Download 486 15.3 Balancing Redox Equations - District 196 e

Section 15.3 Balancing Redox Equations.
15.3 Balancing Redox Equations
Mass
conservation
For redox reactions, like any other type of chemical reaction, mass is conserved. The
number and type of atoms in the reactants must equal to the number and type of atoms in
the products.
Charge
conservation
In addition to mass conservation we must also check to make sure that charge is
conserved. We do this by keeping track of the electrons that are exchanged between
various atoms and make certain that they balance. The number of electrons associated
with the reactants must be equal to the number of electrons associated with the products.
Also, the total increase in the oxidation number of some atoms must be equal to the total
decrease of the oxidation numbers of some other atoms.
Balance mass
and charge
The things that we learned in the previous section about oxidation numbers and redox
reactions are now the tools that we use to balance redox reactions.
Methods for
balancing
redox reactions
Since balancing redox reactions requires both the balance of mass and charge, the
procedure is often more complicated than balancing non-redox reactions.
There are two methods for balancing redox reactions.
1.
2.
The oxidation number method;
The oxidation reaction and reduction reaction method combination. This is
also called thehalf-reaction method.
Both of these methods are based on the same fundamental principles and they simply
represent structured applications of these procedures.
Balancing by
Inspection
In some cases it is possible to balance a redox reaction by the inspection method. These
“easy to balance” reactions are usually the ones that do not occur in aqueous solutions.
Consider the unbalanced reaction of carbon with iron oxide. This reaction takes place in a
furnace where iron oxide (Fe2O3) reacts with carbon resulting in pure iron and carbon
dioxide as a by product.
C + Fe2O3 → Fe + CO2
This is a redox reaction. Carbon is oxidized and iron is reduced. The oxidation number of
oxygen does not change in this reaction. By balancing the mass of the three elements that
take part in this reaction we obtain the balanced equation.
3C + 2Fe2O3 → 4Fe + 3CO2
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A NATURAL APPROACH TO CHEMISTRY
The oxidation number method
Most redox reactions can’t be balanced with the inspection method. For such reactions
we use the oxidation number method.
Balancing
equations
using oxidation
numbers
The basic rule that drives this method is charge balance.
Increase in oxidation number for the oxidized atoms
equals
Decrease in oxidation number for the reduced atoms
1) Assign
oxidation
numbers
The basic steps for balancing a redox reaction with the oxidation number method are:
1.
Assignment of oxidation numbers to all atoms
• Use the rules for assigning the oxidation numbers
• Write the oxidation number for each element below it
2) Identify
oxidation and
reduction
2.
Identify and label the atoms that are oxidized and the atoms that are reduced.
• Increase in oxidation number (loss of electrons) means oxidation
• Decrease in oxidation number (gain of electrons) means reduction
3) Adjust
coefficients to
equalize
oxidation and
reduction
4) Check mass
balance
3.
Adjust the appropriate coefficients in the chemical equation so that the total
increase in oxidation number (total loss of electrons) is equal to the total
decrease in the oxidation number (total gain of electrons).
4.
Check for overall mass balance.
• Check to see that the number of atoms is the same on both sides of the
equation. If not make the necessary changes to the coefficients.
When we apply these steps in order, the resulting balanced equation should satisfy both
mass and charge conservation.
Each step has a well defined set of rules. However, applying all of the steps together can
be challenging! Balancing redox reactions requires practice in order to become familiar
with it.
A NATURAL APPROACH TO CHEMISTRY
487
Section 15.3 Balancing Redox Equations.
Example: balancing a complex redox reaction
Using the oxidation number method, balance the equation
HNO3(aq) + Cu2O(s) → Cu(NO3)2(aq) + NO(g) + H2O(l)
Solve:
Follow the steps for balancing redox equations
Step 1. Assign the oxidation numbers to each element
H N O3
+ Cu 2 O → Cu (N O3 ) 2
+1 +5 -2
+1
-2
+ N O + H2 O
+2 +5 -2
+2 -2
+1 -2
Green numbers under elements are the oxidation numbers
Step 2. Identify the atoms that are reduced and oxidized
H N O3
+ Cu 2 O → Cu (N O3 ) 2
+1 +5 -2
+1
-2
+ N O + H2 O
+2 +5 -2
+2 -2
+1 -2
Oxidation
Reduction
• The oxidation number of N decreases in NO and so N is reduced.
