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WRITING REDOX EQUATIONS : HALF-EQUATION METHOD
Comparing different methods for balancing redox equations:
If the objective is simply to balance an equation, and the equation can be balanced easily BY
INSPECTION, then it may be balanced that way. Balancing by inspection is described on
pages 18 to 23. Most of the examples given involve only exchanges of ions, and only the
reaction of acids and metals on page 20, and combustion reactions on page 21, are
examples of redox. Among the redox equations listed for balancing on page 29, some can
be balanced quite easily by inspection.
The OXIDATION-STATE METHOD, described in pages 28 - 29, is a useful and quick method
for balancing redox equations that cannot be balanced by inspection. It does not have a
good theoretical basis, and occasionally it may be difficult to apply.
The HALF-EQUATION METHOD has a strong theoretical basis, and is therefore a more
important method, even though not always the quickest method to balance an equation. It
is important that students know how to use half-equations, since they have other
applications besides as a way of balancing equations, notably in situations involving cells
and electrolysis.
Some background theory:
On page 25, it was stated that oxidation is a loss of one or more electrons by an atom, ion,
or molecule, whereas reduction is a gain of one or more electrons by an atom, ion or
molecule.
There are two ways to remember which is loss, which is gain of electrons.
a) When an atom is oxidised, its oxidation state becomes more positive; it increases.
When an atom is reduced, its oxidation state becomes less positive or more
negative, that is, it is reduced. Gain of (negative) electrons effectively makes the
atom less positive or more negative, that is, gain of electrons reduces the atom.
The same idea applies, of course, to molecules and ions: reduction involves gain
of electrons.
b) Think of a simple example of an element combining with oxygen, for example,
magnesium plus oxygen forms magnesium oxide.
It should be clear that electrons have been transferred from magnesium to oxygen.
Magnesium has lost electrons in the process of combining with oxygen, that is,
while being oxidised. Oxygen has been reduced by magnesium by gaining electrons
from it.
If there is ever any hesitation about whether oxidation is a loss or a gain of
electrons, use this simple example (or any like it) as a way of remembering.
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The equations written to show electrons (written as e-) are called half-equations.
Note that it is not usual to write a half equation with "- e-”, as
Such an equation is properly written with a positive sign before the electrons, as
Every redox reaction is comprised of two separate components:
1) an oxidation half-reaction, which concerns the atom, molecule or ion that is
oxidised, that is, which loses electrons, and
2) a reduction half-reaction, which concerns the atom, molecule or ion that is
reduced, that is, which gains electrons.
Both of these must be involved, and each can be expressed by a half-equation. For example,
consider the reaction between zinc and sulfuric acid, that produces hydrogen gas and zinc
sulfate in solution:
Written without the sulfate spectator ions, the equation is:
The reaction is redox, because zinc is oxidised, while hydrogen ions are reduced. The
equation can be separated into two half equations:
OXIDATION HALF-EQUATION:
REDUCTION HALF-EQUATION:
Any redox equation can be separated into two half-equations. For example,
If the second half-equation is simplified, the two half-equations are
It should be noticed that electrons are always shown in a redox half-equation, but that the
electrons cancel out when the two half-equations are added together. The "simplified" halfequations shown above cannot be added together unless the second one has been multiplied
by three, so that the electron numbers are the same in both half-equations.
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It should also be noticed that when the half-equations are properly added together, it may
be necessary to simplify the equation with respect to water and hydrogen ions:
How to write balanced redox equations:
EXAMPLE:
Manganese dioxide + hydrochloric acid produces manganese(II) ions + chlorine gas
.
Step one: write the equation in unbalanced form:
Step two: separate the unbalanced equation into two unbalanced half-equations:
Step three: balance either or both equations for any elements other than oxygen or
hydrogen. (NB this is a step often overlooked by students! Be careful!).
Step four: balance oxygen by adding water:
Step five: Balance hydrogen by adding H+ :
Step six: Add electrons to balance charge:
NB the electrons should be on the left-hand-side of one equation and on the right-hand-side
of the other. If this is not so, then either the substances shown on the left of the halfequations will not react with each other in the way that the half-equations show, or else an
error has been made in writing and balancing the half-equations.
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Step seven: If the electrons numbers in the two equations are not equal, multiply one or
both equations by factors so that the numbers of electrons become equal. This
step is not necessary in the example being used, since both equations have 2e-.
A second example below shows the application of step seven.
Step eight: Add the two equations, so that the electrons are cancelled out.
Step nine: Simplify the equation so that water and H+ each appear in only one place:
Step ten: Add spectator ions if required: in this example, chloride ions.
Second example:
Potassium hydroxide solution heated with iodine produces potassium iodate and iodide.
Step one:
Iodine is reduced from oxidation state 0 to -1, and is also oxidised from
oxidation state 0 to +5. This is a redox reaction.
Step two:
Step three:
Step four:
Step five: :
Step six:
Step seven:
Step eight:
NB spectator ions are not written.
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Step nine: Simplify the equation:
Step ten: Add spectator ions if required: in this example, potassium ions.
This equation is correct and balanced, but it does not agree with the original
unbalanced equation at step one, which has KOH as a reactant.
If 6OH- is added to each side of the equation, then 6KOH appears on the left,
and 6HOH (or 6H2O) on the right. The water cancels from left and right, leaving
3H2O on the right.
A redox reaction of this kind, in which the same reactant is both oxidised and reduced
simultaneously (it oxidises and reduces itself) is called disproportionation.
For practice exercises, the equations on page 29 can be balanced by the half equation
method.
Further examples for balancing:
1. Silver metal + nitric acid give silver nitrate + nitric oxide (NO)
2. Calcium hydroxide + chlorine give calcium chloride + calcium hypochlorite
3. Copper + (concentrated) sulfuric acid give copper sulfate + sulfur dioxide
4. Nitric acid + iodide ions give nitric oxide + iodine
5. Nitric acid + sulfide ions give nitrogen dioxide + sulfur
6. Iron(III) chloride + iodide ions give iron(II) chloride + iodine.
7. Thiosulfate ions + iodine give tetrathionate ions (S4O62-) + iodide ions.
8. Ammonium ions + nitrite ions give nitrogen gas + water
9. Manganese dioxide + hydrochloric acid give manganese(II) ions + chlorine gas
10. Potassium dichromate + sulfur dioxide give chromium(III) sulfate + potassium sulfate