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BALANCING EQUATIONS WITHOUT USING HALF-REACTIONS
A simple alternative to writing half-reactions and taking the difference between
them is the following stepwise procedure.
1. Write down only the oxidizing and reducing agents on the left side of
the equation, and their reduced and oxidized forms on the right. If the
oxidized and reduced forms of a given agent differ in the number of
atoms of the element responsible for electron exchange, make these
numbers equal by using a preliminary integer as a coefficient for the
form with the smaller number of atoms.
2. Determine, for the elements responsible for electron exchange, the
number of e~ gained by the oxidizing agent and the number of e~ lost
by the reducing agent, as they go to their respective products.
3. Multiply each of these numbers by integers that will give the same (and
smallest possible) number of e~ lost as gained. Use these integers as
coefficients for the oxidizing agent (and its corresponding product) and
the reducing agent (and its corresponding product) as written down in
step 1. These integers must be multiplied by any preliminary coefficients used in step 1.
4. Determine the sum of the ionic charges on each side of the equation. If
they are not equal, calculate the number of H
+ ions or OH~ ions that
must be added to one side or the other in order to make the sums equal.
If the reaction conditions are acidic, you must use H
+ ; if basic, use
OH^. You cannot have H
+ on one side of the equation and OH~ on the
other; both sides are in the same container with the same acidity.
5. If necessary, balance the H'sor the O's by adding the proper number of
H 2 O molecules to whichever side of the equation needs them. The
equation will now be balanced.
PROBLEM:
Balance the equation for the reaction between Cr 2 O?~ (from Na 2 Cr 2 O 7 ) and Br~
(from NaBr) in acid solution.
SOLUTION:
We follow the steps just outlined.
(1) Write down the oxidizing and reducing agents, and their corresponding
reduced and oxidized forms as products:
Cr 2 O 7
2 - + 2Br -> 2Cr
i+ + Br 2
BALANCING EQUATIONS WITHOUT USING HALF-REACTIONS
A simple alternative to writing half-reactions and taking the difference between
them is the following stepwise procedure.
1. Write down only the oxidizing and reducing agents on the left side of
the equation, and their reduced and oxidized forms on the right. If the
oxidized and reduced forms of a given agent differ in the number of
atoms of the element responsible for electron exchange, make these
numbers equal by using a preliminary integer as a coefficient for the
form with the smaller number of atoms.
2. Determine, for the elements responsible for electron exchange, the
number of e~ gained by the oxidizing agent and the number of e~ lost
by the reducing agent, as they go to their respective products.
3. Multiply each of these numbers by integers that will give the same (and
smallest possible) number of e~ lost as gained. Use these integers as
coefficients for the oxidizing agent (and its corresponding product) and
the reducing agent (and its corresponding product) as written down in
step 1. These integers must be multiplied by any preliminary coefficients used in step 1.
4. Determine the sum of the ionic charges on each side of the equation. If
they are not equal, calculate the number of H
+ ions or OH~ ions that
must be added to one side or the other in order to make the sums equal.
If the reaction conditions are acidic, you must use H
+ ; if basic, use
OH^. You cannot have H
+ on one side of the equation and OH~ on the
other; both sides are in the same container with the same acidity.
5. If necessary, balance the H'sor the O's by adding the proper number of
H 2 O molecules to whichever side of the equation needs them. The
equation will now be balanced.
PROBLEM:
Balance the equation for the reaction between Cr 2 O?~ (from Na 2 Cr 2 O 7 ) and Br~
(from NaBr) in acid solution.
SOLUTION:
We follow the steps just outlined.
(1) Write down the oxidizing and reducing agents, and their corresponding
reduced and oxidized forms as products:
Cr 2 O 7
2 - + 2Br -> 2Cr
i+ + Br 2