• The oxidation number of Cu increases and so Cu is oxidized.
• The oxidation numbers of the other atoms do not change.
Step 3. First we balance all atoms whose oxidation numbers have changed
H N O3
+1 +5 -2
+ Cu 2 O → 2 Cu (N O3 )2
+1
-2
+ N O + H2 O
+2 +5 -2
+2 -2
+1 -2
Oxidation
Reduction
• N atoms whose oxidation number changes are already balanced
• To balance Cu atoms we adjust the coefficients — multiply by 2.
• The total increase in oxidation number is 2: Cu looses 2 electrons.
The decrease in oxidation number of N is 3: N gains 3 electrons.
• Balance the number of electrons lost and gained. Multiply the species
that contain Cu with +3 and the species that contain N with +2
2 H N O3 + 3 Cu 2 O → 6 Cu (N O3 )2
+1 +5 -2
+1
-2
+ 2N O + H2 O
+2 +5 -2
+2 -2
+1 -2
-2(3) = -6
+3(2)= + 6
Step 4. Balance mass.
• Add 12 HNO3 to the left side to balance N.
• Add 6 O and 12 H to the right side to balance oxygen and hydrogen.
• This is done by adding 6H2O to the right side to balance H and O.
14 HNO3 + 3 Cu 2 O → 6 Cu(N O3 ) 2
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+ 2 NO + 7 H 2 O
A NATURAL APPROACH TO CHEMISTRY
Half-Reactions
Redox
reactions in
aqueous
solutions
Many interesting and useful redox reactions occur in aqueous solutions. The procedure
for balancing the equations for aqueous reactions is based on the fundamental principles
described earlier but with an important difference. The difference is the separation of the
complete reaction into two half-reactions. One half-reaction involves the oxidized
elements and the other involves the reduced elements. By separating the oxidation from
the reduction and writing them separately we make it easier to count electrons and
establish charge balance.
Identifying
half-reactions
To see where to split the reaction into half reactions, we need to
know the oxidation numbers of all the elements involved. Once
these are known we can see which are oxidized and which are
reduced. Once we know the oxidation and the reduction parts
we can separate the complete reaction into half-reactions. When
a zinc nail is placed in copper sulfate (CuSO4) solution the
overall reaction can be explained with the half-reactions.
Why halfreactions are
useful
With half-reactions, it is very direct to see the transfer of
electrons. The number of electrons given up by the oxidation
half-reaction must be equal to the number of electrons received
by the reduction half-reaction.
Find the half-reactions of the reaction Zn(s) + CuSO4(aq) → ZnSO4 (aq)+ Cu(s)
Given:
The rules for assigning oxidation numbers.
Relationships: In solution, CuSO4 and ZnSO4 dissociate as follows:
Solve:
• CuSO4 into Cu2+ and SO42- ions
• ZnSO4 into Zn2+ and SO42- ions
• The reaction may now be writen as
• Zn(s) + Cu2+ + SO42- → Zn2+ + SO42- + Cu
• Zn is oxidized and Cu is reduced
• SO42- is neither oxidized or reduced. It is called a spectator ion.
The half-reactions are:
• Zn(s) → Zn2+ + 2e- - oxidation
• Cu2+ + 2e- → Cu - reduction
half-reactions - the oxidation and the reduction parts of a redox reaction.
A NATURAL APPROACH TO CHEMISTRY
489
Section 15.3 Balancing Redox Equations.
The half-reactions method
Steps in the
half-reactions
method
Redox reaction equations can be balanced using the 2 half reactions. The method is
called the half-reactions method for balancing redox reaction equations.
Step 1. Write the complete unbalanced reaction showing explicitly all ions
Step 2. Identify which elements are oxidized and which are reduced. Find the
spectator ions (oxidation number does not change).
Step 3. Write down the two unbalanced half-reactions.
Step 4. Balance mass with elements other than oxygen and hydrogen. Balance
oxygen by adding H2O then balance hydrogen by adding H+.
Step 5. Balance the charge for both half-reactions.
Step 6. Make the number of electrons in both reactions equal by adjusting the
coefficients.
Step 7. Add the two half-reactions and check that both mass and charge balance.
An example
redox reaction
equation:
The plating of copper into zinc in a solution of copper sulfate provides a good example of
the half-reactions method.
