Honors Chemistry I
Hints for Balancing Chemical Equations
I. The trial-and-error method (sight inspection method):
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The key point is to remember that there must be the same number of each of the
elements (atoms) on each side of the equation.
Look for metals (cations) that exist in only one reactant and also in only one product form.
Avoid balancing hydrogen and oxygen until the end. These are usually the most abundant
elements found in chemical reactions.
Look to see if any elements are in their diatomic state. Coefficients for compounds that
contain these elements will have to be even. For example, if H2 gas is produced from
HCl, the coefficient for the HCl will be an even multiple of the coefficient in front of the
H2 molecule.
If an element, such as sulfur, exist only in the form of a polyatomic ion, like the sulfate
ion, count the entire species as one item rather than splitting up the ion into its elements.
For acid base neutralizations, it might be best to think of water as H-OH; that way when
you count, see how many of each of the hydrogen ions and hydroxide ions you have.
II. Algebraic Method:
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Because there must be the same number of each element on both sides of the equation, the
quantity of each substance can be written in an algebraic form.
In place of coefficients, substitute an alphabetic variable (a, b, c, etc.).
List each substance and write an equation to show many atoms of this substance are on
each side of the equation.
Choose any one of the variables and give it a numeric value. In choosing the variable and
the number, remember that you want the lowest whole number ratio in the final balanced
equation. Using a "1" for one of the variable is often helpful.
Solve for all the variables, even if it means doing substitutions within other equations.
Once you have solved for all of the variables, multiply by the least common multiple to
get all of the coefficients in the smallest whole-number ratio.
Note: While the algebraic method for balancing equations will always work, it is a long and
tedious process. You will not have enough time on quizzes and tests to use this method
for every equation. You should use it only when it appears the equation cannot be
balanced by simpler methods.
Balancing Redox Reactions
A. Change-in-Oxidation-State Method
The advantage of this method is that it allows you to quickly isolate the elements that will make balancing these
equations complex.
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Write down the oxidation state of each element above the symbol of the element. Note: I said the oxidation state - not the total
charge on a particular element. For example, the charge on sodium in Na2SO4 is +1 (not +2).
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Identify which elements have lost electrons and which have gained electrons. Use a line to connect the element on the left side of
the equation to that same element on the right side. You should have at least two lines. The substance that loses electrons has
been oxidized and it is called the reducing agent; the substance that has gained electrons has been reduced and it is called the
oxidizing agent.. [Be sure you understand why they are named as they are!]
3.
Determine the least common multiple of the electrons transferred. Multiply each half of the oxidation-reduction process by a
number that will yield this number of electrons. The multiplier becomes the coefficient before the compound involved in the redox
reaction.
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Diatomic substances typically cause the most headaches. When you encounter a reaction that yields a diatomic molecule, such as
O2, H2, or one of the halogens, you will need to put an even coefficient before the reactant compound that contains one of these
elements.
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Once you have balanced the elements that are actually involved in the electron transfer, balance all other elements. Typically these
will only be spectator ions.
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Be sure that all coefficients are reduced to their least common denominator.
B. Half-Cell Method:
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Identify the elements that are reduced and those that are oxidized. Write them in a half-reaction form. If the element is part
of an ion, write the entire ion (not the entire compound!) in the equation.
Balance the number of atoms of these elements on both sides of the equation.
If the elements are part of an ion, balance the other elements. How you balance the other elements (specifically, oxygen and
hydrogen) will depend upon the type of reaction medium you have - either acidic, basic, or neutral.
I. In acidic solutions:
a. To the side lacking in oxygen atoms, add as many waters as you need oxygens.
For example,
NO3 -------> NO(g) add 2 waters to the right-hand side.
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b. Then to balance the hydrogens, add hydrogen ions (H ) to the side opposite where the water was added.
II. In basic (or neutral) solutions:
a. To the side with an excess of oxygen atoms, add as many waters as you have extra oxygens.
For example,
NO3 -------> NO(g) add 2 waters to the left-hand side.
b. Then to balance the hydrogens, add hydroxide ions (OH ) to the side opposite where the water was
added.
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Once the number of atoms are balanced in each half-reaction, the charges must be balanced by adding electrons to the more
positive side. Both sides do not have to add up to zero. Rather the charges on each side have to be equal to each other.
Find the least common multiple for the number of electrons transferred. For example, if two electrons are lost in the
oxidation half-reaction, yet three electrons are gained in the reduction half-reaction, the L.C.M. would be six. Multiply each
half-reaction by the multiplier that will give you this L.C.M.
Add together the two half-reaction. Reduce all coefficients to their least common denominator.