atom in the substance by a set of rules.  We then define an oxidation-reduction reaction as one in which one or more atoms change oxidation numbers, signaling that there was a transfer of electrons.

Rules for Assigning Oxidation Numbers         

1 The oxidation number of an atom in an element is zero.
2 The oxidation number of an atom in a monatomic ion is equal to the charge of the ion.
3 The oxidation number of oxygen is -2 in all oxides, it is -1 in peroxides.
4 The oxidation number of hydrogen is +1 in all compounds except hydrides(binary compounds of a metal and hydrogen) where it is -1.
5 The oxidation number of fluorine is -1 in all of its compounds.   The other halogens(Cl, Br, I) have an oxidation number of -1 in binary compounds, except when the other element is oxygen or another halogen above it in the periodic table.
6 The sum of the oxidation numbers of the atoms in a compound is zero. monatomic ion is equal to the charge of the ion.  The sum of the oxidation numbers of the atoms in a polyatomic ion equals the charge on the ion.

Balancing Oxidation-Reduction Equations

Oxidation-reduction equations are often too dificult to balance by the inspection method(trial and error) we have used up to this point.  There are two systematic methods for balancing redox equations. One is the half-reaction or ion-electron method and the second is the oxidation-state method.  In this summary, I am only going to cover the half-reaction method.  First we will list the rules, and then we will go through a couple of examples in detail.

Rules for balancing oxidation-reductions by the half-reaction method

1 Write a skeleton equation or half-reaction that includes those reactants and products that contain the elements undergoing a change in oxidation number.
2 Write a half-reaction equation for the oxidizing agent, with the element undergoing a reduction on each side of the equation.  The element should not be written as a free element or ion unless it really exists as such.  It should be written a part of a real molecular or ionic species.
3 Write another half-reaction equation for the reducing agent, with the element undergoing an increase in oxidation number on each side.
4 Balance each half-reaction as to number of atoms of each element.   In neutral or acidic solution, H2O and H+ may be used for balancing oxygen and hydrogen atoms.  The oxygen atoms are balanced first.  For each excess oxygen atom on one side of an equation, balance is obtained by adding one   H2O to the other side.  Then H+ is used to balance hydrogens.  Note that O2 and H2 are not used to balance the oxygen and hydrogen atoms unless they are actually reactants or products.

If the solution is alkaline, OH- and H2O may be used.   For each excess oxygen on one side of an equation, balanced is obtained by adding one H2O to the same side and 2 OH- to the other side.  If hydrogen is still unbalanced after this is done, balance is obtained by adding one OH- for each excess hydrogen on the same side as the excess and one H2O to the other side.  If both oxygen and hydrogen are in excess on the same side of the half-reaction, an OH- can be written on the other side for each paired excess of H and O.

5 Balance each half-reaction as to the number of charges by adding electrons to the the left or right side of the half-reaction.  If the preceding rules have been followed carefully, it will be found that electrons must be added to the left side of the half-reaction for the oxidizing agent and to the right side of the half-reaction for the reducing agent.
6 Multiply each half-reaction by a number chosen so that it makes the number of electrons the same in each half-reaction.
7 Add the two half-reactions resulting from the multiplications.   In the resultant equation, cancel any terms common to both sides.  All electrons must cancel.
8 Transform the net ionic equation that results form step 8 into a molecular equation.  This is done by adding to each side of the equation equal numbers of the ions which do not undergo electron transfer (spectator  ions) but which are present along with the reactive components in neutral chemical substances.

As our first example lets balance the following oxidation-reduction equation:

K2Cr2O7 + HCl ==> KCl + CrCl3 + H2O + Cl2

First we balance the half-reaction for the oxidizing agent as follows. Note that we have an acidic solution because of HCl.

Cr2O72- ==> Cr3+

Cr2O72- ==> 2 Cr3+

Cr2O72- ==> 2 Cr3+ + 7 H2O

14 H+ + Cr2O72- ==> 2 Cr3+ + 7 H2O

14 H+ + Cr2O72- + 6 e- ==> 2 Cr3+ + 7 H2O

Next we balance the half-reaction for the reducing agent.

Cl- ==> Cl2

2 Cl- ==> Cl2

2 Cl- ==> Cl2 + 2 e-

Now we multiply each half-reaction by numbers chosen to make the electrons cancel when added.

1 x [14 H+ + Cr2O72- + 6 e- ==> 2 Cr3+ + 7 H2O]

3 x [                     2 Cl- ==> Cl2 + 2 e- ]


14 H+ + Cr2O72 + 6 Cl- ==> 2 Cr3+ + 7 H2O +  Cl2

The 14 H+ was added as 14 HCl, and 6 of the 14 chloride ions were oxidized.  To each side of the equation 8 more Cl- are added to represent those Cl- which were not oxidized.  Similarly, 2 K+ can be added to each side to show that Cr2O72- came from K2Cr2O7. So our balanced molecular equation is

K2Cr2O7 + 14 HCl ==> 2 KCl + 2 CrCl3 + 7 H2O + 3 Cl2

For our second example we will balance the following equation, which is in an alkaline solution.

Zn + NaNO3 + NaOH ==> Na2ZnO2 + NH3 + H2O

The balancing of the half-reaction for the oxidizing agent goes as follows.

NO3- ==> NH3

NO3- + 6 H2O ==> NH3 + 9  OH-

NO3- + 6 H2O   + 8 e- ==> NH3 + 9  OH-

The balancing of the half-reaction for the reducing agent is as follows.

Zn ==> ZnO22-

4 OH- +  Zn ==> ZnO22- + 2 H2O  

4 OH- +  Zn ==> ZnO22- + 2 H2O  + 2 e-

Now we multiply each half-reaction by numbers chosen to make the electrons cancel when added.

1 x [NO3- + 6 H2O   + 8 e- ==> NH3 + 9  OH-]

4 x [ 4 OH- +  Zn ==> ZnO22- + 2 H2O  + 2 e- ]


NO3- + 6 H2O + 4 Zn + 16 OH-  ==> NH3 + 9  OH- + 4 ZnO22- + 8 H2O

Canceling terms that are common to both sides we have.

NO3- + 4 Zn + 7 OH-  ==> NH3 + 4 ZnO22- + 2 H2O

Convert the net ionic equation to a molecular equation by adding ions.

4 Zn + NaNO3 + 7 NaOH ==> 4 Na2ZnO2 + NH3 + 2 H2O







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