Step 1:
Write the unbalanced equation showing all ions
Zn(s) + CuSO4(aq) → ZnSO4 (aq)+ Cu(s).
Zn(s) + Cu2+ + SO42- → Zn2+ + SO42- + Cu
Step 2:
Identify oxidation, reduction and spectator ions.
Step 3:
Write down the two unbalanced half-reactions
Oxidation: Zn (s) → Zn2+ (aq)
Reduction: Cu2+(aq) → Cu (s)
Step 4:
Balance the mass for both half-reactions. It is already balanced.
Zn (s) → Zn2+ (aq)
Cu2+(aq) → Cu (s)
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A NATURAL APPROACH TO CHEMISTRY
Balancing the charge in half-reactions
Why charge
balancing is
necessary
It is possible, even likely, that mass will balance for each half reactions, but one halfreaction uses more electrons that the other one yields. In a fully balanced electrochemical
reaction the electrons given up in the oxidation half-reaction are the same number as the
electrons used in the reduction half-reaction.
Step 5:
Balance the charge for both half-reactions. Add electrons to balance changes in oxidation
numbers
Step 6:
Step 7:
•
Zn(s) → Zn2+ (aq) + 2e- : Add 2 electrons to the right side.
•
Cu2+ (aq) + 2e- → Cu(s) : Add 2 electrons to the left side.
Make the number of electrons in both reactions equal by adjusting the coefficients.
•
Zn(s) → Zn2+ (aq) + 2e-
•
Cu2+(aq) + 2e- → Cu(s)
Add the two half-reactions
•
•
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) :
Mass and charge are balanced.
Return the spectator ions and adjust coefficients to maintain charge neutrality.
Zn(s) + Cu2+(aq) + SO42- (aq) → Zn2+ (aq)+ SO42- (aq) + Cu(s)
Step 8:
What the half
reactions tell us
Simplify equation and check for mass and charge balance.
•
Reactants: Zn, Cu2+, SO42- , 0 net charge.
•
Products: Zn2+, Cu, SO42- , 0 net charge.
•
Mass and charge are balanced
At the end of the analysis we could write the final balanced equation as given below.
However, this does not tell us directly that 2 electrons were transferred. If this reaction
were part of a battery or biochemical process, the fact that 2 electrons were exchanged
might create an electrical current!
Zn(s) + CuSO4(aq) → ZnSO4 (aq)+ Cu(s).
A NATURAL APPROACH TO CHEMISTRY
491
Section 15.3 Balancing Redox Equations.
A redox reaction with chlorine
Chlorine dioxide (ClO2) is used for water treatment. The OH- indicates that the reaction
occurs in a basic solution. for which an element that appears in the reactants is both
oxidized and reduced.
Using the half-reactions method, balance the equation
ClO2 + OH - → ClO2- + ClO3Solve:
Follow the steps for balancing redox equations.
Step 1. Write the complete unbalanced reaction showing explicitly all ions
•
ClO2 + OH - → ClO2- + ClO3-
Step 2. Identify which elements are oxidized and which are reduced
•
ClO2 is oxidized to ClO3- . Cl goes from +4 to +5: Cl is oxidized
ClO2 is reduced to ClO2- . Cl goes from +4 to +3: Cl is reduced
Note that Cl is both oxidized and reduced in this reaction.
Step 3. Write down the two half-reactions
•
•
ClO2 → ClO3- : Oxidation half-reaction
•
ClO2 → ClO2- : Reduction half-reaction
Step 4. Balance the mass for both half-reactions.
a. Cl is balanced in both reactions.
b. Balance O and H of the oxidation half-reaction:
Balance O by adding H2O and OH -: ClO2 + 2OH - → ClO3- + H2O
•
O and H of the reduction half-reaction are balanced: ClO2 → ClO2-
Step 5. Balance the charge for both half-reactions. Add electrons to balance
changes in oxidation numbers
•
ClO2 + 2OH - → ClO3- + H2O + e- : Add electron on the right side
•
ClO2 + e- → ClO2- : Add electron on the left side
Step 6. Number of electrons is equal in both reactions.
Step 7. Add the two half-reactions
•
2ClO2 + 2OH - → ClO3- + ClO2- + H2O, No spectator ions
Step 8. Simplify equation, if needed, and check for mass and charge balance.
•
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2ClO2 + 2OH - → ClO3- + ClO2- + H2O
A NATURAL APPROACH TO CHEMISTRY